Evaluate $\int_0^{\infty } \log \left(\frac{a^2}{x^2}+1\right) \log \left(\frac{b^2}{x^2}+1\right) \log \left(\frac{c^2}{x^2}+1\right) \, dx$
We have for $a,b,c>0$
- $\scriptsize \int_0^{\infty } \log \left(\frac{a^2}{x^2}+1\right) \log \left(\frac{b^2}{x^2}+1\right) \log \left(\frac{c^2}{x^2}+1\right) \, dx=\Re\left(2 c \pi \log ^2(-c)+2 c \pi (-\log (-c)+\log (c)-2) \log (-c)+2 a \pi \log (-c)+2 c \pi \log ^2(c)-2 (a-c) \pi \log \left(1-\frac{a}{a-c}\right)+2 (a-c) \pi \log \left(1-\frac{a+b}{a-c}\right)+2 c \pi \log (b) \log \left(1-\frac{b}{c}\right)-2 c \pi \log \left(1-\frac{b}{c}\right)-2 c \pi \log (a) \log \left(-\frac{c}{a-c}\right)+2 c \pi \log (a+b) \log \left(-\frac{b+c}{a-c}\right)-2 (a+b) \pi \log (-b-c)+2 (b-c) \pi (\log (b-c)-1)-2 c \pi \log (b) \log \left(\frac{b+c}{c}\right)+2 c \pi \log \left(\frac{b+c}{c}\right)+2 c \pi (\log (-c)-1)+2 c \pi (\log (-c)-\log (c)+2)-2 c \pi \log (b-c) (\log (c)-1)+6 c \pi (\log (c)-1)+2 a \pi \log (c)-4 c \pi \log (c)-2 a \pi \log (a) \log (c-a)+2 a \pi \log (a+b) \log (c-a)+2 c \pi \log (a) \log \left(\frac{c}{a+c}\right)-2 c \pi \log (a+b) \log \left(\frac{c-b}{a+c}\right)-2 (a+c) \pi (\log (a+c)-1)-2 a \pi \log (a) \log (a+c)+2 a \pi \log (a+b) \log (a+c)+2 \pi (\log (b-c) c-c-b \log (b-c)) \log (a+c)-2 c \pi (\log (-c)-1) \log (a+c)-2 c \pi (\log (c)-1) \log (a+c)+2 a \pi \log (a+c)-2 b \pi \log (b) \log (c-b)+2 b \pi \log (a+b) \log (c-b)-2 (a+b) \pi \log (c-b)-2 b \pi \log (b) \log (b+c)+2 b \pi \log (a+b) \log (b+c)-2 c \pi (\log (c)-1) \log (b+c)+2 \pi (\log (a+c) c-c+a \log (a+c)) \log (b+c)+2 b \pi \log (b+c)-2 (a+c) \pi \log \left(1-\frac{a}{a+c}\right)+2 (a+c) \pi \log \left(1-\frac{a+b}{a+c}\right)-\frac{2}{3} \pi \left(-3 a \log ^2(a)-3 a \log (-a) \log (a)-3 a \log \left(-\frac{b}{a-b}\right) \log (a)+3 b \log \left(-\frac{b}{a-b}\right) \log (a)+3 a \log (b) \log (a)-3 b \log (b) \log (a)+3 a \log (a+b) \log (a)+a \pi ^2-3 b \log ^2(b)+3 a-3 b-3 a \log \left(\frac{1}{a}\right)-3 a \log \left(-\frac{1}{a}\right)+6 a \log \left(-\frac{b}{a}\right)-6 a \log (-b)-6 b \log (-b)+6 b \log (b)-3 b \log (-b) \log (b)+3 a \log (b) \log \left(\frac{a}{a+b}\right)+3 b \log (b) \log \left(\frac{a}{a+b}\right)+3 a \log (-a) \log (a+b)+3 b \log (-b) \log (a+b)+3 b \log (b) \log (a+b)-6 a \text{Li}_2\left(\frac{a+b}{a}\right)-3 (a-b) \text{Li}_2\left(\frac{a}{a-b}\right)+3 (a+b) \text{Li}_2\left(\frac{b}{a+b}\right)+3 a i \pi \right)-2 a \pi \left(\log \left(\frac{c}{a}+1\right) \log (-c)+\text{Li}_2\left(-\frac{c}{a}\right)\right)-2 a \pi \left(\log \left(\frac{c}{a}+1\right) \log (c)+\text{Li}_2\left(-\frac{c}{a}\right)\right)-2 a \pi \left(\log \left(1-\frac{a+c}{a-b}\right) \log (a+c)+\text{Li}_2\left(\frac{a+c}{a-b}\right)\right)+2 b \pi \left(\log \left(1-\frac{a+c}{a-b}\right) \log (a+c)+\text{Li}_2\left(\frac{a+c}{a-b}\right)\right)+2 b \pi \left(\log \left(1-\frac{c}{b}\right) \log (c)+\text{Li}_2\left(\frac{c}{b}\right)\right)-2 b \pi \left(\log \left(\frac{c}{b}+1\right) \log (c)+\text{Li}_2\left(-\frac{c}{b}\right)\right)+2 a \pi \left(\log \left(1-\frac{b-c}{a+b}\right) \log (b-c)+\text{Li}_2\left(\frac{b-c}{a+b}\right)\right)+2 b \pi \left(\log \left(1-\frac{b-c}{a+b}\right) \log (b-c)+\text{Li}_2\left(\frac{b-c}{a+b}\right)\right)+2 (a-c) \pi \text{Li}_2\left(\frac{a}{a-c}\right)-2 (a-c) \pi \text{Li}_2\left(\frac{a+b}{a-c}\right)+2 c \pi \text{Li}_2\left(\frac{b}{c}\right)-2 c \pi \text{Li}_2\left(-\frac{b}{c}\right)+2 (a+c) \pi \text{Li}_2\left(\frac{a}{a+c}\right)-2 (a+c) \pi \text{Li}_2\left(\frac{a+b}{a+c}\right)\right)$
For instance, taking $(a,b,c)=(1,2,3)$
- $\small \int_0^{\infty } \log \left(\frac{1}{x^2}+1\right) \log \left(\frac{4}{x^2}+1\right) \log \left(\frac{9}{x^2}+1\right) \, dx=4 \pi \Re( \text{Li}_2(3))+2 \pi \Re\left(\text{Li}_2(4)+2 \text{Li}_2\left(\frac{3}{2}\right)\right)+2 \pi \text{Li}_2(-4)-2 \pi \text{Li}_2(-3)-4 \pi \text{Li}_2\left(-\frac{1}{2}\right)+6 \pi \text{Li}_2\left(-\frac{1}{3}\right)-6 \pi \text{Li}_2\left(-\frac{2}{3}\right)+8 \pi \text{Li}_2\left(\frac{1}{4}\right)-8 \pi \text{Li}_2\left(\frac{3}{4}\right)-\frac{5 \pi ^3}{6}+8 \pi \log ^2(2)+12 \pi \log ^2(3)-18 \pi \log (3) \log (2)+10 \pi \log (5) \log (2)$
Define the function $\mathcal{I}:\mathbb{R}_{>0}^{3}\rightarrow\mathbb{R}$ via the improper integral
$$\mathcal{I}{\left(a,b,c\right)}:=\int_{0}^{\infty}\mathrm{d}x\,\ln{\left(1+\frac{a^{2}}{x^{2}}\right)}\ln{\left(1+\frac{b^{2}}{x^{2}}\right)}\ln{\left(1+\frac{c^{2}}{x^{2}}\right)}.\tag{1}$$
Our objective is to derive a closed-form expression for $\mathcal{I}$ in terms of polylogarithms and elementary functions.
This function $\mathcal{I}$ has two important properties that follow almost immediately from definition $(1)$. First, it is symmetric under any permutation of its three parameters. Second, it obeys the scaling relation
$$\mathcal{I}{\left(a,b,c\right)}=d\,\mathcal{I}{\left(d^{-1}a,d^{-1}b,d^{-1}c\right)};~~~\small{\left(a,b,c,d\right)\in\mathbb{R}_{>0}^{4}}.$$
Consider the following derivative, which holds for fixed but arbitrary $x\in\mathbb{R}_{>0}$:
$$\frac{\partial}{\partial y}\ln{\left(1+\frac{y^{2}}{x^{2}}\right)}=\frac{2y}{x^{2}}\cdot\frac{1}{\left(1+\frac{y^{2}}{x^{2}}\right)}=\frac{2y}{\left(x^{2}+y^{2}\right)}.$$
Integrating both sides, we obtain the following integral representation:
$$\ln{\left(1+\frac{a^{2}}{x^{2}}\right)}=\int_{0}^{a}\mathrm{d}y\,\frac{2y}{\left(x^{2}+y^{2}\right)};~~~\small{a\in\mathbb{R}_{>0}\land x\in\mathbb{R}_{>0}}.\tag{2}$$
Suppose $\left(a,b,c\right)\in\mathbb{R}_{>0}^{3}$. Using the integral representation in $(4)$ above to rewrite $\mathcal{I}$ as a multiple integral and changing the order of integration, we find that $\mathcal{I}$ can be expressed as a triple integral as follows:
$$\begin{align} \mathcal{I}{\left(a,b,c\right)} &=\int_{0}^{\infty}\mathrm{d}x\,\ln{\left(1+\frac{a^{2}}{x^{2}}\right)}\ln{\left(1+\frac{b^{2}}{x^{2}}\right)}\ln{\left(1+\frac{c^{2}}{x^{2}}\right)}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{a}\mathrm{d}p\,\frac{2p}{\left(x^{2}+p^{2}\right)}\int_{0}^{b}\mathrm{d}q\,\frac{2q}{\left(x^{2}+q^{2}\right)}\int_{0}^{c}\mathrm{d}r\,\frac{2r}{\left(x^{2}+r^{2}\right)}\\ &=\int_{0}^{\infty}\mathrm{d}x\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{8pqr}{\left(x^{2}+p^{2}\right)\left(x^{2}+q^{2}\right)\left(x^{2}+r^{2}\right)}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\int_{0}^{\infty}\mathrm{d}x\,\frac{8pqr}{\left(x^{2}+p^{2}\right)\left(x^{2}+q^{2}\right)\left(x^{2}+r^{2}\right)}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(-1\right)8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}\frac{\left(q^{2}-r^{2}\right)}{\left(x^{2}+p^{2}\right)}\\ &~~~~~+\frac{\left(r^{2}-p^{2}\right)}{\left(x^{2}+q^{2}\right)}+\frac{\left(p^{2}-q^{2}\right)}{\left(x^{2}+r^{2}\right)}\bigg{]}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{\left(-1\right)8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(q^{2}-r^{2}\right)}{\left(x^{2}+p^{2}\right)}\\ &~~~~~+\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(r^{2}-p^{2}\right)}{\left(x^{2}+q^{2}\right)}+\int_{0}^{\infty}\mathrm{d}x\,\frac{\left(p^{2}-q^{2}\right)}{\left(x^{2}+r^{2}\right)}\bigg{]}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{\left(-1\right)8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}\frac{\left(q^{2}-r^{2}\right)}{p}\\ &~~~~~+\frac{\left(r^{2}-p^{2}\right)}{q}+\frac{\left(p^{2}-q^{2}\right)}{r}\bigg{]}\int_{0}^{\infty}\mathrm{d}y\,\frac{1}{\left(y^{2}+1\right)}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{8pqr}{\left(p^{2}-q^{2}\right)\left(q^{2}-r^{2}\right)\left(r^{2}-p^{2}\right)}\bigg{[}-\frac{\left(q^{2}-r^{2}\right)}{p}\\ &~~~~~-\frac{\left(r^{2}-p^{2}\right)}{q}-\frac{\left(p^{2}-q^{2}\right)}{r}\bigg{]}\frac{\pi}{2}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{c}\mathrm{d}r\,\frac{4\pi\left(p+q+r\right)}{\left(p+q\right)\left(q+r\right)\left(r+p\right)}.\tag{3}\\ \end{align}$$
Suppose $\left(a,b\right)\in\mathbb{R}_{>0}^{2}$, and consider the case where the third parameter of $\mathcal{I}$ is set equal to one. The integrand of our triple integral is a rational function, so at least one of the three integrations can be done in elementary terms. We find
$$\begin{align} \mathcal{I}{\left(a,b,1\right)} &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\int_{0}^{1}\mathrm{d}r\,\frac{4\pi\left(p+q+r\right)}{\left(p+q\right)\left(q+r\right)\left(r+p\right)}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\int_{0}^{1}\mathrm{d}r\,\frac{\left(p-q\right)\left(p+q+r\right)}{\left(q+r\right)\left(r+p\right)}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\int_{0}^{1}\mathrm{d}r\,\left[\frac{p}{\left(q+r\right)}-\frac{q}{\left(r+p\right)}\right]\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\left[p\int_{0}^{1}\mathrm{d}r\,\frac{1}{\left(q+r\right)}-q\int_{0}^{1}\mathrm{d}r\,\frac{1}{\left(r+p\right)}\right]\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{4\pi}{\left(p-q\right)\left(p+q\right)}\left[p\ln{\left(\frac{1+q}{q}\right)}-q\ln{\left(\frac{1+p}{p}\right)}\right]\\ &=2\pi\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{2}{\left(p-q\right)\left(p+q\right)}\left[p\ln{\left(\frac{1+q}{q}\right)}-q\ln{\left(\frac{1+p}{p}\right)}\right]\\ &=2\pi\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\bigg{[}\frac{\ln{\left(\frac{1+q}{q}\right)}+\ln{\left(\frac{1+p}{p}\right)}}{\left(p+q\right)}+\frac{\ln{\left(\frac{1+q}{q}\right)}-\ln{\left(\frac{1+p}{p}\right)}}{\left(p-q\right)}\bigg{]}\\ &=2\pi\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\bigg{[}\frac{\ln{\left(\frac{p}{1+p}\right)}-\ln{\left(\frac{q}{1+q}\right)}}{\left(p-q\right)}-\frac{\ln{\left(\frac{p}{1+p}\right)}+\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\bigg{]}.\tag{4}\\ \end{align}$$
Set $A:=\frac{a}{1+a}\in\left(0,1\right)\land B:=\frac{b}{1+b}\in\left(0,1\right)$. Continuing from the double integral in the last line of $(4)$ above,
$$\begin{align} \frac{\mathcal{I}{\left(a,b,1\right)}}{2\pi} &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\bigg{[}\frac{\ln{\left(\frac{p}{1+p}\right)}-\ln{\left(\frac{q}{1+q}\right)}}{\left(p-q\right)}-\frac{\ln{\left(\frac{p}{1+p}\right)}+\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\bigg{]}\\ &=\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}-\ln{\left(\frac{q}{1+q}\right)}}{\left(p-q\right)}-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}+\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\\ &=\int_{0}^{\frac{a}{1+a}}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{\frac{b}{1+b}}\mathrm{d}y\,\frac{1}{\left(1-y\right)^{2}}\cdot\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(\frac{x}{1-x}-\frac{y}{1-y}\right)};~~~\small{\left[p=\frac{x}{1-x}\land q=\frac{y}{1-y}\right]}\\ &~~~~~-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}}{\left(p+q\right)}-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\\ &=\int_{0}^{A}\mathrm{d}x\int_{0}^{B}\mathrm{d}y\,\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-x\right)\left(x-y\right)\left(1-y\right)}\\ &~~~~~-\int_{0}^{a}\mathrm{d}p\int_{0}^{b}\mathrm{d}q\,\frac{\ln{\left(\frac{p}{1+p}\right)}}{\left(p+q\right)}-\int_{0}^{b}\mathrm{d}q\int_{0}^{a}\mathrm{d}p\,\frac{\ln{\left(\frac{q}{1+q}\right)}}{\left(p+q\right)}\\ &=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\frac{\left(1-x\right)}{\left(x-y\right)\left(1-y\right)}\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]\\ &~~~~~-\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{p}{1+p}\right)}\int_{0}^{b}\mathrm{d}q\,\frac{1}{\left(p+q\right)}-\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{q}{1+q}\right)}\int_{0}^{a}\mathrm{d}p\,\frac{1}{\left(p+q\right)}\\ &=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\left[\frac{1}{\left(x-y\right)}-\frac{1}{\left(1-y\right)}\right]\left[\ln{\left(x\right)}-\ln{\left(y\right)}\right]\\ &~~~~~-\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{p}{1+p}\right)}\ln{\left(\frac{p+b}{p}\right)}\\ &~~~~~-\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{q}{1+q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\left[\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(x-y\right)}-\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-y\right)}\right]\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}\\ &~~~~~+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}.\\ \end{align}$$
Given $0<x<1\land0<B<1$,
$$\begin{align} \int_{0}^{B}\mathrm{d}y\,\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(x-y\right)} &=\int_{0}^{\frac{B}{x}}\mathrm{d}t\,x\frac{\ln{\left(x\right)}-\ln{\left(xt\right)}}{\left(x-xt\right)};~~~\small{\left[y=xt\right]}\\ &=\int_{0}^{\frac{B}{x}}\mathrm{d}t\,\frac{(-1)\ln{\left(t\right)}}{\left(1-t\right)}\\ &=\int_{1-\frac{B}{x}}^{1}\mathrm{d}u\,\frac{(-1)\ln{\left(1-u\right)}}{u};~~~\small{\left[t=1-u\right]}\\ &=\operatorname{Li}_{2}{\left(1\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)},\\ \end{align}$$
and
$$\begin{align} \int_{0}^{B}\mathrm{d}y\,\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-y\right)} &=\int_{1-B}^{1}\mathrm{d}t\,\frac{\ln{\left(x\right)}-\ln{\left(1-t\right)}}{t};~~~\small{\left[y=1-t\right]}\\ &=\operatorname{Li}_{2}{\left(1\right)}-\operatorname{Li}_{2}{\left(1-B\right)}-\ln{\left(1-B\right)}\ln{\left(x\right)}.\\ \end{align}$$
Then,
$$\begin{align} \frac{\mathcal{I}{\left(a,b,1\right)}}{2\pi} &=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\int_{0}^{B}\mathrm{d}y\,\left[\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(x-y\right)}-\frac{\ln{\left(x\right)}-\ln{\left(y\right)}}{\left(1-y\right)}\right]\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\int_{0}^{A}\mathrm{d}x\,\frac{1}{\left(1-x\right)^{2}}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)}+\ln{\left(1-B\right)}\ln{\left(x\right)}\right]\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\left(\frac{A}{1-A}\right)\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\ &~~~~~-\lim_{x\to0}\left(\frac{x}{1-x}\right)\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)}+\ln{\left(1-B\right)}\ln{\left(x\right)}\right]\\ &~~~~~-\int_{0}^{A}\mathrm{d}x\,\left(\frac{x}{1-x}\right)\frac{d}{dx}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{x}\right)}+\ln{\left(1-B\right)}\ln{\left(x\right)}\right];~~~\small{I.B.P.s}\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\left(\frac{A}{1-A}\right)\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\ &~~~~~+\int_{0}^{A}\mathrm{d}x\,\left[\frac{B\ln{\left(\frac{B}{x}\right)}}{\left(1-x\right)\left(B-x\right)}-\frac{\ln{\left(1-B\right)}}{\left(1-x\right)}\right]\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\ &~~~~~-\frac{B}{1-B}\int_{0}^{A}\mathrm{d}x\,\frac{\left(1-B\right)\ln{\left(\frac{x}{B}\right)}}{\left(1-x\right)\left(B-x\right)}-\int_{0}^{A}\mathrm{d}x\,\frac{\ln{\left(1-B\right)}}{\left(1-x\right)}\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\ &~~~~~-\frac{B}{1-B}\int_{0}^{A}\mathrm{d}x\,\left[\frac{\ln{\left(\frac{x}{B}\right)}}{\left(B-x\right)}-\frac{\ln{\left(\frac{x}{B}\right)}}{\left(1-x\right)}\right]\\ &~~~~~+\ln{\left(1-A\right)}\ln{\left(1-B\right)}\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(1-B\right)}\ln{\left(A\right)}\right]\\ &~~~~~+\frac{B}{1-B}\left[-\int_{0}^{A}\mathrm{d}x\,\frac{\ln{\left(\frac{x}{B}\right)}}{\left(B-x\right)}-\int_{0}^{A}\mathrm{d}x\,\frac{\ln{\left(B\right)}-\ln{\left(x\right)}}{\left(1-x\right)}\right]\\ &~~~~~+\ln{\left(1-A\right)}\ln{\left(1-B\right)}\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\frac{A}{1-A}\left[\operatorname{Li}_{2}{\left(1-B\right)}-\operatorname{Li}_{2}{\left(1-\frac{B}{A}\right)}+\ln{\left(A\right)}\ln{\left(1-B\right)}\right]\\ &~~~~~+\frac{B}{1-B}\left[\operatorname{Li}_{2}{\left(1-A\right)}-\operatorname{Li}_{2}{\left(1-\frac{A}{B}\right)}+\ln{\left(B\right)}\ln{\left(1-A\right)}\right]\\ &~~~~~+\ln{\left(1-A\right)}\ln{\left(1-B\right)}\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{p+b}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)},\\ \end{align}$$
and then,
$$\begin{align} \frac{\mathcal{I}{\left(a,b,1\right)}}{2\pi} &=\ln{\left(1+a\right)}\ln{\left(1+b\right)}\\ &~~~~~+a\left[\operatorname{Li}_{2}{\left(\frac{1}{1+b}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{a(1+b)}\right)}-\ln{\left(\frac{a}{1+a}\right)}\ln{\left(1+b\right)}\right]\\ &~~~~~+b\left[\operatorname{Li}_{2}{\left(\frac{1}{1+a}\right)}-\operatorname{Li}_{2}{\left(\frac{b-a}{b(1+a)}\right)}-\ln{\left(\frac{b}{1+b}\right)}\ln{\left(1+a\right)}\right]\\ &~~~~~+\int_{0}^{a}\mathrm{d}p\,\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{b+p}{p}\right)}+\int_{0}^{b}\mathrm{d}q\,\ln{\left(\frac{1+q}{q}\right)}\ln{\left(\frac{a+q}{q}\right)}\\ &=\ln{\left(1+a\right)}\ln{\left(1+b\right)}\\ &~~~~~+a\left[\operatorname{Li}_{2}{\left(\frac{1}{1+b}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{a(1+b)}\right)}-\ln{\left(\frac{a}{1+a}\right)}\ln{\left(1+b\right)}\right]\\ &~~~~~+b\left[\operatorname{Li}_{2}{\left(\frac{1}{1+a}\right)}-\operatorname{Li}_{2}{\left(\frac{b-a}{b(1+a)}\right)}-\ln{\left(\frac{b}{1+b}\right)}\ln{\left(1+a\right)}\right]\\ &~~~~~+a\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+ax}{ax}\right)}\ln{\left(\frac{b+ax}{ax}\right)}\\ &~~~~~+b\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+bx}{bx}\right)}\ln{\left(\frac{a+bx}{bx}\right)}\\ &=\ln{\left(1+a\right)}\ln{\left(1+b\right)}\\ &~~~~~+a\left[\operatorname{Li}_{2}{\left(\frac{1}{1+b}\right)}-\operatorname{Li}_{2}{\left(\frac{a-b}{a(1+b)}\right)}-\ln{\left(\frac{a}{1+a}\right)}\ln{\left(1+b\right)}\right]\\ &~~~~~+b\left[\operatorname{Li}_{2}{\left(\frac{1}{1+a}\right)}-\operatorname{Li}_{2}{\left(\frac{b-a}{b(1+a)}\right)}-\ln{\left(\frac{b}{1+b}\right)}\ln{\left(1+a\right)}\right]\\ &~~~~~+a\,\mathcal{J}{\left(a,\frac{a}{b}\right)}\\ &~~~~~+b\,\mathcal{J}{\left(b,\frac{b}{a}\right)},\\ \end{align}$$
where in the last line above we've defined another function $\mathcal{J}:\mathbb{R}_{>0}\rightarrow\mathbb{R}$ via the integral
$$\mathcal{J}{\left(p,q\right)}:=\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+px}{px}\right)}\ln{\left(\frac{1+qx}{qx}\right)}.$$
For $p>0\land q>0$, we obtain
$$\begin{align} \mathcal{J}{\left(p,q\right)} &=\int_{0}^{1}\mathrm{d}x\,\ln{\left(\frac{1+px}{px}\right)}\ln{\left(\frac{1+qx}{qx}\right)}\\ &=\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{1+q}{q}\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1+qx}{qx}\right)}}{\left(1+px\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(\frac{1+px}{px}\right)}}{\left(1+qx\right)};~~~\small{I.B.P.s}\\ &=\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{1+q}{q}\right)}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+qx\right)}}{\left(1+px\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+px\right)}}{\left(1+qx\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{\left(1+px\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(x\right)}}{\left(1+qx\right)}\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(q\right)}}{\left(1+px\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(p\right)}}{\left(1+qx\right)}\\ &=\ln{\left(\frac{1+p}{p}\right)}\ln{\left(\frac{1+q}{q}\right)}\\ &~~~~~+\frac{\operatorname{Li}_{2}{\left(\frac{p-q}{p+1}\right)}-\operatorname{Li}_{2}{\left(\frac{p}{1+p}\right)}-\operatorname{Li}_{2}{\left(-q\right)}}{p}+\frac{\operatorname{Li}_{2}{\left(\frac{q-p}{q+1}\right)}-\operatorname{Li}_{2}{\left(\frac{q}{1+q}\right)}-\operatorname{Li}_{2}{\left(-p\right)}}{q}\\ &~~~~~-\frac{\operatorname{Li}_{2}{\left(-p\right)}}{p}-\frac{\operatorname{Li}_{2}{\left(-q\right)}}{q}\\ &~~~~~-\frac{\ln{\left(q\right)}\ln{\left(1+p\right)}}{p}-\frac{\ln{\left(p\right)}\ln{\left(1+q\right)}}{q}.\\ \end{align}$$
And with that, our general evaluation of $\mathcal{I}$ is in principle complete, though the final expression is too cumbersome to bother writing out explicitly.
Cheers!