Challenging Integral: $\int_0^\infty\frac{\ln(2+x)\operatorname{Li}_2(-x)}{x(2+x)}dx$

First we start with subbing $ x=\frac{y}{2+y}\Longrightarrow y=\frac{2x}{1-x}\Longrightarrow dy=\frac{2}{(1-x)^2}$

$$\mathcal{I}=\int_0^\infty\frac{\ln(2+y)\operatorname{Li}_2(-y)}{y(2+y)}dy=\frac12\int_0^1\frac{\ln\left(\frac2{1-x}\right)\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx$$

$$=\frac12\ln2\int_0^1\frac{\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx-\frac12\int_0^1\frac{\ln(1-x)\operatorname{Li}_2\left(\frac{2x}{x-1}\right)}{x}dx$$

$$=\frac12\ln2\cdot\mathcal{I}_1+\frac12\mathcal{I}_2$$

---

Evaluating the first integral $\mathcal{I}_1$:

By integration by parts we get

$$\mathcal{I}_1=\int_0^1\frac{\ln(x)\ln\left(\frac{1+x}{1-x}\right)}{x(1-x)}dx=\int_0^1\frac{\ln(x)\ln(1+x)}{x(1-x)}dx-\int_0^1\frac{\ln(x)\ln(1-x)}{x(1-x)}dx\\=\mathcal{A}-\mathcal{B}$$

$$\mathcal{A}=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\frac{x^{n-1}\ln(x)}{1-x}dx=-\sum_{n=1}^\infty \frac{(-1)^n}{n}[H_{n-1}^{(2)}-\zeta(2)]$$

$$=-\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{n}+\sum_{n-1}^\infty\frac{(-1)^n}{n^3}+\ln2\sum_{n=1}^\infty\frac{(-1)^n}{n}$$

$$=-\left(\frac12\ln2\zeta(2)-\zeta(3)\right)-\frac34\zeta(3)-\ln2\zeta(2)$$

$$=\frac14\zeta(3)-\frac32\ln2\zeta(2)$$

By symmetry, we have

$$\mathcal{B}=2\int_0^1\frac{\ln x\ln(1-x)}{x}dx=-\sum_{n=1}^\infty\frac1n\int_0^1x^{n-1}\ln xdx=2\sum_{n=1}^\infty\frac{1}{n^3}=2\zeta(3)$$

Collect $\mathcal{A}$ and $\mathcal{B}$ we get

$$\mathcal{I}_1=-\frac74\zeta(3)-\frac32\ln2\zeta(2)$$

---

Evaluating the second integral $\mathcal{I}_2$:

Use Landen's identity $-\operatorname{Li}_2(z)=\operatorname{Li}_2\left(\frac{z}{z-1}\right)+\ln^2(1-z)$

Set $z=\frac{2x}{x-1}$ we get $-\operatorname{Li}2\left(\frac{2x}{x-1}\right)=\operatorname{Li}2\left(\frac{2x}{x+1}\right)+\frac12\ln^2\left(\frac{1-x}{1+x}\right)$

Then we can write

$$\mathcal{I}_2=\int_0^1\frac{\ln(1-x)\operatorname{Li}_2\left(\frac{2x}{x+1}\right)}{x}dx+\frac12\int_0^1\frac{\ln(1-x)\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx\\=\mathcal{J}+\frac12\mathcal{K}$$

Apply integration by parts for $\mathcal{J}$ we have

$$\mathcal{J}=-\frac52\zeta(4)+\int_0^1\frac{\operatorname{Li}_2(x)\ln\left(\frac{1+x}{1-x}\right)}{x(1+x)}dx$$

$$=-\frac52\zeta(4)+\int_0^1\frac{\operatorname{Li}_2(x)\ln(1+x)}{x(1+x)}dx-\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{x(1+x)}dx\\=-\frac52\zeta(4)+\mathcal{J}_1-\mathcal{J}_2$$

$$\mathcal{J}_1=\int_0^1\frac{\operatorname{Li}_2(x)\ln(1+x)}{x(1+x)}dx=-\sum_{n=1}^\infty (-1)^nH_n\int_0^1 x^{n-1}\operatorname{Li}_2(x)dx$$

$$=-\sum_{n=1}^\infty (-1)^nH_n\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)=\sum_{n=1}^\infty \frac{(-1)^nH_n^2}{n^2}-\zeta(2)\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}$$

You can find here:

$$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^2}{n^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{41}{16}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$

Substitute this result along with $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n}=\frac12\ln^22-\frac12\zeta(2)$ we get

$$\boxed{\mathcal{J}_1=2\operatorname{Li}_4\left(\frac12\right)-\frac{21}{16}\zeta(4)+\frac74\ln2\zeta(3)-\ln^22\zeta(2)+\frac1{12}\ln^42}$$

For $\mathcal{J}_2$, break the denominator first

$$\mathcal{J}_2=\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{x}dx-\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{1+x}dx$$

$$=-\frac54\zeta(4)-\int_0^1\frac{\operatorname{Li}_2(x)\ln(1-x)}{1+x}dx$$

Now set $1-x\to x$ then use the reflection formula $\operatorname{Li}_2(1-x)=\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)$

$$\mathcal{J}_2=-\frac54\zeta(4)-\int_0^1\frac{\ln x}{2-x}\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)dx$$

write $\frac1{2-x}=\sum_{n=1}^\infty \frac1{2^n}x^{n-1}$ to get

$$\mathcal{J}_2=-\frac54\zeta(4)-\sum_{n-1}^\infty \frac1{2^n}\int_0^1\ln x\ x^{n-1}\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)dx$$

$$=-\frac54\zeta(4)-\sum_{n-1}^\infty \frac1{2^n}\frac{\partial}{\partial n}\int_0^1x^{n-1}\left(\zeta(2)-\ln x\ln(1-x)-\operatorname{Li}_2(x)\right)dx, \quad IBP$$

$$=-\frac54\zeta(4)-\sum_{n-1}^\infty \frac1{2^n}\frac{\partial}{\partial n}\left(\frac{\zeta(2)}{n}-\frac{H_n^{(2)}}{n}\right)$$

$$=-\frac54\zeta(4)-\sum_{n=1}^\infty\frac{1}{2^n}\left(\frac{H_n^{(2)}}{n^2}+\frac{2H_n^{(3)}}{n}-\frac{\zeta(2)}{n^2}-\frac{2\zeta(3)}{n}\right)$$

The first sum is already proved here

$$\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42$$

For the second sum, we can just set $x=\frac12$ in the generating function to get

$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n2^n}=\operatorname{Li_4}\left(\frac12\right)-\frac{5}{16}\zeta(4)+\frac78\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac{1}{24}\ln^42$$

Collect these two results along with $\operatorname{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^22$ we get

$$\boxed{\mathcal{J}_2=-3\operatorname{Li_4}\left(\frac12\right)+\frac{9}{16}\zeta(4)+\frac14\ln^22\zeta(2)-\frac{1}{8}\ln^42}$$

Now collect the result of $\mathcal{J}_1$ and $\mathcal{J}_2$ we get

$$\boxed{\mathcal{J}=5\operatorname{Li_4}\left(\frac12\right)-\frac{35}{8}\zeta(4)+\frac74\ln2\zeta(3)-\frac54\ln^22\zeta(2)+\frac{5}{24}\ln^42}$$

---

The integral $\mathcal{K}$ is already calculated here

$$\boxed{\mathcal{K}=-4\operatorname{Li}_4\left(\frac12\right)-\frac{41}{8}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac1{6}\ln^42}$$

Thus

$$\mathcal{I}_2=\mathcal{J}+\frac12\mathcal{K}=3\operatorname{Li}_4\left(\frac12\right)-\frac{111}{16}\zeta(4)-\frac34\ln^22\zeta(2)+\frac1{8}\ln^42$$

and finally

$$\mathcal{I}=\frac12\ln2\cdot\mathcal{I}_1+\frac12\mathcal{I}_2=\frac32\operatorname{Li}_4\left(\frac12\right)-\frac{111}{32}\zeta(4)-\frac78\ln2\zeta(3)-\frac98\ln^22\zeta(2)+\frac1{16}\ln^42$$