How to evaluate $\int_{0}^{1}\frac{\arctan x}{x} \log{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\mathrm dx$

From here we have that $$\frac12 \int_0^1 \frac{\arctan x \ln(1+x^2)}{x} dx =\frac13 \int_0^1 \frac{\arctan x \ln(1+x)}{x}dx$$ $$\Rightarrow I=\int_{0}^{1}\frac{\arctan x}{x} \ln{\left(\frac{1+ x}{\sqrt{1+x^2}}\right)} dx=\frac23 \int_{0}^{1}\frac{\arctan x \ln(1+x)}{x} dx$$ I have encountered this integral too last year and asked it on AoPS, you can take a look at Knas solution from there, giving: $$I=\begin{align}2\Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\text{G}\ln 2-\frac{3}{64}\pi^3-\frac{1}{16}\pi\ln^2 2\end{align}$$

From here , we have $\ \displaystyle \ 3\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=0$

or $\ I=\displaystyle\int_{0}^{1}\frac{\arctan x}{x} \ln{\left(\frac{ 1+ x}{\sqrt{1+x^2}}\right)}\ dx=\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx$

using $\ \displaystyle\arctan x\ln(1+x^2)=-2\sum_{n=0}^{\infty}\frac{(-1)^n H_{2n}} {2n+1}x^{2n+1}$ ( proved here) , we get \begin{align} I&=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1x^{2n}\ dx\\ &=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}\\ &=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}\\ &=-2\Im\sum_{n=1}^\infty\frac{i^nH_n}{n^2}+\frac{\pi^3}{16} \end{align} using the generating function with $x=i$ $$\sum_{n=1}^\infty\frac{x^nH_n}{n^2}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$ we get $\qquad\displaystyle\Im\sum_{n=1}^\infty\frac{i^nHn}{n^2}=-\frac{\pi}{16}\ln^22-\frac12G\ln2-\Im\operatorname{Li}_3(1-i)$

Plugging this result, we get $\quad\boxed{\displaystyle I=\frac{\pi^3}{16}+\frac{\pi}{8}\ln^22+G\ln2+2\Im\operatorname{Li}_3(1-i)}$