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

I tried to integrate by parts, but no way so far, help me, thanks.
 A: 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:
$$I=\begin{align}2\Im\left(\text{Li}_3\left(\frac{1+i}{2}\right)\right)+\text{G}\ln 2-\frac{3}{64}\pi^3-\frac{1}{16}\pi\ln^2 2\end{align}$$
A: 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 $\quad\boxed{\displaystyle I=\frac{\pi^3}{16}+\frac{\pi}{8}\ln^22+G\ln2+2\Im\operatorname{Li}_3(1-i)}$
A: different approach to evaluate $\displaystyle\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx$ :
from here , we have $\displaystyle\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1-x)}{x}\ dx=\frac{\pi^3}{16}\tag{1}$
and from here , we have $\displaystyle \ 3\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx-2\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=0\tag{2}$
by combining $(1)$ and $(2)$, we obtain that $\displaystyle\int_0^1\frac{\arctan x\ln(1+x)}{x}\ dx=3\int_0^1\frac{\arctan x\ln(1-x)}{x}\ dx+\frac{3\pi^3}{32}\tag{3}$
we have
\begin{align}
\int_0^1 \frac{\arctan x\ln(1-x)}{x}\ dx&=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1 x^{2n}\ln(1-x)\ dx\\
&=-\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}=-\text{Im}\sum_{n=1}^\infty\frac{i^nH_n}{n^2}\\
\end{align}
and 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 $\ \displaystyle\int_0^1 \frac{\arctan x\ln(1-x)}{x}\ dx=\frac{\pi}{16}\ln^22+\frac12G\ln2+\text{Im}\operatorname{Li}_3(1-i)\tag{4}$
plugging $(4)$ in $(3)$, we get $$\int_0^1 \frac{\arctan x\ln(1+x)}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2+3\text{Im}\operatorname{Li}_3(1-i)$$
