Evaluating $\int_0^1\frac{\arctan x\ln\left(\frac{2x^2}{1+x^2}\right)}{1-x}dx$ Here is a nice problem proposed by Cornel Valean

$$
I=\int_0^1\frac{\arctan\left(x\right)}{1-x}\,
\ln\left(\frac{2x^2}{1+x^2}\right)\,\mathrm{d}x =
-\frac{\pi}{16}\ln^{2}\left(2\right) -
\frac{11}{192}\,\pi^{3} +
2\Im\left\{%
\text{Li}_{3}\left(\frac{1 + \mathrm{i}}{2}\right)\right\}
$$

My Trial: By subbing $x=\frac{1-t}{1+t}$ we have
$$I=\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln\left(\frac{(1-x)^2}{1+x^2}\right)}{x(1+x)}dx$$
$$=2\underbrace{\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln(1-x)}{x(1+x)}dx}_{x\to (1-x)/(1+x)}-\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln(1+x^2)}{x(1+x)}dx$$
$$=2\int_0^1\frac{\arctan x\ln(\frac{2x}{1+x})}{1-x}dx-\int_0^1\frac{\left(\frac{\pi}{4}-\arctan x\right)\ln(1+x^2)}{x(1+x)}dx$$
and got stuck here. Any idea? thanks.
 A: Update:  the problem and solution will be part of a new paper soon.

A solution by Cornel Ioan Valean
Let's denote the main integral by $\mathcal{I}$, and then we have
$$\mathcal{I=}\int_0^1\frac{(\pi/4-\arctan((1-x)/(1+x)))\log\left(\frac{2x^2}{1+x^2}\right)}{1-x}\textrm{d}x$$
$$=\underbrace{\frac{\pi}{4}\int_0^1\frac{\log\left(\frac{2x^2}{1+x^2}\right)}{1-x}\textrm{d}x}_{\displaystyle J}- \underbrace{\int_0^1\frac{\arctan((1-x)/(1+x))\log\left(\frac{2x^2}{1+x^2}\right)}{1-x}\textrm{d}x}_{\displaystyle 
K}. \tag1$$
The integral $J$ easily reduces to known integrals. If we integrate by parts, we get

$$J=\frac{\pi}{2}\underbrace{\int_0^1\frac{\log (1-x)}{x}\textrm{d}x}_{\displaystyle -\pi^2/6}-\frac{\pi}{2}\underbrace{\int_0^1\frac{x \log (1-x)}{1+x^2}\textrm{d}x}_{\displaystyle 1/8 (\log^2(2)-5\pi^2/12 )}=-\frac{\pi}{16}\log^2(2)-\frac{11}{192}\pi^3,\tag2$$

where the last integral also appears in the book, (Almost) Impossible Integrals, Sums, and Series, page $8$.
For the integral $K$, a bit of magic will be necessary. The first key observation is that
$$K=\Im \biggr\{\int_0^1\frac{\log^2(x (1 + x)/(1 + x^2) + i x (1 - x)/(1 + x^2))}{1-x}\textrm{d}x\biggr\}.$$
Now, we may consider the generalization
$$G(a)=\int_0^1\frac{\displaystyle\log^2\left(\frac{ (1+a) x}{1 + a x}\right)}{1- x}\textrm{d}x,$$
and make the variable change $\displaystyle x\mapsto \frac{1-x}{1+a x}$ that leads to
$$G(a)=\int_0^1 \frac{\log^2(1-x)}{x}\textrm{d}x-a\int_0^1\frac{\displaystyle\log^2(1-x)}{1+ a x}\textrm{d}x,$$
and letting the variable change $x\mapsto 1-x$ in both integrals, we finally get
$$G(a)=\int_0^1 \frac{\log^2(x)}{1-x}\textrm{d}x-\frac{a}{1+a}\int_0^1\frac{\displaystyle\log^2(x)}{1 -a/(1+a) x}\textrm{d}x=2 \zeta(3)-2\operatorname{Li}_3\left(\frac{a}{1+a}\right),$$
where in the calculations we also needed the integral, $\displaystyle \int_0^1 \frac{a \log^2(x)}{1-a x}\textrm{d}x=2\operatorname{Li}_3(a)$, that appears in a generalized form in the same book, (Almost) Impossible Integrals, Sums, and Series, page $4$.
A first note: The variable change $\displaystyle x\mapsto \frac{x}{1+a-ax}$ would work more directly, and no need for a second variable change.
Then, based on the previous result we make the second key observation,
$$K=\Im \{G(i)\}.$$
Thus,

$$\small K=\Im \biggr \{\int_0^1\frac{\log^2(x (1 + x)/(1 + x^2) + i x (1 - x)/(1 + x^2))}{1-x}\textrm{d}x \biggr \}=2 \Im \biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}. \tag3 $$

At last, combining $(1)$, $(2)$, and $(3)$, we conclude that

$$\mathcal{I}=-\frac{\pi}{16}\log^2(2)-\frac{11}{192}\pi^3+2 \Im \biggr\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\biggr\}.$$

End of story
A second note: no software needed for calculating such integrals, or far more advanced ones alike.
Another nice example of an integral calculated by similar means
$$\int_0^1 \frac{1}{x(1+x)}\left(12 \log \left(\frac{(1-x)^2}{1+x^2}\right) \arctan^2(x)-\log ^3\left(\frac{(1-x)^2}{1+x^2}\right)\right) \textrm{d}x$$
$$=\frac{2043 }{64}\zeta (4)+\frac{15}{8} \log ^2(2)\zeta (2)-\frac{1}{2} \log ^4(2)-15 \operatorname{Li}_4\left(\frac{1}{2}\right).$$
