Calculating $\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx$ 
How to prove in a simpe way that
  $$\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx=\frac{13}{8}\ln2\zeta(2)-\frac{33}{32}\zeta(3)\ ?$$
  where $\operatorname{Li}_2$ is the dilogarithm function.

If we integrate by parts, the integral boils down to
$$\int_0^1\left(\frac{1}{x}-\frac{2x}{1+x^2}\right)\left[2\ln(1-x)\ln(1+x)-\ln(1+x)\ln(1+x^2)\right]dx$$
and that seems long and complicated, any other ideas? 
 A: $$I:=\int_0^1\frac{1}{1+x}\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)dx\overset{\large x\to\frac{1-x}{1+x}}=\int_0^1 \frac{1}{1+x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx$$
$$J:=\int_0^1 \frac{1}{1-x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx$$
$$I=\frac12\left((I+J)+(I-J)\right)=\boxed{\int_0^1 \frac{1}{1+x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx=\frac{13}{8}\zeta(2)\ln 2-\frac{33}{32}\zeta(3)}$$
$$J=\frac12\left((I+J)-(I-J)\right)=\boxed{\int_0^1 \frac{1}{1-x}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx=\frac58\zeta(2)\ln 2+\frac{19}{32}\zeta(3)}$$

$$I-J=-2\int_0^1\frac{x}{1-x^2}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx\overset{\large \frac{1-x^2}{1+x^2}\to x}=\color{blue}{\int_0^1 \frac{\operatorname{Li}_2(x)}{1+x}dx}-\color{red}{\int_0^1 \frac{\operatorname{Li}_2(x)}{x}dx}$$
$$\overset{\color{blue}{IBP}}=\color{blue}{\zeta(2)\ln 2+\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx}-\color{red}{\sum_{n=1}^\infty \frac{1}{n^2}\int_0^1 x^{n-1}dx}=\boxed{\zeta(2)\ln 2-\frac{13}{8}\zeta(3)}$$
See here for the first integral.

$$I+J=2\int_0^1 \frac{1}{1-x^2}\operatorname{Li}_2\left(\frac{1-x^2}{1+x^2}\right)dx\overset{\large \frac{1-x^2}{1+x^2}\to x}=\int_0^1 \frac{\operatorname{Li}_2(x)}{x\sqrt{1-x^2}}dx$$
Now we're going to use the following result:
$$\int_0^1 \frac{\ln t}{t-\frac{1}{x}}dt=- \sum_{n=1}^\infty x^n\int_0^1 t^{n-1}\ln t\,dt=-\sum_{n=1}^\infty x^n \left(\frac{1}{n}\right)'=\sum_{n=1}^\infty \frac{x^n}{n^2}=\operatorname{Li}_2(x)$$
$$\Rightarrow I+J=- \int_0^1\int_0^1 \frac{\ln t}{(1-tx)\sqrt{1-x^2}}dx dt\overset{x\to \sin x}=- \int_0^1\ln t\int_0^\frac{\pi}{2} \frac{1}{1-t\sin x}dxdt$$
$$=-\int_0^1\ln t \  \frac{\frac{\pi}{2}+\arcsin t}{\sqrt{1-t^2}} dt\overset{t=\sin x}=-\int_0^\frac{\pi}{2}\left(\frac{\pi}{2}+x\right)\ln (\sin x) dx=\boxed{\frac94\zeta(2)\ln 2 -\frac7{16}\zeta(3)}$$
See here and here for the above integrals.
Also to prove a leftover integral from above we will consider: $$f(a)=\int_0^\pi \frac{dx}{1+\sin a\sin x}\overset{\tan\frac{x}{2}=y}=\int_0^\infty \frac{dy}{1+y^2+2\sin a y}$$
$$=\int_0^\infty \frac{dy}{(\sin a+y)^2+\cos^2  a}=\frac{1}{\cos a}\arctan\left(\frac{\sin a+y}{\cos a}\right)\bigg|_0^\infty=\frac{\frac{\pi}{2}-a}{\cos a}$$
$$\Rightarrow \int_0^\frac{\pi}{2}\frac{1}{1-t\sin x}dx=\frac12 f(-\arcsin t)=\frac{\frac{\pi}{2}+\arcsin t}{\sqrt{1-t^2}}$$
