Finding $\int_{1}^{\infty} \frac{1}{1+x^2} \frac{\operatorname{Li}_2\left ( \frac{1-x}{2} \right ) }{\pi^2+\ln^2\left(\frac{x-1}{2}\right)}\text{d}x$ Prove the integral
$$\int_{1}^{\infty} \frac{1}{1+x^2}
\frac{\operatorname{Li}_2\left ( \frac{1-x}{2}  \right ) }{
\pi^2+\ln^2\left ( \frac{x-1}{2}  \right ) }\text{d}x
=\frac{96C\ln2+7\pi^3}{12(\pi^2+4\ln^2(2))} 
-\frac{\pi}{24}(3+4\pi)$$
where $\operatorname{Li}_2$ is dilogarithm and $C$ is Catalan's constant.
(I checked it high precision. So I believe that it's absolutely true.) 
How we prove that? I tried to use functional equations of $\operatorname{Li}_2$ but I get nothing useful. Any suggestion will be appreciated.
 A: $$I=2\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\pi ^2+\ln ^2x}}\mathrm{d}x$$
Consider
$$f\left( z \right) =\frac{\mathrm{Li}_2\left( z \right)}{1+\left( 2z-1 \right) ^2}\frac{1}{\ln z}$$
Use a key-shaped contour with $0$ and $1$ as the keyholes:
contour
When $x<0$, we have
$$\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln x+\pi \mathrm{i}}}\mathrm{d}x-\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln x-\pi \mathrm{i}}}\mathrm{d}x\\=-2\pi \mathrm{i}\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln ^2x+\pi ^2}}\mathrm{d}x$$
When $x>1$, we have
$$\int_1^{\infty}{\frac{1}{1+\left( 2x-1 \right) ^2}\frac{1}{\ln x}\left( \frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( t-1 \right)}{t}}\mathrm{d}t+\pi \mathrm{i}\ln x \right)}\mathrm{d}x
\\
-\int_1^{\infty}{\frac{1}{1+\left( 2x-1 \right) ^2}\frac{1}{\ln x}\left( \frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( t-1 \right)}{t}}\mathrm{d}t-\pi \mathrm{i}\ln x \right)}\mathrm{d}x
\\
=2\pi \mathrm{i}\int_1^{\infty}{\frac{1}{1+\left( 2x-1 \right) ^2}}\mathrm{d}x=\frac{\pi ^2\mathrm{i}}{4}$$
This is because
$$\mathrm{Li}_2\left( x \right) =-\int_0^x{\frac{\ln \left( 1-t \right)}{t}}\mathrm{d}t=\frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( 1-t \right)}{t}}\mathrm{d}t
\\
=\frac{\pi ^2}{6}-\int_1^x{\frac{\ln \left( t-1 \right)}{t}}\mathrm{d}t\pm \pi \mathrm{i}\ln x\,\,\left( x>1 \right) $$
Then, use the residue theorem, we get
$$-2\pi \mathrm{i}\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\ln ^2x+\pi ^2}}\mathrm{d}x+\frac{\pi ^2\mathrm{i}}{4}=2\pi \mathrm{i}\left[ \mathrm{Res}\left[ f\left( z \right) ,1 \right] +\mathrm{Res}\left[ f\left( z \right) ,\frac{1\pm \mathrm{i}}{2} \right] \right] $$
And
$$\begin{align*}&\mathrm{Res}\left[ f\left( z \right) ,1 \right] =\frac{\pi ^2}{12}
\\
&\mathrm{Res}\left[ f\left( z \right) ,\frac{1-\mathrm{i}}{2} \right] =\frac{\mathrm{i}}{4}\frac{\mathrm{Li}_2\left( \frac{1-\mathrm{i}}{2} \right)}{\ln \left( \frac{1-\mathrm{i}}{2} \right)}=\frac{\mathrm{i}}{4}\frac{\frac{5\pi ^2}{96}-\frac{\ln ^22}{8}+\mathrm{i}\left( \frac{\pi \ln 2}{8}-C \right)}{-\frac{\ln 2}{2}-\frac{\pi \mathrm{i}}{4}}
\\
&\mathrm{Res}\left[ f\left( z \right) ,\frac{1+\mathrm{i}}{2} \right] =-\frac{\mathrm{i}}{4}\frac{\mathrm{Li}_2\left( \frac{1+\mathrm{i}}{2} \right)}{\ln \left( \frac{1+\mathrm{i}}{2} \right)}=-\frac{\mathrm{i}}{4}\frac{\frac{5\pi ^2}{96}-\frac{\ln ^22}{8}-\mathrm{i}\left( \frac{\pi \ln 2}{8}-C \right)}{-\frac{\ln 2}{2}+\frac{\pi \mathrm{i}}{4}}\end{align*}$$
So
$$\int_0^{\infty}{\frac{\mathrm{Li}_2\left( -x \right)}{1+\left( 2x+1 \right) ^2}\frac{1}{\pi ^2+\ln ^2x}}\mathrm{d}x=\frac{7\pi ^3}{24\left( \pi ^2+4\ln ^22 \right)}+\frac{4C\ln 2}{\pi ^2+4\ln ^22}-\frac{\pi ^2}{12}-\frac{\pi}{16}$$
Finally
$$\boxed{I=\frac{7\pi ^3}{12\left( \pi ^2+4\ln ^22 \right)}+\frac{8C\ln 2}{\pi ^2+4\ln ^22}-\frac{\pi ^2}{6}-\frac{\pi}{8}}$$
