Compute $ \int_0^1\frac{\ln^2(1+x)}{1+x^2}\, dx$ How to prove 

$$I=\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx=4\Im\operatorname{Li}_3(1+i)-\frac{7\pi^3}{64}-\frac{3\pi}{16}\ln^22-2\ln2\ G$$

Where $ \operatorname{Li}_3(x)$ is the the trilogarithm function and $G$ is the Catalan constant.
Variant approaches are appreciated.
 A: I proved here in Eq $(1)$:
\begin{align}
\int_0^1\frac{\ln^2(1+x)}{1+x^2}\ dx&=\Im\operatorname{Li}_3(1+i)-\frac{\pi^3}{32}+\overset{\text{IBP}}{\int_0^1\frac{\ln x\ln(1+x)}{1+x^2}\ dx}\\
&=\Im\operatorname{Li}_3(1+i)-\frac{\pi^3}{32}-\int_0^1\frac{\ln x\tan^{-1}x}{1+x}\ dx-\int_0^1\frac{\ln(1+x)\tan^{-1}x}{x}\ dx\\
\end{align}
FDP beautifully calculated here the first integral: $\displaystyle\int_0^1\frac{\ln x\tan^{-1}x}{1+x}\ dx=\frac12G\ln2-\frac{\pi^3}{64}$
and I managed here to find the second integral: $$\displaystyle\int_0^1 \frac{\ln(1+x)\tan^{-1}x}{x}\ dx=\frac{3\pi^3}{32}+\frac{3\pi}{16}\ln^22+\frac32G\ln2-3\text{Im}\operatorname{Li}_3(1+i)$$
Plugging the results of the two integrals, we get the closed form of the original integral. 
A: A hint.
With 

$$\frac{1}{n!}\int\frac{(\ln(x+z))^n}{x-a}\,dx =\sum\limits_{k=0}^n\frac{(\ln(x+z))^k}{k!}(-1)^{n-k+1}\text{Li}_{n-k+1}\left(\frac{x+z}{a+z}\right) + C$$

follows the calculation for:

$$\int\frac{(\ln(1+x))^2}{1+x^2}\,dx = \Im\int\frac{(\ln(x+z))^n}{x-a}\,dx|_{(n,z,a)=(2,1,i)}$$

For the definite integral it's left to simplify $~\displaystyle \Im\,\text{Li}_1(\frac{2}{1+i})~$, $~\displaystyle \Im\,\text{Li}_2(\frac{2}{1+i})~$, $~\displaystyle \Im\,\text{Li}_3(\frac{2}{1+i})~$ 
and $~\displaystyle \Im\,\text{Li}_3(\frac{1}{1+i})~$ .
