Logarithmic integral $\int_0^{1}\frac{\ln x\ln (1-x^{2})}{1+x^{2}}\mathrm dx$ 
Evaluate the integral in a closed-form :
$$I=\int_0^{1}\frac{\ln x\ln (1-x^{2})}{1+x^{2}}\mathrm dx$$

My attempt :
After put $x=\tan y$ we obtain:
$$I=\int_0^{\frac{π}{4}}\ln (\tan x)\ln (1-\tan^{2} x)dx$$
$$1-\tan^{2} x=(1+\tan x)(1-\tan x)$$
$$\ln (1+\tan x)=\ln(\sin x+\cos x)-\ln(\cos x)=\ln (\cos (\frac{π}{4}-x))-\ln (\cos x)$$
Also:
$$\ln (1-\tan x)=\ln(\cos x-\sin x)-\ln(\cos x)=\ln (\cos (\frac{π}{4}+x))-\ln (\cos x)$$
So:
$$\ln (\tan x)\ln (1-\tan^{2} x)$$
$$=(\ln (\sin x)-\ln(\cos x)(\ln(\cos (\frac{π}{4}-x))-\ln (\cos x))(\ln(\cos (\frac{π}{4}+x))-\ln(\cos x))$$
Now I have many integrals. How can I evaluate them? Let me know if anyone has other ideas.
 A: As pointed out by Zacky the solution may be found here given by pisco. Adopting the notation they use there we see that
\begin{align*}
I&=\int_0^1\frac{\log(x)\log(1-x^2)}{1+x^2}\mathrm dx\\
&=\int_0^1\frac{\log(x)\log(1-x)}{1+x^2}\mathrm dx+\int_0^1\frac{\log(x)\log(1+x)}{1+x^2}\mathrm dx\\
&=I_{ab}+I_{ac}
\end{align*}
In the end of their post they get that
\begin{align*}
I_{ab}&=\Im\left(\operatorname{Li}_3\left(\frac{1+i}2\right)\right)-\frac{\pi^3}{128}-\frac\pi{32}\log^2(2)\\
I_{ac}&=-3\Im\left(\operatorname{Li}_3\left(\frac{1+i}2\right)\right)+\frac{11\pi^3}{128}+\frac{3\pi}{32}\log^2(2)-2G\log(2)
\end{align*}
Thus,

$$\small\therefore~I~=~\int_0^1\frac{\log(x)\log(1-x^2)}{1+x^2}\mathrm dx~=~-2\Im\left(\operatorname{Li}_3\left(\frac{1+i}2\right)\right)+\frac{5\pi^3}{64}+\frac\pi{16}\log^2(2)-2G\log(2)$$

A: \begin{align}
\int_0^1\frac{\ln x\ln(1-x^2)}{1+x^2}\ dx=\int_0^1\frac{\ln x\ln(1-x^4)}{1+x^2}\ dx-\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx
\end{align}
The first integral was nicely done here by Bennett Gardiner.
$$\int_0^1\frac{\ln x\ln(1-x^4)}{1+x^2}\ dx=\frac{\pi^3}{16}-3G\ln2$$
and I managed to calculate the second integral here.
$$\int_0^1\frac{\ln x\ln(1+x^2)}{1+x^2}\ dx=\frac{3\pi^3}{32}+\frac{\pi}8\ln^22-\ln2~G-2\text{Im}\operatorname{Li_3}(1+i)$$
Thus $$\int_0^1\frac{\ln x\ln(1-x^2)}{1+x^2}\ dx=2\text{Im}\operatorname{Li_3}(1+i)-\frac{\pi^3}{32}-\frac{\pi}8\ln^22-2\ln2~G$$
A: Presented below is a self-contained evaluation. Let $J=\int_0^1 \frac{\ln^2(1+x)}{1+x^2}dx$
\begin{align}
&\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx
\overset{x\to\frac{1-x}{1+x}}= J
-2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx-2G\ln2+\frac{\pi^3}{16}\\
&\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}dx
\overset{(0,1)+\overset{x\to 1/x}{(1,\infty)}} = 2J -2 \int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx+\frac{\pi^3}{16}
\end{align}
Eliminate $J$ to get
$$\int_0^1 \frac{\ln x\ln(1+x)}{1+x^2}dx
=\frac12\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}dx-K-2G\ln2+\frac{\pi^3}{32}
\tag1$$
where $K=\int_0^1 \frac{\ln^2(1-x)}{1+x^2}dx =2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right)
$. Also
$$\int_0^1 \frac{\ln x\ln(1-x)}{1+x^2}dx
=\frac12 K-\frac12\int_0^1\frac{\ln\frac x{1-x}}{1+x^2}dx +\frac12\int_0^1\frac{\ln^2x}{1+x^2}dx\tag2
$$
Then, (1) + (2)
$$
\int_0^1 \frac{\ln x\ln(1-x^2)}{1+x^2}dx 
=-2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right)-2G\ln2+\frac{\pi^3}{16} +\frac12 A \tag3$$
where
\begin{align}
A =& \int_0^1 \underset{t=1-x}{\frac{\ln^2(1-x)}{1+x^2}}dx- \int_0^1\underset{t=\frac x{1-x}}{\frac{\ln\frac x{1-x}}{1+x^2}}dx +\underset{t=1+x}{
\int_0^\infty \frac{\ln^2(1+x)}{1+x^2}}dx \\
 =& \int_0^\infty {\frac{4t \ln^2 t}{4+t^4}}dt
\overset{t^2=2u}= \frac14 \int_0^\infty \frac{\ln^2(2u)}{1+u^2}du
 = \frac{\pi}{8}\ln^22+\frac{\pi^3}{32}
\end{align}
Plug into (3) to obtain
$$
\int_0^1 \frac{\ln x\ln(1-x^2)}{1+x^2}dx  =-2\text{Im}\>\text{Li}_3\left(\frac{1+i}2\right) -2G\ln2
+\frac{\pi}{16}\ln^22+\frac{5\pi^3}{64}$$
