Logarithmic integral $ \int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}\ \mathrm{d}x $ I found this integral weeks ago.
$$ \int_0^1 \dfrac{x\ln(x)\ln(1+x)}{1+x^2}\ \mathrm{d}x $$
I tried to solve this integral using various series representation and ended up with a complicated double series which I have asked here. How can I solve this integral?

Let $K= \int_0^1 \frac{\ln x\ln(1-x)}{1+x} dx$ and note
\begin{align}
I&= \int_0^1 \dfrac{x\ln x\ln(1-x^2)}{1+x^2} dx
\overset{x^2\to x}
=\frac14K \\
J &= \int_0^1 \dfrac{x\ln x\ln\frac{1-x}{1+x}}{1+x^2} dx
\overset{x\to \frac{1-x}{1+x}}=\int_0^1 \dfrac{\ln x\ln\frac{1-x}{1+x}}{1+x} dx -J\\
&= \frac12K-\frac12 \int_0^1 \dfrac{\ln x\ln(1+x)}{1+x} \overset{ibp}{dx }=  \frac12K+\frac14 \int_0^1 \underset {= \frac14\zeta(3)}{\frac{\ln^2(1+x)}x }dx
\end{align}
Then
\begin{align}
&\int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}dx=\frac12 (I-J)
=-\frac1{32}\zeta(3)-\frac18K \tag1\\
\overset{ibp}=& -\frac1{32}\zeta(3)-\frac18\int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx+\frac18\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx
\end{align}
where
\begin{align}\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx
 =&\frac14\int_0^1 \frac{\ln^2(1-\overset{\to x}{x^2})}xdx -\frac14\int_0^1 \frac{\ln^2\frac{\overset{\to x}{1-x}}{1+x}}xdx\\
 =&\frac18\int_0^1 \frac{\ln^2(1-x)}xdx -\frac12\int_0^1 \frac{\ln^2 x}{1-x^2}dx\\
=&\frac18\cdot 2\zeta(3) -\frac12\cdot \frac74\zeta(3)
=-\frac58\zeta(3)
\end{align}
\begin{align}\int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx
 =&\int_0^1 \ln(1+x)\>d\left( \int_0^x \frac{\ln t}{1-t}dt \right)_{t=xy}\\
 =& \ln2\int_0^1 \frac{\ln t}{1-t}dt
-\int_0^1\int_0^1 \frac{x\ln x+\overset{y\leftrightarrow x}{x \ln y}}{(1+x)(1-xy)}
dxdy \\
=& -\frac{\pi^2}{6}\ln2 
-\int_0^1\int_0^1 \frac{\ln x}{1-xy} -\frac{\ln x}{(1+x)(1+y)} dydx \\
= & -\frac{\pi^2}{6}\ln2 
+\frac12 \int_0^1\frac{\ln^2x}{1-x} dx +\ln2\int_0^1 \frac{\ln x}{1+x}dx\\
= & -\frac{\pi^2}{6}\ln2 +\frac12\cdot 2\zeta(3)-\frac{\pi^2}{12}\ln2
= -\frac{\pi^2}4 \ln2 +\zeta(3)
\end{align}
Plug into (1) to obtain
$$ \int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}dx= -\frac{15}{64}\zeta(3) + \frac{\pi^2}{32}\ln2$$
