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?
 A: Here is a magical solution where no advanced results are used.
$$\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx$$
$$=-\underbrace{\int _0^{\infty }\frac{\ln \left(x\right)\ln \left(1+x\right)}{x\left(1+x^2\right)}\:dx}_{J}+\underbrace{\int _0^1\frac{\ln \left(x\right)\ln \left(1+x\right)}{x\left(1+x^2\right)}\:dx}_{I}+\int _0^1\frac{x\ln ^2\left(x\right)}{1+x^2}\:dx.$$

Let's try differentiation under the integral sign for $I$.
$$I=\int _0^1\frac{\ln \left(x\right)\ln \left(1+x\right)}{x\left(1+x^2\right)}\:dx,\:I\left(a\right)=\int _0^1\frac{\ln \left(x\right)\ln \left(1+ax\right)}{x\left(1+x^2\right)}\:dx$$
$$I'\left(a\right)=-G\frac{1}{1+a^2}+\frac{a\operatorname{Li}_2\left(-a\right)}{1+a^2}+\frac{1}{8}\zeta \left(2\right)\frac{a}{1+a^2}$$
$$I=-G\int _0^1\frac{1}{1+a^2}\:da+\underbrace{\int _0^1\frac{a\operatorname{Li}_2\left(-a\right)}{1+a^2}\:da}_{\mathcal{A}}+\frac{1}{8}\zeta \left(2\right)\int _0^1\frac{a}{1+a^2}\:da$$
$$=-\frac{\pi }{4}G+\int _0^1\frac{\operatorname{Li}_2\left(-a\right)}{a}\:da-\frac{\pi }{4}\int _0^1\frac{\ln \left(t\right)}{1+t^2}\:dt+\frac{1}{2}\ln \left(2\right)\int _0^1\frac{t\ln \left(t\right)}{1+t^2}\:dt$$
$$-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t\right)}{1+t^2}\:dt+\frac{1}{16}\ln \left(2\right)\zeta \left(2\right)$$
$$=-\frac{3}{4}\zeta \left(3\right)-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t\right)}{1+t^2}\:dt.$$
Note that for $\mathcal{A}$ I made use of the well-known representation $\int _0^1\frac{a\ln \left(t\right)}{1+at}\:dt=\operatorname{Li}_2\left(-a\right)$ and interchanged the order of integration.

With the same technique and parameter we can prove that $J$ is:
$$J=\int _0^{\infty }\frac{\ln \left(x\right)\ln \left(1+x\right)}{x\left(1+x^2\right)}\:dx=-\frac{3}{4}\zeta \left(2\right)\int _0^1\frac{a}{1+a^2}\:da-\frac{1}{2}\int _0^1\frac{a\ln ^2\left(a\right)}{1+a^2}\:da$$
$$=-\frac{3}{32}\zeta \left(3\right)-\frac{3}{8}\ln \left(2\right)\zeta \left(2\right).$$

Plugging these results into the original expression we obtain:
$$\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx=\frac{3}{32}\zeta \left(3\right)+\frac{3}{8}\ln \left(2\right)\zeta \left(2\right)-\frac{3}{4}\zeta \left(3\right)-\int _0^1\frac{t\ln \left(t\right)\ln \left(1+t\right)}{1+t^2}\:dt$$
$$+\int _0^1\frac{x\ln ^2\left(x\right)}{1+x^2}\:dx.$$
Therefore:
$$\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx=-\frac{15}{64}\zeta \left(3\right)+\frac{3}{16}\ln \left(2\right)\zeta \left(2\right).$$
A: 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
$$ \int_0^1 \frac{x\ln x\ln(1+x)}{1+x^2}dx= -\frac{15}{64}\zeta(3) + \frac{\pi^2}{32}\ln2$$
A: IBP:
$$\frac3{16}\sum_{K\ge1}{\frac1{k^2}\sum_{u\ge1}\frac {(-1)^{u+1}}u{}-\frac5{64}}\sum_{k\ge1}\sum_{u\ge1}\frac{(u-1)!(k-1)!}{(k+1)!}H_{k+u} = \frac3{16}\zeta(2)\ln(2)-\frac{15}{64}\zeta(3)$$
