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?

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:
$$\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx=-\frac{15}{64}\zeta \left(3\right)+\frac{3}{16}\ln \left(2\right)\zeta \left(2\right).$$
