Closed-form of $\int\limits_0^1\left(\frac{\left(x^2+1\right)\arcsin(x)}{\sqrt{1-x^2}}+2x\ln\left(x^2+1\right)\right)\frac{\ln x}{x^3+x}\,dx$ I've conjectured the following closed-form:
$$
I = \int\limits_0^1\left(\frac{\left(x^2+1\right)\arcsin(x)}{\sqrt{1-x^2}}+2x\ln\left(x^2+1\right)\right)\frac{\ln x}{x^3+x}\,dx = -2\,G\,\ln2,
$$
where $G$ is Catalan's constant.
Numerically
$$ I \approx -1.2697979381877088371491554851603117320986537271546606092465\dots$$
How to prove it?

Apply the obvious substitution $x\mapsto\sin{x}$ to the first integral
$$I=\int^\frac{\pi}{2}\limits_0\frac{x\ln(\sin{x})}{\sin{x}}\ {\rm d}x+\int\limits^1_0\frac{2\ln{x}\ln(1+x^2)}{1+x^2}\ {\rm d}x$$
The latter integral has been addressed here and is equivalent to
\begin{align}
\int\limits^1_0\frac{2\ln{x}\ln(1+x^2)}{1+x^2}\ {\rm d}x=4\Im{\rm Li}_3(1-i)-2\mathbf{G}\ln{2}+\frac{3\pi^3}{16}+\frac{\pi}{4}\ln^2{2}
\end{align}
As for the first integral,
\begin{align}
\int\limits^\frac{\pi}{2}_0\frac{x\ln(\sin{x})}{\sin{x}}\ {\rm d}x
&=2\int\limits^1_0\frac{\arctan{x}\ln\left(\frac{2x}{1+x^2}\right)}{x}\ {\rm d}x\\
&=2{\rm Ti}_2(1)\ln{2}+2\int\limits^1_0\frac{\arctan{x}\ln{x}}{x}\ {\rm d}x-2\int\limits^1_0\frac{\arctan{x}\ln(1+x^2)}{x}\ {\rm d}x\\
&=2\mathbf{G}\ln{2}-2\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^3}+2\int\limits^1_0\frac{\ln{x}\ln(1+x^2)}{1+x^2}\ {\rm d}x+4\int\limits^1_0\frac{x\arctan{x}\ln{x}}{1+x^2}\ {\rm d}x
\end{align}
and the integral
\begin{align}
4\int\limits^1_0\frac{x\arctan{x}\ln{x}}{1+x^2}\ {\rm d}x=-8\Im{\rm Li}_3(1-i)-\frac{5\pi^3}{16}-\frac{\pi}{2}\ln^2{2}
\end{align}
has also been established in the link above. (Both integrals were covered in the evaluation of $\mathscr{J}_2$.) Hence the closed form is indeed
\begin{align}
I=-2\mathbf{G}\ln{2}
\end{align}
