TITLE: Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$
QUESTION [39 upvotes]: Here is a challenging one maybe some would like a go at.
Show that:
$$\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx=\frac{-9\pi^{4}}{256}+\frac{\pi^{3}}{32}\ln2+\frac{\pi^{2}}{6}G-\frac{1}{1536}\left[\psi_{3}\left(\frac34\right)-\psi_{3}\left(\frac14\right)\right]$$

REPLY [2 votes]: Following the same idea of @Cody we have
$$I=\int_{0}^{1}\frac{x\ln^{3}x}{(1-x)(1+x^2)}\ dx+\int_{0}^{1}\frac{\ln^{2}x\ln(1+x)}{1+x^2}\ dx=K+J$$

\begin{align}
K&=\int_{0}^{1}\frac{x\ln^{3}x}{(1-x)(x^{2}+1)}\ dx\\
&=\frac12\int_0^1\frac{x\ln^3x}{1+x^2}\ dx-\frac12\int_0^1\frac{\ln^3x}{1+x^2}\ dx+\frac12\int_0^1\frac{\ln^3x}{1-x}\ dx\\
&=\frac12\left(-\frac{21}{64}\zeta(4)\right)-\frac12\left(-6\beta(4)\right)+\frac12\left(-6\zeta(4)\right)\\
&=-\frac{405}{128}\zeta(4)+3\beta(4)
\end{align}

The integral $J$ is evaluated here in two methods 
$$J=\int_0^1\frac{\ln^2x\ln(1+x)}{1+x^2}\ dx=\frac{\pi^3}{32}\ln2+\zeta(2)G-2\beta(4)$$
Combining the results of $K$ and $J$ we have 
$$I=-\frac{405}{128}\zeta(4)+\frac{\pi^3}{32}\ln2+\zeta(2)G+\beta(4)$$
Substituting $\beta(4)=\frac1{768}\left(\psi_3(1/4)-8\pi^4\right)$ along with $\zeta(4)=\frac{\pi^4}{90}$ and $\zeta(2)=\frac{\pi^2}{6}$ we get

$$I=\frac{\pi^2}{6}G+\frac{\pi^{3}}{32}\ln2-\frac{35}{768}\pi^4+\frac{1}{768}\psi_{3}(1/4)$$