real analysis - Evaluating $int_0^1 frac{log x log left(1-x^4 right)}{1+x^2}dx$

I am trying to prove that

$$\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx = \frac{\pi^3}{16}-3G\log 2 \tag{1}$$

where $G$ is Catalan's Constant.

I was able to express it in terms of Euler Sums but it does not seem to be of any use.

$$\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx = \frac{1}{16}\sum_{n=1}^\infty \frac{\psi_1 \left(\frac{1}{4}+n \right)-\psi_1 \left(\frac{3}{4}+n \right)}{n} \tag{2}$$

Here $\psi_n(z)$ denotes the polygamma function.

Can you help me solve this problem?

Answer

I tried substitutions and the differentiation w.r.t a paramater trick like the other posters. Another partial result, or a trail of breadcrumbs to follow, is the following. We try a series expansion,
$$

\frac{\log\left(1-x^4\right)}{1+x^2} = \displaystyle \sum_{k=1}^{\infty} x^{4k}\left(x^{2} -1\right)H_k,
$$
where $H_k$ are the Harmonic numbers. Then
\begin{align}
\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}\ \mathrm{d}x &=\displaystyle \sum_{k=1}^{\infty}\, H_k\int_0^1 x^{4k}\left(x^{2} -1\right)\log x \ \mathrm{d}x \\
&=\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+1)^2}-\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+3)^2}.
\end{align}
These sums look very similar to the ones evaluated in this post, in which they are transformed into alternating sums. Using the same techniques, or perhaps working back from the answers, we can hopefully show that
$$
\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+1)^2} = -G\left(\frac{\pi}{4}+\frac{\log 8}{2} \right) +\frac{7}{4}\zeta(3) +\frac{\pi^3}{32} - \frac{\pi^2}{16}\log 8,

$$
$$
\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+3)^2} = -G\left(\frac{\pi}{4}-\frac{\log 8}{2} \right) +\frac{7}{4}\zeta(3) -\frac{\pi^3}{32} - \frac{\pi^2}{16}\log 8,
$$
Subtracting the second from the first gives us
$$
\frac{\pi^3}{16}-G\log 8.
$$