Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ 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.
\begin{align}
&\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x
\\[3mm] = &\
\frac{1}{16}\sum_{n = 1}^{\infty}
\frac{\psi_{1}\left(1/4 + n\right) -
\psi_{1}\left(3/4 + n\right)}{n} \tag{2}
\end{align}
Here $\psi_{n}\left(z\right)$ denotes the polygamma function.
Can you help me solve this problem $?$.

The following is a proof of the formula $$S= \sum_{k=1}^{\infty} \frac{H_{k}}{ (k+a)^{2}}= \left(\gamma + \psi(a) \right) \psi_{1}(a) - \frac{\psi_{2}(a)}{2} \, , \quad a >0.$$ 
This formula is mentioned in a comment under Bennett Gardiner's answer. 
(For $a=0$, the right side of the equation should be interpreted as a limit).
$$ \begin{align}  S &= \sum_{k=1}^{\infty} \frac{H_{k}}{(k+a)^{2}} \\ &= \sum_{k=1}^{\infty} \frac{1}{(k+a)^{2}} \sum_{n=1}^{k} \frac{1}{n}  \\& = \sum_{n=1}^{\infty} \frac{1}{n} \sum_{k=n}^{\infty}  \frac{1}{(k+a)^2} \\ &= \sum_{n=1}^{\infty} \frac{\psi_{1}(a+n)}{n}  \\ &= - \sum_{n=1}^{\infty} \frac{1}{n} \int_{0}^{1} \frac{x^{a+n-1} \ln x}{1-x} \, dx \tag{1} \\ &= - \int_{0}^{1}  \frac{x^{a-1} \ln x}{1-x} \sum_{n=1}^{\infty} \frac{x^{n}}{n} \, dx \\ &= \int_{0}^{1} \frac{x^{a-1} \ln x \ln(1-x)}{1-x} \, dx \\ &= \lim_{b \to 0^{+}} \frac{\partial }{\partial a \, \partial b} B(a,b) \\ &= \small  \lim_{b \to 0^{+}} \frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)} \left( \psi(a) \psi(b) - \psi(a)\psi(a+b) - \psi(b) \psi(a+b) + \psi^{2}(a+b) - \psi_{1}(a+b) \right) \tag{2} \\ &=  \lim_{b \to 0^{+}} \frac{\Gamma(a)}{\Gamma(a+b)} \left( \frac{1}{b} - \gamma + \mathcal{O}(b) \right)\left( \left( \gamma \psi_{1}(a) + \psi(a) \psi_{1} (a)  - \frac{\psi_{2}(a)}{2} \right)b + \mathcal{O}(b^{2}) \right) \\ &= \left(\gamma + \psi(a) \right) \psi_{1}(a) -  \frac{\psi_{2}(a)}{2} \end{align}$$

$(1)$ https://en.wikipedia.org/wiki/Trigamma_function#Calculation 
$(2)$ http://mathworld.wolfram.com/BetaFunction.html (26)
