How to compute $\sum_{n=1}^\infty \frac{H_{2n}^2}{n^2}$?

Question

where $H_n$ denotes the harmonic number.
I can't see $$\sum_{n\geq 1} \frac{1}{n^2}\left(\int_0^1 \frac{1-x^{2n}}{1-x}\ \mathrm{d}x\right)^2$$ be of any assistance; even $$-\sum_{n\geq 1}H_{2n}^2\int_0^1 x^{n-1}\log{x}$$ does not seem like it would be of any assistance unless we know of a nice closed form/ generating function for $\sum_{n\geq 1} x^{n-1}H_{2n }^2$ which I have high doubts about. Evidently, I really do not even know where to start. I know how to compute $$\sum_{n\geq1}\frac{H_n^2}{n}$$ and $$\sum_{k\geq 1}\frac{H_{2n}}{n^2}$$ but the desired sum is a mystery. Thank you!

Answer 1

$$\sum_{n=1}^\infty\frac{H_{2n}^2}{n^2}=4\sum_{n=1}^\infty\frac{H_{2n}^2}{(2n)^2}$$

now use $2\sum_{n=1}^\infty f(2n)=\sum_{n=1}^\infty f(n)+\sum_{n=1}^\infty (-1)^nf(n)$

$$\Longrightarrow \sum_{n=1}^\infty\frac{H_{2n}^2}{n^2}=2\sum_{n=1}^\infty\frac{H_{n}^2}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_{n}^2}{n^2}$$

where

$$\sum_{n=1}^\infty\frac{H_n^2}{n^2}=\frac{17}4\zeta(4)$$

and

$$\sum_{n=1}^{\infty}\frac{(-1)^nH_n^2}{n^2}=2\operatorname{Li}_4\left(\frac12\right)-\frac{41}{16}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$