Challenging sum: Compute $\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}$ Prove that 

$$S=\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3)$$
   where $H_n^{(m)}=\sum_{k=1}^n\frac1{k^m}$ is the n$th$ generalized harmonic number of order $m$ and $\zeta$ is the Riemann zeta function.

This problem is proposed by Cornel Valean and can be found here. 
Here is how I managed to find the integral representation:
We have $\int_0^1 x^{2n-1}\ln(1-x)\ dx=-\frac{H_{2n}}{2n}$ , then we can write
$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=-\frac12\int_0^1\frac{\ln(1-x)}{x}\sum_{n=1}^\infty\frac{H_n^{(2)}}{n}(x^2)^n\ dx\\=\small{-\frac12\int_0^1\frac{\ln(1-x)}{x}\left(\operatorname{Li}_3(x^2)+2\operatorname{Li}_3(1-x^2)-\ln(1-x^2)\operatorname{Li}_2(1-x^2)-\zeta(2)\ln(1-x^2)-2\zeta(3)\right)\ dx}$$
So any idea how to crack this integral or different approach? 
Thanks.

UPDATE:
This result was mentioned by @nospoon here in equation $(3)$. He didn't post the solution but he provided the idea.
 A: Update: the details may be found in the preprint The evaluation of a special harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$
The magical way by Cornel Ioan Valean
By the Cauchy product, we have $\operatorname{Li}_2(x^2) \log(1-x^2)= 3\sum _{n=1}^{\infty } \frac{x^{2 n}}{n^3}-2\sum _{n=1}^{\infty } x^{2n}\frac{H_n}{n^2}-\sum _{n=1}^{\infty } x^{2n}\frac{H_n^{(2)}}{n}$, and if we multiply both sides by $\log(1-x)/x$, and integrate from $x=0$ to $x=1$, using that $\int_0^1 x^{n-1}\log(1-x)\textrm{d}x=-H_n/n$, and doing all the reductions, we arrive at 
$$2\sum _{n=1}^{\infty } \frac{H_{2 n} H_n^{(2)}}{(2 n)^2}-12\sum _{n=1}^{\infty } \frac{H_n}{n^4}+12\sum _{n=1}^{\infty }(-1)^{n-1} \frac{H_n}{n^4}+\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^3}$$
$$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log \left(1-x^2\right) \log (1-x)}{x} \textrm{d}x$$
$$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log (1+x) \log (1-x)}{x}\textrm{d}x+2 \int_0^1 \frac{\text{Li}_2(-x) \log ^2(1-x)}{x} \textrm{d}x\\+2 \int_0^1 \frac{\text{Li}_2(x) \log ^2(1-x)}{x} \textrm{d}x$$
$$=\int_0^1 \frac{\text{Li}_2\left(x^2\right) \log (1+x) \log (1-x)}{x} \textrm{d}x+2 \sum _{n=1}^{\infty } \frac{H_n^2}{n^3}-2 \sum _{n=1}^{\infty } \frac{(-1)^{n-1}H_n^2}{n^3}+2 \sum _{n=1}^{\infty } \frac{H_n^{(2)}}{n^3}\\-2 \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_n^{(2)}}{n^3},$$
where the last integral is given here Two very advanced harmonic series of weight $5$, and all the last resulting harmonic series are given in the book (Almost) Impossible Integrals, Sums, and Series. The reduction to the last series has been achieved by using the identity, $\displaystyle \int_0^1 x^{n-1}\log^2(1-x)\textrm{d}x=\frac{H_n^2+H_n^{(2)}}{n}$. The series $\sum _{n=1}^{\infty } \frac{H_n H_{2 n}}{n^3}$ maybe found calculated in the paper On the calculation of two essential harmonicseries with a weight 5 structure, involving harmonic numbers of the type H_{2n} by Cornel Ioan Valean.
Thus, we have 

$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n)^2}=\frac{101}{64}\zeta(5)-\frac5{16}\zeta(2)\zeta(3).$$

All the details will appear in a new paper.
