real analysis - Advanced Sum: Compute $sum_{n=1}^inftyfrac{H_{2n}H_n^{(2)}}{(2n+1)^2}$

How to prove

$$\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}= \\ \small{\frac43\ln^32\zeta(2)-\frac72\ln^22\zeta(3)-\frac{21}{16}\zeta(2)\zeta(3)+\frac{713}{64}\zeta(5)-\frac4{15}\ln^52-8\ln2\operatorname{Li}_4\left(\frac12\right)-8\operatorname{Li}_5\left(\frac12\right)}$$

where $H_n^{(q)}=\sum_{k=1}^n\frac{1}{n^q}$ is the harmonic number, $\operatorname{Li}_r(x)=\sum_{n=1}^\infty\frac{x^n}{n^r}$ is the polylogarithm function and $\zeta$ is the Riemann zeta function.

This problem is proposed by Cornel with no solution submitted.

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My trial

By applying integration by parts we have

$$\int_0^1 x^{2n}(\operatorname{Li}_2(x)-\zeta(2))\ dx=-\frac{H_{2n}}{(2n+1)^2}-\frac{1}{(2n+1)^3}$$

now multiply both sides by $H_n^{(2)}$ then sum both sides from $n=1$ to $\infty$ we get

$$\int_0^1(\operatorname{Li}_2(x)-\zeta(2))\sum_{n=1}^\infty H_n^{(2)}x^{2n}\ dx=-\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}-\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^3}$$

$$\int_0^1\frac{(\operatorname{Li}_2(x)-\zeta(2))\operatorname{Li}_2(x^2)}{1-x^2}\ dx=-\sum_{n=1}^\infty\frac{H_{2n}H_n^{(2)}}{(2n+1)^2}-\color{blue}{\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^3}}$$

I managed to find the blue sum using Abel's summation. As for the integral, I tried integration by parts but still resistant.

QUESTION

Any idea how to crack the integral or a different approach to find the target sum?

Thanks.