challenging sum $\sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}$ How to prove that

\begin{align}
\sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}=\frac13\ln^42-2\ln^22\zeta(2)+7\ln2\zeta(3)-\frac{121}{16}\zeta(4)+8\operatorname{Li}_4\left(\frac12\right)
\end{align}
  where $H_n^{(m)}=1+\frac1{2^m}+\frac1{3^m}+...+\frac1{n^m}$ is the $n$th harmonic number of order $m$.

This problem was proposed by Cornel Valean.
Here is integral expression of the sum $\displaystyle -\int_0^1\frac{\ln x\operatorname{Li}_2(x^2)}{1-x^2}\ dx\quad $.
 A: Different approach:
\begin{align}
S&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{(2n+1)^2}\\
&=\sum_{n=1}^\infty H_n^{(2)}\int_0^1-x^{2n}\ln x\ dx\\
&=-\int_0^1\ln x\sum_{n=1}^\infty(x^2)^nH_n^{(2)}\\
&=-\int_0^1\frac{\ln x\operatorname{Li}_2(x^2)}{1-x^2}\ dx,\quad \operatorname{Li}_2(x^2)=2\operatorname{Li}_2(x)+2\operatorname{Li}_2(-x)\\
&=-2\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x^2}\ dx-2\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1-x^2}\ dx\\
&=-\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x}-\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1+x}-\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1-x}-\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1+x}\ dx\\
&=-I_1-I_2-I_3-I_4
\end{align}

\begin{align}
I_1&=\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1-x}\ dx\\
&=\sum_{n=1}^\infty H_n^{(2)}\int_0^1 x^n\ln x\ dx\\
&=-\sum_{n=1}^\infty \frac{H_n^{(2)}}{(n+1)^2}\\
&=-\sum_{n=1}^\infty \frac{H_n^{(2)}}{n^2}+\zeta(4)
\end{align}

\begin{align}
I_2&=\int_0^1\frac{\ln x\operatorname{Li}_2(x)}{1+x}\ dx\\
&=-\sum_{n=1}^\infty (-1)^n\int_0^1\ x^{n-1}\ln x\operatorname{Li}_2(x)\ dx\\
&=-\sum_{n=1}^\infty (-1)^n\left(\frac{H_n^{(2)}}{n^2}+\frac{2H_n}{n^3}-\frac{2\zeta(2)}{n^2}\right)
\end{align}

\begin{align}
I_3&=\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1-x}\ dx\\
&=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1\frac{x^n \ln x}{1-x}\ dx\\
&=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\left(H_n^{(2)}-\zeta(2)\right)
\end{align}

\begin{align}
I_4&=\int_0^1\frac{\ln x\operatorname{Li}_2(-x)}{1+x}\ dx\\
&=-\sum_{n=1}^\infty (-1)^nH_n^{(2)}\int_0^1x^n\ln x\ dx\\
&=-\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{(n+1)^2}\\
&=\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{n^2}+\frac78\zeta(4)
\end{align}

Combine the four integrals we get 
$$S=\frac98\zeta(4)+2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}-\sum_{n=1}^\infty(-1)^n\frac{H_n^{(2)}}{n^2}$$
Plugging the two sums we get 

$$S=\frac13\ln^42-2\ln^22\zeta(2)+7\ln2\zeta(3)-\frac{121}{16}\zeta(4)+8\operatorname{Li}_4\left(\frac12\right)
$$

