Interesting Harmonic Sum $\sum_{k\geq 1}\frac{(-1)^{k-1}}{k^2}H_k^{(2)}$

A related problem. You can have the following identity

$$\sum_{k=1}^{\infty}(-1)^{k-1} \frac{H_k^{(2)}}{k^2} = \frac{37}{16}\zeta(4)+2\sum_{k=1}^{\infty}(-1)^k \frac{H_k}{k^3}\sim 0.7843781621 .$$

we know that $$\int_0^1x^{n-1}\ln(1-x)\ dx=-\frac{H_n}{n}$$ differentiate both sides with respect to $n$ $$\int_0^1x^{n-1}\ln(1-x)\ln x\ dx=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\zeta(2)}{n} $$ multiply both sides by $\frac{(-1)^n}{n}$ then take the sum, we get \begin{align} \sum_{n=1}^\infty\frac{(-1)^n H_n}{n^3}+\sum_{n=1}^\infty\frac{(-1)^n H_n^{(2)}}{n^2}-\underbrace{\zeta(2)\sum_{n=1}^\infty\frac{(-1)^n}{n^2}}_{-\frac54\zeta(4)}&=\int_0^1\frac{\ln(1-x)\ln x}{x}\sum_{n=1}^\infty\frac{(-x)^n}{n}\ dx\\ &=-\int_0^1\frac{\ln(1-x)\ln(1+x)\ln x}{x}\ dx \end{align} I was able here to prove $$\int_0^1\frac{\ln(1-x)\ln(1+x)\ln x}{x}\ dx=\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}+\frac34\sum_{n=1}^\infty\frac{H_n}{n^3}+\frac18\zeta(4)$$ which follows that$$\sum_{n=1}^\infty\frac{(-1)^n H_n^{(2)}}{n^2}=-2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\frac34\sum_{n=1}^\infty\frac{H_n}{n^3}-\frac{11}8\zeta(4)$$ substituting the well known results: $$\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)$$ $$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}=2\operatorname{Li}_4\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac1{12}\ln^42$$ finally we get$$\sum_{n=1}^\infty\frac{(-1)^n H_n^{(2)}}{n^2}=-4\operatorname{Li}_4\left(\frac12\right)-\frac{51}{16}\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac16\ln^42$$