integration - Challenging Logarithmic Integral $int_0^1frac{ln^3(1-x)ln(1+x)}{x}dx$

Challenging Integral:

\begin{align}
I=\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx&=6\operatorname{Li}_5\left(\frac12\right)+6\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{81}{16}\zeta(5)-\frac{21}{8}\zeta(2)\zeta(3)\\&\quad+\frac{21}8\ln^22\zeta(3)-\ln^32\zeta(2)+\frac15\ln^52
\end{align}

The way I computed this integral is really long as it's based on values of tough alternating Euler sums which themselves long to calculate. I hope we can find other approaches that save us such tedious calculations. Any way, here is my approach:

Using the identity from this solution: $\displaystyle\int_0^1 x^{n-1}\ln^3(1-x)\ dx=-\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$

Multiplying both sides by $\frac{(-1)^{n-1}}{n}$ then summing both sides from $n=1$ to $n=\infty$, gives:
\begin{align}
I&=\int_0^1\frac{\ln^3(1-x)}{x}\sum_{n=1}^\infty-\frac{(-x)^{n}}{n}dx=\int_0^1\frac{\ln^3(1-x)\ln(1+x)}{x}dx\\
&=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}+3\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}+2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}
\end{align}

We have:
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}&=-6\operatorname{Li}_5\left(\frac12\right)-6\ln2\operatorname{Li}_4\left(\frac12\right)+\ln^32\zeta(2)-\frac{21}{8}\ln^22\zeta(3)\\&\quad+\frac{27}{16}\zeta(2)\zeta(3)+\frac94\zeta(5)-\frac15\ln^52
\end{align}

\begin{align}
\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}&=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac23\ln^32\zeta(2)+\frac74\ln^22\zeta(3)\\&\quad-\frac{15}{16}\zeta(2)\zeta(3)-\frac{23}8\zeta(5)+\frac2{15}\ln^52
\end{align}

$$\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}=\frac{21}{32}\zeta(5)-\frac34\zeta(2)\zeta(3)$$

The proof of the first and second sum can be found here and the third sum can be found here.

By substituting these three sums,we get the closed form of $I$.

---

Other try is by using the rule: ( see here)
$$\int_0^1 \frac{\ln^a(1-x)\ln(1+x)}{x}dx=(-1)^a a! \sum_{n=1}^\infty\frac{H_n^{(a+1)}}{n2^n}$$

We get $\quad\displaystyle I=-6\sum_{n=1}^\infty\frac{H_n^{(4)}}{n2^n}\quad$ and this sum is really hard to crack and I think I made it more complicated this way. All approaches are appreciated.

By the way, the last sum was proposed by Cornel last year on his FB page here but he has not revealed his solution yet.

Thanks.