What is a closed form for ${\large\int}_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx$? Some time ago I asked How to find $\displaystyle{\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$.
Thanks to great effort of several MSE users, we now know that
\begin{align}
\int_0^1\frac{\ln^3(1+x)\,\ln x}xdx=&\,\frac{\pi^2}3\ln^32-\frac25\ln^52+\frac{\pi^2}2\zeta(3)+\frac{99}{16}\zeta(5)\\&\,-\frac{21}4\zeta(3)\ln^22-12\operatorname{Li}_4\!\left(\tfrac12\right)\ln2-12\operatorname{Li}_5\!\left(\tfrac12\right)\tag1
\end{align}
Now, a natural follow-up to that question is to bump the power of the logarithm and to ask:

Question: What is a closed form for the next integral?
  $$I=\int_0^1\frac{\ln^3(1+x)\,\ln^2x}xdx\tag2$$

I think it is likely that $I$ has a closed form, because there are several very similar integrals having known closed forms:
$$\int_0^1\frac{\ln^2(1+x)\,\ln^2x}xdx=\frac{\pi^2\,\zeta(3)}3-\frac{29\,\zeta(5)}8\tag3$$

$$\int_0^1\frac{\ln^3(1-x)\,\ln^2x}xdx=12\zeta^2(3)-\frac{23\pi^6}{1260}\tag4$$

\begin{align}
\int_0^1\frac{\ln^3(1+x)\,\ln^2x}{x^2}dx=&\,\frac{3\zeta(3)}2+2\pi^2\zeta(3)+\frac{3\zeta(5)}2-\frac{21\zeta(3)}2\ln^22\\&\,-\frac{63\zeta(3)}2\ln2+\frac{23\pi^4}{60}-\frac{4\ln^52}5-\frac{3\ln^42}2\\&\,-4\ln^32+\frac{2\pi^2}3\ln^32+\frac{3\pi^2}2\ln^22-24\operatorname{Li}_5\!\left(\tfrac12\right)\\&\,-36\operatorname{Li}_4\!\left(\tfrac12\right)-24\operatorname{Li}_4\!\left(\tfrac12\right)\ln2\tag5
\end{align}

partial solution
using the following identity: ( I can provide the proof if needed)
$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^2-H_n^{(2)}\right)$$
replace $x$ with $-x$ , then multiply both sides by $\ln^3x$ and integrate from $0$ to $1$, we have
\begin{align}
I&=\int_0^1\frac{\ln^2(1+x)\ln^3x}{1+x}\ dx=\sum_{n=1}^\infty (-1)^n\left(H_n^2-H_n^{(2)}\right)\int_0^1x^n\ln^3x\ dx\\
&=-6\sum_{n=1}^\infty \frac{(-1)^n}{(n+1)^4}\left(H_n^2-H_n^{(2)}\right)=6\sum_{n=1}^\infty \frac{(-1)^n}{n^4}\left(H_{n-1}^2-H_{n-1}^{(2)}\right)\\
&=6\sum_{n=1}^\infty \frac{(-1)^n}{n^4}\left(H_{n}^2-H_{n}^{(2)}-2\frac{H_n}{n}+\frac2{n^2}\right)\\
&=6\left(\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^4}-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^4}-2\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}-\frac{31}{16}\zeta(6)\right)
\end{align}
