I will be using the following results:
$$2\sum^\infty_{n=1}\frac{H_n}{n^q}=(q+2)\zeta(q+1)-\sum^{q-2}_{j=1}\zeta(j+1)\zeta(q-j)\tag1$$
$$\sum^\infty_{n=1}\frac{H_n}{n^22^n}=\zeta(3)-\frac{\pi^2}{12}\ln{2}\tag2$$
$$\sum^\infty_{n=1}\frac{H_n}{n^32^n}={\rm Li}_4\left(\tfrac{1}{2}\right)+\frac{\pi^4}{720}-\frac{1}{8}\zeta(3)\ln{2}+\frac{1}{24}\ln^4{2}\tag3$$
\begin{align}
\sum^\infty_{n=1}\frac{H_n}{n^42^n}
=&2{\rm Li}_5\left(\tfrac{1}{2}\right)+\frac{1}{32}\zeta(5)+{\rm Li}_4\left(\tfrac{1}{2}\right)\ln{2}-\frac{\pi^4}{720}\ln{2}+\frac{1}{2}\zeta(3)\ln^2{2}\\&-\frac{\pi^2}{12}\zeta(3)-\frac{\pi^2}{36}\ln^3{2}+\frac{1}{40}\ln^5{2}\tag4
\end{align}
Proofs of $(1)$, $(2)$ and $(4)$ can be found here, here and here respectively. Unfortunately, there has not been a mathematically sound proof of $(3)$ on MSE as of now.
Using $\mathcal{I}$ to denote the integral in question,
\begin{align}
\mathcal{I}
&=-\int^1_0\frac{\ln^3{x}\ln^2(1+x)}{1+x}{\rm d}x\\
&=-\int^2_1\frac{\ln^2{x}\ln^3(x-1)}{x}{\rm d}x\\
&=\underbrace{-\int^1_\frac{1}{2}\frac{\ln^2{x}\ln^3(1-x)}{x}{\rm d}x}_{\mathcal{I}_1}\underbrace{+3\int^1_{\frac{1}{2}}\frac{\ln^3{x}\ln^2(1-x)}{x}}_{\mathcal{I}_2}\underbrace{-3\int^1_{\frac{1}{2}}\frac{\ln^4{x}\ln(1-x)}{x}{\rm d}x}_{\mathcal{I}_3}-\frac{1}{6}\ln^6{2}
\end{align}
For $\mathcal{I}_1$, integration by parts gives
$$\mathcal{I}_1=\frac{1}{3}\ln^6{2}-\int^1_\frac{1}{2}\frac{\ln^3{x}\ln^2(1-x)}{1-x}{\rm d}x$$
On the other hand, $x\mapsto1-x$ yields
$$\mathcal{I}_1=-\int^\frac{1}{2}_0\frac{\ln^3{x}\ln^2(1-x)}{1-x}{\rm d}x$$
Combining these two equalities, we have
\begin{align}
\mathcal{I}_1
&=\frac{1}{6}\ln^6{2}-\frac{1}{2}\int^1_0\frac{\ln^3{x}\ln^2(1-x)}{1-x}{\rm d}x\\
&=\frac{1}{6}\ln^6{2}-\frac{1}{2}\frac{\partial^5\beta}{\partial a^3\partial b^2}(1,0^+)\\
&=\frac{1}{6}\ln^6{2}-\frac{1}{2}\left[\frac{1}{b}+\mathcal{O}(1)\right]\left[\left(12\zeta^2(3)-\frac{23\pi^6}{1260}\right)b+\mathcal{O}(b^2)\right]_{b=0}\\
&=\frac{23\pi^6}{2520}-6\zeta^2(3)+\frac{1}{6}\ln^6{2}
\end{align}
Even with the help of Wolfram Alpha, evaluating that fifth derivative was horribly unpleasant to say the least. As for $\mathcal{I}_2$,
\begin{align}
\mathcal{I}_2
=&6\sum^\infty_{n=1}\frac{H_n}{n+1}\int^1_\frac{1}{2}x^n\ln^3{x}\ {\rm d}x\\
=&6\sum^\infty_{n=1}\frac{H_n}{n+1}\frac{\partial^3}{\partial n^3}\left(\frac{1}{n+1}-\frac{1}{(n+1)2^{n+1}}\right)\\
=&\color{#E2062C}{-\sum^\infty_{n=1}\frac{36H_n}{(n+1)^5}}+\color{#FF4F00}{\sum^\infty_{n=1}\frac{36H_n}{(n+1)^52^{n+1}}}+\color{#00A000}{\sum^\infty_{n=1}\frac{36\ln{2}H_n}{(n+1)^42^{n+1}}}+\color{#21ABCD}{\sum^\infty_{n=1}\frac{18\ln^2{2}H_n}{(n+1)^32^{n+1}}}\\&+\color{#6F00FF}{\sum^\infty_{n=1}\frac{6\ln^3{2}H_n}{(n+1)^22^{n+1}}}\\
=&\color{#E2062C}{-\frac{\pi^6}{35}+18\zeta^2(3)}+\color{#FF4F00}{\sum^\infty_{n=1}\frac{36H_n}{n^52^{n}}-36{\rm Li}_6\left(\tfrac{1}{2}\right)}+\color{#00A000}{36{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}+\frac{9}{8}\zeta(5)\ln{2}}\\
&+\color{#00A000}{36{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}-\frac{\pi^4}{20}\ln^2{2}+18\zeta(3)\ln^3{2}-3\pi^2\zeta(3)\ln{2}-\pi^2\ln^4{2}+\frac{9}{10}\ln^6{2}}\\
&+\color{#21ABCD}{\frac{\pi^4}{40}\ln^2{2}-\frac{9}{4}\zeta(3)\ln^3{2}+\frac{3}{4}\ln^6{2}}+\color{#6F00FF}{\frac{3}{4}\zeta(3)\ln^3{2}-\ln^6{2}}\\
=&\sum^\infty_{n=1}\frac{36H_n}{n^52^{n}}-36{\rm Li}_6\left(\tfrac{1}{2}\right)-\frac{\pi^6}{35}+36{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}+\frac{9}{8}\zeta(5)\ln{2}+36{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}\\
&-\frac{\pi^4}{40}\ln^2{2}+18\zeta^2(3)-3\pi^2\zeta(3)\ln{2}+\frac{33}{2}\zeta(3)\ln^3{2}-\pi^2\ln^4{2}+\frac{13}{20}\ln^6{2}
\end{align}
For $\mathcal{I}_3$,
\begin{align}
\mathcal{I}_3
=&3\sum^\infty_{n=1}\frac{1}{n}\int^1_\frac{1}{2}x^{n-1}\ln^4{x}\ {\rm d}x\\
=&3\sum^\infty_{n=1}\frac{1}{n}\frac{\partial^4}{\partial n^4}\left(\frac{1}{n}-\frac{1}{n2^n}\right)\\
=&\sum^\infty_{n=1}\left(\frac{72}{n^6}-\frac{72}{n^62^n}-\frac{72\ln{2}}{n^52^n}-\frac{36\ln^2{2}}{n^42^n}-\frac{12\ln^3{2}}{n^32^n}-\frac{3\ln^4{2}}{n^22^n}\right)\\
=&-72{\rm Li}_6\left(\tfrac{1}{2}\right)+\frac{8\pi^6}{105}-72{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}-36{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}\\&-\frac{21}{2}\zeta(3)\ln^3{2}+\frac{3\pi^2}{4}\ln^4{2}-\frac{1}{2}\ln^6{2}
\end{align}
Thus
\begin{align}
\color{#BF00FF}{\mathcal{I}
=}&\color{#BF00FF}{36\sum^\infty_{n=1}\frac{H_n}{n^52^n}-108{\rm Li}_6\left(\tfrac{1}{2}\right)+\frac{143\pi^6}{2520}-36{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}+\frac{9}{8}\zeta(5)\ln{2}-\frac{\pi^4}{40}\ln^2{2}}\\&\color{#BF00FF}{+12\zeta^2(3)-3\pi^2\zeta(3)\ln{2}+6\zeta(3)\ln^3{2}-\frac{\pi^2}{4}\ln^4{2}+\frac{3}{20}\ln^6{2}}
\end{align}
We note that
\begin{align}
\zeta(\bar{5},1)
=&\frac{1}{24}\int^1_0\frac{\ln^4{x}\ln(1+x)}{1+x}{\rm d}x\\
=&\frac{1}{24}\int^2_1\frac{\ln{x}\ln^4(x-1)}{x}{\rm d}x\\
=&-\frac{1}{24}\int^1_\frac{1}{2}\frac{\ln{x}\ln^4(1-x)}{x}{\rm d}x+\frac{1}{6}\int^1_\frac{1}{2}\frac{\ln^2{x}\ln^3(1-x)}{x}{\rm d}x-\frac{1}{4}\int^1_\frac{1}{2}\frac{\ln^3{x}\ln^2(1-x)}{x}{\rm d}x\\
&+\frac{1}{6}\int^1_\frac{1}{2}\frac{\ln^4{x}\ln(1-x)}{x}{\rm d}x+\frac{1}{144}\ln^6{2}\\
=&\underbrace{-\frac{1}{24}\int^\frac{1}{2}_0\frac{\ln^4{x}\ln(1-x)}{1-x}{\rm d}x}_{\mathcal{J}}-3\sum^\infty_{n=1}\frac{H_n}{n^52^n}+7{\rm Li}_6\left(\tfrac{1}{2}\right)-\frac{17\pi^6}{5040}+{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}\\
&-\frac{3}{32}\zeta(5)\ln{2}-{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}+\frac{\pi^4}{480}\ln^2{2}-\frac{1}{2}\zeta^2(3)+\frac{\pi^2}{4}\zeta(3)\ln{2}-\frac{19}{24}\zeta(3)\ln^3{2}\\
&+\frac{\pi^2}{24}\ln^4{2}-\frac{17}{360}\ln^6{2}
\end{align}
since we have already derived the values of the last three integrals. For the remaining integral,
\begin{align}
\mathcal{J}
=&\frac{1}{24}\sum^\infty_{n=1}H_n\frac{\partial^4}{\partial n^4}\left(\frac{1}{(n+1)2^{n+1}}\right)\\
=&\sum^\infty_{n=1}\frac{H_n}{(n+1)^52^{n+1}}+\sum^\infty_{n=1}\frac{\ln{2}H_n}{(n+1)^42^{n+1}}+\sum^\infty_{n=1}\frac{\ln^2{2}H_n}{2(n+1)^32^{n+1}}+\sum^\infty_{n=1}\frac{\ln^3{2}H_n}{6(n+1)^22^{n+1}}\\
&+\sum^\infty_{n=1}\frac{\ln^4{2}H_n}{24(n+1)2^{n+1}}\\
=&\sum^\infty_{n=1}\frac{H_n}{n^52^n}-{\rm Li}_6\left(\tfrac{1}{2}\right)+{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}+\frac{1}{32}\zeta(5)\ln{2}+{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}-\frac{\pi^4}{720}\ln^2{2}\\
&+\frac{1}{2}\zeta(3)\ln^3{2}-\frac{\pi^2}{12}\zeta(3)\ln{2}-\frac{\pi^2}{36}\ln^4{2}+\frac{1}{40}\ln^6{2}+\frac{\pi^4}{1440}\ln^2{2}-\frac{1}{16}\zeta(3)\ln^3{2}\\&+\frac{1}{48}\ln^6{2}+\frac{1}{48}\zeta(3)\ln^3{2}-\frac{1}{36}\ln^6{2}+\frac{1}{48}\ln^6{2}\\
=&\sum^\infty_{n=1}\frac{H_n}{n^52^n}-{\rm Li}_6\left(\tfrac{1}{2}\right)+{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}+\frac{1}{32}\zeta(5)\ln{2}+{\rm Li}_4\left(\tfrac{1}{2}\right)\ln^2{2}-\frac{\pi^4}{1440}\ln^2{2}\\
&+\frac{11}{24}\zeta(3)\ln^3{2}-\frac{\pi^2}{12}\zeta(3)\ln{2}-\frac{\pi^2}{36}\ln^4{2}+\frac{7}{180}\ln^6{2}\\
\end{align}
Hence we can express $\zeta(\bar{5},1)$ as
\begin{align}
\zeta(\bar{5},1)
=&-2\sum^\infty_{n=1}\frac{H_n}{n^52^n}+6{\rm Li}_6\left(\tfrac{1}{2}\right)-\frac{17\pi^6}{5040}+2{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}-\frac{1}{16}\zeta(5)\ln{2}+\frac{\pi^4}{720}\ln^2{2}\\
&-\frac{1}{2}\zeta^2(3)-\frac{1}{3}\zeta(3)\ln^3{2}+\frac{\pi^2}{6}\zeta(3)\ln{2}+\frac{\pi^2}{72}\ln^4{2}-\frac{1}{120}\ln^6{2}
\end{align}
This implies that
\begin{align}
\sum^\infty_{n=1}\frac{H_n}{n^52^n}
=&3{\rm Li}_6\left(\tfrac{1}{2}\right)-\frac{1}{2}\zeta(\bar{5},1)-\frac{17\pi^6}{10080}+{\rm Li}_5\left(\tfrac{1}{2}\right)\ln{2}-\frac{1}{32}\zeta(5)\ln{2}+\frac{\pi^4}{1440}\ln^2{2}\\
&-\frac{1}{4}\zeta^2(3)-\frac{1}{6}\zeta(3)\ln^3{2}+\frac{\pi^2}{12}\zeta(3)\ln{2}+\frac{\pi^2}{144}\ln^4{2}-\frac{1}{240}\ln^6{2}
\end{align}
Plucking this back into the original integral, we get another form in terms of $\zeta(\bar{5},1)$
\begin{align}
\color{#BF00FF}{\mathcal{I}
=}&\color{#BF00FF}{-\frac{\pi^6}{252}-18\zeta(\bar{5},1)+3\zeta^2(3)}
\end{align}
This is as close to a "closed form" as I can get. The sheer number of cancellations involved in the last step makes me think that my answer could be roundabout and inefficient. Note that no known simple closed form for $\zeta(\bar{5},1)$ exists, implying that closed forms for higher power integrals are unlikely to exist as well.
Following the same approach as the given link, we have $\displaystyle\operatorname{arctanh}(x)=\frac12\ln\left(\frac{1+x}{1-x}\right)$, and with the same substitution $\displaystyle t=\frac{1+x}{1-x}$ we get
$$I(a,b)=-\frac1{2^{a-1}}\int_0^1\frac{(1+t)^{b-2}}{(1-t)^b}\ln^a(t)~\mathrm dt$$
Using generalized binomial expansion theorem, we can expand this as
$$I(a,b)=-\frac1{2^{a-1}(b-1)!}\sum_{j=0}^{b-2}\binom{b-2}j\sum_{k=0}^\infty(k+1)\dots(k+b-1)\int_0^1t^{k+j}\ln^a(t)~\mathrm dt$$
The integral at the end can be viewed as the $a$th derivative of $\displaystyle\int_0^1t^u~\mathrm dt=\frac1{u+1}$ at $u=k+j$, which is given by $\displaystyle\frac{(-1)^aa!}{(k+j+1)^{a+1}}$, which leaves us with the series
$$I(a,b)=\frac{(-1)^{a+1}a!}{2^{a-1}(b-1)!}\sum_{j=0}^{b-2}\binom{b-2}j\sum_{k=0}^\infty\frac{(k+1)\dots(k+b-1)}{(k+j+1)^{a+1}}$$
which boils the problem down to computing the last series. See that we have
$$\sum_{k=0}^\infty\frac{(k+1)\dots(k+b-1)}{(k+j+1)^{a+1}}=\sum_{k=j+1}^\infty\frac{P_{b,j}(k)}{k^{a+1}}=\sum_{m=0}^{b-1}\alpha_{b,j,m}\zeta(a-m+1)-\sum_{k=1}^j\frac{P_{b,j}(k)}{k^{a+1}}$$
for a polynomial $\displaystyle P_{b,j}(k)=\sum_{m=0}^{b-1}\alpha_{b,j,m}k^m$. This leaves us with the solution given by
$$I(a,b)=\frac{(-1)^{a+1}a!}{2^{a-1}(b-1)!}\sum_{j=0}^{b-2}\binom{b-2}j\left[\sum_{m=0}^{b-1}\alpha_{b,j,m}\zeta(a-m+1)-\sum_{k=1}^j\frac{P_{b,j}(k)}{k^{a+1}}\right]$$
The rest is just multiplying out $(k-j)\dots(k+b-j-2)$ to extract the coefficients of $P_{b,j}$. These are Stirling numbers of the first kind.
Best Answer
Following the same approach here: $$I(a)=\int_0^1\frac{\ln x\ln^a(1-x)}{x^3}dx=\int_0^1\frac{\ln(1-x)\ln^ax}{(1-x)^3}dx\\\overset{IBP}{=}\frac12\int_0^1\frac{\ln^ax}{(1-x)^3}dx-\frac12a\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2x}dx$$.
The first integral:
$$\int_0^1\frac{\ln^ax}{(1-x)^3}dx=\frac12\sum_{n=2}^\infty n(n-1)\int_0^1 x^{n-2}\ln^ax\ dx\\=\frac12(-1)^aa!\sum_{n=2}^\infty\frac{n(n-1)}{(n-1)^{a+1}}=\frac12(-1)^aa!\sum_{n=1}^\infty\frac{n+1}{n^a}\\=\boxed{\frac12(-1)^aa!(\zeta(a-1)+\zeta(a))}$$
The second integral:
Use the partial fraction decomposition: $\frac1{(1-x)^2x}=\frac1x+\frac1{1-x}+\frac1{(1-x)^2}$
$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2x}dx\\=\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{x}dx+\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{1-x}dx+\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2}dx$$
where the first integral:
$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{x}dx=-\sum_{n=1}^\infty\frac1n\int_0^1 x^{n-1}\ln^{a-1}x\ dx\\=(-1)^a(a-1)!\sum_{n=1}^\infty\frac1{n^{a+1}}=(-1)^a(a-1)!\zeta(a+1)$$
and the second integral
$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{1-x}dx=-\sum_{n=1}^\infty H_n\int_0^1x^n\ln^{a-1}x\ dx\\=(-1)^a(a-1)!\sum_{n=1}^\infty\frac{H_n}{(n+1)^a}\\=(-1)^a(a-1)!\left(\sum_{n=1}^\infty\frac{H_n}{n^a}-\zeta(a+1)\right)$$
As for the third integral, its already calculated in the link I mentioned above which shows
$$\int_0^1\frac{\ln x\ln^b(1-x)}{x^2}dx=\int_0^1\frac{\ln(1-x)\ln^bx}{(1-x)^2}dx=(-1)^b b!\left(\zeta(b)-\sum_{n=1}^\infty\frac{H_n}{n^b}\right)$$
then we can write
$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2}dx=(-1)^a (a-1)!\left(\sum_{n=1}^\infty\frac{H_n}{n^{a-1}}-\zeta(a-1)\right)$$
Collect the three sub-integrals to get
$$\int_0^1\frac{\ln(1-x)\ln^{a-1}x}{(1-x)^2x}dx=\boxed{(-1)^a(a-1)!\left(\sum_{n=1}^\infty\frac{H_n}{n^a}+\sum_{n=1}^\infty\frac{H_n}{n^{a-1}}-\zeta(a-1)\right)}$$
Combine the boxed results we get
where $$\sum_{n=1}^\infty \frac{H_n}{n^a}=\left(1+\frac a2\right)\zeta(a+1)-\frac12\sum_{k=1}^{a-2}\zeta(k+1)\zeta(a-k)$$ is the generalized Euler identity.