Voy a estar usando los siguientes resultados:
$$2\sum^\infty_{n=1}\frac{H_n}{n^q}=(q+2)\zeta(q+1)-\sum^{q-2}_{j=1}\zeta(j+1)\zeta(q-j)\tag1$$
$$\sum^\infty_{n=1}\frac{H_n}{n^22^n}=\zeta(3)-\frac{\pi^2}{12}\ln{2}\tag2$$
$$\sum^\infty_{n=1}\frac{H_n}{n^32^n}={\rm Li}_4\left(\tfrac{1}{2}\right)+\frac{\pi^4}{720}-\frac{1}{8}\zeta(3)\ln{2}+\frac{1}{24}\ln^4{2}\tag3$$
\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}
Pruebas de $(1)$, $(2)$ y $(4)$ se puede encontrar aquí, aquí y aquí respectivamente. Por desgracia, no ha sido un matemáticamente prueba de sonido de $(3)$ sobre el MSE como la de ahora.
El uso de $\mathcal{I}$ para denotar la integral en cuestión,
\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}
Para $\mathcal{I}_1$, integración por partes da
$$\mathcal{I}_1=\frac{1}{3}\ln^6{2}-\int^1_\frac{1}{2}\frac{\ln^3{x}\ln^2(1-x)}{1-x}{\rm d}x$$
Por otro lado, $x\mapsto1-x$ rendimientos
$$\mathcal{I}_1=-\int^\frac{1}{2}_0\frac{\ln^3{x}\ln^2(1-x)}{1-x}{\rm d}x$$
La combinación de estas dos igualdades, tenemos
\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}
Incluso con la ayuda de Wolfram Alpha, la evaluación de la quinta derivado era horriblemente desagradable por decir lo menos. Como para $\mathcal{I}_2$,
\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}
Para $\mathcal{I}_3$,
\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}
Así
\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}
Tomamos nota de que
\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}
como ya hemos obtuvieron los valores de los últimos tres integrales. Para el resto de la 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}
Por lo tanto podemos expresar $\zeta(\bar{5},1)$
\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}
Esto implica que
\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}
Depilación a la original de la integral, obtenemos otra forma, en términos de $\zeta(\bar{5},1)$
\begin{align}
\color{#BF00FF}{\mathcal{I}
=}&\color{#BF00FF}{-\frac{\pi^6}{252}-18\zeta(\bar{5},1)+3\zeta^2(3)}
\end{align}
Esto es lo más cercano a una "forma cerrada" como la puedo conseguir. El gran número de cancelaciones que intervienen en el último paso me hace pensar que mi respuesta podría ser rotonda e ineficiente. Tenga en cuenta que no se conocen simple forma cerrada para $\zeta(\bar{5},1)$ existe, lo que implica que las formas cerradas para mayor potencia de las integrales es raro existir.
