Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^5 2^n}$ Given the nth harmonic number $ H_n = \sum_{j=1}^{n} \frac{1}{j}$, we get from this post that apparently,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^k}z^n= S_{k-1,2}(z) + \rm{Li}_{\,k+1}(z)$$
for $-1\leq z\leq 1$, and with Nielsen generalized polylogarithm $S_{n,p}(z)$ and polylogarithm $\rm{Li}_n(z)$. Hence for small $k$,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2\, 2^n}= S_{1,2}\big(\tfrac12\big)+\rm{Li}_3\big(\tfrac12\big)$$
$$\sum_{n=1}^{\infty}\frac{H_n}{n^3\, 2^n}= S_{2,2}\big(\tfrac12\big)+\rm{Li}_4\big(\tfrac12\big)$$
$$\sum_{n=1}^{\infty}\frac{H_n}{n^4\, 2^n}= S_{3,2}\big(\tfrac12\big)+\rm{Li}_5\big(\tfrac12\big)$$
and so on. Explicitly, given $a=\ln 2$,
$$S_{1,2}\big(\tfrac12\big) +\tfrac1{6}a^3-\tfrac18 \zeta(3)=0 $$
$$S_{2,2}\big(\tfrac12\big) +\tfrac1{168}a^4+\tfrac17a^2\,\rm{Li}_2\big(\tfrac12\big)+\tfrac17a\,\rm{Li}_3\big(\tfrac12\big)-\tfrac18\zeta(4) = 0$$
which are discussed in this and this post. And by yours truly, 
$$S_{3,2}\big(\tfrac12\big) -A+B  = 0$$
$$A = \tfrac{41}{840}a^5+\tfrac5{21}a^3\,\rm{Li}_2\big(\tfrac12\big)+\tfrac47a^2\,\rm{Li}_3\big(\tfrac12\big)+a\,\rm{Li}_4\big(\tfrac12\big) + \rm{Li}_5\big(\tfrac12\big) $$
$$B=\tfrac12\zeta(2)\zeta(3)+\tfrac18a\,\zeta(4)-\tfrac1{32}\zeta(5)$$

Q: What, however, is the explicit evaluation in ordinary polylogs of the next steps, namely $S_{4,2}\big(\tfrac12\big)$ and $S_{5,2}\big(\tfrac12\big)$?

P.S. Try as I might, they resist being evaluated and there are indications these higher order integrals may not be expressible by ordinary polylogs.
 A: Let $I$ denotes $\int_0^1\frac{\ln^4(1+x)\ln x}{x}\ dx$
I proved here 

\begin{align}
I&=-120\operatorname{Li}_6\left(\frac12\right)-72\ln2\operatorname{Li}_5\left(\frac12\right)-24\ln^22\operatorname{Li}_4\left(\frac12\right)+78\zeta(6)+\frac34\ln2\zeta(5)\\
&\quad-\frac32\ln^22\zeta(4)-3\ln^32\zeta(3)+2\ln^42\zeta(2)+12\zeta^2(3)-12\ln2\zeta(2)\zeta(3)\\
&\quad-\frac{17}{30}\ln^62+24\sum_{n=1}^\infty\frac{H_n}{n^52^n}\tag{1}
\end{align}

On the other hand and by applying integration by parts we have
\begin{align}
I=-2\int_0^1\frac{\ln^3(1+x)\ln^2x}{1+x}\ dx
\end{align}
You can find here the following identity
$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
Replacing $x$ with $-x$ yields
$$-\frac{\ln^3(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^nx^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$ 
Now we can write
\begin{align}
I&=2\sum_{n=1}^\infty (-1)^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\int_0^1 x^n \ln^2x\ dx\\
&=-2\sum_{n=1}^\infty (-1)^n\left(H_{n-1}^3-3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)}\right)\int_0^1 x^{n-1} \ln^2x\ dx\\
&=-2\sum_{n=1}^\infty (-1)^n\left(H_{n}^3-3H_{n}H_{n}^{(2)}+2H_{n}^{(3)}+\frac{6H_n}{n^3}-\frac{3H_n^2}{n}+\frac{3H_n^{(2)}}{n}-\frac{6}{n^2}\right)\left(\frac{2}{n^3}\right)\\
\end{align}
Then 

\begin{align}
I&=12\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^3}-4\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^3}-8\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^3}-24\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}\\
&\quad+12\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^4}-12\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^4}+24\sum_{n=1}^\infty\frac{(-1)^n}{n^6}\tag{2}
\end{align}


We can see from $(1)$ and $(2)$ that our target sum is related to $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$.
I am not sure if the alternating sums in $(2)$ have closed form or not but I am sure that $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$ has no closed form because the power of the denominator is odd more than 3. So the target sum has no closed form unless $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}$ cancels out with other sums somehow which I doubt.
