Is there a nice way to represent $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n+m+1}$? Here, $H_n$ denotes the harmonic number. More colloquially, is there any way to represent $$\int_0^1 x^{n-1}\log^2\left(1+x\right)\ \mathrm{d}x$$ in a nice way? The latter is corollary to the original question; namely,
$$\int_0^1 x^{n-1}\log^2\left(1+x\right)\ \mathrm{d}x = \frac{\log^2(2)}{n}-\frac{2}{n}\sum_{k=1}^\infty \frac{(-1)^{k+1}H_{\ell}}{n+k+1}...$$
which I do not know how to simplify. Thank you!

Mathematica choked when I gave it the general expression (at least the way I tried).  Here's how the sum works out for some small values of $m$.
\begin{gather}
m = 0 : \tfrac{1}{2}\log^2(2) \\
m = 1 : -1 + 2 \log(2) - \tfrac{1}{2}\log^2(2) \\
m = 2 : \tfrac{5}{4} - 2 \log(2) + \tfrac{1}{2}\log^2(2)   \\
m = 3 : - \tfrac{55}{36} + \tfrac{8}{3} \log(2) - \tfrac{1}{2}\log^2(2)  \\
m = 4 : \tfrac{241}{144} -\tfrac{8}{3} \log(2)-\tfrac{1}{2}\log^2(2)  \\
m = 5 : - \tfrac{6589}{3600} + \tfrac{46}{15} \log(2) + \tfrac{1}{2}\log^2(2) 
\end{gather}
