Is there a nice way to represent $\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{n+m+1}$?

Question

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!

Answer 1

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}