Is there a closed form for the integral $\int_0^1 x^n \log^m (1-x) \, {\rm d}x$?

Change variables $x=1-t$: $$ I=\int_0^1 x^n \log^m (1-x) \, {\rm d}x=\int_0^1dt (1-t)^n\log^m t=\sum_{k=0}^n {n\choose k}(-1)^{n-k}\int_0^1 dt \ t^{n-k}\log^m t\ . $$ Now change variable $t=\exp(z)$ and get $$ I=\sum_{k=0}^n {n\choose k}(-1)^{n-k}\int_{-\infty}^0 dz\ e^{(n-k+1)z}z^m= \boxed{\Gamma (m+1)\sum_{k=0}^n {n\choose k}\frac{(-1)^{n-k+m} }{ (n+1-k)^{m+1}}}\ , $$ which is a finite sum.

Another closed form follows by differentiating the beta function multiple times and applying the Faà di Bruno's formula.

Claim. For positive integers $m$ and $n$, $$ \mathcal{J}_{n,m} := \int_0^1 x^{n-1}\log^m (1-x) \, \mathrm{d}x = (-1)^m \frac{m!}{n} \sum_{\alpha\in I_m} \prod_{k=1}^m \frac{1}{\alpha_k!} \bigg(\frac{H_n^{(k)}}{k}\bigg)^{\alpha_k} \tag{1} $$ where $\alpha$ runs over the set of indices $$I_m = \{(\alpha_1,\cdots,\alpha_m)\in\Bbb{N}_0^m : 1\cdot\alpha_1+\cdots+m\cdot\alpha_m=m\}.$$

This formula gives an almost explicit formula for $\mathcal{J}_{n,m}$ in terms of polynomial of $H_n^{(1)}, \cdots, H_n^{(n)}$ at the expense of introducing certain combinatorial object, namely $I_m$.

Proof. Notice that

$$ \int_0^1 x^{n-1}(1-x)^s \, \mathrm{d}x = \frac{(n-1)!}{(s+1)\cdots(s+n)} = (n-1)!\exp\left(-\sum_{j=1}^n \log(s+j) \right). $$

Letting $f(s) = -\sum_{j=1}^n \log(s+j) $ and applying the Faà di Bruno's formula, we have

$$ \mathcal{J}_{n,m} = (n-1)!e^{f(0)} \sum_{\alpha \in I_m} m! \prod_{k=1}^{m} \frac{1}{\alpha_k !} \bigg( \frac{f^{(k)}(0)}{k!} \bigg)^{\alpha_k}. \tag{2}$$

Plugging $f(0) = -\log n!$ and

$$ f^{(k)}(0) = \sum_{j=1}^n (-1)^k (k-1)! (s+j)^{-k} \bigg|_{s=0} = (-1)^k (k-1)! H_n^{(k)} $$

into $\text{(2)}$ and simplifying the resulting expression yields $\text{(1)}$.