Denote $(q;q)_n=(1-q)(1-q^2)\cdots(1-q^n)$.

The below two identities are known. \begin{align*}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}}{(1-q^n)\,(q;q)_n} &=\sum_{n=1}^{\infty}\frac{q^n}{1-q^n}, \\ \sum_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}}{(1-q^n)^2\,(q;q)_n} &=\frac12\sum_{n=1}^{\infty}\frac{(n+1)q^n}{1-q^n}+\frac12\left(\sum_{n=1}^{\infty}\frac{q^n}{1-q^n}\right)^2. \end{align*}

QUESTION. Is there a similar expression for the following? $$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}}{(1-q^n)^m\,(q;q)_n}.$$

Remark. Let's at least try out this for small values of $m$, say $m=3$ or $m=4$.