Denote $(q;q)_n=(1-q)(1-q^2)\cdots(1-q^n)$.
The below twothree 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*}\begin{align*} \sum_{n=1}^{\infty}\frac{(-1)^{n-1}q^{\binom{n+1}2}}{(q;q)_n} &=1-\sum_{n\in\mathbb{Z}}(-1)^nq^{\frac{n(3n+1)}2}, \\ \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 $m=4$.