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

The below three identities are known.
\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 $m=4$.