Does the following series have a close form? $$\sum_{n=1}^{\infty}\frac{n^3q^{\frac{3n}{2}}}{\left(1-q^n\right)^3}$$ The final result may be related with combinations of Jacobi theta function, Dedekind function $\eta(\tau)$ and Eisenstein series.
This kind of series appear when we compute the following kind of integral: $\mathcal{I}=\int_0^1 d\mathfrak{a}_1\int_0^1 d\mathfrak{a}_2\wp\left(\mathfrak{a}_1-\mathfrak{a}_2+\frac{\tau}{2}|\tau\right)\wp\left(2\mathfrak{a}_1+\mathfrak{a}_2+\frac{\tau}{2}|\tau\right)\wp\left(2\mathfrak{a}_2+\mathfrak{a}_1+\frac{\tau}{2}|\tau\right)$ where the Weierstrass p function $\wp(z|\tau)$ is defined by: $$ \wp(z|\tau) = \ \frac{1}{z^2} + \sum_{(m,n) \ne (0,0)} \left[\frac{1}{(z - m - n \tau)^2} - \frac{1}{(m + n \tau)^2}\right] $$ By using its Fourier expansion $$\ 4\pi^2 \mathbb{E}_2(\tau) - 2 \pi^2 i \sum_{n \in \mathbb{Z}-\{0\}}\frac{n q^{- \frac{n}{2}}}{\sin n \pi \tau}e^{2\pi i nz}$$ we can transform this integral into: $$\mathcal{I}=64\pi^6\mathbb{E}_2(\tau)^3-128\pi^6\sum_{n=1}^{\infty}\frac{n^3q^{\frac{3n}{2}}}{\left(1-q^n\right)^3}$$ where $\mathbb{E}_2(\tau)$ represents the Eisenstein Series.
$$\sum_{n=1}^{\infty}\frac{n^3q^{\frac{3n}{2}}}{\left(1-q^n\right)^3}=-\frac{1}{4}\left(2 \mathbb{E}_2(\tau)^3-30\mathbb{E}_2(\tau)\mathbb{E}_4(\tau)+70\mathbb{E}_6(\tau)\right)+\frac{1}{16}\left(2 \mathbb{E}_2\left(\frac{\tau}{2}\right)^3-30\mathbb{E}_2\left(\frac{\tau}{2}\right)\mathbb{E}_4\left(\frac{\tau}{2}\right)+70\mathbb{E}_6\left(\frac{\tau}{2}\right)\right)\notag\\ -\frac{3}{8}\mathbb{E}_4\left(\frac{\tau}{2}\right)+\frac{3}{8}\mathbb{E}_4\left(\tau\right)$$