How to integrate $\int_0^{\infty} \frac{x^3}{e^{\frac{x-\mu}{kT}} + 1} dx$

Question

So far, I've tried changing the variable to $y = x-\mu$, but doing this makes the lower limit become $-\mu$ instead of $0$: \begin{equation} \int_{-\mu}^{\infty}\frac{y^3 +3y^2\mu+3y\mu^2 + \mu^3}{e^{\frac{y}{kT}} + 1}dy \end{equation}

so I can't use the Riemann zeta integral to solve it:

\begin{equation} \int_{0}^{\infty} \frac{x^{\nu -1}}{e^{ax} + 1}dx = \frac{1-2^{1-\nu}}{a^{\nu}}\Gamma(\nu)\zeta(\nu) \end{equation}

Does anyone know how could I solve it? Or is there any different way to do it, with geometric series? (I haven't found an analogous to the Riemann zeta integral with lower limit...).

Thanks.