Inverse Laplace transform of $\frac{1}{s^b-c}$. I am trying to find the inverse Laplace transform of $$F(s) = \frac{1}{(1+a\,s)^b-c}$$ where $a$, $b$, and $c$ are positive real numbers.
For $c=0$, we can use the following: $$\mathcal L^{-1}\left\{\frac1{s^b}\right\}=\frac{x^{b-1}}{\Gamma(b)}, \qquad for\quad b>0$$ Then we
have 
$$\mathcal L^{-1}\left\{\frac1{(1+a s)^b}\right\}
=\frac1{a^b}\mathcal L^{-1}\left\{\frac1{(\frac1{a}+s)^b}\right\}
=\frac{e^{-\frac{x}{a}}}{a^b}\mathcal L^{-1}\left\{\frac1{s^b}\right\}
=\frac{e^{-\frac{x}{a}}}{a^b}\frac{x^{b-1}}{\Gamma(b)}.$$
Now, what do I do with $c\neq 0\,$?
This simplifies the question to: 

What is the inverse Laplace transform of $F(s)$ given by:
  $$\frac{1}{s^b-c}?$$

 A: Reduction to $c=1$ is just "time scaling":
$$\mathcal{L}^{-1}\left\{\frac{1}{s^b-1}\right\}=f(x)\implies\mathcal{L}^{-1}\left\{\frac{1}{s^b-c}\right\}=c^{1/b-1}f(c^{1/b}x).$$
For the former, there's also a "series path": around $s=+\infty$,
$$\frac{1}{s^b-1}=\sum_{n=1}^{\infty}s^{-nb}\implies\mathcal{L}^{-1}\left\{\frac{1}{s^b-1}\right\}=\sum_{n=1}^{\infty}\frac{x^{nb-1}}{\Gamma(nb)}$$
as you know already, leading to the Mittag-Leffler function.
A: Not a complete solution, merely reducing the number of parameters in the integrand by one.
\begin{align*}
\frac{1}{2\pi\mathrm{i}}&\int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty}
\frac{1}{s^b-c} \mathrm{e}^{st} \,\mathrm{d}s   \\
&= \frac{1}{2\pi\mathrm{i}}
\frac{1}{c} \int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty} \frac{1}{(s^b)/c -1} \mathrm{e}^{st} \,\mathrm{d}s  \\
&= \frac{1}{2\pi\mathrm{i}}
\frac{1}{c} \int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty} \frac{1}{(s/c^{1/b})^b -1} \mathrm{e}^{st} \,\mathrm{d}s  \\
&= \frac{1}{2\pi\mathrm{i}}\frac{1}{c} \int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty} \frac{1}{(c^{1/b}\hat{s}/c^{1/b})^b -1} \mathrm{e}^{c^{1/b}\hat{s}t} \,c^{1/b}\mathrm{d}\hat{s}  \\
&= \frac{1}{2\pi\mathrm{i}}\frac{c^{1/b}}{c} \int_{\gamma - \mathrm{i}\infty}^{\gamma + \mathrm{i}\infty} \frac{1}{\hat{s}^b -1} \mathrm{e}^{\hat{s}\hat{t}} \,\mathrm{d}\hat{s}  \text{,}
\end{align*}
where $\hat{t} = c^{1/b} t$.
The singularities of $\frac{1}{\hat{s}^b -1}$ are on the unit circle.  (There's also a branch cut for almost all choices of $b$ which we may choose to run along the negative real axis).  So any $\gamma > 1$ suffices.
