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: $$\frac{1}{s^b-c}?$$