Finding the $(n-1)$th derivative of $f(x)=\frac{e^{bx}}{(x+a)^n}$ I'm interested in calculating the quantity $f(n,x)=\frac{d^{n-1}}{dx^{n-1}}\frac{e^{bx}}{(x+a)^n}$, where $a$ and $b$ are positive real constants. $n$ is a natural number.
I've tried and failed to find any kind of recursion relation which might help in finding a closed form for $f(n)$. Does anybody have any ideas on how to calculate it? I suspect that there might be a nice closed form in terms of Legendre polynomials, but maybe I'm being too hopeful.

HINT:
Using Leibniz' Rule, we have
$$\frac{d^{n-1}}{dx^{n-1}}\frac{e^{ix}}{(x+a)^n}=\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{d^k(x+a)^{-n}}{dx^k}\frac{d^{n-k}e^{ix}}{dx^{n-k}}$$
