Partial fractions on $(cx^2+dx+e)^n$ If I have 
$$\frac{ax+b}{(cx^2+dx+e)^n}$$
with real coefficients and $(cx^2+dx+e)$ has complex roots, what does
$$\frac{ax+b}{[c(x-\alpha)(x-\alpha^*)]^n}$$
turn into, in terms of partial fractions?

Based on differentiating this answer w.r.t $x$, you have:
$$ \frac{1}{ \left( x-\mu \right)  ^
{1+n} \left( x-\nu \right) ^
{1+n} }=\sum _{m=0}^{n}{2\,n-m\choose n} \left( -\nu+\mu
 \right) ^{m-1-2\,n} \left( {\frac { \left( -1 \right) ^{n+m}}{
 \left( -x+\nu \right) ^{m+1}}}-{\frac { \left( -1 \right) ^{n}}{
 \left( -x+\mu \right) ^{m+1}}} \right) $$
which, if you add up terms backwards $(m\rightarrow n-m)$, is equivalent to:
$$\sum _{m=0}^{n}{n+m\choose n} \frac{1}{\left( -\nu+\mu \right) ^{1+n+m}}\left( {\frac {
 \left( -1 \right) ^{-m}}{ \left( -x+\nu \right) ^{n-m+1}}}-{\frac {
 \left( -1 \right) ^{n}}{ \left( -x+\mu \right) ^{n-m+1}}} \right) 
 $$
It is simple to multiply by the numerator in your question if you so wish.
