Partial fractions on $(cx^2+dx+e)^n$

Question

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?

Answer 1

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.