I am trying to find the first 2-3 terms of the asymptotic expansion in terms of 1/ρ of the elliptic integral \begin{equation} I_n(\rho)=\int_0^\frac{h_2}{\rho}\frac{t^{2n}/h_2^{2n}}{(E_n(t))^2\sqrt{1-t^2}\sqrt{1-\frac{h_3^2}{h_2^2}t^2}}dt. \end{equation}
where E is a polynomial and $E^2$ can be written as \begin{equation} (E_n(t))^2=\sum_{r}c_rt^{2r}, \end{equation}
Using binomial theorem i deduced that
\begin{equation}I_n(\rho)=\int_0^{h_2/\rho}\left[\frac{t^{2n}/h_2^{2n}}{(E_n(t))^2 }\sum_{n=0}^{+\infty}\sum_{k=0}^n(-1)^n\binom{-\frac{1}{2}}{n-k}\binom{-\frac{1}{2}}{k}\frac{h_3^{2k}}{h_2^{2k+1}}t^{2n}\right] dt .\end{equation}
Because i can t find a closed form series, is it OK to expand $1/E_n^2(t)$ in terms of $1/t$ and then integrate term by term? i.e is it right to take Taylor expansion for \begin{equation} \frac{1}{(E_n(t))^2}=\frac{1}{c_0+c_1t^2+c_1t^4+\cdots} \end{equation} which is \begin{equation} \frac{1}{(E_n(t))^2}=\frac{1}{c_0}-\frac{c_1}{c_0^2} t^2+\frac{c_1^2-c_0 c_2}{c_0^3}t^4+O(t^6) \end{equation} and then after multiplicating, integrate term by term?
Thank you in advance!