I require to evaluate the following indefinite integration:
$$\int\left(\frac{A}{r^2}+\frac{B}{r^3}+\frac{C}{r^4}+\frac{D}{r^5}+\frac{E}{r^6} \right)\left(1+\frac{F}{r^2}+\frac{G}{r^3}\right)^{-1}dr$$ which can be re-expressed using $\epsilon=\frac{F}{r^2}+\frac{G}{r^3}$ as $$\int\left(\frac{A}{r^2}+\frac{B}{r^3}+\frac{C}{r^4}+\frac{D}{r^5}+\frac{E}{r^6} \right)\left(1+\epsilon\right)^{-1}dr\\ \qquad\qquad=\int\left(\frac{A}{r^2}+\frac{B}{r^3}+\frac{C}{r^4}+\frac{D}{r^5}+\frac{E}{r^6} \right)\left(1-\epsilon+\epsilon^2-\epsilon^3+\dots\right)dr$$
I am trying to solve the integral by expanding the denominator in binomial series rather than the standard process of evaluating the roots of the denominator because the roots (one real root and two complex roots) are extremely complicated to deal with as in this question.
However, to get the approximately correct value of the integral, I require to expand the series up to $\epsilon^{11}$ and then Mathematica yields a series expansion of the integral that contains around $40$ terms. Though this method gives approximately the correct curves of the integral, the method seems to be too much complicated.
I am asking the question to know if any simpler method is possible to solve this integral.
A related question is How to integrate a binomial expression without expanding it before?
In the first place, you should rewrite as
$$\int\frac{A+Bs+Cs^2+Ds^3+Es^4}{F+Gs^2+s^3}ds.$$
(Notice the displacement of the coefficients at the denominator.) Then by long division you will get rid of the terms $Ds^3+Es^4$.
Next, knowing the real root, factor the denominator as $(s+a)(s^2+bs+c)$ and decompose as
$$\frac{A'+B's+C's^2}{(s+a)(s^2+bs+c)}=\frac{\alpha(2Gs+3s^2)}{F+Gs^2+s^3}+\frac{\beta}{s+a}+\frac\gamma{s^2+bs+c}.$$
All three terms are relatively easy, and the computation involves only real numbers.