Simplifying indefinite integral (possibly in terms of elliptic integrals) I would like to simplify the following integral
$$\int ds \frac{s^n}{\sqrt{1-s^2}}\frac{1}{\sqrt{-(s-s_1)(s-s_2)(s-s_3)}}$$.
for $n=0, 1$, and $2$. Here, $s_1$, $s_2$, and $s_3$ are constants (they come from the roots of a complicated cubic polynomial). I'm pretty sure that this integral can be rewritten in terms of elliptic integrals for each $n$, but I'm not sure how to proceed. Any help would be greatly appreciated!

I will write here a (very) partial answer to your problem.
First of all, I am going to consider only one case, that is $n = 0$.
Even in this case, it seems like no expression exists for a "primitive" of that integral.
However if we make a deal (a bad one) we can manage to find an expression. The deal is: one of the three roots must be zero. If that happens, then we have a very nasty expression for the integral.
I will use $a, b, c$ instead of $s_1, s_2, s_3$ to avoid confusion and to write clear.
In the first case, we need to assume $c = 0$ to get an expression for the integral, as said.
$$\int \frac{\text{d}s}{\sqrt{1-s^2}}\frac{1}{\sqrt{-s^2(s-a)(s-b)}} = -\frac{\color{blue}{2 (s-1) s (b-s) \sqrt{\frac{(s+1) (a-b)}{(b+1) (a-s)}} \left(b F\left(\sin ^{-1}\left(\sqrt{\frac{(a-1) (b-s)}{(b-1) (a-s)}}\right)|\frac{(a+1) (b-1)}{(a-1) (b+1)}\right)+(a-b) \Pi \left(\frac{a (b-1)}{(a-1) b};\sin ^{-1}\left(\sqrt{\frac{(a-1) (b-s)}{(b-1) (a-s)}}\right)|\frac{(a+1) (b-1)}{(a-1) (b+1)}\right)\right)}}{\color{red}{a (b-1) b \sqrt{1-s^2} \sqrt{\frac{(a-1) (s-1) (a-b) (b-s)}{(b-1)^2 (a-s)^2}} \sqrt{s^2 (a-s) (s-b)}}}$$
Where $F$ denotes the complete Elliptic integral of the first kind and $\Pi$ denotes the complete elliptic integral of the third kind.

Second case: $n = 1$.
Here we get, assuming $c = 0$ as before.
$$\int \frac{s}{\sqrt{1-s^2}}\frac{1}{\sqrt{-s^2(s-a)(s-b)}}\text{d}s = \frac{\color{blue}{2 \sqrt{\frac{(s-1) (a-b)}{(b-1) (a-s)}} \sqrt{\frac{(s+1) (a-b)}{(b+1) (a-s)}} \sqrt{-s^2 (s-a) (s-b)} F\left(\sin ^{-1}\left(\sqrt{\frac{(a+1) (b-s)}{(b+1) (a-s)}}\right)|\frac{(a-1) (b+1)}{(a+1) (b-1)}\right)}}{\color{red}{s \sqrt{1-s^2} (a-b) \sqrt{\frac{(a+1) (b-s)}{(b+1) (a-s)}}}}$$
With the same meaning of $F$ as before.
Again I did not find any expression for $c\neq 0$.

Third Case: $n = 2$. Always assuming $c = 0$ we have
$$\int \frac{s^2}{\sqrt{1-s^2}}\frac{1}{\sqrt{-s^2(s-a)(s-b)}}\text{d}s = -\frac{\color{blue}{2 s (s+1) (b-s) \sqrt{\frac{(s-1) (a-b)}{(b-1) (a-s)}} \left(a F\left(\sin ^{-1}\left(\sqrt{\frac{(a+1) (b-s)}{(b+1) (a-s)}}\right)|\frac{(a-1) (b+1)}{(a+1) (b-1)}\right)+(b-a) \Pi \left(\frac{b+1}{a+1};\sin ^{-1}\left(\sqrt{\frac{(a+1) (b-s)}{(b+1) (a-s)}}\right)|\frac{(a-1) (b+1)}{(a+1) (b-1)}\right)\right)}}{\color{red}{(b+1) \sqrt{1-s^2} \sqrt{\frac{(a+1) (s+1) (a-b) (b-s)}{(b+1)^2 (a-s)^2}} \sqrt{s^2 (a-s) (s-b)}}}$$
Same meaning of $F$ and $\Pi$.
Again, no expression found for $c\neq 0$.
I apologise for thise answer which does not cover the general cases you presented here. However this could be a start.
Otherwise, you need numerical integration.
Postface
I coloured the terms because the writing is rather nasty, and confusing too. In each of the expressions, the results come into a form of fraction: the $\color{blue}{BLUE}$ part is the numerator and the $\color{red}{RED}$ part is the denominator.
