How to calculate integral $\int \frac{dx}{\sqrt{(x^2-B_1^2)(x^2-B_2^2)}}$

Question

How to calculate the integral $$\int \frac{dx}{\sqrt{(x^2-B_1^2)(x^2-B_2^2)}}, $$ where $B_1,B_2 \in \mathbb{R}$ are constants.

I think it involves elliptical integrals and more so, that the integral is the form $sn(x)$. I consulted the book Hand Book of Elliptical Integrals for Engineers and Scientis (Byrd and Friedman) but found nothing to help me.

Answer 1

Let $B_1^2 = a$ and $B_2^2 = b$. Then tables (or software) give:

$$\frac{\sqrt{1-\frac{x^2}{a}} \sqrt{1-\frac{x^2}{b}} F\left(\sin ^{-1}\left(\sqrt{\frac{1}{a}} x\right)|\frac{a}{b}\right)}{\sqrt{\frac{1}{a}} \sqrt{\left(a-x^2\right) \left(b-x^2\right)}}$$

Where $F$ denotes an elliptic integral.

Answer 2

my idea is

$\frac{1}{\sqrt{(x^2-a^2)(x^2-b^2)}}=\frac{1}{\sqrt{(x^2-b^2+b^2-a^2)(x^2-b^2)}}=\frac{1}{\sqrt{(x^2-b^2)^2 + (x^2-b^2)(b^2-a^2)}}$, then if you choose $t=x^2-b^2$ then $dt=2xdx$ and $x=\sqrt{t+b^2}$ or equivalent $\frac{dt}{2\sqrt{t+b^2}}=dx$ so we need to resolve $\frac{1}{2} \int \frac{dt}{t \sqrt{t+b^2}}$, again if we put $u=\sqrt{t+b^2}$ then $du=\frac{dt}{2\sqrt{t+b^2}}$ and $t=u^2-b^2$ or equivalent $2udu=dt$ and your new integral is $\int \frac{u}{u^2-b^2}= \int \frac{u-b+b}{(u-b)(u+b)}= \int \frac{du}{u+b} +\int \frac{b}{u^2-b^2}= ...$ and backup with your initial substitution