Solve $\int\frac{a^2-x^2}{\sqrt{(a^2-x^2)^2-e^2}}dx$

Question

Solve $\int\frac{a^2-x^2}{\sqrt{(a^2-x^2)^2-e^2}}dx$

First I thought of adding and subtracting $e^2$ in the numerator but then realized there was a whole square of $a^2-x^2$ in the denominator, so, dropped this idea.

Then I tried substituting $a^2-x^2=t$ but it gave me $-2xdx=dt\implies dx=-\frac{dt}{2\sqrt{a^-t}}$, thus the integration became $$-\int\frac{tdt}{2\sqrt{a^2-t}\sqrt{t^2-e^2}}$$

It's not appearing very helpful either. Any hint please?

There is a closed form but it requires elliptic integrals of the first and second kinds. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Jul 06 '21 at 10:39
In fact, there is nothing special about this particular form, the elliptic integral method works for integrals of the form $$\int\frac{x^2-A}{\sqrt{x^4-Bx^2+C}},dx$$ for constants $A,B,C$. — ə̷̶̸͇̘̜́̍͗̂̄︣͟, Jul 06 '21 at 10:42

score 2 · Answer 1 · answered Jul 06 '21 at 10:49

Using $$(a^2-x^2)^2-e^2 = (a^2-x^2+e)(a^2-x^2-e)$$ and $$a^2-x^2 = \frac{1}{2}\big((a^2-x^2+e)+(a^2-x^2-e)\big) $$ we can write this integral as $$ \frac12\int\sqrt{\frac{a^2-x^2+e}{a^2-x^2-e}}dx + \frac12\int\sqrt{\frac{a^2-x^2-e}{a^2-x^2+e}}dx$$ and by appropriate rescaling of variable $x$ these integrals can be expressed using the incomplete elliptic integral of the second kind: $$ E(x;k) = \int_0^x \sqrt{\frac{1-k^2t^2}{1-t^2}}dt $$

Solve $\int\frac{a^2-x^2}{\sqrt{(a^2-x^2)^2-e^2}}dx$

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