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

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:
$$ E(x;k) = \int_0^x \sqrt{\frac{1-k^2t^2}{1-t^2}}dt $$
