# Indefinite Integral $\int \frac{dx}{\sqrt {ax^4-bx^2}}$

I am trying to Integrate $$I=\int \frac{dx}{\sqrt {ax^4-bx^2}}, \qquad a,b\in \mathbb{R}.$$ Thanks. I tried to do $x=\sin \phi$ $$\int \frac{\cos \phi\, d\phi}{\sqrt{a\sin^4 \phi-b\sin^2 \phi}}=\int \frac{\cot \phi \, b\phi}{\sqrt{a\sin^2\phi-b}}$$ but get stuck here. Mathematica gives a closed form result
$$I=-\frac{x\sqrt{ax^2-2b}}{\sqrt b \sqrt{ax^4-bx^2}}\tan^{-1}\bigg(\frac{\sqrt{2b}}{\sqrt{ax^2-2b}}\bigg).$$

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Shouldn't you have $$I=-\dfrac{x\sqrt{cx^2-2d}}{\sqrt{d}\sqrt{cx^4-dx^2}}\tan^{-1}\left(\dfrac{\sqr‌​t{2d}}{\sqrt{cx^2-2d}}\right)?$$ –  Sanath Devalapurkar May 9 at 3:22
The denominator can be rewritten as $x\sqrt{ax^2-b}$. An obvious substitution would be $x=\sqrt{\dfrac ba}\cdot\cosh t$ –  Lucian May 9 at 3:38

Hint: try $x= \sqrt{\frac{b}{a}}\tan{\theta}$.
You have been given two possible changes of variable. A key point was also mentioned by Lucian for a rewrite of the denominator. After all of that, you should arrive to something much simpler that what Mathematica gave you (you did not simplify its result) since $$I=\int \frac{dx}{\sqrt {ax^4-bx^2}}=-\frac{\tan ^{-1}\left(\frac{\sqrt{b}}{\sqrt{a x^2-b}}\right)}{\sqrt{b}}$$