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

Question

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
$$ I=-\frac{x\sqrt{ax^2-2b}}{\sqrt b \sqrt{ax^4-bx^2}}\tan^{-1}\bigg(\frac{\sqrt{2b}}{\sqrt{ax^2-2b}}\bigg). $$

Answer 1

Hint: try $x= \sqrt{\frac{b}{a}}\tan{\theta}$.

Answer 2

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 $$I=\int \frac{dx}{\sqrt {ax^4-bx^2}}=-\frac{\tan ^{-1}\left(\frac{\sqrt{b}}{\sqrt{a x^2-b}}\right)}{\sqrt{b}}$$

Answer 3

$$ \begin{aligned} \int\frac{\mathrm{d}x}{\sqrt{ax^4 - bx^2}}&=\int\frac{\mathrm{d}x}{\sqrt{bx^4\left(\frac{a}{b} - \frac{1}{x^2}\right)}}\\ &=\frac{1}{\sqrt{b}}\int\frac{1}{\sqrt{\frac{a}{b} - \left(\frac{1}{x}\right)^2}}\frac{1}{x^2}\,\mathrm{d}x \end{aligned} $$ Now, set $u=\dfrac{1}{x}$ and $\mathrm{d}u=-\dfrac{1}{x^2}\,\mathrm{d}x$: $$ I=\frac{1}{\sqrt{b}} \int-\frac{1}{\sqrt{\left(\sqrt{\frac{a}{b}}\right)^{\!2} - u^2}}\,\mathrm{d}u $$ We have the integral $$ \int-\frac{\mathrm{d}u}{\sqrt{\alpha^2 - u^2}}=\arccos\left(\frac{u}{\alpha}\right)+C $$ Then, $$ \begin{aligned} I&=\frac{1}{\sqrt{b}}\arccos\left(u\sqrt{\frac{b}{a}}\right)+C\\ &=\frac{1}{\sqrt{b}}\arccos\left(\frac{1}{x}\sqrt{\frac{b}{a}}\right)+C \end{aligned} $$

Answer 4

$\displaystyle\int\dfrac{dx}{\sqrt{ax^4-bx^2}}=\displaystyle\int\dfrac{dx}{x\sqrt{ax^2-b}}=\dfrac{1}{\sqrt{a}}\displaystyle\int\dfrac{dx}{x\sqrt{x^2-\dfrac{b}{a}}}$

$\therefore x^2-\dfrac{b}{a}=t^2 \implies x\ dx=t\ dt$

$\therefore \displaystyle\int\dfrac{dx}{x\sqrt{x^2-\dfrac{b}{a}}}=\displaystyle\int\dfrac{dt}{\left(t^2+\dfrac{b}{a}\right)}$