$\int \frac{dx}{\sqrt{|(x-a_1)(x-a_2)(x-a_3)|}}$

Question

I need to find the integrals $$\int _{-\infty}^{a_1}\frac{dx}{\sqrt{|(x-a_1)(x-a_2)(x-a_3)|}},$$ $$\int _{a_2}^{a_3}\frac{dx}{\sqrt{|(x-a_1)(x-a_2)(x-a_3)|}}$$ $a_1,a_2,a_3 \in \mathbb{R}: a_1<a_2<a_3.$

I tried to solve it using different substitution such as $u^2=t-a_3$ and $tan(\psi) =u/\sqrt{a_3-a_1}$. But it doesn't work...

Thank you for any help

Answer 1

For the first integral, substitute $ x = a_3 + (a_1 - a_3)/ y^2$. For the second integral, substitute $x = a_3 + (a_2 - a_3)y^2$. In both cases, the integral you get will reduce to $$ \int_0^1\frac{dy}{\sqrt{(1-y^2)\left(1 - \frac{a_3-a_2}{a_3 - a_1}y^2\right)}} = \frac{2}{\sqrt{a_3-a_1}}K\left(\sqrt{\frac{a_3-a_2}{a_3-a_1}}\right). $$

Answer 2

Elliptic integrals.
If $a_1 < a_2 < a_3$,

$$ \int_{-\infty }^{{a_1}}\!{\frac {1}{\sqrt { \left( {a_1}-x \right) \left( {a_2}-x \right) \left( {a_3}-x \right) }}} \,{\rm d}x = {\frac {2}{\sqrt {{a_3}-{a_1}}}{\rm K} \left( \sqrt {{\frac {{a_3}-{a_2}}{{a_3}-{a_1}}}} \right) } $$ and $$ \int_{{a_2}}^{{a_3}}\!{\frac {1}{\sqrt { \left( {a_1}-x \right) \left( {a_2}-x \right) \left( {a_3}-x \right) }}} \,{\rm d}x={\frac {2}{\sqrt {{a_3}-{a_1}}}{\rm K} \left( \sqrt {{\frac {{a_3}-{a_2}}{{a_3}-{a_1}}}} \right) } $$