Integrate: $$\int\frac{1}{\sqrt{x(x-9)(x-5)}}\,dx$$
I did some substitutions, but it seems not to be the right path to follow. Some hints?
Noticing that $x(x-9)(x-5) =x((x-7)^2-4)$ we have: $$\int\frac{1}{\sqrt{x(x-9)(x-5)}}\,dx\,\,= \int\frac{1}{\sqrt{(y+7)(y^2-4)}}\,dy\,\,\,=\int\frac{1}{\sqrt{7+2\cosh(t)}}\,dt\,\,= \int\frac{2}{\sqrt{z^4+7z^2+1}}\,dz\,\,= \int\frac{2}{\sqrt{\left(z^2+\frac{7}{2}\right)^2- \frac{45}{4}}}\,dz\,\,=\,\,??? $$
Substitutions are: $y=x-7\,\,\,;y=2\cosh(t)\,\,\,;z=e^\frac{t}{2}$
In this case, there is a cubic expression in the square root, which, unfortunately, means elliptic integral, and related, elliptic functions.
Consider the Weierstrass' elliptic function, which is the inverse of the following integral:
$$u = \int_y^\infty \frac {ds} {\sqrt{4s^3 - g_2s -g_3}}$$
where $g_2, g_3$ are constants. Your integral can be readily expressed in this form by a direct substitution. Then determining the actual values of $g_2, g_3$, you can calculate the fundamental parallelogram of the elliptic function. After that, you can use the formulae that express the inverse of Weierstrass' elliptic function in terms of incomplete elliptic integrals.
Or, try Wolfram Alpha.