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: $y=x-7\,\,\,;y=2\cosh(t)\,\,\,;z=e^\frac{t}{2}$

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    $\begingroup$ I am afraid that this will lead to some elliptic integral. $\endgroup$ – Claude Leibovici Aug 19 '18 at 6:47
  • $\begingroup$ @ClaudeLeibovici the strangest thing is that I have taken it from an old exam of Mathematical Analysis 1, given in the uni I am attending, and elliptic integrals are in M.A.2 $\endgroup$ – Arcticmonkey Aug 19 '18 at 7:07
  • $\begingroup$ May be a typo in the integral ? Who knows ? The result from WA is still worse that what I was thinking about. $\endgroup$ – Claude Leibovici Aug 19 '18 at 7:09
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    $\begingroup$ Who wants to bet that the problem is $\int\frac{dx}{\sqrt{(x-9)(x-5)}}$ ? $\endgroup$ – Claude Leibovici Aug 19 '18 at 7:22
  • $\begingroup$ @ClaudeLeibovici A typo is possible. I'm writing an e-mail to the professor, who wrote it $\endgroup$ – Arcticmonkey Aug 19 '18 at 7:24

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.


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