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Please refer to my previous question (How to solve this integral using Gamma Functions). Now there is a slight change in the question.

$$ \int_{-c}^c \sqrt{b - \frac{c^6}{x^6-c^6}}dx$$

I know that final answer includes gamma function entries like ($\frac{\Gamma(2/3)\Gamma(1/2)}{\Gamma(1/6)})$.

I tried to use the same method as used previously, but couldn't get desired the result.

J. M. ain't a mathematician
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Coderzz
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  • Well, a first step could be to set $u=x/c$ to get rid of the $c$ in the integral. You can also use parity to get an integral from $0$ to $1$. What else have you tried? – mickep Jul 31 '17 at 06:00

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Doing the same as JJacquelin in his answer to your previous question, using $x=ct$ we have $$I=\int \sqrt{b - \frac{c^6}{x^6-c^6}}\,dx=c \int \sqrt{b-\frac{1}{ t^6-1}}\,dt$$ that is to say $$I=\sqrt{b+1}\, c \,t F_1\left(\frac{1}{6};\frac{1}{2},-\frac{1}{2};\frac{7}{6};t^6,\frac{b t^6}{b+1}\right)$$ where appears the Appell hypergeometric function of two variables.

Integrating between $0$ and $1$ and doubling the result $$J=\int_{-c}^c \sqrt{b - \frac{c^6}{x^6-c^6}}dx=2c\sqrt{b+1 } \frac{ \Gamma \left(\frac{7}{6}\right) \,\Gamma \left(\frac{1}{2}\right) }{\Gamma \left(\frac{2}{3}\right)}\,_2F_1\left(-\frac{1}{2},\frac{1}{6};\frac{2}{3};\frac{b}{b+1}\right)$$

Edit

From a computational point of view, it is interesting to notice that $$\, _2F_1\left(-\frac{1}{2},\frac{1}{6};\frac{2}{3};u\right)=1-\frac{u}{8}-\frac{7 u^2}{320}-\frac{91 u^3}{10240}-\frac{1729 u^4}{360448}-\frac{8645 u^5}{2883584}+O\left(u^6\right)$$ which would make the converge quite fast even for large values of $b$.

Claude Leibovici
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  • Have you looked into whether or not your hypergeometric result can be further reduced to elliptic integrals. possibly through a cubic transformation or some such? – David H Jul 31 '17 at 09:36
  • @DavidH. No, I did not. Any underlying idea in your question ? I should be interested to look at something else. – Claude Leibovici Jul 31 '17 at 09:40
  • Well, the integral in question is very similar to the one asked about here. It's possible that the substitutions used there could be adapted to the problem here... – David H Jul 31 '17 at 09:54