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QUESTION: I want to integrate the following expression: $$\frac{dx}{\sqrt{c_1x^{-2s}-1}}$$ Where $s \in \mathbb{R}-{0}$.

My Attempt:

I could only rearrange and create the following: $$\frac{dx}{\sqrt{c_1x^{-2s}-1}}$$ $$=\frac{x^s \, dx}{\sqrt{c_1-x^{2s}}}$$ $$=\frac{x}{s}\cdot \frac{d(x^s)}{\sqrt{c^2-(x^{s})^2}}$$ where $c_1=c^2$ Integrating by parts, we get that $$=\frac{x}{s}\cdot \arcsin \left(\frac{x^s}{c}\right)-\frac{1}{s}\int \arcsin \left(\frac{x^s}{c}\right) dx + c_2$$

If you need to know where this came from, then I must add that it was obtained from the equation $(2)$ as in this link by putting $n(x,y)=x^s$.

I cannot proceed any further. Help is needed. Any kind of closed form evaluation is welcome.

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    You need the absolute value for $x^s$ in the numerator. – 2017-02-22
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    This certainly won't be elementary: even the case $s=2$ requires elliptic integrals. – 2017-02-22
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    this integral leads to a hypergeometric series – 2017-02-22
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    @Chappers Any closed form is welcome. Can you tell me if I can express this in terms of any non elementary function/ standard function? – 2017-02-22

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Suppose $s>0$ (similar calculation can be done for $s<0$, but will need different endpoints to start with). Consider $$ I = \int_0^x \frac{t^s}{c} (1-t^{2s}/c^2)^{-1/2} \, dt, $$ which is one antiderivative of your function. Setting $u=(t/x)^{2s}$ changes the limits to $0$ and $1$, and $du = 2s t^{2s-1} x^{-2s} \, dt $, so $$ I = \frac{x^{s+1}}{2sc} \int_0^1 u^{1/(2s)-1/2} (1-x^{2s}/c^2 u)^{-1/2} \, dt, $$ which is in the form of the hypergeometric integral $$ F(a,b;c;z) = \frac{1}{B(b,c-b)}\int_0^1 u^{b-1} (1-u)^{c-b-1} (1-zu)^{-a} \, du. $$ Thus $$ I = \frac{x^{s+1}}{(s+1)c} F\left( \frac{1}{2} , \frac{1}{2}+\frac{1}{2s} ; \frac{3}{2}+\frac{1}{2s} ; \frac{x^{2s}}{c^2} \right), $$ which, as they say, is better'n'nowt.

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    (+1) Strange saying, I haven't ever heard it. Does it stand for *now, it is not better than before at all*? – 2017-02-22
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    For $s<0$, do the limits go from $-x$ to $0$? – 2017-02-22
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    @JackD'Aurizio It's Northern English slang. "Better than nothing". I only know _nowt_ means nothing because _Coronation Street_ popularised it. – 2017-02-22
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    Another thing: How did $2sc$ become $(s+1)c$ in the denominator in the last line? – 2017-02-22
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    @SchrodingersCat The transformation only sends $0 \mapsto 0$ if $s>0$. For $s<0$, you need to start at $\infty$ instead, so it gets mapped to $0$. – 2017-02-22
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    @SchrodingersCat It's the beta function: $B(1/2+1/(2s),1) = 2s/(1+s) $, and you have to multiply by that to get from the integral to the $F$. – 2017-02-22