Integral of $\int\frac{dx}{(x^4-1)^{3/2}}$ Can $$\int\frac{dx}{(x^4-1)^{3/2}}$$ be represented in terms of elementary integrals?
I have tried multiple substitutions, such as $x^2=\sec(t)$ or $\sqrt{x^4-1}=t$, but nothing seems to be working. Please provide some insight.
 A: This will be a way, but I warn you that what will we obtain in the end will need a huge computational help the same.
Starting to write the integrand function as
$$S = \int\left(x^4\left(1 - \frac{1}{x^4}\right)\right)^{-3/2}\ \text{d}x$$
Arranging a  bit
$$S = \int x^{-6} \left(1 - \frac{1}{x^4}\right)^{-3/2}\ \text{d}x$$
Now we can use the Binomial Series for the expression in the brackets: 
$$S = \int x^{-6}\sum_{k = 0}^{+\infty}\binom{-3/2}{k}\left(-\frac{1}{x^4}\right)^k\ \text{d}x$$
Arranging again
$$S = \sum_{k = 0}^{+\infty}(-1)^k \binom{-3/2}{k}\int x^{-6-4k}\ \text{d}x$$
Integration is trivial and you'll obtain in the end the series
$$S = \sum_{k = 0}^{+\infty}(-1)^k \binom{-3/2}{k}\frac{x^{-5-4k}}{-5-4k}$$
The series does converge to a so called hypergeometric function, which is:
$$S = -\frac{_2F_1\left(\frac{5}{4},\ \frac{3}{2},\ \frac{9}{4},\ \frac{1}{x^4}\right)}{5x^5}$$
The plot of this function is

More on hypergeometric functions:
http://mathworld.wolfram.com/HypergeometricFunction.html
https://en.wikipedia.org/wiki/Hypergeometric_function
Otherwise you can always refer to the Jacobi elliptic integral as mentioned.
A: From formula 260.78 of Byrd/Friedman, we make the substitution $x=\mathrm{nc}\left(u\mid\frac12\right)$, where $\mathrm{nc}\left(u\mid m\right)$ is a Jacobi elliptic function with parameter $m$. This yields the integral
$$\int\frac1{\mathrm{nc}^4\left(u\mid\frac12\right)-1}\mathrm du$$
Skipping the messy algebra (but if you want to do it yourself, use formulae 320.02 and 361.61 from Byrd/Friedman after a partial fraction decomposition, then undo the substitution), we finally obtain the result
$$-\frac{1}{2\sqrt{2}}F\left(\cos^{-1}\left(\frac1{x}\right)\mid\frac12\right)-\frac{x}{2\sqrt{x^4-1}}$$
where $F(\phi\mid m)$ is the incomplete elliptic integral of the first kind with amplitude $\phi$ and modulus $m$.
A: Case $1$: $|x^4|\leq1$
Then $\int\dfrac{dx}{(x^4-1)^\frac{3}{2}}$
$=\int\dfrac{i}{(1-x^4)^\frac{3}{2}}dx$
$=\int\sum\limits_{n=0}^\infty\dfrac{i(2n+1)!x^{4n}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\infty\dfrac{i(2n+1)!x^{4n+1}}{4^n(n!)^2(4n+1)}+C$
Case $2$: $|x^4|\geq1$
Then $\int\dfrac{dx}{(x^4-1)^\frac{3}{2}}$
$=\int\dfrac{dx}{x^6\left(1-\dfrac{1}{x^4}\right)^\frac{3}{2}}$
$=\int\sum\limits_{n=0}^\infty\dfrac{(2n+1)!x^{-4n-6}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(2n+1)!x^{-4n-5}}{4^n(n!)^2(-4n-5)}+C$
$=-\sum\limits_{n=0}^\infty\dfrac{(2n+1)!}{4^n(n!)^2(4n+5)x^{4n+5}}+C$
