Integral of $\int\frac{dx}{(x^4-1)^{3/2}}$

Question

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.

Answer 1

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.

Answer 2

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 $m$.

Answer 3

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 $\int\dfrac{dx}{(x^4-1)^\frac{3}{2}}$

$=\int\dfrac{dx}{x^6\left(1-\dfrac{1}{x^4}\right)^\frac{3}{2}}$

$=\int\sum\limits_{n=0}^\infty\dfrac{(2n+1)!x^{-4n-6}}{4^n(n!)^2}dx$

$=\sum\limits_{n=0}^\infty\dfrac{(2n+1)!x^{-4n-5}}{4^n(n!)^2(-4n-5)}+C$

$=-\sum\limits_{n=0}^\infty\dfrac{(2n+1)!}{4^n(n!)^2(4n+5)x^{4n+5}}+C$