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.

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 $\int\dfrac{dx}{(x^4-1)^\frac{3}{2}}$
$=\int\dfrac{dx}{x^6\left(1-\dfrac{1}{x^4}\right)^\frac{3}{2}}$
$=\int\sum\limits_{n=0}^\infty\dfrac{(2n+1)!x^{-4n-6}}{4^n(n!)^2}dx$
$=\sum\limits_{n=0}^\infty\dfrac{(2n+1)!x^{-4n-5}}{4^n(n!)^2(-4n-5)}+C$
$=-\sum\limits_{n=0}^\infty\dfrac{(2n+1)!}{4^n(n!)^2(4n+5)x^{4n+5}}+C$
