The indefinite integration of $\frac{1}{\sqrt{n^4-1}}$ I need the indefinite integral:
$$\int\frac{1}{\sqrt{n^4-1}}dn $$
I know it has a relation with the $\tanh^{-1}$ function, but can't find a proper substitution. 
 A: Introduce variables $c, s$ and $\theta$ such that
$$\cos\theta = c = \frac{1}{n}\quad\text{ and }\quad \sin\theta = s = \sqrt{1-c^2} = \frac{\sqrt{n^2-1}}{n}.$$
We have
$$\begin{align}
\int \frac{dn}{\sqrt{n^4-1}} 
&= - \int \frac{dc}{c^2\sqrt{\frac{1}{c^4}-1}}
= - \int \frac{dc}{\sqrt{1-c^4}}\\
&= - \frac{1}{\sqrt{2}} \int \frac{dc}{\sqrt{(1-c^2)(1 - \frac12(1-c^2))}}\\
&= \frac{1}{\sqrt{2}}\int \frac{ds}{\sqrt{(1-s^2)(1-\frac12 s^2)}}\\
&= \frac{1}{\sqrt{2}}\int \frac{d\theta}{\sqrt{1-\frac12 \sin^2\theta}}\\
\end{align}$$
The integrals in last two lines are in the Jacobi's form and regular form of the incomplete elliptic integral of the first kind for modulus $m = \frac12$:
$$
\begin{align}
F(x;m)     &= \int_0^x \frac{dt}{\sqrt{(1-t^2)(1-mt^2)}}\\
F(\phi\mid m) &= \int_0^\phi \frac{d\theta}{\sqrt{1-m\sin^2\theta}} = F(\sin\phi;m)
\end{align}
$$
As a result
$$\int_1^x \frac{dn}{\sqrt{n^4-1}}
= \frac{1}{\sqrt{2}}F\left(\frac{\sqrt{x^2-1}}{x};\frac12\right)
= \frac{1}{\sqrt{2}}F\left(\cos^{-1}\frac{1}{x}\bigg|\frac12\right)
$$
On WolframAlpha, the regular form of the incomplete elliptic integral can be accessed with the function EllipticF. The integral can be evaluated using following expression (for real $x > 1$):
$$\bf 1/\text{Sqrt}[2]*\text{EllipticF}[\text{ArcCos}[1/x],1/2]$$
A: According to Maple, it is an elliptic integral,
$$
\int\frac{dn}{\sqrt{n^4-1}} = F(-in,i)
$$
Here $i^2=-1$, a complex number.
A: Case $1$: $|n^4|\geq1$
Then $\int\dfrac{1}{\sqrt{n^4-1}}dn$
$=\int\dfrac{1}{n^2\sqrt{1-\dfrac{1}{n^4}}}dn$
$=\int\dfrac{1}{n^2}\sum\limits_{k=0}^\infty\dfrac{(2k)!n^{-4k}}{4^k(k!)^2}dn$
$=\int\sum\limits_{k=0}^\infty\dfrac{(2k)!n^{-4k-2}}{4^k(k!)^2}dn$
$=\sum\limits_{k=0}^\infty\dfrac{(2k)!n^{-4k-1}}{4^k(k!)^2(-4k-1)}+C$
$=-\sum\limits_{k=0}^\infty\dfrac{(2k)!}{4^k(k!)^2(4k+1)n^{4k+1}}+C$
Case $2$: $|n^4|\leq1$
Then $\int\dfrac{1}{\sqrt{n^4-1}}dn$
$=\int\dfrac{1}{i\sqrt{1-n^4}}dn$
$=-\int\sum\limits_{k=0}^\infty\dfrac{i(2k)!n^{4k}}{4^k(k!)^2}dn$
$=-\sum\limits_{k=0}^\infty\dfrac{i(2k)!n^{4k+1}}{4^k(k!)^2(4k+1)}+C$
A: Put n=tanx, after making substitution you will be left with a cosine function which is easy to integrate
