The indefinite integration of $\frac{1}{\sqrt{n^4-1}}$

Question

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.

Answer 1

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.

Answer 2

Put n=tanx, after making substitution you will be left with a cosine function which is easy to integrate

Answer 3

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 $x > 1$): $$\bf 1/\text{Sqrt}[2]*\text{EllipticF}[\text{ArcCos}[1/x],1/2]$$

Answer 4

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 $\int\dfrac{1}{\sqrt{n^4-1}}dn$

$=\int\dfrac{1}{i\sqrt{1-n^4}}dn$

$=-\int\sum\limits_{k=0}^\infty\dfrac{i(2k)!n^{4k}}{4^k(k!)^2}dn$

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