Find $\int\frac1{\sqrt{1+x^4}}\,dx$

Question

$$ \text{Find } \int \frac 1 {\sqrt{1+x^4}} \, dx$$

Let $x^2=\tan u$ $\implies 2x \,dx=\sec^2 u \,du$ $\implies dx=\dfrac{\sec^2 u}{2\sqrt{\tan u}}\,du$

$$= \int \frac{\sec^2 u}{2\sec u\sqrt{\tan u}} \, du $$ $$= \int \frac{\sec u}{2\sqrt{\tan u}} \, du $$

I am unsure how to continue..

Answer 1

Let $$I = \int\frac{1}{\sqrt{1+x^4}}dx\;,$$ Now put $x^2=\tan t\;,$ Then $2xdx = \sec^2 tdt$

So $$I = \frac{1}{2}\int\frac{\sec^2 t}{\sec t \sqrt{\tan t}}dt = \frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{\sin 2 t}}dt$$

Now put $\displaystyle 2t=\frac{\pi}{2}-2\theta\;,$ Then $dt = -d\theta\;,$ So we get $$I = -\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{\cos 2\theta}}d\theta$$

So $$I = -\frac{1}{\sqrt{2}}\int\frac{1}{\sqrt{1-2\sin^2 \theta}}d\theta = -\frac{1}{\sqrt{2}}\int^{\theta}_{0}\frac{1}{\sqrt{1-2\sin^2 u}}du = -\frac{1}{\sqrt{2}}F\left(\theta\mid 2\right)+\mathcal{C}$$

$$ = -\frac{1}{\sqrt{2}}F\left(\frac{\frac{\pi}{4}-t}{2}\mid 2\right) +\mathcal{C}= -\frac{1}{\sqrt{2}}F\left(\frac{\frac{\pi}{4}-\tan^{-1}x^2}{2}\mid 2\right)+\mathcal{C}$$

Using elliptical integral of first kind, $$F(\phi\mid k^2) = \int^{\phi}_{0}\frac{1}{\sqrt{1-k^2\sin^2\theta}}d\theta$$

Answer 2

the integral can't expressed by a known elelmentary function, Mathematica says this here $$-\sqrt[4]{-1} F\left(\left.i \sinh ^{-1}\left(\sqrt[4]{-1} x\right)\right|-1\right)$$

Answer 3

$$\displaystyle \int{\dfrac{1}{\sqrt{1-m^4}}dm} = F(\arcsin(m)|-1)$$

Let $x = (-1)^{1/4}m$

$$-\int{\dfrac{1}{(-1)^{1/4}\sqrt{1+x^4}}dx} = F(\arcsin(x/(-1)^{1/4})|-1) $$

$$ \int{\dfrac{1}{\sqrt{1+x^4}}dx} =(-1)^{1/4} F(\sin^{-1}(x/(-1)^{1/4})|-1) = (-1)^{1/4} F(-\arcsin((-1)^{3/4} x)| -1 ) $$

Where $F(x|y)$ is the first kind of Elliptic integral function.

EDIT: Sorry for my carelessness, I didn't appraise my steps about $dx$ substitution, the equivalence can check from Wolframalpha.

UPGRADE: To show the equivalence of my answer and the answer given by WA,

By $F(-x|y) = -F(x|y)$, $$(-1)^{1/4} F(-\arcsin((-1)^{3/4} x)| -1 ) =-(-1)^{1/4} F(\arcsin((-1)^{3/4} x)| -1 ) $$

By $arcsin(ix) = iarcsinh(x)$

$$-(-1)^{1/4} F(\arcsin((-1)^{3/4} x)| -1 ) = -(-1)^{1/4} F(i \times arcsinh((-1)^{1/4} x)| -1 )$$

Everything is done.

Answer 4

For any real number of $x$ ,

When $|x|\leq1$ ,

$\int\dfrac{1}{\sqrt{1+x^4}}dx$

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

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

When $|x|\geq1$ ,

$\int\dfrac{1}{\sqrt{1+x^4}}dx$

$=\int\dfrac{1}{x^2\sqrt{1+\dfrac{1}{x^4}}}dx$

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

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

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