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

Question

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

My Try: I used parts as follows:

$$I=x{\sqrt{x^4+1}}-4 \int \frac{x^4 dx}{\sqrt{x^4+1}}$$ $\implies$

$$I=\frac{x}{\sqrt{x^4+1}}-4I+4 \int \frac{dx}{\sqrt{x^4+1}}$$ $\implies$

$$5I=\frac{x}{\sqrt{x^4+1}}+4J$$ where

$$J=\int \frac{dx}{\sqrt{x^4+1}}$$

any clue here?

Answer 1

$$\int \sqrt{x^4+1}\:dx$$ This integral cannot be written on a closed form with a finite number of elementary functions.

Either you express it on the form of infinite series, or you need a special function.

A more general problem is $\quad\int (x^a+1)^b\:dx$

In the general case, it involves a special function called "Hypergeometric 2F1" : $$\int (x^a+1)^b\:dx = x\:\:_2\text{F}_1\left(\frac{1}{a}\:,\:-b\:;\:1+\frac{1}{a}\:;\:-x^a\right)$$ These kind of function has numerous particular cases, especially when $a$ and/or $b$ are integer or rational numbers.

In the case $a=4$ and $b=\frac{1}{2}$ the hypergeometric function reduces into a function of lower level which involves a simpler kind of function called Elliptic Integral. But it isn't simpler in fact because they are complex numbers in the formula. So, the general form : $$\int \sqrt{x^4+1}\:dx = x\:\:_2\text{F}_1\left(\frac{1}{4}\:,\:-\frac{1}{2}\:;\:\frac{5}{4}\:;\:-x^4\right)$$ might be preferable. If not, look for series expansion (depending on the limits of the integral). In practice, numerical integration could be the most efficient way.

http://mathworld.wolfram.com/HypergeometricFunction.html

http://mathworld.wolfram.com/EllipticIntegral.html

A paper for general public about the use of special functions : https://fr.scribd.com/doc/14623310/Safari-on-the-country-of-the-Special-Functions-Safari-au-pays-des-fonctions-speciales

Answer 2

When $|x|\leq1$ ,

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

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

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

When $|x|\geq1$ ,

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

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

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

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

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^{3-4n}}{4^n(n!)^2(1-2n)(3-4n)}+C$

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!}{4^n(n!)^2(2n-1)(4n-3)x^{4n-3}}+C$

Answer 3

To get

$$I = \int\sqrt{x^4+1}\,\mathrm{d}x=\int\frac{x^4+1}{\sqrt{x^4+1}}\,\mathrm{d}x.$$

First use partial fractions to get

$$ I=x{\sqrt{x^4+1}}-2\int x \mathrm{d}\sqrt{x^4+1}=x{\sqrt{x^4+1}}-2 \int \frac{x^4+1-1}{\sqrt{x^4+1}}\mathrm{d}x $$

and rewrite as

$$ I=x{\sqrt{x^4+1}}-2I+2 J,\quad \mathrm{where}\quad J=\int \frac{1}{\sqrt{x^4+1}}\mathrm{d}x $$

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let $t^4=-x^4$, solved

$$ \begin{aligned} t &= e^{\pi i / 4} x\\ x &= e^{-\pi i / 4} t\\ 1-x^4&=1 - t^4= (1-t^2)(1- i^2t^2)\\ \end{aligned} $$

So

$$ J= \int \frac{1}{\sqrt{1+x^4}}\mathrm{d}x = \int \frac{e^{-\pi i/4}}{\sqrt{1-t^4}}\mathrm{d}x=e^{-\pi i/4}F(\arcsin t; i^2) $$

where F(x;k) is Elliptic Integral of the First Kind.

$$ F(x;k^2)=\int _{0}^{\sin x}{\frac {{\rm {d}}t}{\sqrt {(1-t^{2})(1-k^{2}t^{2})}}} $$

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Then

$$ \begin{aligned} I &=\frac{x}{3}{\sqrt{x^4+1}}+\frac{2}{3}J\\ &=\frac{x}{3}{\sqrt{x^4+1}}+\frac{2e^{-\pi i/4}}{3}F(\arcsin e^{\pi i/4} x; -1)\\ \end{aligned} $$

Finally, take the derivation and verify that the result is correct

D[x Sqrt[x^4 + 1] / 3 + 2 Exp[-Pi I / 4] EllipticF[ArcSin[Exp[Pi I / 4] x], -1] / 3, x] // FullSimplify