Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$ Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$
$\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$
So Integral is $\displaystyle \int_{0}^{\infty}\frac{\sqrt{x-1}}{\sqrt{x^5-1}}dx$
now i did not understand how can i solve it
Help me
Thanks

The answer in terms of elliptic integrals turns out to be very simple – but it requires a very sneaky trick.
As pointed out in Jack's answer, the integrals over $[0,1]$ and $[1,\infty]$ are the same, so
$$I=\int_0^\infty\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx=2\int_0^1\frac1{\sqrt{x^4+x^3+x^2+x+1}}\,dx$$
The sneaky trick is to substitute $u=x+\frac1x$, which gives
$$I=2\int_2^\infty\frac1{\sqrt{(u^2-4)(u^2+u-1)}}\,du$$
This changes all the denominator quartic's roots from complex to real, which will make the sequel easier. Furthermore, the roots are very simple: $-2,-\varphi,1/\varphi,2$ where $\varphi$ is the golden ratio. We can now apply Byrd and Friedman 258.00 directly, which says that if $d<c<b<a<y$,
$$\int_a^y\frac1{\sqrt{(t-a)(t-b)(t-c)(t-d)}}\,dt=gF(\psi,m)\\
\text{where }g=\frac2{\sqrt{(a-c)(b-d)}},\psi=\sin^{-1}\sqrt{\frac{(b-d)(y-a)}{(a-d)(y-b)}},m=\frac{(b-c)(a-d)}{(a-c)(b-d)}$$
($m=k^2$ is the parameter, as used by Mathematica and mpmath.) We have $d=-2,c=-\varphi,b=1/\varphi,a=2$, but since $y=\infty$ we have to take the limit of the expression for $\psi$ as $y\to\infty$, which therefore becomes
$$\psi=\sin^{-1}\sqrt{\frac{b-d}{a-d}}=\sin^{-1}\sqrt{\frac{1/\varphi+2}4}=\sin^{-1}\frac\varphi2=\frac{3\pi}{10}$$
The other key numbers follow:
$$g=\frac2{\sqrt{(2+\varphi)(1/\varphi+2)}}=\frac2{\sqrt{4\varphi+3}}$$
$$m=\frac{4(1/\varphi+\varphi)}{(2+\varphi)(1/\varphi+2)}=\frac{4(2\varphi-1)}{4\varphi+3}=8\varphi-12$$
Finally we have the answer in terms of elliptic integrals:
$$\boxed{I=\frac4{\sqrt{4\varphi+3}}F\left(\frac{3\pi}{10},8\varphi-12\right)}$$

#!/usr/bin/env python3
from mpmath import *
mp.dps = 150

print(quad(lambda t: 1/sqrt(polyval((1,1,1,1,1),t)), [0, inf]))

g = 4/sqrt(4*phi+3)
m = 8*phi-12
print(g*ellipf(3*pi/10, m))


P.S. A certain Dr. Sonnhard Graubner gave an answer here, but it's incorrect.
