TITLE: How to simplify $\int{\sqrt[4]{1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}}dx}$?
QUESTION [18 upvotes]: I have been asked about the following integral:
$$\int{\sqrt[4]{1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}}dx}$$
I think this is a joke of bad taste. I have tried every elementary method of integration which i know, also i tried integrating using Maple but as i suspected, the integrad doesn't have an anti derivative. Any ideas?

REPLY [4 votes]: Hint:
Let $\text{arcsec}x=t, x=\sec t$
Using Principal values, $0\le t\le\pi,t\ne\dfrac\pi2$
$\sin t=\sqrt{\left(1-\dfrac1x\right)^2}=\dfrac{\sqrt{x^2-1}}{|x|}$ as $\sin t\ge0$
$\tan t=\sin t\sec t=?$
$$1-8x^2+8x^4-4x\sqrt{x^2-1}+8x^3\sqrt{x^2-1}=\dfrac{\cos^4t-8\cos^2t+8-4\sin t\cos^2t+8\sin t}{\cos^4t}$$
Now $\cos^4t-8\cos^2t+8-4\sin t\cos^2t+8\sin t$
$=(1-\sin^2t)^2-8(1-\sin^2t)+8-4\sin t(1-\sin^2t)+8\sin t$
$=\sin^4t+4\sin^3t+6\sin^2t+4\sin t+1$
$=(\sin t+1)^4$
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