How to simplify $\int{\sqrt[4]{1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}}dx}$? 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?   
 A: Hint:
$(x+\sqrt{x^2-1})^4 = x^4+4x^3\sqrt{x^2-1}+6x^2(x^2-1)+4x(x^2-1)\sqrt{x^2-1}+(x^2-1)^2 = 8x^4-8x^2+1-4x\sqrt{x^2-1}+8x^3\sqrt{x^2-1}$
A: Use $$\sqrt[4]{1-8x^2+8x^4-4x\sqrt{x^2-1}+8x^3\sqrt{x^{2}-1}}=\left|x+\sqrt{x^2-1}\right|.$$
A: 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 $\cos^4t-8\cos^2t+8-4\sin t\cos^2t+8\sin t$
$=(1-\sin^2t)^2-8(1-\sin^2t)+8-4\sin t(1-\sin^2t)+8\sin t$
$=\sin^4t+4\sin^3t+6\sin^2t+4\sin t+1$
$=(\sin t+1)^4$
A: $$\frac{1}{16}{{\left( \sqrt{x-1}+\sqrt{x+1} \right)}^{8}}=1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}$$
So the integral becomes
$$\begin{align}
  & =\int{\sqrt[4]{\frac{1}{16}{{\left( \sqrt{x-1}+\sqrt{x+1} \right)}^{8}}}dx} \\ 
 & =\frac{1}{2}\int{{{\left( \sqrt{x-1}+\sqrt{x+1} \right)}^{2}}dx} \\ 
 & =\int{\left( x+\sqrt{{{x}^{2}}-1} \right)dx} \\ 
\end{align}$$
