How to simplify $\int{\sqrt[4]{1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}}dx}$?

Use $$\sqrt[4]{1-8x^2+8x^4-4x\sqrt{x^2-1}+8x^3\sqrt{x^{2}-1}}=\left|x+\sqrt{x^2-1}\right|.$$

Hint:

$(x+\sqrt{x^2-1})^4 = x^4+4x^3\sqrt{x^2-1}+6x^2(x^2-1)+4x(x^2-1)\sqrt{x^2-1}+(x^2-1)^2 = 8x^4-8x^2+1-4x\sqrt{x^2-1}+8x^3\sqrt{x^2-1}$

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 $\cos^4t-8\cos^2t+8-4\sin t\cos^2t+8\sin t$

$=(1-\sin^2t)^2-8(1-\sin^2t)+8-4\sin t(1-\sin^2t)+8\sin t$

$=\sin^4t+4\sin^3t+6\sin^2t+4\sin t+1$

$=(\sin t+1)^4$