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

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?

Answer 1

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

Answer 2

Hint:

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

Answer 3

$$\frac{1}{16}{{\left( \sqrt{x-1}+\sqrt{x+1} \right)}^{8}}=1-8{{x}^{2}}+8{{x}^{4}}-4x\sqrt{{{x}^{2}}-1}+8{{x}^{3}}\sqrt{{{x}^{2}}-1}$$ So the integral becomes

$$\begin{align} & =\int{\sqrt[4]{\frac{1}{16}{{\left( \sqrt{x-1}+\sqrt{x+1} \right)}^{8}}}dx} \\ & =\frac{1}{2}\int{{{\left( \sqrt{x-1}+\sqrt{x+1} \right)}^{2}}dx} \\ & =\int{\left( x+\sqrt{{{x}^{2}}-1} \right)dx} \\ \end{align}$$

Answer 4

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

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

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

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

Answer 5

First of all, in order to deal with such ‘ugly’ integrand, $$ E(x)=\sqrt[4]{1-8 x^2+8 x^4-4 x \sqrt{x^2-1}+8 x^3 \sqrt{x^2-1}}, $$ where $|x|\ge 1$,

I put $x=\cosh \theta$ into $E(x)$ and then simplify. $$ \begin{aligned} E(x)= & \sqrt[4]{ 1-8 \cosh ^2 \theta+8 \cosh ^4 \theta-4 \cosh \theta \sinh \theta +8 \cosh ^3 \theta \sinh \theta }\\ = & \sqrt[4]{ 1+8 \cosh ^2 \theta\left(\cosh ^2 \theta-1\right)+4 \cosh \theta \sinh \theta \left(2 \cosh ^2 \theta-1\right)}\\=& \sqrt[4]{ 1+2 \sinh ^2 2 \theta+2 \sinh (2 \theta) \cosh (2 \theta)}\\=& \sqrt[4]{[\cosh (2 \theta)+\sinh (2 \theta)]^2}\\=&e^\theta\\ \textrm{ OR } E(x) =&x+\sqrt{x^2-1} \end{aligned} $$

Then $$I= \int e^\theta \sinh \theta d\theta = \frac 14\left(e^{2\theta}-2\theta\right) +C= \frac{1}{2}\left[x^2+ x \sqrt{x^2+1}-\ln \left(x+\sqrt{x^2-1}\right)\right]+C$$