<s>[INST] Evaluation of $\int x^8\sqrt{x^2+1}\ dx$ 
Evaluation of $\displaystyle \int x^8\sqrt{x^2+1}\ dx$

$\bf{My\: Try::}$ Put $x=\tan \theta\;,$ Then $dx = \sec^2 \theta d \theta$
So $$I = \int \sec ^{3}\theta \cdot \tan^{8}\theta d \theta=\int \frac{\sin^{8}\theta}{\cos^{5}\theta }d\theta$$
Now How can I solve it after that, Help Required, Thanks
 [/INST] Apply the reduction formula
$$\int x^n\sqrt{1+x^2}dx= I_n=\frac{x^{n-1}}{n+2}(1+x^2)^{3/2}-\frac{n-1}{n+2}I_{n-2}
$$
4 times to reduce the integral
\begin{align}
\int x^8\sqrt{1+x^2}\ dx
=7\left(\frac{x^7}{70}-\frac{x^5 }{80}+\frac{x^3 }{96}-\frac{x}{128}\right)(1+x^2)^{3/2}+\frac7{128}I_0
\end{align}
where
$
I_0=\int \sqrt{1+x^2}\ dx
=\frac{x}{2}(1+x^2)^{1/2}+\frac{1}{2}\sinh^{-1}x
$
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