Integrate $\int \frac {x^4}{\sqrt {x^2-9}} \,dx$ Integrate $\int \dfrac {x^4}{\sqrt {x^2-9}} dx$
My Attempt:
Let $x=3\sec (\theta )$
$$dx=3\sec (\theta).\tan (\theta).d\theta$$
Then,
$$=\int \dfrac {x^4}{\sqrt {x^2-9}}$$
$$=\int \dfrac {81. \sec^4 (\theta)}{\sqrt {(3\sec (\theta))^2 - 9}} 3\sec (\theta).\tan (\theta).d\theta $$
$$=\int \dfrac {81 \sec^5 (\theta). 3\tan (\theta).d\theta}{3\tan (\theta)}$$
$$=81\int \sec^5 (\theta) d\theta$$

For $x\ge 3$ let substitute $\begin{cases}x=3\cosh(u) & u\ge 0\\3\sinh(u)=\sqrt{x^2-9}\end{cases}$
$\begin{align}\displaystyle \int \dfrac{x^4}{\sqrt{x^2-9}}\mathop{dx}&=81\int \cosh(u)^4\mathop{du}\\&=\dfrac{81}8\bigg(2\cosh(u)^3\sinh(u)+3\cosh(u)\sinh(u)+3u\bigg)+C\\&=\frac 14x^3\sqrt{x^2-9}+\dfrac{27}{8}x\sqrt{x^2-9}+\dfrac{243}8\operatorname{argch}(\frac x3)+C\end{align}$
For $x\le-3$ just reverse the sign.
Your choice of substitution leads to $\sec^5(x)$ which is difficult to linearise. 
With $\cosh^4(u)=\frac 18\cosh(4u)+\frac 12\cosh(2u)+\frac 38$ we can integrate and apply the inverse transformation to go back to powers of $(\cosh,\sinh)$, I wrote only the resulting anti-derivative.
