Find the integral of $\tan^4(x) \sec(x)$ I want to know if there is a shorter way to find integral of $\tan^4(x) \sec(x)$
without using reduction formula of $\sec(x)$
because it's really takes a long time. 
Thanks all 
 A: Write it as
$$
\int\frac{\sin^4x}{\cos^5x}\,dx=
\int\frac{\sin^4{x}}{\cos^6x}\cos x\,dx=
\int\frac{\sin^4{x}}{(1-\sin^2x)^3}\cos x\,dx
$$
Can you see an obvious substitution?
A: $$I=\int\tan^4x\sec xdx=\int (\tan^3x)\sec x\tan xdx$$
$$=\tan^3x\int \sec x\tan xdx-\int\left(\frac{d(\tan^3x)}{dx}\int \sec x\tan xdx\right)dx$$
$$=\tan^3x\sec x-3\int\tan^2x\sec^2x\cdot \sec xdx$$
$$=\tan^3x\sec x-3\int\tan^2x(1+\tan^2x)\cdot \sec xdx$$
$$\implies I=\tan^3x\sec x-3I-3\int\tan^2x\cdot\sec xdx$$
$$\text{Firstly,}\int\tan^2x\cdot\sec xdx=\tan x\int \sec x\tan xdx-\int\left(\frac{d(\tan x)}{dx}\cdot\int \sec x\tan xdx\right)dx$$
$$\implies\int\tan^2x\cdot\sec xdx=\tan x\sec x-\int\sec^3x dx\ \ \ \ (1)$$
$$\text{Again,}\int\tan^2x\cdot\sec xdx=\int(\sec^2x-1)\sec xdx$$
$$\implies\int\tan^2x\cdot\sec xdx=\int\sec^3x dx-\ln|\sec x+\tan x|\ \ \ \ (2)$$
Solve $(1),(2)$ for $\displaystyle \int\tan^2x\cdot\sec xdx$
We don't need to calculate $\displaystyle\int\sec^3x$
A: What about $(\tan x)^3 \cdot \dfrac{1}{(\cos x)^2} \cdot \sin x$ and make integration by parts? 
