What is the step-by-step way to do the integral of $\int \tan^4 u \sec^3 u \,du$?

Question

$$\int \tan^4 u~ \sec^3 u\, du$$

The 2006 MIT OpenCourseWare notes "Single Variable Calculus" (PDF link via mit.edu) arrives at this integral (with a multiplied constant) in Example 2, but it did not show the complete way to solve it. WolframAlpha knows the answer, but it is not free to use.

The lecture note link actually provided some context about a family of integrals: $\int \sec^n x\tan^m x dx$. However, the lecture note only solved two cases: 1st, when $n$ is even; 2nd, when $m$ is odd. The integral does not belong to either case.

I tried leveraging the following results, but failed to get the solution.

$\sec^2 x = 1 + \tan^2 x$

$\int \sec^2 x dx = \tan x + c$

$\int \sec x \tan x dx = \sec x + c$

$\int \tan x dx = -\log(\cos x) + c$

$\int \sec x dx = \log (\tan x + \sec x) + c$

Answer 1

Start writing $$I=\int\tan^4\left(u\right)\sec^3\left(u\right)\,du=\int\sec^3\left(u\right)\left(\tan^2\left(u\right)\right)^2\,du$$ $$I=\int\sec^3\left(u\right)\left(\sec^2\left(u\right)-1\right)^2\,du=\int\left(\sec^7\left(u\right)-2\sec^5\left(u\right)+\sec^3\left(u\right)\right)\mathrm{d}u$$ Now, use the reduction formula

$$\int\sec^n\left(u\right)\,du=\frac{n-2}{n-1}\,\int\sec^{n-2}\left(u\right)\,du+\frac 1 {n-1}\sec^{n-2}\left(u\right)\,\tan(u)$$ starting with $$\int\sec\left(u\right)\,du=\log\left(\tan\left(u\right)+\sec\left(u\right)\right)$$

Answer 2

Bioche's rules suggest to use the substitution $$s=\sin u,\quad \mathrm ds=\cos u\,\mathrm du$$ so you obtain the integral of a rational function of $s$: $$\int \tan^4 u \sec^3 u\,\mathrm du=\int\frac{\sin^4u\,\mathrm du}{\cos^7u}= \int\frac{s^4\,\mathrm ds}{\cos^8u}=\int\frac{s^4\,\mathrm ds}{(1-s^2)^4},$$ which you can determine using partial fractions.