<s>[INST] Ideas on the ways to integrate $\int \tan^2( x)\sec^3( x) dx$ I would proceed by thus , let $y = [\sec (x)]^2 $
then 
$$dy = 2 \cdot \sec(x) \cdot \sec(x) \cdot \tan(x) \cdot dx = 2 \cdot ( \sec (x))^2 \cdot \tan(x) \cdot dx $$
so, 
$$
2 \tan^2(x) \sec^2 (x) dx = \sec(x) \cdot \tan(x) \cdot dy = y(y-1)^\frac{1}{2} \cdot dy
$$ 
since $$\sec(x) = y^{\frac{1}{2}}$$ and by considering
positive square roots only $\tan y = ( \sec^2(x) - 1)^{1/2} = (y - 1)^{1/2}$. Thus the substitution $y = \sec^2 x$ yields
$$
2 \int \tan^2 (x) \sec^3(x) dx = \int (y(y - 1) )^{1/2} dy
$$
and this later form  can be reduced to  the  standard form  $\int(z^2 - a^2)^{1/2} dz$ since 
$$
y(y-1)=(y-(1/2))^2 - (1/2)^2 . 
$$
What  are the  other ways to integrate this expression, except for the substitution
$\tan^2(x)^2=\sec (x)^2 -1$ which gives 
$$
\sec^5(x) - \sec^3 (x)
$$
in the integrand which I quite don't like.
 [/INST] We can do this by intgration by parts
$ I=\int tan^2 x \cdot sec^3x \space dx$
$=\int (sec^2 x-1)\cdot sec^3 x \space dx$
$=\int sec^5 x \space dx-\int sec^3 x \space dx$
$=sec^3 x \cdot tan x-\int 3sec^2 x \cdot sec x tan x \cdot tan x \space dx-\int sec^3 x \space dx  $
$= sec^3x \cdot tanx-3I-I_1$
$ or, 4I= sec^3x \cdot tanx-I_1$
We can now find out the second integral 
$I_1=\int sec^3 x ~dx=\int sec x ~sec^2x~dx=sec x~tan x-\int sec x~(sec^2 x-1)dx=sec x~tanx-I_1+\int sec x~dx$
$or, 2I_1=secx~tan x+ln|sec x+tan x|$
Now back to original problem:
$I={1 \over 4}sec^3x\cdot tanx-{1 \over 8}(sec x~tan x+ln|sec x+tan x|)+c$
</s>