Integrate tan x . tan 2x . tan 3x dx?

Question

This problem has a very simple solution.
A hint - think of trigonometrical simplification first.

tan3x = tan (2x + x )
= (tan 2x + tan x )/( 1 - tanx tan2x )
tan3x - tanx tan2x tan3x = tan2x + tanx

tanx tan2x tan3x = tan3x -tan2x -tanx

Now integrate

Answer 1

2/3 * ln|cosx| + 1/2 * ln|cos2x| - 1/3 * ln|2cos2x - 1|

The idea is to rewrite this expression in the form sinx / f(cosx). That way we can make the sub u = cosx, du = -sinx*dx.

2sinA*sinB = cos(A-B) - cos(A+B)
2cosA*cosB = cos(A-B) + cos(A+B)
from that
tanA*tanB = [cos(A-B) - cos(A+B)] / [cos(A-B) - cos(A+B)]

tan3x*tanx = [cos2x - cos4x] / [cos2x + cos4x]
tanx*tan2x*tan3x = [cos2x - cos4x] / [cos2x + cos4x] * sin2x / cos2x
= [cos2x + 1 - 2cos^2 (2x)] / {[cos2x - 1 + 2cos^2 (2x)]*[cos2x]} * sin2x * dx
Now let's make the sub u = cos2x, du = -2sin2x * dx

The conversion yields
-1/2 * (1 + u - 2u^2) / [u*(2u^2 + u - 1)] * du
= -1/2 * (1 + u - 2u^2) / [u*(2u - 1)*(u + 1)] * du
= [ 1/(2u) - (2/3)/(2u-1) + (1/3)/(u+1) ] * du

The integral is:
1/2 * ln(u) - 1/3 * ln(2u-1) + 1/3 * ln(u+1)
= 1/2 * ln(cos2x) - 1/3 * ln(2cos2x-1) + 1/3 * ln(cos2x+1) + C1
= 1/2 * ln(cos2x) - 1/3 * ln(2cos2x-1) + 1/3 * ln(2cos^2 x) + C1
= 1/2 * ln(cos2x) - 1/3 * ln(2cos2x-1) + 1/3 * ln(cosx)^2 - 2/3*ln2 + C1
= 1/2 * ln(cos2x) - 1/3 * ln(2cos2x-1) + 2/3 * ln(cosx) + C2

**EDIT**
This solution actually differentiates to give the original expression. And it's also consistent with Mathematica. I'd be very interested to see the simplified solution.

**EDIT**
I wish I could give you a thumbs up for that approach.

Answer 2

Integrate Tan 2x