$\displaystyle \int tan(x)tan(2x)tan(3x)dx$
Smoke..
Just mow I came back home and I started to solve your integral...
I have discovered a very interesting result
believe it or not (tanx)(tan2x)(tan3x) = tan(3x)-tan(2x)-tanx
indeed :
tan(3x) = (tanx+tan(2x))/(1-tanxtan(2x))
this implies : tan(3x)(1-tanxtan(2x))=tanx+tan(2x) and after all simplifications we obtain : tan(3x)-tan(2x)-tan(x) = tanxtan(2x)tan(3x)
therefore your integral now is simple to calculate:
since integral(tanx)dx = -ln(cosx)
integral(tan(2x))dx = (-1/2)ln(cos(2x))
and Integral (tan(3x))dx = (-1/3)ln(cos(3x))
you can verify all these and check the calculations..
MINOAS