How to integrate tan ⁡ ( x ) tan ⁡ ( 2 x ) tan ⁡ ( 3 x )?

Question

How to integrate   tan  ⁡  (  x  )  tan  ⁡  (  2  x  )  tan  ⁡  (  3  x  )?

Perman7z · Accepted Answer

Since  tan  ⁡  3  x  =  tan  ⁡  (  x  +  2  x  )  =                    tan        ⁡        x        +        tan        ⁡        2        x                    1        −        tan        ⁡        x        tan        ⁡        2        x              ,we have  tan  ⁡  3  x  −  tan  ⁡  x  tan  ⁡  2  x  tan  ⁡  3  x  =  tan  ⁡  x  +  tan  ⁡  2  x  ,i.e.  tan  ⁡  x  tan  ⁡  2  x  tan  ⁡  3  x  =  tan  ⁡  3  x  −  tan  ⁡  x  −  tan  ⁡  2  x  .For   a  ≠  0 we have  ∫  tan  ⁡  (  a  x  )    d  x  =  −            1      a        ln  ⁡      |    cos  ⁡  (  a  x  )      |    +  c  ,and therefore                    ∫        tan        ⁡        x        tan        ⁡        2        x        tan        ⁡        3        x                d        x                            =                    ∫        (        tan        ⁡        3        x        −        tan        ⁡        x        −        tan        ⁡        2        x        )                d        x                                          =                    ln        ⁡                  |                cos        ⁡        x                  |                +                              1            2                          ln        ⁡                  |                cos        ⁡        2        x                  |                −                              1            3                          ln        ⁡                  |                cos        ⁡        3        x                  |                +        c

orlovskihmw · Accepted Answer

In addition to the other answers, we can also do like this:
Use

\tan 3 x = \frac{3 \cos^{2} x \sin x - \sin^{3} x}{\cos^{3} x - 3 \cos x \sin^{2} x}

,

\tan 2 x = \frac{2 \cos x \sin x}{\cos^{2} x - \sin^{2} x}

and substitute

u = \cos x

to get

I = \int \tan x \tan 2 x \tan 3 x d x = \int - \frac{2 (4 \cos^{4} x - 5 \cos^{2} x + 1)}{\cos x (2 \cos^{2} x - 1) (4 \cos^{2} x - 3)} \sin x d x = 2 \int \frac{4 u^{4} - 5 u^{2} + 1}{u (2 u^{2} - 1) (4 u^{2} - 3)} d u = - \frac{8}{3} \int \frac{u}{4 u^{2} - 3} d u + \int \frac{2 u}{2 u^{2} - 1} d u + \int \frac{2}{3 u} d u

How to integrate tan ⁡

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