Integrand size = 24, antiderivative size = 144 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^m \, dx=-\frac {2 (a+i a \tan (c+d x))^m}{d m (2+m)}+\frac {\operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^m}{2 d m}+\frac {\tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (2+m)}-\frac {m (a+i a \tan (c+d x))^{1+m}}{a d \left (2+3 m+m^2\right )} \]
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Time = 0.24 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3641, 3673, 3608, 3562, 70} \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^m \, dx=\frac {(a+i a \tan (c+d x))^m \operatorname {Hypergeometric2F1}\left (1,m,m+1,\frac {1}{2} (i \tan (c+d x)+1)\right )}{2 d m}-\frac {m (a+i a \tan (c+d x))^{m+1}}{a d \left (m^2+3 m+2\right )}+\frac {\tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac {2 (a+i a \tan (c+d x))^m}{d m (m+2)} \]
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Rule 70
Rule 3562
Rule 3608
Rule 3641
Rule 3673
Rubi steps \begin{align*} \text {integral}& = \frac {\tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (2+m)}-\frac {\int \tan (c+d x) (a+i a \tan (c+d x))^m (2 a+i a m \tan (c+d x)) \, dx}{a (2+m)} \\ & = \frac {\tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (2+m)}-\frac {m (a+i a \tan (c+d x))^{1+m}}{a d \left (2+3 m+m^2\right )}-\frac {\int (a+i a \tan (c+d x))^m (-i a m+2 a \tan (c+d x)) \, dx}{a (2+m)} \\ & = -\frac {2 (a+i a \tan (c+d x))^m}{d m (2+m)}+\frac {\tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (2+m)}-\frac {m (a+i a \tan (c+d x))^{1+m}}{a d \left (2+3 m+m^2\right )}+i \int (a+i a \tan (c+d x))^m \, dx \\ & = -\frac {2 (a+i a \tan (c+d x))^m}{d m (2+m)}+\frac {\tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (2+m)}-\frac {m (a+i a \tan (c+d x))^{1+m}}{a d \left (2+3 m+m^2\right )}+\frac {a \text {Subst}\left (\int \frac {(a+x)^{-1+m}}{a-x} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {2 (a+i a \tan (c+d x))^m}{d m (2+m)}+\frac {\operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (c+d x))\right ) (a+i a \tan (c+d x))^m}{2 d m}+\frac {\tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (2+m)}-\frac {m (a+i a \tan (c+d x))^{1+m}}{a d \left (2+3 m+m^2\right )} \\ \end{align*}
Time = 0.79 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.72 \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^m \, dx=\frac {(a+i a \tan (c+d x))^m \left (-2 \left (2+2 m+m^2\right )+\left (2+3 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (1,m,1+m,\frac {1}{2} (1+i \tan (c+d x))\right )-2 i m^2 \tan (c+d x)+2 m (1+m) \tan ^2(c+d x)\right )}{2 d m (1+m) (2+m)} \]
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\[\int \left (\tan ^{3}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{m}d x\]
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\[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^m \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )^{3} \,d x } \]
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\[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^m \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{m} \tan ^{3}{\left (c + d x \right )}\, dx \]
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\[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^m \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )^{3} \,d x } \]
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\[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^m \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )^{3} \,d x } \]
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Timed out. \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^m \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^m \,d x \]
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