Optimal. Leaf size=144 \[ \frac{(a+i a \tan (c+d x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (c+d x)+1)\right )}{2 d m}-\frac{m (a+i a \tan (c+d x))^{m+1}}{a d \left (m^2+3 m+2\right )}+\frac{\tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac{2 (a+i a \tan (c+d x))^m}{d m (m+2)} \]
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Rubi [A] time = 0.191903, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3560, 3592, 3527, 3481, 68} \[ \frac{(a+i a \tan (c+d x))^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (i \tan (c+d x)+1)\right )}{2 d m}-\frac{m (a+i a \tan (c+d x))^{m+1}}{a d \left (m^2+3 m+2\right )}+\frac{\tan ^2(c+d x) (a+i a \tan (c+d x))^m}{d (m+2)}-\frac{2 (a+i a \tan (c+d x))^m}{d m (m+2)} \]
Antiderivative was successfully verified.
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Rule 3560
Rule 3592
Rule 3527
Rule 3481
Rule 68
Rubi steps
\begin{align*} \int \tan ^3(c+d x) (a+i a \tan (c+d x))^m \, dx &=\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 \operatorname{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{\, _2F_1\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*}
Mathematica [F] time = 6.54585, size = 0, normalized size = 0. \[ \int \tan ^3(c+d x) (a+i a \tan (c+d x))^m \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.622, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{3} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m}{\left (i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m} \tan \left (d x + c\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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