Optimal. Leaf size=74 \[ -\frac{\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{3 i \tan (c+d x)}{2 a d}-\frac{\log (\cos (c+d x))}{a d}+\frac{3 i x}{2 a} \]
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Rubi [A] time = 0.0679656, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3550, 3525, 3475} \[ -\frac{\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))}-\frac{3 i \tan (c+d x)}{2 a d}-\frac{\log (\cos (c+d x))}{a d}+\frac{3 i x}{2 a} \]
Antiderivative was successfully verified.
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Rule 3550
Rule 3525
Rule 3475
Rubi steps
\begin{align*} \int \frac{\tan ^3(c+d x)}{a+i a \tan (c+d x)} \, dx &=-\frac{\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \tan (c+d x) (2 a-3 i a \tan (c+d x)) \, dx}{2 a^2}\\ &=\frac{3 i x}{2 a}-\frac{3 i \tan (c+d x)}{2 a d}-\frac{\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))}+\frac{\int \tan (c+d x) \, dx}{a}\\ &=\frac{3 i x}{2 a}-\frac{\log (\cos (c+d x))}{a d}-\frac{3 i \tan (c+d x)}{2 a d}-\frac{\tan ^2(c+d x)}{2 d (a+i a \tan (c+d x))}\\ \end{align*}
Mathematica [B] time = 1.05023, size = 174, normalized size = 2.35 \[ -\frac{i \cos (c) \sec (c+d x) (\cos (d x)+i \sin (d x)) \left (-4 d x \tan ^2(c)+(-4-4 i \tan (c)) \tan ^{-1}(\tan (d x))-2 i d x \tan (c)+4 d x \sec ^2(c)-i \tan (c) \sin (2 d x)-2 i \log \left (\cos ^2(c+d x)\right )+(\tan (c)+i) \cos (2 d x)+4 \sec (c) \sin (d x) \sec (c+d x)+2 \tan (c) \log \left (\cos ^2(c+d x)\right )+4 i \tan (c) \sec (c) \sin (d x) \sec (c+d x)-6 d x+\sin (2 d x)\right )}{4 d (a+i a \tan (c+d x))} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.022, size = 73, normalized size = 1. \begin{align*}{\frac{-i\tan \left ( dx+c \right ) }{ad}}-{\frac{{\frac{i}{2}}}{ad \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{5\,\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{4\,ad}}-{\frac{\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{4\,ad}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.28772, size = 278, normalized size = 3.76 \begin{align*} \frac{10 i \, d x e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (10 i \, d x + 9\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 4 \,{\left (e^{\left (4 i \, d x + 4 i \, c\right )} + e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 1}{4 \,{\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + a d e^{\left (2 i \, d x + 2 i \, c\right )}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 1.01913, size = 121, normalized size = 1.64 \begin{align*} \begin{cases} \frac{e^{- 2 i c} e^{- 2 i d x}}{4 a d} & \text{for}\: 4 a d e^{2 i c} \neq 0 \\x \left (\frac{\left (5 i e^{2 i c} - i\right ) e^{- 2 i c}}{2 a} - \frac{5 i}{2 a}\right ) & \text{otherwise} \end{cases} + \frac{5 i x}{2 a} - \frac{\log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a d} + \frac{2 e^{- 2 i c}}{a d \left (e^{2 i d x} + e^{- 2 i c}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.80923, size = 95, normalized size = 1.28 \begin{align*} -\frac{\frac{\log \left (\tan \left (d x + c\right ) + i\right )}{a} - \frac{5 \, \log \left (-i \, \tan \left (d x + c\right ) - 1\right )}{a} + \frac{4 i \, \tan \left (d x + c\right )}{a} + \frac{5 \, \tan \left (d x + c\right ) - 3 i}{a{\left (\tan \left (d x + c\right ) - i\right )}}}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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