Optimal. Leaf size=119 \[ \frac{7 i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{i \log (\cos (c+d x))}{a^3 d}-\frac{7 x}{8 a^3}-\frac{\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.241476, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.292, Rules used = {3558, 3595, 3589, 3475, 12, 3526, 8} \[ \frac{7 i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{i \log (\cos (c+d x))}{a^3 d}-\frac{7 x}{8 a^3}-\frac{\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3558
Rule 3595
Rule 3589
Rule 3475
Rule 12
Rule 3526
Rule 8
Rubi steps
\begin{align*} \int \frac{\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=-\frac{\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}-\frac{\int \frac{\tan ^2(c+d x) (-3 a+6 i a \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=-\frac{\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\tan (c+d x) \left (-18 i a^2-24 a^2 \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=-\frac{\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}-\frac{i \int \frac{42 a^3 \tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{24 a^5}+\frac{i \int \tan (c+d x) \, dx}{a^3}\\ &=-\frac{i \log (\cos (c+d x))}{a^3 d}-\frac{\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}-\frac{(7 i) \int \frac{\tan (c+d x)}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=-\frac{i \log (\cos (c+d x))}{a^3 d}-\frac{\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac{7 i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac{7 \int 1 \, dx}{8 a^3}\\ &=-\frac{7 x}{8 a^3}-\frac{i \log (\cos (c+d x))}{a^3 d}-\frac{\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac{7 i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}
Mathematica [A] time = 0.417045, size = 118, normalized size = 0.99 \[ \frac{\sec ^3(c+d x) (-81 i \sin (c+d x)+84 d x \sin (3 (c+d x))+2 i \sin (3 (c+d x))-51 \cos (c+d x)+\cos (3 (c+d x)) (96 \log (\cos (c+d x))-84 i d x-2)+96 i \sin (3 (c+d x)) \log (\cos (c+d x)))}{96 a^3 d (\tan (c+d x)-i)^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.026, size = 98, normalized size = 0.8 \begin{align*}{\frac{{\frac{7\,i}{8}}}{d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}+{\frac{{\frac{15\,i}{16}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{d{a}^{3}}}-{\frac{1}{6\,d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}+{\frac{17}{8\,d{a}^{3} \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{i}{16}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{d{a}^{3}}} \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 = 2.2894, size = 246, normalized size = 2.07 \begin{align*} -\frac{{\left (180 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 96 i \, e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 66 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 3.77616, size = 134, normalized size = 1.13 \begin{align*} - \frac{\left (\begin{cases} 15 x e^{6 i c} - \frac{11 i e^{4 i c} e^{- 2 i d x}}{2 d} + \frac{5 i e^{2 i c} e^{- 4 i d x}}{4 d} - \frac{i e^{- 6 i d x}}{6 d} & \text{for}\: d \neq 0 \\x \left (15 e^{6 i c} - 11 e^{4 i c} + 5 e^{2 i c} - 1\right ) & \text{otherwise} \end{cases}\right ) e^{- 6 i c}}{8 a^{3}} - \frac{i \log{\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 2.40082, size = 108, normalized size = 0.91 \begin{align*} -\frac{-\frac{90 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} - \frac{6 i \, \log \left (-i \, \tan \left (d x + c\right ) + 1\right )}{a^{3}} + \frac{165 i \, \tan \left (d x + c\right )^{3} + 291 \, \tan \left (d x + c\right )^{2} - 171 i \, \tan \left (d x + c\right ) - 29}{a^{3}{\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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