Optimal. Leaf size=124 \[ \frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {2 \tan ^2(c+d x)}{a^2 d}-\frac {15 i \tan (c+d x)}{4 a^2 d}-\frac {4 \log (\cos (c+d x))}{a^2 d}+\frac {15 i x}{4 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.19, antiderivative size = 124, normalized size of antiderivative = 1.00, 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 = {3558, 3595, 3528, 3525, 3475} \[ \frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {2 \tan ^2(c+d x)}{a^2 d}-\frac {15 i \tan (c+d x)}{4 a^2 d}-\frac {4 \log (\cos (c+d x))}{a^2 d}+\frac {15 i x}{4 a^2}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3475
Rule 3525
Rule 3528
Rule 3558
Rule 3595
Rubi steps
\begin {align*} \int \frac {\tan ^5(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan ^3(c+d x) (-4 a+6 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \tan ^2(c+d x) \left (-30 i a^2-32 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=-\frac {2 \tan ^2(c+d x)}{a^2 d}+\frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \tan (c+d x) \left (32 a^2-30 i a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac {15 i x}{4 a^2}-\frac {15 i \tan (c+d x)}{4 a^2 d}-\frac {2 \tan ^2(c+d x)}{a^2 d}+\frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {4 \int \tan (c+d x) \, dx}{a^2}\\ &=\frac {15 i x}{4 a^2}-\frac {4 \log (\cos (c+d x))}{a^2 d}-\frac {15 i \tan (c+d x)}{4 a^2 d}-\frac {2 \tan ^2(c+d x)}{a^2 d}+\frac {5 i \tan ^3(c+d x)}{4 a^2 d (1+i \tan (c+d x))}-\frac {\tan ^4(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end {align*}
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Mathematica [B] time = 1.69, size = 300, normalized size = 2.42 \[ \frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (-128 i d x \sin ^2(c)+60 d x \sin (2 c)-\sin (2 c) \sin (4 d x)-64 d x \tan (c)-i \sin (2 c) \cos (4 d x)-16 \sec (c) \cos (2 c-d x) \sec (c+d x)+16 \sec (c) \cos (2 c+d x) \sec (c+d x)+8 i \sin (2 c) \sec ^2(c+d x)-16 i \sec (c) \sin (2 c-d x) \sec (c+d x)+16 i \sec (c) \sin (2 c+d x) \sec (c+d x)+32 i \sin (2 c) \log \left (\cos ^2(c+d x)\right )+64 (\sin (2 c)-i \cos (2 c)) \tan ^{-1}(\tan (d x))+\cos (2 c) \left (-64 d x \tan (c)+8 \sec ^2(c+d x)+32 \log \left (\cos ^2(c+d x)\right )-60 i d x-i \sin (4 d x)+\cos (4 d x)\right )+64 i d x+16 i \sin (2 d x)-16 \cos (2 d x)\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.45, size = 151, normalized size = 1.22 \[ \frac {124 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} - 8 \, {\left (-31 i \, d x - 6\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (124 i \, d x + 95\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 64 \, {\left (e^{\left (8 i \, d x + 8 i \, c\right )} + 2 \, e^{\left (6 i \, d x + 6 i \, c\right )} + e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 14 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1}{16 \, {\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 2 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 13.78, size = 98, normalized size = 0.79 \[ \frac {\frac {2 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {62 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} - \frac {8 \, {\left (a^{2} \tan \left (d x + c\right )^{2} + 4 i \, a^{2} \tan \left (d x + c\right )\right )}}{a^{4}} - \frac {93 \, \tan \left (d x + c\right )^{2} - 150 i \, \tan \left (d x + c\right ) - 61}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 108, normalized size = 0.87 \[ -\frac {2 i \tan \left (d x +c \right )}{d \,a^{2}}-\frac {\tan ^{2}\left (d x +c \right )}{2 a^{2} d}+\frac {\ln \left (\tan \left (d x +c \right )+i\right )}{8 d \,a^{2}}-\frac {9 i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )}+\frac {1}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {31 \ln \left (\tan \left (d x +c \right )-i\right )}{8 d \,a^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.03, size = 114, normalized size = 0.92 \[ \frac {31\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{8\,a^2\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,2{}\mathrm {i}}{a^2\,d}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,a^2\,d}+\frac {\frac {9\,\mathrm {tan}\left (c+d\,x\right )}{4\,a^2}-\frac {2{}\mathrm {i}}{a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 4.92, size = 218, normalized size = 1.76 \[ \frac {- 2 e^{2 i c} e^{2 i d x} - 4}{- a^{2} d e^{4 i c} e^{4 i d x} - 2 a^{2} d e^{2 i c} e^{2 i d x} - a^{2} d} + \begin {cases} \frac {\left (16 a^{2} d e^{4 i c} e^{- 2 i d x} - a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{16 a^{4} d^{2}} & \text {for}\: 16 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {i \left (31 e^{4 i c} - 8 e^{2 i c} + 1\right ) e^{- 4 i c}}{4 a^{2}} - \frac {31 i}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {31 i x}{4 a^{2}} - \frac {4 \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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