Optimal. Leaf size=104 \[ \frac {i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {9 \tan (c+d x)}{4 a^2 d}+\frac {2 i \log (\cos (c+d x))}{a^2 d}+\frac {9 x}{4 a^2}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.16, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3558, 3595, 3525, 3475} \[ \frac {i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {9 \tan (c+d x)}{4 a^2 d}+\frac {2 i \log (\cos (c+d x))}{a^2 d}+\frac {9 x}{4 a^2}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 3475
Rule 3525
Rule 3558
Rule 3595
Rubi steps
\begin {align*} \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^2} \, dx &=-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {\int \frac {\tan ^2(c+d x) (-3 a+5 i a \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \tan (c+d x) \left (-16 i a^2-18 a^2 \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac {9 x}{4 a^2}-\frac {9 \tan (c+d x)}{4 a^2 d}+\frac {i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}-\frac {(2 i) \int \tan (c+d x) \, dx}{a^2}\\ &=\frac {9 x}{4 a^2}+\frac {2 i \log (\cos (c+d x))}{a^2 d}-\frac {9 \tan (c+d x)}{4 a^2 d}+\frac {i \tan ^2(c+d x)}{a^2 d (1+i \tan (c+d x))}-\frac {\tan ^3(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end {align*}
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Mathematica [B] time = 1.53, size = 273, normalized size = 2.62 \[ -\frac {\sec ^2(c+d x) (\cos (d x)+i \sin (d x))^2 \left (64 d x \sin ^2(c)+36 i d x \sin (2 c)-i \sin (2 c) \sin (4 d x)-32 i d x \tan (c)+\sin (2 c) \cos (4 d x)-8 i \sec (c) \cos (2 c-d x) \sec (c+d x)+8 i \sec (c) \cos (2 c+d x) \sec (c+d x)+8 \sec (c) \sin (2 c-d x) \sec (c+d x)-8 \sec (c) \sin (2 c+d x) \sec (c+d x)-16 \sin (2 c) \log \left (\cos ^2(c+d x)\right )+32 (\cos (2 c)+i \sin (2 c)) \tan ^{-1}(\tan (d x))+\cos (2 c) \left (-32 i d x \tan (c)+16 i \log \left (\cos ^2(c+d x)\right )+36 d x+\sin (4 d x)+i \cos (4 d x)\right )-32 d x-12 \sin (2 d x)-12 i \cos (2 d x)\right )}{16 a^2 d (\tan (c+d x)-i)^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.42, size = 111, normalized size = 1.07 \[ \frac {68 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (68 \, d x - 44 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (32 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 32 i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 11 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{16 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 6.81, size = 79, normalized size = 0.76 \[ -\frac {-\frac {2 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {34 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {16 \, \tan \left (d x + c\right )}{a^{2}} + \frac {-51 i \, \tan \left (d x + c\right )^{2} - 74 \, \tan \left (d x + c\right ) + 27 i}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{16 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.13, size = 93, normalized size = 0.89 \[ -\frac {\tan \left (d x +c \right )}{a^{2} d}+\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{8 d \,a^{2}}-\frac {17 i \ln \left (\tan \left (d x +c \right )-i\right )}{8 d \,a^{2}}-\frac {i}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {7}{4 d \,a^{2} \left (\tan \left (d x +c \right )-i\right )} \]
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 = 100, normalized size = 0.96 \[ -\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,17{}\mathrm {i}}{8\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,a^2\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )}{a^2\,d}-\frac {\frac {3}{2\,a^2}+\frac {\mathrm {tan}\left (c+d\,x\right )\,7{}\mathrm {i}}{4\,a^2}}{d\,\left ({\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.53, size = 180, normalized size = 1.73 \[ \begin {cases} \frac {\left (- 48 i a^{2} d e^{4 i c} e^{- 2 i d x} + 4 i a^{2} d e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{64 a^{4} d^{2}} & \text {for}\: 64 a^{4} d^{2} e^{6 i c} \neq 0 \\x \left (\frac {\left (17 e^{4 i c} - 6 e^{2 i c} + 1\right ) e^{- 4 i c}}{4 a^{2}} - \frac {17}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {2 i}{- a^{2} d e^{2 i c} e^{2 i d x} - a^{2} d} + \frac {17 x}{4 a^{2}} + \frac {2 i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{2} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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