Integrand size = 24, antiderivative size = 119 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {7 x}{8 a^3}-\frac {i \log (\cos (c+d x))}{a^3 d}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2}+\frac {7 i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )} \]
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Time = 0.30 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3639, 3676, 3670, 3556, 12, 3607, 8} \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\frac {7 i}{8 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {i \log (\cos (c+d x))}{a^3 d}-\frac {7 x}{8 a^3}-\frac {\tan ^3(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac {3 i \tan ^2(c+d x)}{8 a d (a+i a \tan (c+d x))^2} \]
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Rule 8
Rule 12
Rule 3556
Rule 3607
Rule 3639
Rule 3670
Rule 3676
Rubi steps \begin{align*} \text {integral}& = -\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*}
Time = 0.50 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.09 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\sec ^3(c+d x) (51 \cos (c+d x)+\cos (3 (c+d x)) (-51+90 \log (i-\tan (c+d x))+6 \log (i+\tan (c+d x)))+i (81 \sin (c+d x)+(-55+90 \log (i-\tan (c+d x))+6 \log (i+\tan (c+d x))) \sin (3 (c+d x))))}{96 a^3 d (-i+\tan (c+d x))^3} \]
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Time = 0.36 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.77
method | result | size |
risch | \(-\frac {15 x}{8 a^{3}}+\frac {11 i {\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{3} d}-\frac {5 i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a^{3} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{3} d}-\frac {2 c}{a^{3} d}-\frac {i \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a^{3} d}\) | \(92\) |
derivativedivides | \(\frac {7 i}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}-\frac {7 \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {1}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {17}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}\) | \(95\) |
default | \(\frac {7 i}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}-\frac {7 \arctan \left (\tan \left (d x +c \right )\right )}{8 d \,a^{3}}-\frac {1}{6 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {17}{8 d \,a^{3} \left (\tan \left (d x +c \right )-i\right )}\) | \(95\) |
norman | \(\frac {-\frac {7 x}{8 a}+\frac {17 \left (\tan ^{5}\left (d x +c \right )\right )}{8 a d}-\frac {21 x \left (\tan ^{2}\left (d x +c \right )\right )}{8 a}-\frac {21 x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}-\frac {7 x \left (\tan ^{6}\left (d x +c \right )\right )}{8 a}+\frac {17 i}{12 a d}+\frac {7 \tan \left (d x +c \right )}{8 a d}+\frac {7 \left (\tan ^{3}\left (d x +c \right )\right )}{3 a d}+\frac {3 i \left (\tan ^{4}\left (d x +c \right )\right )}{d a}+\frac {15 i \left (\tan ^{2}\left (d x +c \right )\right )}{4 d a}}{a^{2} \left (1+\tan ^{2}\left (d x +c \right )\right )^{3}}+\frac {i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d \,a^{3}}\) | \(176\) |
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Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.65 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {{\left (180 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} + 96 i \, e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) - 66 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{3} d} \]
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Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.55 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\begin {cases} \frac {\left (16896 i a^{6} d^{2} e^{10 i c} e^{- 2 i d x} - 3840 i a^{6} d^{2} e^{8 i c} e^{- 4 i d x} + 512 i a^{6} d^{2} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{24576 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{12 i c} \neq 0 \\x \left (\frac {\left (- 15 e^{6 i c} + 11 e^{4 i c} - 5 e^{2 i c} + 1\right ) e^{- 6 i c}}{8 a^{3}} + \frac {15}{8 a^{3}}\right ) & \text {otherwise} \end {cases} - \frac {15 x}{8 a^{3}} - \frac {i \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{a^{3} d} \]
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Exception generated. \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 1.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.66 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {-\frac {6 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{3}} - \frac {90 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{3}} + \frac {165 i \, \tan \left (d x + c\right )^{3} + 291 \, \tan \left (d x + c\right )^{2} - 171 i \, \tan \left (d x + c\right ) - 29}{a^{3} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{96 \, d} \]
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Time = 4.86 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.92 \[ \int \frac {\tan ^4(c+d x)}{(a+i a \tan (c+d x))^3} \, dx=-\frac {\frac {27\,\mathrm {tan}\left (c+d\,x\right )}{8\,a^3}-\frac {17{}\mathrm {i}}{12\,a^3}+\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,17{}\mathrm {i}}{8\,a^3}}{d\,\left (-{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (c+d\,x\right )}^2+\mathrm {tan}\left (c+d\,x\right )\,3{}\mathrm {i}+1\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,15{}\mathrm {i}}{16\,a^3\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,a^3\,d} \]
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