Integrand size = 22, antiderivative size = 110 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {i x}{16 a^4}-\frac {1}{8 d (a+i a \tan (c+d x))^4}+\frac {1}{12 a d (a+i a \tan (c+d x))^3}+\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {1}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \]
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Time = 0.09 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3607, 3560, 8} \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {1}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {i x}{16 a^4}+\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {1}{12 a d (a+i a \tan (c+d x))^3}-\frac {1}{8 d (a+i a \tan (c+d x))^4} \]
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Rule 8
Rule 3560
Rule 3607
Rubi steps \begin{align*} \text {integral}& = -\frac {1}{8 d (a+i a \tan (c+d x))^4}-\frac {i \int \frac {1}{(a+i a \tan (c+d x))^3} \, dx}{2 a} \\ & = -\frac {1}{8 d (a+i a \tan (c+d x))^4}+\frac {1}{12 a d (a+i a \tan (c+d x))^3}-\frac {i \int \frac {1}{(a+i a \tan (c+d x))^2} \, dx}{4 a^2} \\ & = -\frac {1}{8 d (a+i a \tan (c+d x))^4}+\frac {1}{12 a d (a+i a \tan (c+d x))^3}+\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}-\frac {i \int \frac {1}{a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = -\frac {1}{8 d (a+i a \tan (c+d x))^4}+\frac {1}{12 a d (a+i a \tan (c+d x))^3}+\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {1}{16 d \left (a^4+i a^4 \tan (c+d x)\right )}-\frac {i \int 1 \, dx}{16 a^4} \\ & = -\frac {i x}{16 a^4}-\frac {1}{8 d (a+i a \tan (c+d x))^4}+\frac {1}{12 a d (a+i a \tan (c+d x))^3}+\frac {1}{16 d \left (a^2+i a^2 \tan (c+d x)\right )^2}+\frac {1}{16 d \left (a^4+i a^4 \tan (c+d x)\right )} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.85 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\sec ^4(c+d x) (16 \cos (2 (c+d x))+(-3-24 i d x) \cos (4 (c+d x))+32 i \sin (2 (c+d x))+3 i \sin (4 (c+d x))+24 d x \sin (4 (c+d x)))}{384 a^4 d (-i+\tan (c+d x))^4} \]
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Time = 0.32 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.55
method | result | size |
risch | \(-\frac {i x}{16 a^{4}}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )}}{16 a^{4} d}-\frac {{\mathrm e}^{-6 i \left (d x +c \right )}}{48 a^{4} d}-\frac {{\mathrm e}^{-8 i \left (d x +c \right )}}{128 a^{4} d}\) | \(60\) |
derivativedivides | \(\frac {i}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {1}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}\) | \(96\) |
default | \(\frac {i}{12 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{3}}-\frac {i}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )}-\frac {1}{8 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{4}}-\frac {1}{16 d \,a^{4} \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {i \arctan \left (\tan \left (d x +c \right )\right )}{16 d \,a^{4}}\) | \(96\) |
norman | \(\frac {-\frac {i x}{16 a}+\frac {1}{12 a d}-\frac {\tan ^{4}\left (d x +c \right )}{4 a d}+\frac {i \tan \left (d x +c \right )}{16 d a}-\frac {53 i \left (\tan ^{3}\left (d x +c \right )\right )}{48 d a}-\frac {11 i \left (\tan ^{5}\left (d x +c \right )\right )}{48 a d}-\frac {i \left (\tan ^{7}\left (d x +c \right )\right )}{16 a d}-\frac {i x \left (\tan ^{2}\left (d x +c \right )\right )}{4 a}-\frac {3 i x \left (\tan ^{4}\left (d x +c \right )\right )}{8 a}-\frac {i x \left (\tan ^{6}\left (d x +c \right )\right )}{4 a}-\frac {i x \left (\tan ^{8}\left (d x +c \right )\right )}{16 a}+\frac {5 \left (\tan ^{2}\left (d x +c \right )\right )}{6 a d}}{\left (1+\tan ^{2}\left (d x +c \right )\right )^{4} a^{3}}\) | \(191\) |
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Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.49 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {{\left (-24 i \, d x e^{\left (8 i \, d x + 8 i \, c\right )} + 24 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3\right )} e^{\left (-8 i \, d x - 8 i \, c\right )}}{384 \, a^{4} d} \]
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Time = 0.29 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.42 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\begin {cases} \frac {\left (6144 a^{8} d^{2} e^{14 i c} e^{- 2 i d x} - 2048 a^{8} d^{2} e^{10 i c} e^{- 6 i d x} - 768 a^{8} d^{2} e^{8 i c} e^{- 8 i d x}\right ) e^{- 16 i c}}{98304 a^{12} d^{3}} & \text {for}\: a^{12} d^{3} e^{16 i c} \neq 0 \\x \left (\frac {\left (- i e^{8 i c} - 2 i e^{6 i c} + 2 i e^{2 i c} + i\right ) e^{- 8 i c}}{16 a^{4}} + \frac {i}{16 a^{4}}\right ) & \text {otherwise} \end {cases} - \frac {i x}{16 a^{4}} \]
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Exception generated. \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\text {Exception raised: RuntimeError} \]
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Time = 0.68 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.80 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=\frac {\frac {12 \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{4}} - \frac {12 \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{4}} + \frac {25 \, \tan \left (d x + c\right )^{4} - 124 i \, \tan \left (d x + c\right )^{3} - 246 \, \tan \left (d x + c\right )^{2} + 252 i \, \tan \left (d x + c\right ) + 57}{a^{4} {\left (\tan \left (d x + c\right ) - i\right )}^{4}}}{384 \, d} \]
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Time = 4.46 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.55 \[ \int \frac {\tan (c+d x)}{(a+i a \tan (c+d x))^4} \, dx=-\frac {x\,1{}\mathrm {i}}{16\,a^4}+\frac {-\frac {{\mathrm {tan}\left (c+d\,x\right )}^3\,1{}\mathrm {i}}{16}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{4}+\frac {\mathrm {tan}\left (c+d\,x\right )\,19{}\mathrm {i}}{48}+\frac {1}{12}}{a^4\,d\,{\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^4} \]
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