Integrand size = 15, antiderivative size = 117 \[ \int (a+i a \tan (c+d x))^5 \, dx=16 a^5 x-\frac {16 i a^5 \log (\cos (c+d x))}{d}-\frac {8 a^5 \tan (c+d x)}{d}+\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d}+\frac {2 i a \left (a^2+i a^2 \tan (c+d x)\right )^2}{d} \]
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Time = 0.10 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3559, 3558, 3556} \[ \int (a+i a \tan (c+d x))^5 \, dx=-\frac {8 a^5 \tan (c+d x)}{d}-\frac {16 i a^5 \log (\cos (c+d x))}{d}+16 a^5 x+\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {2 i a \left (a^2+i a^2 \tan (c+d x)\right )^2}{d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d} \]
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Rule 3556
Rule 3558
Rule 3559
Rubi steps \begin{align*} \text {integral}& = \frac {i a (a+i a \tan (c+d x))^4}{4 d}+(2 a) \int (a+i a \tan (c+d x))^4 \, dx \\ & = \frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d}+\left (4 a^2\right ) \int (a+i a \tan (c+d x))^3 \, dx \\ & = \frac {2 i a^3 (a+i a \tan (c+d x))^2}{d}+\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d}+\left (8 a^3\right ) \int (a+i a \tan (c+d x))^2 \, dx \\ & = 16 a^5 x-\frac {8 a^5 \tan (c+d x)}{d}+\frac {2 i a^3 (a+i a \tan (c+d x))^2}{d}+\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d}+\left (16 i a^5\right ) \int \tan (c+d x) \, dx \\ & = 16 a^5 x-\frac {16 i a^5 \log (\cos (c+d x))}{d}-\frac {8 a^5 \tan (c+d x)}{d}+\frac {2 i a^3 (a+i a \tan (c+d x))^2}{d}+\frac {2 i a^2 (a+i a \tan (c+d x))^3}{3 d}+\frac {i a (a+i a \tan (c+d x))^4}{4 d} \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.61 \[ \int (a+i a \tan (c+d x))^5 \, dx=\frac {a^5 \left (35 i+192 i \log (i+\tan (c+d x))-180 \tan (c+d x)-66 i \tan ^2(c+d x)+20 \tan ^3(c+d x)+3 i \tan ^4(c+d x)\right )}{12 d} \]
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Time = 0.15 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.62
method | result | size |
derivativedivides | \(\frac {a^{5} \left (-15 \tan \left (d x +c \right )+\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {11 i \left (\tan ^{2}\left (d x +c \right )\right )}{2}+8 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+16 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(72\) |
default | \(\frac {a^{5} \left (-15 \tan \left (d x +c \right )+\frac {i \left (\tan ^{4}\left (d x +c \right )\right )}{4}+\frac {5 \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {11 i \left (\tan ^{2}\left (d x +c \right )\right )}{2}+8 i \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+16 \arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(72\) |
parallelrisch | \(\frac {3 i a^{5} \left (\tan ^{4}\left (d x +c \right )\right )-66 i a^{5} \left (\tan ^{2}\left (d x +c \right )\right )+20 \left (\tan ^{3}\left (d x +c \right )\right ) a^{5}+96 i a^{5} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+192 a^{5} x d -180 a^{5} \tan \left (d x +c \right )}{12 d}\) | \(83\) |
risch | \(-\frac {32 a^{5} c}{d}-\frac {4 i a^{5} \left (48 \,{\mathrm e}^{6 i \left (d x +c \right )}+108 \,{\mathrm e}^{4 i \left (d x +c \right )}+88 \,{\mathrm e}^{2 i \left (d x +c \right )}+25\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {16 i a^{5} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(89\) |
norman | \(16 a^{5} x -\frac {15 a^{5} \tan \left (d x +c \right )}{d}+\frac {5 a^{5} \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}-\frac {11 i a^{5} \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {i a^{5} \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}+\frac {8 i a^{5} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{d}\) | \(92\) |
parts | \(a^{5} x -\frac {10 i a^{5} \left (\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {i a^{5} \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {5 i a^{5} \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {10 a^{5} \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {5 a^{5} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(161\) |
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Time = 0.25 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.51 \[ \int (a+i a \tan (c+d x))^5 \, dx=-\frac {4 \, {\left (48 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 108 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 88 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + 25 i \, a^{5} + 12 \, {\left (i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{5}\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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Time = 0.25 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.52 \[ \int (a+i a \tan (c+d x))^5 \, dx=- \frac {16 i a^{5} \log {\left (e^{2 i d x} + e^{- 2 i c} \right )}}{d} + \frac {- 192 i a^{5} e^{6 i c} e^{6 i d x} - 432 i a^{5} e^{4 i c} e^{4 i d x} - 352 i a^{5} e^{2 i c} e^{2 i d x} - 100 i a^{5}}{3 d e^{8 i c} e^{8 i d x} + 12 d e^{6 i c} e^{6 i d x} + 18 d e^{4 i c} e^{4 i d x} + 12 d e^{2 i c} e^{2 i d x} + 3 d} \]
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Time = 0.46 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.41 \[ \int (a+i a \tan (c+d x))^5 \, dx=a^{5} x + \frac {5 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{5}}{3 \, d} + \frac {10 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{5}}{d} + \frac {i \, a^{5} {\left (\frac {4 \, \sin \left (d x + c\right )^{2} - 3}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 2 \, \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{4 \, d} + \frac {5 i \, a^{5} {\left (\frac {1}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right )^{2} - 1\right )\right )}}{d} + \frac {5 i \, a^{5} \log \left (\sec \left (d x + c\right )\right )}{d} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (99) = 198\).
Time = 0.40 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.90 \[ \int (a+i a \tan (c+d x))^5 \, dx=-\frac {4 \, {\left (12 i \, a^{5} e^{\left (8 i \, d x + 8 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 72 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 48 i \, a^{5} e^{\left (6 i \, d x + 6 i \, c\right )} + 108 i \, a^{5} e^{\left (4 i \, d x + 4 i \, c\right )} + 88 i \, a^{5} e^{\left (2 i \, d x + 2 i \, c\right )} + 12 i \, a^{5} \log \left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right ) + 25 i \, a^{5}\right )}}{3 \, {\left (d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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Time = 4.40 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.62 \[ \int (a+i a \tan (c+d x))^5 \, dx=\frac {a^5\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,16{}\mathrm {i}-15\,a^5\,\mathrm {tan}\left (c+d\,x\right )-\frac {a^5\,{\mathrm {tan}\left (c+d\,x\right )}^2\,11{}\mathrm {i}}{2}+\frac {5\,a^5\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+\frac {a^5\,{\mathrm {tan}\left (c+d\,x\right )}^4\,1{}\mathrm {i}}{4}}{d} \]
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