Integrand size = 22, antiderivative size = 99 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {5 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 i a^3 \sec (c+d x)}{2 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}+\frac {5 i \sec (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d} \]
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Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {3579, 3567, 3855} \[ \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {5 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 i a^3 \sec (c+d x)}{2 d}+\frac {5 i \sec (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d} \]
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Rule 3567
Rule 3579
Rule 3855
Rubi steps \begin{align*} \text {integral}& = \frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}+\frac {1}{3} (5 a) \int \sec (c+d x) (a+i a \tan (c+d x))^2 \, dx \\ & = \frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}+\frac {5 i \sec (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+\frac {1}{2} \left (5 a^2\right ) \int \sec (c+d x) (a+i a \tan (c+d x)) \, dx \\ & = \frac {5 i a^3 \sec (c+d x)}{2 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}+\frac {5 i \sec (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d}+\frac {1}{2} \left (5 a^3\right ) \int \sec (c+d x) \, dx \\ & = \frac {5 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 i a^3 \sec (c+d x)}{2 d}+\frac {i a \sec (c+d x) (a+i a \tan (c+d x))^2}{3 d}+\frac {5 i \sec (c+d x) \left (a^3+i a^3 \tan (c+d x)\right )}{6 d} \\ \end{align*}
Time = 1.48 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.94 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 (\cos (3 d x)+i \sin (3 d x)) \left (60 \text {arctanh}\left (\sin (c)+\cos (c) \tan \left (\frac {d x}{2}\right )\right )+i \sec ^3(c+d x) (20+24 \cos (2 (c+d x))+9 i \sin (2 (c+d x)))\right )}{12 d (\cos (d x)+i \sin (d x))^3} \]
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Time = 1.83 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.01
method | result | size |
risch | \(\frac {i a^{3} \left (33 \,{\mathrm e}^{5 i \left (d x +c \right )}+40 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(100\) |
derivativedivides | \(\frac {-i a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )-3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {3 i a^{3}}{\cos \left (d x +c \right )}+a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(147\) |
default | \(\frac {-i a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )-3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\sin \left (d x +c \right )}{2}-\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+\frac {3 i a^{3}}{\cos \left (d x +c \right )}+a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(147\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (81) = 162\).
Time = 0.24 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.04 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {66 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} + 80 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} + 30 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + 15 \, {\left (a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 15 \, {\left (a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3}\right )} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{6 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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\[ \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx=- i a^{3} \left (\int i \sec {\left (c + d x \right )}\, dx + \int \left (- 3 \tan {\left (c + d x \right )} \sec {\left (c + d x \right )}\right )\, dx + \int \tan ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \left (- 3 i \tan ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\right )\, dx\right ) \]
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Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.10 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {9 \, a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} + \log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + \frac {36 i \, a^{3}}{\cos \left (d x + c\right )} + \frac {4 i \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{3}}{\cos \left (d x + c\right )^{3}}}{12 \, d} \]
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Time = 0.48 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.26 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {15 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right ) - 15 \, a^{3} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right ) - \frac {2 \, {\left (9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 48 i \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 22 i \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
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Time = 6.25 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.37 \[ \int \sec (c+d x) (a+i a \tan (c+d x))^3 \, dx=\frac {5\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,6{}\mathrm {i}-a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,16{}\mathrm {i}-3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a^3\,22{}\mathrm {i}}{3}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
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