Integrand size = 28, antiderivative size = 150 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{33 a^2 d e^2}+\frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{33 a^2 d e \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]
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Time = 0.15 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3581, 3854, 3856, 2720} \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\frac {4 i e^2}{11 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{7/2}}+\frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{33 a^2 d e^2}+\frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{33 a^2 d e \sqrt {e \sec (c+d x)}} \]
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Rule 2720
Rule 3581
Rule 3854
Rule 3856
Rubi steps \begin{align*} \text {integral}& = \frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (7 e^2\right ) \int \frac {1}{(e \sec (c+d x))^{7/2}} \, dx}{11 a^2} \\ & = \frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {5 \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{11 a^2} \\ & = \frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{33 a^2 d e \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {5 \int \sqrt {e \sec (c+d x)} \, dx}{33 a^2 e^2} \\ & = \frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{33 a^2 d e \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (5 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{33 a^2 e^2} \\ & = \frac {10 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{33 a^2 d e^2}+\frac {2 e \sin (c+d x)}{11 a^2 d (e \sec (c+d x))^{5/2}}+\frac {10 \sin (c+d x)}{33 a^2 d e \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{11 d (e \sec (c+d x))^{7/2} \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}
Time = 1.54 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=-\frac {\sec ^4(c+d x) \left (28 i+24 i \cos (2 (c+d x))-4 i \cos (4 (c+d x))+40 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))-6 \sin (2 (c+d x))+7 \sin (4 (c+d x))\right )}{132 a^2 d (e \sec (c+d x))^{3/2} (-i+\tan (c+d x))^2} \]
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Time = 8.66 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.27
method | result | size |
default | \(\frac {\frac {4 i \left (\cos ^{5}\left (d x +c \right )\right )}{11}+\frac {4 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{11}-\frac {10 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}{33}+\frac {2 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{11}-\frac {10 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}}{33}+\frac {10 \sin \left (d x +c \right )}{33}}{a^{2} d \sqrt {e \sec \left (d x +c \right )}\, e}\) | \(190\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\frac {{\left (\sqrt {2} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-11 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 30 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 56 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 18 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} - 80 i \, \sqrt {2} \sqrt {e} e^{\left (6 i \, d x + 6 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{264 \, a^{2} d e^{2}} \]
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\[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=- \frac {\int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (c + d x \right )} - 2 i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan {\left (c + d x \right )} - \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx}{a^{2}} \]
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Exception generated. \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
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\[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\int { \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2} \, dx=\int \frac {1}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]
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