Integrand size = 28, antiderivative size = 138 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=-\frac {14 a^2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 i a^2 (e \sec (c+d x))^{3/2}}{15 d}+\frac {14 a^2 e \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \]
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Time = 0.19 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3579, 3567, 3853, 3856, 2719} \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=-\frac {14 a^2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 i a^2 (e \sec (c+d x))^{3/2}}{15 d}+\frac {14 a^2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 d}+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d} \]
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Rule 2719
Rule 3567
Rule 3579
Rule 3853
Rule 3856
Rubi steps \begin{align*} \text {integral}& = \frac {2 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac {1}{5} (7 a) \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x)) \, dx \\ & = \frac {14 i a^2 (e \sec (c+d x))^{3/2}}{15 d}+\frac {2 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac {1}{5} \left (7 a^2\right ) \int (e \sec (c+d x))^{3/2} \, dx \\ & = \frac {14 i a^2 (e \sec (c+d x))^{3/2}}{15 d}+\frac {14 a^2 e \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}-\frac {1}{5} \left (7 a^2 e^2\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx \\ & = \frac {14 i a^2 (e \sec (c+d x))^{3/2}}{15 d}+\frac {14 a^2 e \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}-\frac {\left (7 a^2 e^2\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = -\frac {14 a^2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {14 i a^2 (e \sec (c+d x))^{3/2}}{15 d}+\frac {14 a^2 e \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 i (e \sec (c+d x))^{3/2} \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 3.30 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.93 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=\frac {(e \sec (c+d x))^{3/2} \left (-\frac {14 i \sqrt {2} \left (3 \sqrt {1+e^{2 i (c+d x)}}-e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}}}+\frac {1}{2} \csc (c) \sec ^{\frac {5}{2}}(c+d x) (\cos (2 c)-i \sin (2 c)) (36 \cos (d x)+27 \cos (2 c+d x)+21 \cos (2 c+3 d x)-20 i \sin (d x)+20 i \sin (2 c+d x))\right ) (a+i a \tan (c+d x))^2}{15 d \sec ^{\frac {7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^2} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 429 vs. \(2 (143 ) = 286\).
Time = 14.59 (sec) , antiderivative size = 430, normalized size of antiderivative = 3.12
method | result | size |
default | \(\frac {2 e \,a^{2} \sqrt {e \sec \left (d x +c \right )}\, \left (21 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{2}\left (d x +c \right )\right )-21 i E\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{2}\left (d x +c \right )\right )+42 i \cos \left (d x +c \right ) F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-42 i \cos \left (d x +c \right ) E\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+21 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-21 i E\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+10 i+21 \sin \left (d x +c \right )+10 i \sec \left (d x +c \right )-3 \tan \left (d x +c \right )-3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}\) | \(430\) |
parts | \(\text {Expression too large to display}\) | \(830\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.20 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (21 i \, a^{2} e e^{\left (5 i \, d x + 5 i \, c\right )} + 16 i \, a^{2} e e^{\left (3 i \, d x + 3 i \, c\right )} + 7 i \, a^{2} e e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 21 \, \sqrt {2} {\left (i \, a^{2} e e^{\left (4 i \, d x + 4 i \, c\right )} + 2 i \, a^{2} e e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{2} e\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
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\[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=- a^{2} \left (\int \left (- \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}\right )\, dx + \int \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (c + d x \right )}\, dx + \int \left (- 2 i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan {\left (c + d x \right )}\right )\, dx\right ) \]
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\[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \,d x } \]
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\[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \,d x } \]
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Timed out. \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2 \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \]
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