Integrand size = 28, antiderivative size = 183 \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=-\frac {22 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )} \]
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Time = 0.17 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3581, 3853, 3856, 2719} \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=-\frac {22 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 e^7 \sin (c+d x) \sqrt {e \sec (c+d x)}}{15 a^2 d}+\frac {22 e^5 \sin (c+d x) (e \sec (c+d x))^{5/2}}{45 a^2 d}+\frac {22 e^3 \sin (c+d x) (e \sec (c+d x))^{9/2}}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )} \]
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Rule 2719
Rule 3581
Rule 3853
Rule 3856
Rubi steps \begin{align*} \text {integral}& = -\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (11 e^2\right ) \int (e \sec (c+d x))^{11/2} \, dx}{7 a^2} \\ & = \frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (11 e^4\right ) \int (e \sec (c+d x))^{7/2} \, dx}{9 a^2} \\ & = \frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (11 e^6\right ) \int (e \sec (c+d x))^{3/2} \, dx}{15 a^2} \\ & = \frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {\left (11 e^8\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{15 a^2} \\ & = \frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {\left (11 e^8\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 a^2 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}} \\ & = -\frac {22 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^2 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{45 a^2 d}+\frac {22 e^3 (e \sec (c+d x))^{9/2} \sin (c+d x)}{63 a^2 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{7 d \left (a^2+i a^2 \tan (c+d x)\right )} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 2.88 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.65 \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=\frac {(e \sec (c+d x))^{15/2} (\cos (d x)+i \sin (d x))^2 \left (\frac {22 i \sqrt {2} e^{3 i c-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}+\frac {1}{56} \csc (c) \sec ^{\frac {9}{2}}(c+d x) (\cos (2 c)+i \sin (2 c)) (1260 \cos (d x)+1050 \cos (2 c+d x)+1078 \cos (2 c+3 d x)+77 \cos (4 c+3 d x)+231 \cos (4 c+5 d x)+720 i \sin (d x)-720 i \sin (2 c+d x))\right )}{45 d \sec ^{\frac {11}{2}}(c+d x) (a+i a \tan (c+d x))^2} \]
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 472 vs. \(2 (185 ) = 370\).
Time = 9.20 (sec) , antiderivative size = 473, normalized size of antiderivative = 2.58
method | result | size |
default | \(-\frac {2 i e^{7} \sqrt {e \sec \left (d x +c \right )}\, \left (-231 \left (\cos ^{2}\left (d x +c \right )\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}}+231 \left (\cos ^{2}\left (d x +c \right )\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}}-462 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\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}}\, \cos \left (d x +c \right )+462 \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}}\, \cos \left (d x +c \right )-231 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, E\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}}+231 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}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+231 i \sin \left (d x +c \right )+77 i \tan \left (d x +c \right )+77 i \tan \left (d x +c \right ) \sec \left (d x +c \right )-35 i \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )-35 i \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )+90 \left (\sec ^{2}\left (d x +c \right )\right )+90 \left (\sec ^{3}\left (d x +c \right )\right )\right )}{315 a^{2} d \left (\cos \left (d x +c \right )+1\right )}\) | \(473\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.40 \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (231 i \, e^{7} e^{\left (9 i \, d x + 9 i \, c\right )} + 1078 i \, e^{7} e^{\left (7 i \, d x + 7 i \, c\right )} + 1980 i \, e^{7} e^{\left (5 i \, d x + 5 i \, c\right )} + 1770 i \, e^{7} e^{\left (3 i \, d x + 3 i \, c\right )} + 77 i \, e^{7} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 231 \, \sqrt {2} {\left (i \, e^{7} e^{\left (8 i \, d x + 8 i \, c\right )} + 4 i \, e^{7} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, e^{7} e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{7} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, e^{7}\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{315 \, {\left (a^{2} d e^{\left (8 i \, d x + 8 i \, c\right )} + 4 \, a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 4 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )}} \]
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Timed out. \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=\text {Timed out} \]
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Exception generated. \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]
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\[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^2} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{15/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]
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