Integrand size = 30, antiderivative size = 165 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {16 i}{35 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {12 i \sqrt {a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}-\frac {32 i \sqrt {a+i a \tan (c+d x)}}{35 a d e^2 \sqrt {e \sec (c+d x)}} \]
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Time = 0.35 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3583, 3578, 3569} \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {32 i \sqrt {a+i a \tan (c+d x)}}{35 a d e^2 \sqrt {e \sec (c+d x)}}+\frac {16 i}{35 d e^2 \sqrt {a+i a \tan (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {12 i \sqrt {a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}+\frac {2 i}{7 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{5/2}} \]
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Rule 3569
Rule 3578
Rule 3583
Rubi steps \begin{align*} \text {integral}& = \frac {2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {6 \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx}{7 a} \\ & = \frac {2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}-\frac {12 i \sqrt {a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}+\frac {24 \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{35 e^2} \\ & = \frac {2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {16 i}{35 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {12 i \sqrt {a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}+\frac {16 \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}} \, dx}{35 a e^2} \\ & = \frac {2 i}{7 d (e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}+\frac {16 i}{35 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {12 i \sqrt {a+i a \tan (c+d x)}}{35 a d (e \sec (c+d x))^{5/2}}-\frac {32 i \sqrt {a+i a \tan (c+d x)}}{35 a d e^2 \sqrt {e \sec (c+d x)}} \\ \end{align*}
Time = 1.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.48 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i (17+\cos (2 (c+d x))+3 i \sec (c+d x) \sin (3 (c+d x))+35 i \tan (c+d x))}{35 d e^2 \sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]
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Time = 10.23 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.42
method | result | size |
default | \(-\frac {2 \left (i \left (\cos ^{2}\left (d x +c \right )\right )-6 \sin \left (d x +c \right ) \cos \left (d x +c \right )+8 i-16 \tan \left (d x +c \right )\right )}{35 d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {e \sec \left (d x +c \right )}\, e^{2}}\) | \(70\) |
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Time = 0.24 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-7 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 112 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 40 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-\frac {7}{2} i \, d x - \frac {7}{2} i \, c\right )}}{140 \, a d e^{3}} \]
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\[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]
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Time = 0.59 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {5 i \, \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 7 i \, \cos \left (\frac {5}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 35 i \, \cos \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) - 105 i \, \cos \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 5 \, \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, \sin \left (\frac {5}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 35 \, \sin \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 105 \, \sin \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right )}{140 \, \sqrt {a} d e^{\frac {5}{2}}} \]
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\[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {5}{2}} \sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]
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Time = 5.35 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(e \sec (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (-\sin \left (c+d\,x\right )-\frac {3\,\sin \left (3\,c+3\,d\,x\right )}{35}+\frac {\cos \left (c+d\,x\right )\,1{}\mathrm {i}}{2}+\frac {\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}}{70}\right )}{d\,e^3\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]
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