Integrand size = 30, antiderivative size = 132 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {8 i a (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {16 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d} \]
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Time = 0.33 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3596, 3578, 3583, 3569} \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 i \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}}{5 d}-\frac {16 i \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{5/2}}{15 d}+\frac {8 i a \sec ^2(c+d x) (e \cos (c+d x))^{5/2}}{15 d \sqrt {a+i a \tan (c+d x)}} \]
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Rule 3569
Rule 3578
Rule 3583
Rule 3596
Rubi steps \begin{align*} \text {integral}& = \left ((e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{5/2}} \, dx \\ & = -\frac {2 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {\left (4 a (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{5 e^2} \\ & = \frac {8 i a (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{5 d}+\frac {\left (8 (e \cos (c+d x))^{5/2} (e \sec (c+d x))^{5/2}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}} \, dx}{15 e^2} \\ & = \frac {8 i a (e \cos (c+d x))^{5/2} \sec ^2(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)}}{5 d}-\frac {16 i (e \cos (c+d x))^{5/2} \sec ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{15 d} \\ \end{align*}
Time = 1.71 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.48 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {i e^2 \sqrt {e \cos (c+d x)} (-15+\cos (2 (c+d x))-4 i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}}{15 d} \]
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Time = 8.99 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.47
method | result | size |
default | \(\frac {2 \left (i \left (\cos ^{2}\left (d x +c \right )\right )+4 \sin \left (d x +c \right ) \cos \left (d x +c \right )-8 i\right ) \sqrt {e \cos \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, e^{2}}{15 d}\) | \(62\) |
risch | \(-\frac {i e^{2} \sqrt {2}\, \sqrt {e \cos \left (d x +c \right )}\, \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (30-2 \cos \left (2 d x +2 c \right )+8 i \sin \left (2 d x +2 c \right )\right )}{30 d}\) | \(74\) |
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Time = 0.26 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.65 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {\sqrt {2} \sqrt {\frac {1}{2}} {\left (-3 i \, e^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 30 i \, e^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e^{2}\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {3}{2} i \, d x - \frac {3}{2} i \, c\right )}}{30 \, d} \]
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Timed out. \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\text {Timed out} \]
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Time = 0.39 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.12 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {{\left (5 i \, e^{2} \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 i \, e^{2} \cos \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) - 30 i \, e^{2} \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 5 \, e^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, e^{2} \sin \left (\frac {5}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 30 \, e^{2} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )} \sqrt {a} \sqrt {e}}{30 \, d} \]
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\[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {5}{2}} \sqrt {i \, a \tan \left (d x + c\right ) + a} \,d x } \]
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Time = 0.84 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.64 \[ \int (e \cos (c+d x))^{5/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {e^2\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+4\,\sin \left (2\,c+2\,d\,x\right )-15{}\mathrm {i}\right )}{15\,d} \]
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