Integrand size = 30, antiderivative size = 169 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{11/2}} \, dx=\frac {32 i a^3 \sqrt {e \sec (c+d x)}}{77 d e^6 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a^2 \sqrt {a+i a \tan (c+d x)}}{77 d e^4 (e \sec (c+d x))^{3/2}}-\frac {12 i a (a+i a \tan (c+d x))^{3/2}}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}} \]
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Time = 0.35 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3578, 3569} \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{11/2}} \, dx=\frac {32 i a^3 \sqrt {e \sec (c+d x)}}{77 d e^6 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a^2 \sqrt {a+i a \tan (c+d x)}}{77 d e^4 (e \sec (c+d x))^{3/2}}-\frac {12 i a (a+i a \tan (c+d x))^{3/2}}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}} \]
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Rule 3569
Rule 3578
Rubi steps \begin{align*} \text {integral}& = -\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}}+\frac {(6 a) \int \frac {(a+i a \tan (c+d x))^{3/2}}{(e \sec (c+d x))^{7/2}} \, dx}{11 e^2} \\ & = -\frac {12 i a (a+i a \tan (c+d x))^{3/2}}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}}+\frac {\left (24 a^2\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{77 e^4} \\ & = -\frac {16 i a^2 \sqrt {a+i a \tan (c+d x)}}{77 d e^4 (e \sec (c+d x))^{3/2}}-\frac {12 i a (a+i a \tan (c+d x))^{3/2}}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}}+\frac {\left (16 a^3\right ) \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{77 e^6} \\ & = \frac {32 i a^3 \sqrt {e \sec (c+d x)}}{77 d e^6 \sqrt {a+i a \tan (c+d x)}}-\frac {16 i a^2 \sqrt {a+i a \tan (c+d x)}}{77 d e^4 (e \sec (c+d x))^{3/2}}-\frac {12 i a (a+i a \tan (c+d x))^{3/2}}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{11 d (e \sec (c+d x))^{11/2}} \\ \end{align*}
Time = 1.60 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.72 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{11/2}} \, dx=\frac {a^2 (-55 i \cos (c+d x)+35 i \cos (3 (c+d x))-22 \sin (c+d x)+42 \sin (3 (c+d x))) (\cos (2 (c+2 d x))+i \sin (2 (c+2 d x))) \sqrt {a+i a \tan (c+d x)}}{154 d e^5 \sqrt {e \sec (c+d x)} (\cos (d x)+i \sin (d x))^2} \]
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Time = 9.82 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.57
method | result | size |
risch | \(-\frac {i a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (7 \,{\mathrm e}^{5 i \left (d x +c \right )}+33 \,{\mathrm e}^{3 i \left (d x +c \right )}+154 i \sin \left (d x +c \right )\right )}{308 e^{5} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\) | \(97\) |
default | \(\frac {2 i \left (\tan \left (d x +c \right )-i\right )^{2} a^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (42 i \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-35 \left (\cos ^{5}\left (d x +c \right )\right )-16 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+40 \left (\cos ^{3}\left (d x +c \right )\right )\right )}{77 d \sqrt {e \sec \left (d x +c \right )}\, e^{5}}\) | \(104\) |
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Time = 0.24 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.59 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{11/2}} \, dx=\frac {{\left (-7 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 40 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 110 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} + 77 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{308 \, d e^{6}} \]
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Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{11/2}} \, dx=\text {Timed out} \]
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Time = 0.84 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.73 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{11/2}} \, dx=\frac {{\left (-7 i \, a^{2} \cos \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) - 33 i \, a^{2} \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 77 i \, a^{2} \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 77 i \, a^{2} \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, a^{2} \sin \left (\frac {11}{2} \, d x + \frac {11}{2} \, c\right ) + 33 \, a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 77 \, a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 77 \, a^{2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{308 \, d e^{\frac {11}{2}}} \]
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\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{11/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}} \,d x } \]
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Time = 6.80 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.79 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{11/2}} \, dx=-\frac {a^2\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (-187\,\sin \left (2\,c+2\,d\,x\right )-40\,\sin \left (4\,c+4\,d\,x\right )-7\,\sin \left (6\,c+6\,d\,x\right )+\cos \left (2\,c+2\,d\,x\right )\,33{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,40{}\mathrm {i}+\cos \left (6\,c+6\,d\,x\right )\,7{}\mathrm {i}\right )}{616\,d\,e^6} \]
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