Integrand size = 28, antiderivative size = 155 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\frac {10 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 d e^6}+\frac {10 a^3 \sin (c+d x)}{77 d e^5 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}} \]
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Time = 0.21 (sec) , antiderivative size = 155, 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 = {3578, 3577, 3854, 3856, 2720} \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\frac {10 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 d e^6}+\frac {10 a^3 \sin (c+d x)}{77 d e^5 \sqrt {e \sec (c+d x)}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}} \]
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Rule 2720
Rule 3577
Rule 3578
Rule 3854
Rule 3856
Rubi steps \begin{align*} \text {integral}& = -\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}+\frac {(5 a) \int \frac {(a+i a \tan (c+d x))^2}{(e \sec (c+d x))^{7/2}} \, dx}{11 e^2} \\ & = -\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}+\frac {\left (15 a^3\right ) \int \frac {1}{(e \sec (c+d x))^{3/2}} \, dx}{77 e^4} \\ & = \frac {10 a^3 \sin (c+d x)}{77 d e^5 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}+\frac {\left (5 a^3\right ) \int \sqrt {e \sec (c+d x)} \, dx}{77 e^6} \\ & = \frac {10 a^3 \sin (c+d x)}{77 d e^5 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}}+\frac {\left (5 a^3 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{77 e^6} \\ & = \frac {10 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{77 d e^6}+\frac {10 a^3 \sin (c+d x)}{77 d e^5 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{11 d (e \sec (c+d x))^{11/2}}-\frac {20 i \left (a^3+i a^3 \tan (c+d x)\right )}{77 d e^2 (e \sec (c+d x))^{7/2}} \\ \end{align*}
Time = 2.11 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\frac {a^3 \sqrt {e \sec (c+d x)} \left (-46 i \cos (c+d x)-22 i \cos (3 (c+d x))-15 \sin (c+d x)+20 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (3 (c+d x))-i \sin (3 (c+d x)))-15 \sin (3 (c+d x))\right ) (\cos (3 (c+2 d x))+i \sin (3 (c+2 d x)))}{154 d e^6 (\cos (d x)+i \sin (d x))^3} \]
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Time = 20.10 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.30
method | result | size |
default | \(-\frac {2 a^{3} \left (28 i \left (\cos ^{5}\left (d x +c \right )\right )-28 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+5 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-11 i \left (\cos ^{3}\left (d x +c \right )\right )+5 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-3 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-5 \sin \left (d x +c \right )\right )}{77 e^{5} d \sqrt {e \sec \left (d x +c \right )}}\) | \(201\) |
risch | \(-\frac {i {\mathrm e}^{i \left (d x +c \right )} \left (7 \,{\mathrm e}^{4 i \left (d x +c \right )}+24 \,{\mathrm e}^{2 i \left (d x +c \right )}+37\right ) a^{3} \sqrt {2}}{308 d \,e^{5} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}+\frac {10 \sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}\, \sqrt {i \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}\, \sqrt {i {\mathrm e}^{i \left (d x +c \right )}}\, F\left (\sqrt {-i \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}, \frac {\sqrt {2}}{2}\right ) a^{3} \sqrt {e \,{\mathrm e}^{i \left (d x +c \right )} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}}{77 d \sqrt {e \,{\mathrm e}^{3 i \left (d x +c \right )}+e \,{\mathrm e}^{i \left (d x +c \right )}}\, e^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}\) | \(260\) |
parts | \(-\frac {2 a^{3} \left (-7 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+15 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+15 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-9 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-15 \sin \left (d x +c \right )\right )}{77 d \sqrt {e \sec \left (d x +c \right )}\, e^{5}}-\frac {2 i a^{3} \left (7 \left (\cos ^{5}\left (d x +c \right )\right )-11 \left (\cos ^{3}\left (d x +c \right )\right )\right )}{77 d \sqrt {e \sec \left (d x +c \right )}\, e^{5}}-\frac {6 i a^{3}}{11 d \left (e \sec \left (d x +c \right )\right )^{\frac {11}{2}}}+\frac {2 a^{3} \left (21 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+10 i F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+10 i \sec \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, F\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}-6 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-10 \sin \left (d x +c \right )\right )}{77 d \sqrt {e \sec \left (d x +c \right )}\, e^{5}}\) | \(420\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\frac {-40 i \, \sqrt {2} a^{3} \sqrt {e} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (-7 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 31 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 61 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 37 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{308 \, d e^{6}} \]
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Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\text {Timed out} \]
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\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}} \,d x } \]
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\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {11}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{11/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{11/2}} \,d x \]
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