Optimal. Leaf size=132 \[ \frac {2 i e^2}{45 d \left (a^3+i a^3 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}+\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{9 a d (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}} \]
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Rubi [A] time = 0.13, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3500, 3502, 3771, 2639} \[ \frac {2 i e^2}{45 d \left (a^3+i a^3 \tan (c+d x)\right ) \sqrt {e \sec (c+d x)}}+\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{9 a d (a+i a \tan (c+d x))^2 \sqrt {e \sec (c+d x)}} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 3500
Rule 3502
Rule 3771
Rubi steps
\begin {align*} \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^3} \, dx &=\frac {4 i e^2}{9 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}+\frac {e^2 \int \frac {1}{\sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))} \, dx}{9 a^2}\\ &=\frac {4 i e^2}{9 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}+\frac {2 i e^2}{45 d \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{15 a^3}\\ &=\frac {4 i e^2}{9 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}+\frac {2 i e^2}{45 d \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {e^2 \int \sqrt {\cos (c+d x)} \, dx}{15 a^3 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {4 i e^2}{9 a d \sqrt {e \sec (c+d x)} (a+i a \tan (c+d x))^2}+\frac {2 i e^2}{45 d \sqrt {e \sec (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}
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Mathematica [C] time = 0.85, size = 140, normalized size = 1.06 \[ -\frac {e^{-i d x} \sec ^2(c+d x) (\cos (d x)+i \sin (d x)) (e \sec (c+d x))^{3/2} \left (6 e^{2 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )+3 i \sin (2 (c+d x))+8 \cos (2 (c+d x))+8\right )}{45 a^3 d (\tan (c+d x)-i)^3} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.84, size = 0, normalized size = 0.00 \[ \frac {{\left (90 \, a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )} {\rm integral}\left (-\frac {i \, \sqrt {2} e \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{15 \, a^{3} d}, x\right ) + \sqrt {2} {\left (12 i \, e e^{\left (6 i \, d x + 6 i \, c\right )} + 23 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, e\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{90 \, a^{3} d} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.34, size = 388, normalized size = 2.94 \[ \frac {2 \left (20 i \left (\cos ^{5}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-3 i \cos \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right ) \sin \left (d x +c \right )-20 \left (\cos ^{6}\left (d x +c \right )\right )+3 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-3 i \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-9 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )+19 \left (\cos ^{4}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right )}{45 a^{3} d \sin \left (d x +c \right )^{5}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \frac {i \int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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