Optimal. Leaf size=178 \[ -\frac {22 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 e^7 \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 a^3 d}+\frac {22 e^5 \sin (c+d x) (e \sec (c+d x))^{5/2}}{15 a^3 d}-\frac {22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2} \]
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Rubi [A] time = 0.17, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3500, 3501, 3768, 3771, 2639} \[ -\frac {22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}+\frac {22 e^7 \sin (c+d x) \sqrt {e \sec (c+d x)}}{5 a^3 d}+\frac {22 e^5 \sin (c+d x) (e \sec (c+d x))^{5/2}}{15 a^3 d}-\frac {22 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 2639
Rule 3500
Rule 3501
Rule 3768
Rule 3771
Rubi steps
\begin {align*} \int \frac {(e \sec (c+d x))^{15/2}}{(a+i a \tan (c+d x))^3} \, dx &=-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}+\frac {\left (11 e^2\right ) \int \frac {(e \sec (c+d x))^{11/2}}{a+i a \tan (c+d x)} \, dx}{3 a^2}\\ &=-\frac {22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}+\frac {\left (11 e^4\right ) \int (e \sec (c+d x))^{7/2} \, dx}{3 a^3}\\ &=-\frac {22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}+\frac {\left (11 e^6\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 a^3}\\ &=-\frac {22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}+\frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^3 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}-\frac {\left (11 e^8\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac {22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}+\frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^3 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}-\frac {\left (11 e^8\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a^3 \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\\ &=-\frac {22 e^8 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^3 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}-\frac {22 i e^4 (e \sec (c+d x))^{7/2}}{21 a^3 d}+\frac {22 e^7 \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 a^3 d}+\frac {22 e^5 (e \sec (c+d x))^{5/2} \sin (c+d x)}{15 a^3 d}-\frac {4 i e^2 (e \sec (c+d x))^{11/2}}{3 a d (a+i a \tan (c+d x))^2}\\ \end {align*}
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Mathematica [C] time = 1.65, size = 128, normalized size = 0.72 \[ -\frac {e^6 (\tan (c+d x)-i) (e \sec (c+d x))^{3/2} \left (77 e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-e^{2 i (c+d x)}\right )-868 \cos (2 (c+d x))+143 i \tan (c+d x)+203 i \sin (3 (c+d x)) \sec (c+d x)-556\right )}{210 a^3 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.47, size = 0, normalized size = 0.00 \[ \frac {\sqrt {2} {\left (-462 i \, e^{7} e^{\left (7 i \, d x + 7 i \, c\right )} - 1694 i \, e^{7} e^{\left (5 i \, d x + 5 i \, c\right )} - 2266 i \, e^{7} e^{\left (3 i \, d x + 3 i \, c\right )} - 1274 i \, e^{7} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 105 \, {\left (a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} {\rm integral}\left (\frac {11 i \, \sqrt {2} e^{7} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{5 \, a^{3} d}, x\right )}{105 \, {\left (a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \]
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 \[ \int \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 1.41, size = 392, normalized size = 2.20 \[ -\frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (231 i \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-231 i \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+231 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-231 i \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+231 \left (\cos ^{4}\left (d x +c \right )\right )+140 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-294 \left (\cos ^{3}\left (d x +c \right )\right )-15 i \sin \left (d x +c \right )+63 \cos \left (d x +c \right )\right ) \left (\frac {e}{\cos \left (d x +c \right )}\right )^{\frac {15}{2}} \left (\cos ^{4}\left (d x +c \right )\right )}{105 a^{3} d \sin \left (d x +c \right )^{5}} \]
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 \[ \int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{15/2}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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