Optimal. Leaf size=164 \[ -\frac {14 \cos ^{\frac {11}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )} \]
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Rubi [A]
time = 0.14, antiderivative size = 164, 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 = {3596, 3581,
3853, 3856, 2719} \begin {gather*} -\frac {14 \cos ^{\frac {11}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \sin (c+d x) \cos ^5(c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \sin (c+d x) \cos ^3(c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d \left (a^2+i a^2 \tan (c+d x)\right ) (e \cos (c+d x))^{11/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 2719
Rule 3581
Rule 3596
Rule 3853
Rule 3856
Rubi steps
\begin {align*} \int \frac {1}{(e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \, dx &=\frac {\int \frac {(e \sec (c+d x))^{11/2}}{(a+i a \tan (c+d x))^2} \, dx}{(e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\\ &=-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (7 e^2\right ) \int (e \sec (c+d x))^{7/2} \, dx}{3 a^2 (e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\\ &=\frac {14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}+\frac {\left (7 e^4\right ) \int (e \sec (c+d x))^{3/2} \, dx}{5 a^2 (e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\\ &=\frac {14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {\left (7 e^6\right ) \int \frac {1}{\sqrt {e \sec (c+d x)}} \, dx}{5 a^2 (e \cos (c+d x))^{11/2} (e \sec (c+d x))^{11/2}}\\ &=\frac {14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}-\frac {\left (7 \cos ^{\frac {11}{2}}(c+d x)\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a^2 (e \cos (c+d x))^{11/2}}\\ &=-\frac {14 \cos ^{\frac {11}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^3(c+d x) \sin (c+d x)}{15 a^2 d (e \cos (c+d x))^{11/2}}+\frac {14 \cos ^5(c+d x) \sin (c+d x)}{5 a^2 d (e \cos (c+d x))^{11/2}}-\frac {4 i \cos ^2(c+d x)}{3 d (e \cos (c+d x))^{11/2} \left (a^2+i a^2 \tan (c+d x)\right )}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in
optimal.
time = 6.10, size = 406, normalized size = 2.48 \begin {gather*} \frac {2 \sqrt {2} e^{3 i c+2 i d x} \cos ^{\frac {7}{2}}(c+d x) \csc (c) \left (-42 \sqrt {2-2 i e^{i (c+d x)}} \sqrt {e^{i (c+d x)} \left (-i+e^{i (c+d x)}\right )} \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\text {ArcSin}\left (\sqrt {-i \cos (c+d x)+\sin (c+d x)}\right )\right |-1\right )+42 \sqrt {2-2 i e^{i (c+d x)}} \sqrt {e^{i (c+d x)} \left (-i+e^{i (c+d x)}\right )} \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\text {ArcSin}\left (\sqrt {-i \cos (c+d x)+\sin (c+d x)}\right )\right |-1\right )-\frac {1}{2} e^{-2 i c} \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )} \left (\left (-1+e^{2 i c}\right ) \left (47+56 e^{2 i (c+d x)}+21 e^{4 i (c+d x)}\right )+7 \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 )\right )\right ) (\cos (2 c)+i \sin (2 c)) (\cos (d x)+i \sin (d x))^2}{15 d \left (1+e^{2 i (c+d x)}\right )^3 (e \cos (c+d x))^{11/2} (a+i a \tan (c+d x))^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 2.56, size = 321, normalized size = 1.96
method | result | size |
default | \(\frac {\frac {112 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {56 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}-\frac {112 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}+\frac {56 \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {8 i \left (\sin ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\frac {24 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}-\frac {14 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{5}-\frac {4 i \sin \left (\frac {d x}{2}+\frac {c}{2}\right )}{3}}{\left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) e +e}\, e^{5} d}\) | \(321\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 0.09, size = 186, normalized size = 1.13 \begin {gather*} -\frac {2 \, {\left (2 \, \sqrt {\frac {1}{2}} {\left (21 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 56 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 47 i \, e^{\left (2 i \, d x + 2 i \, c\right )}\right )} \sqrt {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 )} + 21 \, {\left (i \, \sqrt {2} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, \sqrt {2} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, \sqrt {2} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{15 \, {\left (a^{2} d e^{\frac {11}{2}} + a^{2} d e^{\left (6 i \, d x + 6 i \, c + \frac {11}{2}\right )} + 3 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c + \frac {11}{2}\right )} + 3 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c + \frac {11}{2}\right )}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {1}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{11/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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