Optimal. Leaf size=163 \[ \frac{486 i a^4 (d \sec (e+f x))^{2/3}}{35 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac{81 i a^3 (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{35 f}+\frac{27 i a^2 (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}}{35 f}+\frac{3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f} \]
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Rubi [A] time = 0.320744, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {3494, 3493} \[ \frac{486 i a^4 (d \sec (e+f x))^{2/3}}{35 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac{81 i a^3 (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{35 f}+\frac{27 i a^2 (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}}{35 f}+\frac{3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f} \]
Antiderivative was successfully verified.
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Rule 3494
Rule 3493
Rubi steps
\begin{align*} \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx &=\frac{3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f}+\frac{1}{5} (9 a) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx\\ &=\frac{27 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{35 f}+\frac{3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f}+\frac{1}{35} \left (108 a^2\right ) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx\\ &=\frac{81 i a^3 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{35 f}+\frac{27 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{35 f}+\frac{3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f}+\frac{1}{35} \left (162 a^3\right ) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3} \, dx\\ &=\frac{486 i a^4 (d \sec (e+f x))^{2/3}}{35 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac{81 i a^3 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{35 f}+\frac{27 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{35 f}+\frac{3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f}\\ \end{align*}
Mathematica [A] time = 1.11004, size = 116, normalized size = 0.71 \[ \frac{3 a^3 (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{5/3} (\sin (e-2 f x)+i \cos (e-2 f x)) (442 \cos (2 (e+f x))+45 i \tan (e+f x)+59 i \sin (3 (e+f x)) \sec (e+f x)+364)}{140 d f (\cos (f x)+i \sin (f x))^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.141, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{{\frac{2}{3}}} \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{{\frac{11}{3}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.26319, size = 1319, normalized size = 8.09 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.19718, size = 389, normalized size = 2.39 \begin{align*} \frac{2 \cdot 2^{\frac{1}{3}}{\left (420 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} + 945 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} + 810 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + 243 i \, a^{3}\right )} \left (\frac{a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}}{35 \,{\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{2}{3}}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{11}{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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