Optimal. Leaf size=69 \[ \frac{12 i \sqrt [6]{2} a^2 \sqrt [3]{d \sec (e+f x)} \text{Hypergeometric2F1}\left (-\frac{7}{6},\frac{1}{6},\frac{7}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{f \sqrt [6]{1+i \tan (e+f x)}} \]
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Rubi [A] time = 0.157643, antiderivative size = 69, normalized size of antiderivative = 1., 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 = {3505, 3523, 70, 69} \[ \frac{12 i \sqrt [6]{2} a^2 \sqrt [3]{d \sec (e+f x)} \text{Hypergeometric2F1}\left (-\frac{7}{6},\frac{1}{6},\frac{7}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{f \sqrt [6]{1+i \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 3505
Rule 3523
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2 \, dx &=\frac{\sqrt [3]{d \sec (e+f x)} \int \sqrt [6]{a-i a \tan (e+f x)} (a+i a \tan (e+f x))^{13/6} \, dx}{\sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}}\\ &=\frac{\left (a^2 \sqrt [3]{d \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{7/6}}{(a-i a x)^{5/6}} \, dx,x,\tan (e+f x)\right )}{f \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}}\\ &=\frac{\left (2 \sqrt [6]{2} a^3 \sqrt [3]{d \sec (e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{i x}{2}\right )^{7/6}}{(a-i a x)^{5/6}} \, dx,x,\tan (e+f x)\right )}{f \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{\frac{a+i a \tan (e+f x)}{a}}}\\ &=\frac{12 i \sqrt [6]{2} a^2 \, _2F_1\left (-\frac{7}{6},\frac{1}{6};\frac{7}{6};\frac{1}{2} (1-i \tan (e+f x))\right ) \sqrt [3]{d \sec (e+f x)}}{f \sqrt [6]{1+i \tan (e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.965874, size = 128, normalized size = 1.86 \[ -\frac{3 a^2 e^{-2 i e} \sqrt [3]{d \sec (e+f x)} (\cos (2 (e+f x))+i \sin (2 (e+f x))) \left (7 i \sqrt [3]{1+e^{2 i (e+f x)}} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (e+f x)}\right )+\sec (e) \sin (f x) \sec (e+f x)+\tan (e)-8 i\right )}{4 f (\cos (f x)+i \sin (f x))^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.132, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{d\sec \left ( fx+e \right ) } \left ( a+ia\tan \left ( fx+e \right ) \right ) ^{2}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{1}{3}}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2^{\frac{1}{3}}{\left (27 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 21 i \, a^{2}\right )} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{1}{3}} e^{\left (\frac{1}{3} i \, f x + \frac{1}{3} i \, e\right )} + 4 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}{\rm integral}\left (-\frac{7 i \cdot 2^{\frac{1}{3}} a^{2} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{1}{3}} e^{\left (-\frac{2}{3} i \, f x - \frac{2}{3} i \, e\right )}}{4 \, f}, x\right )}{4 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int \sqrt [3]{d \sec{\left (e + f x \right )}}\, dx + \int - \sqrt [3]{d \sec{\left (e + f x \right )}} \tan ^{2}{\left (e + f x \right )}\, dx + \int 2 i \sqrt [3]{d \sec{\left (e + f x \right )}} \tan{\left (e + f x \right )}\, dx\right ) \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{1}{3}}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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