Optimal. Leaf size=81 \[ \frac{3 i (1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{11}{6},\frac{7}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{2^{5/6} f (a+i a \tan (e+f x))} \]
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Rubi [A] time = 0.169389, antiderivative size = 81, 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} \[ \frac{3 i (1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{11}{6},\frac{7}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{2^{5/6} f (a+i a \tan (e+f x))} \]
Antiderivative was successfully verified.
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Rule 3505
Rule 3523
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \frac{\sqrt [3]{d \sec (e+f x)}}{a+i a \tan (e+f x)} \, dx &=\frac{\sqrt [3]{d \sec (e+f x)} \int \frac{\sqrt [6]{a-i a \tan (e+f x)}}{(a+i a \tan (e+f x))^{5/6}} \, 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{1}{(a-i a x)^{5/6} (a+i a x)^{11/6}} \, 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 (a \sqrt [3]{d \sec (e+f x)} \left (\frac{a+i a \tan (e+f x)}{a}\right )^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{\left (\frac{1}{2}+\frac{i x}{2}\right )^{11/6} (a-i a x)^{5/6}} \, dx,x,\tan (e+f x)\right )}{2\ 2^{5/6} f \sqrt [6]{a-i a \tan (e+f x)} (a+i a \tan (e+f x))}\\ &=\frac{3 i \, _2F_1\left (\frac{1}{6},\frac{11}{6};\frac{7}{6};\frac{1}{2} (1-i \tan (e+f x))\right ) \sqrt [3]{d \sec (e+f x)} (1+i \tan (e+f x))^{5/6}}{2^{5/6} f (a+i a \tan (e+f x))}\\ \end{align*}
Mathematica [A] time = 0.42943, size = 103, normalized size = 1.27 \[ -\frac{3 i e^{-2 i (e+f x)} \left (4 e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (e+f x)}\right )-e^{2 i (e+f x)}-1\right ) \sqrt [3]{d \sec (e+f x)}}{10 a f} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.154, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{a+ia\tan \left ( fx+e \right ) }\sqrt [3]{d\sec \left ( fx+e \right ) }}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \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{{\left (10 \, a f e^{\left (2 i \, f x + 2 i \, e\right )}{\rm integral}\left (-\frac{2 i \cdot 2^{\frac{1}{3}} \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 )}}{5 \, a f}, x\right ) + 2^{\frac{1}{3}} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{1}{3}}{\left (3 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i\right )} e^{\left (\frac{1}{3} i \, f x + \frac{1}{3} i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{10 \, a f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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 \frac{\left (d \sec \left (f x + e\right )\right )^{\frac{1}{3}}}{i \, a \tan \left (f x + e\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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