Optimal. Leaf size=71 \[ \frac{12 i 2^{5/6} a^2 (d \sec (e+f x))^{5/3} \text{Hypergeometric2F1}\left (-\frac{11}{6},\frac{5}{6},\frac{11}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{5 f (1+i \tan (e+f x))^{5/6}} \]
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Rubi [A] time = 0.176646, antiderivative size = 71, 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} \[ \frac{12 i 2^{5/6} a^2 (d \sec (e+f x))^{5/3} \text{Hypergeometric2F1}\left (-\frac{11}{6},\frac{5}{6},\frac{11}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{5 f (1+i \tan (e+f x))^{5/6}} \]
Antiderivative was successfully verified.
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Rule 3505
Rule 3523
Rule 70
Rule 69
Rubi steps
\begin{align*} \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx &=\frac{(d \sec (e+f x))^{5/3} \int (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{17/6} \, dx}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\\ &=\frac{\left (a^2 (d \sec (e+f x))^{5/3}\right ) \operatorname{Subst}\left (\int \frac{(a+i a x)^{11/6}}{\sqrt [6]{a-i a x}} \, dx,x,\tan (e+f x)\right )}{f (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\\ &=\frac{\left (2\ 2^{5/6} a^3 (d \sec (e+f x))^{5/3}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{i x}{2}\right )^{11/6}}{\sqrt [6]{a-i a x}} \, dx,x,\tan (e+f x)\right )}{f (a-i a \tan (e+f x))^{5/6} \left (\frac{a+i a \tan (e+f x)}{a}\right )^{5/6}}\\ &=\frac{12 i 2^{5/6} a^2 \, _2F_1\left (-\frac{11}{6},\frac{5}{6};\frac{11}{6};\frac{1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{5/3}}{5 f (1+i \tan (e+f x))^{5/6}}\\ \end{align*}
Mathematica [B] time = 2.71808, size = 267, normalized size = 3.76 \[ \frac{(a+i a \tan (e+f x))^2 (d \sec (e+f x))^{5/3} \left (\frac{3}{4} \csc (e) (\cos (2 e)-i \sin (2 e)) \sec ^{\frac{8}{3}}(e+f x) (64 i \sin (2 e+f x)+75 \cos (2 e+f x)+55 \cos (2 e+3 f x)-64 i \sin (f x)+90 \cos (f x))-\frac{33 i 2^{2/3} \left (5 \sqrt [3]{1+e^{2 i (e+f x)}}-\left (-1+e^{2 i e}\right ) e^{2 i f x} \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},-e^{2 i (e+f x)}\right )\right )}{\left (-1+e^{2 i e}\right ) \sqrt [3]{\frac{e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt [3]{1+e^{2 i (e+f x)}}}\right )}{80 f \sec ^{\frac{11}{3}}(e+f x) (\cos (f x)+i \sin (f x))^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.132, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{{\frac{5}{3}}} \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{5}{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{2}{3}}{\left (-165 i \, a^{2} d e^{\left (5 i \, f x + 5 i \, e\right )} - 78 i \, a^{2} d e^{\left (3 i \, f x + 3 i \, e\right )} - 33 i \, a^{2} d e^{\left (i \, f x + i \, e\right )}\right )} \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}} e^{\left (\frac{2}{3} i \, f x + \frac{2}{3} i \, e\right )} + 80 \,{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}{\rm integral}\left (\frac{11 i \cdot 2^{\frac{2}{3}} a^{2} d \left (\frac{d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac{2}{3}} e^{\left (\frac{2}{3} i \, f x + \frac{2}{3} i \, e\right )}}{16 \, f}, x\right )}{80 \,{\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, 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(-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{5}{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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