Optimal. Leaf size=69 \[ \frac{6 i 2^{5/6} a (d \sec (e+f x))^{5/3} \text{Hypergeometric2F1}\left (-\frac{5}{6},\frac{5}{6},\frac{11}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{5 f (1+i \tan (e+f x))^{5/6}} \]
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Rubi [A] time = 0.166016, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3505, 3523, 70, 69} \[ \frac{6 i 2^{5/6} a (d \sec (e+f x))^{5/3} \text{Hypergeometric2F1}\left (-\frac{5}{6},\frac{5}{6},\frac{11}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{5 f (1+i \tan (e+f x))^{5/6}} \]
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)) \, 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))^{11/6} \, 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)^{5/6}}{\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^{5/6} a^2 (d \sec (e+f x))^{5/3}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{1}{2}+\frac{i x}{2}\right )^{5/6}}{\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{6 i 2^{5/6} a \, _2F_1\left (-\frac{5}{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 [A] time = 1.2436, size = 104, normalized size = 1.51 \[ \frac{3 a d e^{-2 i f x} (d \sec (e+f x))^{2/3} (\cos (e+3 f x)+i \sin (e+3 f x)) \left (i \left (1+e^{2 i (e+f x)}\right )^{2/3} \text{Hypergeometric2F1}\left (\frac{2}{3},\frac{5}{6},\frac{11}{6},-e^{2 i (e+f x)}\right )+2 \tan (e+f x)-3 i\right )}{10 f} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.109, 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 ) \, 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 )}\,{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 (-15 i \, a d e^{\left (3 i \, f x + 3 i \, e\right )} - 3 i \, a 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 )} + 10 \,{\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}{\rm integral}\left (\frac{i \cdot 2^{\frac{2}{3}} a 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 )}}{2 \, f}, x\right )}{10 \,{\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(-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 )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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