Optimal. Leaf size=69 \[ -\frac{3 i \sqrt [6]{2} a (1+i \tan (e+f x))^{5/6} \text{Hypergeometric2F1}\left (-\frac{5}{6},\frac{5}{6},\frac{1}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{5 f (d \sec (e+f x))^{5/3}} \]
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Rubi [A] time = 0.152235, 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} \[ -\frac{3 i \sqrt [6]{2} a (1+i \tan (e+f x))^{5/6} \text{Hypergeometric2F1}\left (-\frac{5}{6},\frac{5}{6},\frac{1}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{5 f (d \sec (e+f x))^{5/3}} \]
Antiderivative was successfully verified.
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Rule 3505
Rule 3523
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \frac{a+i a \tan (e+f x)}{(d \sec (e+f x))^{5/3}} \, dx &=\frac{\left ((a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}\right ) \int \frac{\sqrt [6]{a+i a \tan (e+f x)}}{(a-i a \tan (e+f x))^{5/6}} \, dx}{(d \sec (e+f x))^{5/3}}\\ &=\frac{\left (a^2 (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}\right ) \operatorname{Subst}\left (\int \frac{1}{(a-i a x)^{11/6} (a+i a x)^{5/6}} \, dx,x,\tan (e+f x)\right )}{f (d \sec (e+f x))^{5/3}}\\ &=\frac{\left (a^2 (a-i a \tan (e+f x))^{5/6} \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 )^{5/6} (a-i a x)^{11/6}} \, dx,x,\tan (e+f x)\right )}{2^{5/6} f (d \sec (e+f x))^{5/3}}\\ &=-\frac{3 i \sqrt [6]{2} a \, _2F_1\left (-\frac{5}{6},\frac{5}{6};\frac{1}{6};\frac{1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{5/6}}{5 f (d \sec (e+f x))^{5/3}}\\ \end{align*}
Mathematica [A] time = 0.443457, size = 106, normalized size = 1.54 \[ -\frac{3 i a e^{i (e+f x)} \left (4 \sqrt [3]{1+e^{2 i (e+f x)}} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (e+f x)}\right )+e^{2 i (e+f x)}+1\right )}{5 d f \left (1+e^{2 i (e+f x)}\right ) (d \sec (e+f x))^{2/3}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.151, size = 0, normalized size = 0. \begin{align*} \int{(a+ia\tan \left ( fx+e \right ) ) \left ( d\sec \left ( fx+e \right ) \right ) ^{-{\frac{5}{3}}}}\, 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 \frac{i \, a \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{3}}}\,{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{10 \, d^{2} f{\rm integral}\left (-\frac{2 i \cdot 2^{\frac{1}{3}} a \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 \, d^{2} f}, x\right ) + 2^{\frac{1}{3}}{\left (-3 i \, a e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, a\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 )}}{10 \, d^{2} f} \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 \left (\int \frac{1}{\left (d \sec{\left (e + f x \right )}\right )^{\frac{5}{3}}}\, dx + \int \frac{i \tan{\left (e + f x \right )}}{\left (d \sec{\left (e + f x \right )}\right )^{\frac{5}{3}}}\, 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 \frac{i \, a \tan \left (f x + e\right ) + a}{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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