Optimal. Leaf size=71 \[ -\frac{3 i (1+i \tan (e+f x))^{5/6} \text{Hypergeometric2F1}\left (-\frac{5}{6},\frac{17}{6},\frac{1}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{10\ 2^{5/6} a f (d \sec (e+f x))^{5/3}} \]
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Rubi [A] time = 0.20907, 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} \[ -\frac{3 i (1+i \tan (e+f x))^{5/6} \text{Hypergeometric2F1}\left (-\frac{5}{6},\frac{17}{6},\frac{1}{6},\frac{1}{2} (1-i \tan (e+f x))\right )}{10\ 2^{5/6} a f (d \sec (e+f x))^{5/3}} \]
Antiderivative was successfully verified.
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Rule 3505
Rule 3523
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \frac{1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))} \, dx &=\frac{\left ((a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}\right ) \int \frac{1}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{11/6}} \, 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)^{17/6}} \, dx,x,\tan (e+f x)\right )}{f (d \sec (e+f x))^{5/3}}\\ &=\frac{\left ((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 )^{17/6} (a-i a x)^{11/6}} \, dx,x,\tan (e+f x)\right )}{4\ 2^{5/6} f (d \sec (e+f x))^{5/3}}\\ &=-\frac{3 i \, _2F_1\left (-\frac{5}{6},\frac{17}{6};\frac{1}{6};\frac{1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{5/6}}{10\ 2^{5/6} a f (d \sec (e+f x))^{5/3}}\\ \end{align*}
Mathematica [A] time = 0.861761, size = 119, normalized size = 1.68 \[ -\frac{3 \sec ^2(e+f x) \left (\frac{128 e^{2 i (e+f x)} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (e+f x)}\right )}{\left (1+e^{2 i (e+f x)}\right )^{2/3}}+16 i \sin (2 (e+f x))+6 \cos (2 (e+f x))-26\right )}{220 a f (\tan (e+f x)-i) (d \sec (e+f x))^{5/3}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.133, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{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(-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 (440 \, a d^{2} f e^{\left (4 i \, f x + 4 i \, e\right )}{\rm integral}\left (-\frac{16 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 )}}{55 \, a d^{2} 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 (-33 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 45 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 93 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 15 i\right )} e^{\left (\frac{1}{3} i \, f x + \frac{1}{3} i \, e\right )}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{440 \, a d^{2} 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{1}{\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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