Optimal. Leaf size=119 \[ \frac{3 d \left (8 a^2-3 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\cos ^2(e+f x)\right )}{16 f \sqrt{\sin ^2(e+f x)}}+\frac{33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f} \]
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Rubi [A] time = 0.152344, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3508, 3486, 3772, 2643} \[ \frac{3 d \left (8 a^2-3 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\cos ^2(e+f x)\right )}{16 f \sqrt{\sin ^2(e+f x)}}+\frac{33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f} \]
Antiderivative was successfully verified.
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Rule 3508
Rule 3486
Rule 3772
Rule 2643
Rubi steps
\begin{align*} \int (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2 \, dx &=\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}+\frac{3}{8} \int (d \sec (e+f x))^{5/3} \left (\frac{8 a^2}{3}-b^2+\frac{11}{3} a b \tan (e+f x)\right ) \, dx\\ &=\frac{33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}+\frac{1}{8} \left (8 a^2-3 b^2\right ) \int (d \sec (e+f x))^{5/3} \, dx\\ &=\frac{33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}+\frac{1}{8} \left (\left (8 a^2-3 b^2\right ) \left (\frac{\cos (e+f x)}{d}\right )^{2/3} (d \sec (e+f x))^{2/3}\right ) \int \frac{1}{\left (\frac{\cos (e+f x)}{d}\right )^{5/3}} \, dx\\ &=\frac{33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac{3 \left (8 a^2-3 b^2\right ) d \, _2F_1\left (-\frac{1}{3},\frac{1}{2};\frac{2}{3};\cos ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \sin (e+f x)}{16 f \sqrt{\sin ^2(e+f x)}}+\frac{3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f}\\ \end{align*}
Mathematica [A] time = 1.73674, size = 108, normalized size = 0.91 \[ \frac{(d \sec (e+f x))^{5/3} \left (-5 \left (8 a^2-3 b^2\right ) \sin (2 (e+f x)) \sqrt [3]{\cos ^2(e+f x)} \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{1}{2},\frac{3}{2},\sin ^2(e+f x)\right )+15 \left (8 a^2-3 b^2\right ) \sin (2 (e+f x))+12 b (16 a+5 b \tan (e+f x))\right )}{160 f} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.147, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{{\frac{5}{3}}} \left ( a+b\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(-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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} d \sec \left (f x + e\right ) \tan \left (f x + e\right )^{2} + 2 \, a b d \sec \left (f x + e\right ) \tan \left (f x + e\right ) + a^{2} d \sec \left (f x + e\right )\right )} \left (d \sec \left (f x + e\right )\right )^{\frac{2}{3}}, x\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 (b \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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