Optimal. Leaf size=119 \[ \frac {3 d \left (8 a^2-3 b^2\right ) \sin (e+f x) (d \sec (e+f x))^{2/3} \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(e+f x)\right )}{16 f \sqrt {\sin ^2(e+f x)}}+\frac {33 a b (d \sec (e+f x))^{5/3}}{40 f}+\frac {3 b (d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}{8 f} \]
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Rubi [A] time = 0.15, antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 2643
Rule 3486
Rule 3508
Rule 3772
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*}
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Mathematica [A] time = 1.71, size = 108, normalized size = 0.91 \[ \frac {(d \sec (e+f x))^{5/3} \left (-5 \left (8 a^2-3 b^2\right ) \sin (2 (e+f x)) \sqrt [3]{\cos ^2(e+f x)} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {3}{2};\sin ^2(e+f x)\right )+15 \left (8 a^2-3 b^2\right ) \sin (2 (e+f x))+12 b (16 a+5 b \tan (e+f x))\right )}{160 f} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b^{2} d \sec \left (f x + e\right ) \tan \left (f x + e\right )^{2} + 2 \, a b d \sec \left (f x + e\right ) \tan \left (f x + e\right ) + a^{2} d \sec \left (f x + e\right )\right )} \left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.68, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +b \tan \left (f x +e \right )\right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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