Optimal. Leaf size=119 \[ -\frac{3 d \left (2 a^2+3 b^2\right ) \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{4}{3},\frac{7}{3},\cos ^2(e+f x)\right )}{16 f \sqrt{\sin ^2(e+f x)} (d \sec (e+f x))^{8/3}}+\frac{3 a b}{10 f (d \sec (e+f x))^{5/3}}-\frac{3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}} \]
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Rubi [A] time = 0.153647, 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} \[ -\frac{3 d \left (2 a^2+3 b^2\right ) \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{4}{3},\frac{7}{3},\cos ^2(e+f x)\right )}{16 f \sqrt{\sin ^2(e+f x)} (d \sec (e+f x))^{8/3}}+\frac{3 a b}{10 f (d \sec (e+f x))^{5/3}}-\frac{3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}} \]
Antiderivative was successfully verified.
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Rule 3508
Rule 3486
Rule 3772
Rule 2643
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/3}} \, dx &=-\frac{3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}}-\frac{3}{2} \int \frac{-\frac{2 a^2}{3}-b^2+\frac{1}{3} a b \tan (e+f x)}{(d \sec (e+f x))^{5/3}} \, dx\\ &=\frac{3 a b}{10 f (d \sec (e+f x))^{5/3}}-\frac{3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}}-\frac{1}{2} \left (-2 a^2-3 b^2\right ) \int \frac{1}{(d \sec (e+f x))^{5/3}} \, dx\\ &=\frac{3 a b}{10 f (d \sec (e+f x))^{5/3}}-\frac{3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}}-\frac{1}{2} \left (\left (-2 a^2-3 b^2\right ) \sqrt [3]{\frac{\cos (e+f x)}{d}} \sqrt [3]{d \sec (e+f x)}\right ) \int \left (\frac{\cos (e+f x)}{d}\right )^{5/3} \, dx\\ &=\frac{3 a b}{10 f (d \sec (e+f x))^{5/3}}-\frac{3 \left (2 a^2+3 b^2\right ) \cos ^3(e+f x) \, _2F_1\left (\frac{1}{2},\frac{4}{3};\frac{7}{3};\cos ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \sin (e+f x)}{16 d^2 f \sqrt{\sin ^2(e+f x)}}-\frac{3 b (a+b \tan (e+f x))}{2 f (d \sec (e+f x))^{5/3}}\\ \end{align*}
Mathematica [A] time = 0.487629, size = 119, normalized size = 1. \[ \frac{\sec ^2(e+f x) \left (2 \left (2 a^2+3 b^2\right ) \cos ^2(e+f x)^{2/3} \tan (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2}{3},\frac{3}{2},\sin ^2(e+f x)\right )+3 a^2 \sin (2 (e+f x))-6 a b \cos (2 (e+f x))-6 a b-3 b^2 \sin (2 (e+f x))\right )}{10 f (d \sec (e+f x))^{5/3}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.142, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{2} \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{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\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*}{\rm integral}\left (\frac{{\left (b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}\right )} \left (d \sec \left (f x + e\right )\right )^{\frac{1}{3}}}{d^{2} \sec \left (f x + e\right )^{2}}, x\right ) \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*} \int \frac{\left (a + b \tan{\left (e + f x \right )}\right )^{2}}{\left (d \sec{\left (e + f x \right )}\right )^{\frac{5}{3}}}\, dx \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{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\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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