Optimal. Leaf size=119 \[ -\frac{3 d \left (2 a^2-3 b^2\right ) \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2}{3},\frac{5}{3},\cos ^2(e+f x)\right )}{8 f \sqrt{\sin ^2(e+f x)} (d \sec (e+f x))^{4/3}}-\frac{15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac{3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}} \]
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Rubi [A] time = 0.142824, 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} \[ -\frac{3 d \left (2 a^2-3 b^2\right ) \sin (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2}{3},\frac{5}{3},\cos ^2(e+f x)\right )}{8 f \sqrt{\sin ^2(e+f x)} (d \sec (e+f x))^{4/3}}-\frac{15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac{3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}} \]
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}{\sqrt [3]{d \sec (e+f x)}} \, dx &=\frac{3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac{3}{2} \int \frac{\frac{2 a^2}{3}-b^2+\frac{5}{3} a b \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx\\ &=-\frac{15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac{3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac{1}{2} \left (2 a^2-3 b^2\right ) \int \frac{1}{\sqrt [3]{d \sec (e+f x)}} \, dx\\ &=-\frac{15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac{3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}+\frac{1}{2} \left (\left (2 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 \sqrt [3]{\frac{\cos (e+f x)}{d}} \, dx\\ &=-\frac{15 a b}{2 f \sqrt [3]{d \sec (e+f x)}}-\frac{3 \left (2 a^2-3 b^2\right ) \cos ^2(e+f x) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(e+f x)\right ) (d \sec (e+f x))^{2/3} \sin (e+f x)}{8 d f \sqrt{\sin ^2(e+f x)}}+\frac{3 b (a+b \tan (e+f x))}{2 f \sqrt [3]{d \sec (e+f x)}}\\ \end{align*}
Mathematica [A] time = 3.91785, size = 209, normalized size = 1.76 \[ \frac{3 d \sin (e+f x) (a+b \tan (e+f x))^2 \left (\frac{\left (\left (2 a^2-3 b^2\right ) \cot (e+f x)+4 a b\right ) \left (\left (2 a^2-3 b^2\right ) \sqrt{\sin ^2(e+f x)} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{2},\frac{5}{6},\sec ^2(e+f x)\right )-4 a b \cos (e+f x) \sqrt{-\tan ^2(e+f x)}\right )}{\sqrt{-\tan ^2(e+f x)} \left (\left (2 a^2-3 b^2\right ) \sqrt{\sin ^2(e+f x)} \cot (e+f x)+4 a b \sin (e+f x)\right )}+b^2\right )}{2 f (d \sec (e+f x))^{4/3} (a \cos (e+f x)+b \sin (e+f x))^2} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.144, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b\tan \left ( fx+e \right ) \right ) ^{2}{\frac{1}{\sqrt [3]{d\sec \left ( fx+e \right ) }}}}\, 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{1}{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{2}{3}}}{d \sec \left (f x + e\right )}, 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}}{\sqrt [3]{d \sec{\left (e + f x \right )}}}\, 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{1}{3}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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