Optimal. Leaf size=143 \[ -\frac{2 d^2 \left (5 a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 d \left (5 a^2-2 b^2\right ) \sin (e+f x) \sqrt{d \sec (e+f x)}}{5 f}+\frac{14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f} \]
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Rubi [A] time = 0.162875, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3508, 3486, 3768, 3771, 2639} \[ -\frac{2 d^2 \left (5 a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 d \left (5 a^2-2 b^2\right ) \sin (e+f x) \sqrt{d \sec (e+f x)}}{5 f}+\frac{14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f} \]
Antiderivative was successfully verified.
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Rule 3508
Rule 3486
Rule 3768
Rule 3771
Rule 2639
Rubi steps
\begin{align*} \int (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))^2 \, dx &=\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}+\frac{2}{5} \int (d \sec (e+f x))^{3/2} \left (\frac{5 a^2}{2}-b^2+\frac{7}{2} a b \tan (e+f x)\right ) \, dx\\ &=\frac{14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}+\frac{1}{5} \left (5 a^2-2 b^2\right ) \int (d \sec (e+f x))^{3/2} \, dx\\ &=\frac{14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac{2 \left (5 a^2-2 b^2\right ) d \sqrt{d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}-\frac{1}{5} \left (\left (5 a^2-2 b^2\right ) d^2\right ) \int \frac{1}{\sqrt{d \sec (e+f x)}} \, dx\\ &=\frac{14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac{2 \left (5 a^2-2 b^2\right ) d \sqrt{d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}-\frac{\left (\left (5 a^2-2 b^2\right ) d^2\right ) \int \sqrt{\cos (e+f x)} \, dx}{5 \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}\\ &=-\frac{2 \left (5 a^2-2 b^2\right ) d^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{14 a b (d \sec (e+f x))^{3/2}}{15 f}+\frac{2 \left (5 a^2-2 b^2\right ) d \sqrt{d \sec (e+f x)} \sin (e+f x)}{5 f}+\frac{2 b (d \sec (e+f x))^{3/2} (a+b \tan (e+f x))}{5 f}\\ \end{align*}
Mathematica [A] time = 0.636578, size = 126, normalized size = 0.88 \[ -\frac{2 d^2 (a+b \tan (e+f x))^2 \left (\left (3 b^2-\frac{15 a^2}{2}\right ) \sin (2 (e+f x))+3 \left (5 a^2-2 b^2\right ) \cos ^{\frac{3}{2}}(e+f x) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )-b (10 a+3 b \tan (e+f x))\right )}{15 f \sqrt{d \sec (e+f x)} (a \cos (e+f x)+b \sin (e+f x))^2} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.296, size = 712, normalized size = 5. \begin{align*} \text{result too large to display} \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 )} \sqrt{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(-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{3}{2}}{\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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