Optimal. Leaf size=145 \[ \frac{2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac{2 a b}{15 f (d \sec (e+f x))^{5/2}}-\frac{2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}} \]
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Rubi [A] time = 0.165958, antiderivative size = 145, 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, 3769, 3771, 2639} \[ \frac{2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac{2 a b}{15 f (d \sec (e+f x))^{5/2}}-\frac{2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 3508
Rule 3486
Rule 3769
Rule 3771
Rule 2639
Rubi steps
\begin{align*} \int \frac{(a+b \tan (e+f x))^2}{(d \sec (e+f x))^{5/2}} \, dx &=-\frac{2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}-\frac{2}{3} \int \frac{-\frac{3 a^2}{2}-b^2-\frac{1}{2} a b \tan (e+f x)}{(d \sec (e+f x))^{5/2}} \, dx\\ &=-\frac{2 a b}{15 f (d \sec (e+f x))^{5/2}}-\frac{2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}-\frac{1}{3} \left (-3 a^2-2 b^2\right ) \int \frac{1}{(d \sec (e+f x))^{5/2}} \, dx\\ &=-\frac{2 a b}{15 f (d \sec (e+f x))^{5/2}}+\frac{2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}+\frac{\left (3 a^2+2 b^2\right ) \int \frac{1}{\sqrt{d \sec (e+f x)}} \, dx}{5 d^2}\\ &=-\frac{2 a b}{15 f (d \sec (e+f x))^{5/2}}+\frac{2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}+\frac{\left (3 a^2+2 b^2\right ) \int \sqrt{\cos (e+f x)} \, dx}{5 d^2 \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}\\ &=-\frac{2 a b}{15 f (d \sec (e+f x))^{5/2}}+\frac{2 \left (3 a^2+2 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{5 d^2 f \sqrt{\cos (e+f x)} \sqrt{d \sec (e+f x)}}+\frac{2 \left (3 a^2+2 b^2\right ) \sin (e+f x)}{15 d f (d \sec (e+f x))^{3/2}}-\frac{2 b (a+b \tan (e+f x))}{3 f (d \sec (e+f x))^{5/2}}\\ \end{align*}
Mathematica [A] time = 0.922147, size = 92, normalized size = 0.63 \[ \frac{\left (6 a^2+4 b^2\right ) E\left (\left .\frac{1}{2} (e+f x)\right |2\right )+2 \cos ^{\frac{3}{2}}(e+f x) \left (\left (a^2-b^2\right ) \sin (e+f x)-2 a b \cos (e+f x)\right )}{5 f \cos ^{\frac{5}{2}}(e+f x) (d \sec (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [C] time = 0.275, size = 670, normalized size = 4.6 \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] 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}{2}}}\,{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 )} \sqrt{d \sec \left (f x + e\right )}}{d^{3} \sec \left (f x + e\right )^{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 \frac{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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