Optimal. Leaf size=130 \[ -\frac{24 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{5 b^2 d^2 f \sqrt{\sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{12 (b \tan (e+f x))^{3/2}}{5 b^3 f (d \sec (e+f x))^{5/2}}-\frac{2}{b f \sqrt{b \tan (e+f x)} (d \sec (e+f x))^{5/2}} \]
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Rubi [A] time = 0.186373, antiderivative size = 130, 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 = {2609, 2612, 2616, 2640, 2639} \[ -\frac{24 E\left (\left .\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{5 b^2 d^2 f \sqrt{\sin (e+f x)} \sqrt{d \sec (e+f x)}}-\frac{12 (b \tan (e+f x))^{3/2}}{5 b^3 f (d \sec (e+f x))^{5/2}}-\frac{2}{b f \sqrt{b \tan (e+f x)} (d \sec (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2609
Rule 2612
Rule 2616
Rule 2640
Rule 2639
Rubi steps
\begin{align*} \int \frac{1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}} \, dx &=-\frac{2}{b f (d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)}}-\frac{6 \int \frac{\sqrt{b \tan (e+f x)}}{(d \sec (e+f x))^{5/2}} \, dx}{b^2}\\ &=-\frac{2}{b f (d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)}}-\frac{12 (b \tan (e+f x))^{3/2}}{5 b^3 f (d \sec (e+f x))^{5/2}}-\frac{12 \int \frac{\sqrt{b \tan (e+f x)}}{\sqrt{d \sec (e+f x)}} \, dx}{5 b^2 d^2}\\ &=-\frac{2}{b f (d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)}}-\frac{12 (b \tan (e+f x))^{3/2}}{5 b^3 f (d \sec (e+f x))^{5/2}}-\frac{\left (12 \sqrt{b \tan (e+f x)}\right ) \int \sqrt{b \sin (e+f x)} \, dx}{5 b^2 d^2 \sqrt{d \sec (e+f x)} \sqrt{b \sin (e+f x)}}\\ &=-\frac{2}{b f (d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)}}-\frac{12 (b \tan (e+f x))^{3/2}}{5 b^3 f (d \sec (e+f x))^{5/2}}-\frac{\left (12 \sqrt{b \tan (e+f x)}\right ) \int \sqrt{\sin (e+f x)} \, dx}{5 b^2 d^2 \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}\\ &=-\frac{2}{b f (d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)}}-\frac{24 E\left (\left .\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )\right |2\right ) \sqrt{b \tan (e+f x)}}{5 b^2 d^2 f \sqrt{d \sec (e+f x)} \sqrt{\sin (e+f x)}}-\frac{12 (b \tan (e+f x))^{3/2}}{5 b^3 f (d \sec (e+f x))^{5/2}}\\ \end{align*}
Mathematica [C] time = 0.656595, size = 81, normalized size = 0.62 \[ \frac{24 \sqrt [4]{-\tan ^2(e+f x)} \, _2F_1\left (-\frac{1}{4},\frac{1}{4};\frac{3}{4};\sec ^2(e+f x)\right )+\cos (2 (e+f x))-11}{5 b d^2 f \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.217, size = 570, normalized size = 4.4 \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{1}{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{3}{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{\sqrt{d \sec \left (f x + e\right )} \sqrt{b \tan \left (f x + e\right )}}{b^{2} d^{3} \sec \left (f x + e\right )^{3} \tan \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(-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{1}{\left (d \sec \left (f x + e\right )\right )^{\frac{5}{2}} \left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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