Optimal. Leaf size=106 \[ -\frac{64 \sqrt{b \tan (e+f x)}}{15 b^3 d^2 f \sqrt{d \sec (e+f x)}}-\frac{16 \sqrt{b \tan (e+f x)}}{15 b^3 f (d \sec (e+f x))^{5/2}}-\frac{2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}} \]
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Rubi [A] time = 0.165777, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2609, 2612, 2605} \[ -\frac{64 \sqrt{b \tan (e+f x)}}{15 b^3 d^2 f \sqrt{d \sec (e+f x)}}-\frac{16 \sqrt{b \tan (e+f x)}}{15 b^3 f (d \sec (e+f x))^{5/2}}-\frac{2}{3 b f (b \tan (e+f x))^{3/2} (d \sec (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2609
Rule 2612
Rule 2605
Rubi steps
\begin{align*} \int \frac{1}{(d \sec (e+f x))^{5/2} (b \tan (e+f x))^{5/2}} \, dx &=-\frac{2}{3 b f (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}-\frac{8 \int \frac{1}{(d \sec (e+f x))^{5/2} \sqrt{b \tan (e+f x)}} \, dx}{3 b^2}\\ &=-\frac{2}{3 b f (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}-\frac{16 \sqrt{b \tan (e+f x)}}{15 b^3 f (d \sec (e+f x))^{5/2}}-\frac{32 \int \frac{1}{\sqrt{d \sec (e+f x)} \sqrt{b \tan (e+f x)}} \, dx}{15 b^2 d^2}\\ &=-\frac{2}{3 b f (d \sec (e+f x))^{5/2} (b \tan (e+f x))^{3/2}}-\frac{16 \sqrt{b \tan (e+f x)}}{15 b^3 f (d \sec (e+f x))^{5/2}}-\frac{64 \sqrt{b \tan (e+f x)}}{15 b^3 d^2 f \sqrt{d \sec (e+f x)}}\\ \end{align*}
Mathematica [A] time = 3.05235, size = 159, normalized size = 1.5 \[ \frac{-6 \sqrt{\frac{1}{\cos (e+f x)+1}} (2 \cos (2 (e+f x))-1) \tan (e+f x)-228 \tan \left (\frac{1}{2} (e+f x)\right ) \sqrt{\sec (e+f x)+1} \sqrt{\sec (e+f x)}+\sqrt{\frac{1}{\cos (e+f x)+1}} (3 \cos (2 (e+f x))-43) \csc (e+f x) \sec (e+f x)}{60 b^2 d^2 f \sqrt{\frac{1}{\cos (e+f x)+1}} \sqrt{b \tan (e+f x)} \sqrt{d \sec (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.156, size = 72, normalized size = 0.7 \begin{align*}{\frac{2\,\sin \left ( fx+e \right ) \left ( 3\, \left ( \cos \left ( fx+e \right ) \right ) ^{4}+24\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-32 \right ) }{15\,f \left ( \cos \left ( fx+e \right ) \right ) ^{5}} \left ({\frac{d}{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}} \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{5}{2}}}} \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{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.93458, size = 213, normalized size = 2.01 \begin{align*} -\frac{2 \,{\left (3 \, \cos \left (f x + e\right )^{5} + 24 \, \cos \left (f x + e\right )^{3} - 32 \, \cos \left (f x + e\right )\right )} \sqrt{\frac{b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt{\frac{d}{\cos \left (f x + e\right )}}}{15 \,{\left (b^{3} d^{3} f \cos \left (f x + e\right )^{2} - b^{3} d^{3} f\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{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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