Optimal. Leaf size=96 \[ \frac {6 b^2 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}} \]
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Rubi [A] time = 0.12, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2610, 2616, 2640, 2639} \[ \frac {6 b^2 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2610
Rule 2616
Rule 2639
Rule 2640
Rubi steps
\begin {align*} \int \frac {(b \tan (e+f x))^{5/2}}{(d \sec (e+f x))^{5/2}} \, dx &=-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}+\frac {\left (3 b^2\right ) \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx}{5 d^2}\\ &=-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}+\frac {\left (3 b^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {b \sin (e+f x)} \, dx}{5 d^2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}\\ &=-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}+\frac {\left (3 b^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {\sin (e+f x)} \, dx}{5 d^2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}\\ &=\frac {6 b^2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{5 d^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}-\frac {2 b (b \tan (e+f x))^{3/2}}{5 f (d \sec (e+f x))^{5/2}}\\ \end {align*}
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Mathematica [C] time = 0.77, size = 81, normalized size = 0.84 \[ \frac {b^3 \left (-6 \sqrt [4]{-\tan ^2(e+f x)} \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {3}{4};\sec ^2(e+f x)\right )+\cos (2 (e+f x))-1\right )}{5 d^2 f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \sec \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )} b^{2} \tan \left (f x + e\right )^{2}}{d^{3} \sec \left (f x + e\right )^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.71, size = 565, normalized size = 5.89 \[ -\frac {\left (6 \cos \left (f x +e \right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticE \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}-3 \cos \left (f x +e \right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}+6 \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticE \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\frac {i \left (-1+\cos \left (f x +e \right )\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )-\left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {2}+4 \cos \left (f x +e \right ) \sqrt {2}-3 \sqrt {2}\right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sqrt {2}}{5 f \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sin \left (f x +e \right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \tan \left (f x + e\right )\right )^{\frac {5}{2}}}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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