Optimal. Leaf size=97 \[ -\frac {2 d^2 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}} \]
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Rubi [A] time = 0.12, antiderivative size = 97, 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 = {2608, 2616, 2640, 2639} \[ -\frac {2 d^2 E\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {\sin (e+f x)} \sqrt {d \sec (e+f x)}}-\frac {2 d^2}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2608
Rule 2616
Rule 2639
Rule 2640
Rubi steps
\begin {align*} \int \frac {(d \sec (e+f x))^{3/2}}{(b \tan (e+f x))^{3/2}} \, dx &=-\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {d^2 \int \frac {\sqrt {b \tan (e+f x)}}{\sqrt {d \sec (e+f x)}} \, dx}{b^2}\\ &=-\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\left (d^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {b \sin (e+f x)} \, dx}{b^2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}\\ &=-\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {\left (d^2 \sqrt {b \tan (e+f x)}\right ) \int \sqrt {\sin (e+f x)} \, dx}{b^2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}\\ &=-\frac {2 d^2}{b f \sqrt {d \sec (e+f x)} \sqrt {b \tan (e+f x)}}-\frac {2 d^2 E\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}\\ \end {align*}
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Mathematica [C] time = 0.63, size = 70, normalized size = 0.72 \[ \frac {2 d^2 \left (\sqrt [4]{-\tan ^2(e+f x)} \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {3}{4};\sec ^2(e+f x)\right )-1\right )}{b f \sqrt {b \tan (e+f x)} \sqrt {d \sec (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d \sec \left (f x + e\right )} \sqrt {b \tan \left (f x + e\right )} d \sec \left (f x + e\right )}{b^{2} \tan \left (f x + e\right )^{2}}, x\right ) \]
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 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.59, size = 535, normalized size = 5.52 \[ -\frac {\left (-2 \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 )}}+\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 )}}-2 \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 )+\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 )+\sqrt {2}\right ) \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right ) \sqrt {2}}{f \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}} \]
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 \[ \int \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (b \tan \left (f x + e\right )\right )^{\frac {3}{2}}}\,{d x} \]
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 \[ \int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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