Optimal. Leaf size=95 \[ -\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {4 F\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}} \]
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Rubi [A]
time = 0.08, antiderivative size = 95, 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 = {2689, 2696,
2721, 2720} \begin {gather*} -\frac {4 \sqrt {\sin (e+f x)} F\left (\left .\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {d \sec (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}}-\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Rule 2689
Rule 2696
Rule 2720
Rule 2721
Rubi steps
\begin {align*} \int \frac {\sqrt {d \sec (e+f x)}}{(b \tan (e+f x))^{5/2}} \, dx &=-\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {2 \int \frac {\sqrt {d \sec (e+f x)}}{\sqrt {b \tan (e+f x)}} \, dx}{3 b^2}\\ &=-\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {\left (2 \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \int \frac {1}{\sqrt {b \sin (e+f x)}} \, dx}{3 b^2 \sqrt {b \tan (e+f x)}}\\ &=-\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {\left (2 \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}\right ) \int \frac {1}{\sqrt {\sin (e+f x)}} \, dx}{3 b^2 \sqrt {b \tan (e+f x)}}\\ &=-\frac {2 \sqrt {d \sec (e+f x)}}{3 b f (b \tan (e+f x))^{3/2}}-\frac {4 F\left (\left .\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )\right |2\right ) \sqrt {d \sec (e+f x)} \sqrt {\sin (e+f x)}}{3 b^2 f \sqrt {b \tan (e+f x)}}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in
optimal.
time = 0.81, size = 70, normalized size = 0.74 \begin {gather*} -\frac {2 \sqrt {d \sec (e+f x)} \left (1+2 \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};\sec ^2(e+f x)\right ) \left (-\tan ^2(e+f x)\right )^{3/4}\right )}{3 b f (b \tan (e+f x))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains complex when optimal does not.
time = 0.36, size = 322, normalized size = 3.39
method | result | size |
default | \(-\frac {\left (2 i \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right ) \cos \left (f x +e \right )+2 i \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (f x +e \right )+\cos \left (f x +e \right ) \sqrt {2}\right ) \sqrt {\frac {d}{\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{3 f \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \cos \left (f x +e \right )^{2}}\) | \(322\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 0.12, size = 147, normalized size = 1.55 \begin {gather*} \frac {2 \, {\left (\sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )^{2} - \sqrt {-2 i \, b d} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - \sqrt {2 i \, b d} {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\rm weierstrassPInverse}\left (4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right )}}{3 \, {\left (b^{3} f \cos \left (f x + e\right )^{2} - b^{3} f\right )}} \end {gather*}
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 \begin {gather*} \int \frac {\sqrt {d \sec {\left (e + f x \right )}}}{\left (b \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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