Optimal. Leaf size=131 \[ \frac {d \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{\sqrt {b} f \sqrt {b \tan (e+f x)}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{\sqrt {b} f \sqrt {b \tan (e+f x)}} \]
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Rubi [A]
time = 0.07, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2696, 2644,
335, 218, 212, 209} \begin {gather*} \frac {d \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \text {ArcTan}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{\sqrt {b} f \sqrt {b \tan (e+f x)}}+\frac {d \sqrt {b \sin (e+f x)} \sqrt {d \sec (e+f x)} \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right )}{\sqrt {b} f \sqrt {b \tan (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 212
Rule 218
Rule 335
Rule 2644
Rule 2696
Rubi steps
\begin {align*} \int \frac {(d \sec (e+f x))^{3/2}}{\sqrt {b \tan (e+f x)}} \, dx &=\frac {\left (d \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \int \frac {\sec (e+f x)}{\sqrt {b \sin (e+f x)}} \, dx}{\sqrt {b \tan (e+f x)}}\\ &=\frac {\left (d \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-\frac {x^2}{b^2}\right )} \, dx,x,b \sin (e+f x)\right )}{b f \sqrt {b \tan (e+f x)}}\\ &=\frac {\left (2 d \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{1-\frac {x^4}{b^2}} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{b f \sqrt {b \tan (e+f x)}}\\ &=\frac {\left (d \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{f \sqrt {b \tan (e+f x)}}+\frac {\left (d \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \sin (e+f x)}\right )}{f \sqrt {b \tan (e+f x)}}\\ &=\frac {d \tan ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{\sqrt {b} f \sqrt {b \tan (e+f x)}}+\frac {d \tanh ^{-1}\left (\frac {\sqrt {b \sin (e+f x)}}{\sqrt {b}}\right ) \sqrt {d \sec (e+f x)} \sqrt {b \sin (e+f x)}}{\sqrt {b} f \sqrt {b \tan (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 5.01, size = 105, normalized size = 0.80 \begin {gather*} -\frac {\left (\text {ArcTan}\left (\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )-\tanh ^{-1}\left (\frac {\sqrt {\sec (e+f x)}}{\sqrt [4]{\tan ^2(e+f x)}}\right )\right ) (d \sec (e+f x))^{3/2} \sqrt {b \tan (e+f x)}}{b f \sec ^{\frac {3}{2}}(e+f x) \sqrt [4]{\tan ^2(e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 0.36, size = 344, normalized size = 2.63
method | result | size |
default | \(\frac {\sqrt {2}\, \left (2 i \EllipticF \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-i \EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+\EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )\right ) \cos \left (f x +e \right ) \sqrt {-\frac {i \left (\cos \left (f x +e \right )-1\right )}{\sin \left (f x +e \right )}}\, \left (\frac {d}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \left (\sin ^{2}\left (f x +e \right )\right ) \sqrt {\frac {i \cos \left (f x +e \right )-i+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {-\frac {i \cos \left (f x +e \right )-i-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}}{2 f \left (\cos \left (f x +e \right )-1\right ) \sqrt {\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}}}\) | \(344\) |
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 [B] Leaf count of result is larger than twice the leaf count of optimal. 350 vs.
\(2 (115) = 230\).
time = 0.57, size = 709, normalized size = 5.41 \begin {gather*} \left [-\frac {2 \, d \sqrt {-\frac {d}{b}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (d \cos \left (f x + e\right )^{2} - {\left (d \cos \left (f x + e\right ) + d\right )} \sin \left (f x + e\right ) - d\right )}}\right ) - d \sqrt {-\frac {d}{b}} \log \left (\frac {d \cos \left (f x + e\right )^{4} - 72 \, d \cos \left (f x + e\right )^{2} + 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} - {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \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}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} + 28 \, {\left (d \cos \left (f x + e\right )^{2} - 2 \, d\right )} \sin \left (f x + e\right ) + 72 \, d}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right )}{8 \, f}, \frac {2 \, d \sqrt {\frac {d}{b}} \arctan \left (\frac {{\left (\cos \left (f x + e\right )^{3} - 5 \, \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) + 4\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) + 4\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {d}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{4 \, {\left (d \cos \left (f x + e\right )^{2} + {\left (d \cos \left (f x + e\right ) + d\right )} \sin \left (f x + e\right ) - d\right )}}\right ) + d \sqrt {\frac {d}{b}} \log \left (\frac {d \cos \left (f x + e\right )^{4} - 72 \, d \cos \left (f x + e\right )^{2} - 8 \, {\left (7 \, \cos \left (f x + e\right )^{3} + {\left (\cos \left (f x + e\right )^{3} - 8 \, \cos \left (f x + e\right )\right )} \sin \left (f x + e\right ) - 8 \, \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}{b}} \sqrt {\frac {d}{\cos \left (f x + e\right )}} - 28 \, {\left (d \cos \left (f x + e\right )^{2} - 2 \, d\right )} \sin \left (f x + e\right ) + 72 \, d}{\cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right ) + 8}\right )}{8 \, 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 {\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\sqrt {b \tan {\left (e + f x \right )}}}\, 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 {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{\sqrt {b\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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