Optimal. Leaf size=58 \[ \frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f} \]
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Rubi [A]
time = 0.03, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2702, 335, 304,
209, 212} \begin {gather*} \frac {\sqrt {b} \text {ArcTan}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 212
Rule 304
Rule 335
Rule 2702
Rubi steps
\begin {align*} \int \csc (e+f x) \sqrt {b \sec (e+f x)} \, dx &=\frac {\text {Subst}\left (\int \frac {\sqrt {x}}{-1+\frac {x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{b f}\\ &=\frac {2 \text {Subst}\left (\int \frac {x^2}{-1+\frac {x^4}{b^2}} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{b f}\\ &=-\frac {b \text {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{f}+\frac {b \text {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{f}\\ &=\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}-\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{f}\\ \end {align*}
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Mathematica [A]
time = 0.26, size = 73, normalized size = 1.26 \begin {gather*} \frac {\left (2 \tan ^{-1}\left (\sqrt {\sec (e+f x)}\right )+\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (1+\sqrt {\sec (e+f x)}\right )\right ) \sqrt {b \sec (e+f x)}}{2 f \sqrt {\sec (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(168\) vs.
\(2(46)=92\).
time = 0.20, size = 169, normalized size = 2.91
method | result | size |
default | \(-\frac {\sqrt {\frac {b}{\cos \left (f x +e \right )}}\, \cos \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right ) \left (\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )-\ln \left (-\frac {2 \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1\right )}{\sin \left (f x +e \right )^{2}}\right )\right )}{2 f \sin \left (f x +e \right )^{2} \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(169\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.49, size = 75, normalized size = 1.29 \begin {gather*} \frac {b {\left (\frac {2 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{\sqrt {b}} + \frac {\log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{\sqrt {b}}\right )}}{2 \, f} \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. 126 vs.
\(2 (48) = 96\).
time = 0.44, size = 265, normalized size = 4.57 \begin {gather*} \left [\frac {2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) + 1\right )}}{2 \, b}\right ) + \sqrt {-b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, f}, -\frac {2 \, \sqrt {b} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}} {\left (\cos \left (f x + e\right ) - 1\right )}}{2 \, \sqrt {b}}\right ) - \sqrt {b} \log \left (\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, 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 \sqrt {b \sec {\left (e + f x \right )}} \csc {\left (e + f x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 4.29, size = 64, normalized size = 1.10 \begin {gather*} \frac {b^{2} {\left (\frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {-b}}\right )}{\sqrt {-b} b} - \frac {\arctan \left (\frac {\sqrt {b \cos \left (f x + e\right )}}{\sqrt {b}}\right )}{b^{\frac {3}{2}}}\right )} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{f} \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 = -0.02 \begin {gather*} \int \frac {\sqrt {\frac {b}{\cos \left (e+f\,x\right )}}}{\sin \left (e+f\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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