Optimal. Leaf size=78 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}+\frac {2}{b f \sqrt {b \sec (e+f x)}} \]
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Rubi [A] time = 0.05, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2622, 325, 329, 298, 203, 206} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}+\frac {2}{b f \sqrt {b \sec (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 203
Rule 206
Rule 298
Rule 325
Rule 329
Rule 2622
Rubi steps
\begin {align*} \int \frac {\csc (e+f x)}{(b \sec (e+f x))^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{x^{3/2} \left (-1+\frac {x^2}{b^2}\right )} \, dx,x,b \sec (e+f x)\right )}{b f}\\ &=\frac {2}{b f \sqrt {b \sec (e+f x)}}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {x}}{-1+\frac {x^2}{b^2}} \, dx,x,b \sec (e+f x)\right )}{b^3 f}\\ &=\frac {2}{b f \sqrt {b \sec (e+f x)}}+\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{-1+\frac {x^4}{b^2}} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{b^3 f}\\ &=\frac {2}{b f \sqrt {b \sec (e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{b-x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{b f}+\frac {\operatorname {Subst}\left (\int \frac {1}{b+x^2} \, dx,x,\sqrt {b \sec (e+f x)}\right )}{b f}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b \sec (e+f x)}}{\sqrt {b}}\right )}{b^{3/2} f}+\frac {2}{b f \sqrt {b \sec (e+f x)}}\\ \end {align*}
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Mathematica [A] time = 2.00, size = 89, normalized size = 1.14 \[ \frac {\sqrt {\sec (e+f x)} \left (\log \left (1-\sqrt {\sec (e+f x)}\right )-\log \left (\sqrt {\sec (e+f x)}+1\right )\right )+2 \sqrt {\sec (e+f x)} \tan ^{-1}\left (\sqrt {\sec (e+f x)}\right )+4}{2 b f \sqrt {b \sec (e+f x)}} \]
Antiderivative was successfully verified.
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fricas [B] time = 1.02, size = 314, normalized size = 4.03 \[ \left [\frac {2 \, \sqrt {-b} \arctan \left (\frac {2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) + b}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) - \sqrt {-b} \log \left (-\frac {b \cos \left (f x + e\right )^{2} + 4 \, {\left (\cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} - 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, b^{2} f}, \frac {2 \, \sqrt {b} \arctan \left (\frac {2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{b \cos \left (f x + e\right ) - b}\right ) + 8 \, \sqrt {\frac {b}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) + \sqrt {b} \log \left (-\frac {b \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {b}{\cos \left (f x + e\right )}} + 6 \, b \cos \left (f x + e\right ) + b}{\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1}\right )}{4 \, b^{2} f}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.17, size = 221, normalized size = 2.83 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (4 \cos \left (f x +e \right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right )}{2 f \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} \left (\frac {b}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.62, size = 89, normalized size = 1.14 \[ \frac {b {\left (\frac {2 \, \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b}}\right )}{b^{\frac {5}{2}}} + \frac {\log \left (-\frac {\sqrt {b} - \sqrt {\frac {b}{\cos \left (f x + e\right )}}}{\sqrt {b} + \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}{b^{\frac {5}{2}}} + \frac {4}{b^{2} \sqrt {\frac {b}{\cos \left (f x + e\right )}}}\right )}}{2 \, f} \]
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 \[ \int \frac {1}{\sin \left (e+f\,x\right )\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \]
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 \[ \int \frac {\csc {\left (e + f x \right )}}{\left (b \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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