Optimal. Leaf size=141 \[ \frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {a \text {ArcTan}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {a \sin (e+f x)}}-\frac {a \tanh ^{-1}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {a \sin (e+f x)}} \]
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Rubi [A]
time = 0.10, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2675, 2681, 12,
2645, 335, 218, 212, 209} \begin {gather*} -\frac {a \sqrt {\cos (e+f x)} \text {ArcTan}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {a \sin (e+f x)}}-\frac {a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)} \tanh ^{-1}\left (\sqrt {\cos (e+f x)}\right )}{b^2 f \sqrt {a \sin (e+f x)}}+\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 209
Rule 212
Rule 218
Rule 335
Rule 2645
Rule 2675
Rule 2681
Rubi steps
\begin {align*} \int \frac {\sqrt {a \sin (e+f x)}}{(b \tan (e+f x))^{3/2}} \, dx &=\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}+\frac {a^2 \int \frac {\sqrt {b \tan (e+f x)}}{(a \sin (e+f x))^{3/2}} \, dx}{b^2}\\ &=\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}+\frac {\left (a^2 \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \int \frac {\csc (e+f x)}{a \sqrt {\cos (e+f x)}} \, dx}{b^2 \sqrt {a \sin (e+f x)}}\\ &=\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}+\frac {\left (a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \int \frac {\csc (e+f x)}{\sqrt {\cos (e+f x)}} \, dx}{b^2 \sqrt {a \sin (e+f x)}}\\ &=\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {\left (a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-x^2\right )} \, dx,x,\cos (e+f x)\right )}{b^2 f \sqrt {a \sin (e+f x)}}\\ &=\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {\left (2 a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\sqrt {\cos (e+f x)}\right )}{b^2 f \sqrt {a \sin (e+f x)}}\\ &=\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {\left (a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {\cos (e+f x)}\right )}{b^2 f \sqrt {a \sin (e+f x)}}-\frac {\left (a \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {\cos (e+f x)}\right )}{b^2 f \sqrt {a \sin (e+f x)}}\\ &=\frac {2 \sqrt {a \sin (e+f x)}}{b f \sqrt {b \tan (e+f x)}}-\frac {a \tan ^{-1}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {a \sin (e+f x)}}-\frac {a \tanh ^{-1}\left (\sqrt {\cos (e+f x)}\right ) \sqrt {\cos (e+f x)} \sqrt {b \tan (e+f x)}}{b^2 f \sqrt {a \sin (e+f x)}}\\ \end {align*}
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Mathematica [A]
time = 0.40, size = 88, normalized size = 0.62 \begin {gather*} \frac {\left (-\text {ArcTan}\left (\sqrt [4]{\cos ^2(e+f x)}\right )-\tanh ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )+2 \sqrt [4]{\cos ^2(e+f x)}\right ) \sqrt {a \sin (e+f x)}}{b f \sqrt [4]{\cos ^2(e+f x)} \sqrt {b \tan (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.34, size = 237, normalized size = 1.68
method | result | size |
default | \(-\frac {\left (\cos \left (f x +e \right )-1\right ) \left (4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )-\ln \left (-\frac {2 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-\left (\cos ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )-2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-1}{\sin \left (f x +e \right )^{2}}\right )+\arctan \left (\frac {1}{2 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\right )+4 \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\right ) \sqrt {a \sin \left (f x +e \right )}}{2 f \sin \left (f x +e \right ) \cos \left (f x +e \right ) \left (\frac {b \sin \left (f x +e \right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sqrt {-\frac {\cos \left (f x +e \right )}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\) | \(237\) |
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. 286 vs.
\(2 (131) = 262\).
time = 0.65, size = 577, normalized size = 4.09 \begin {gather*} \left [\frac {2 \, b \sqrt {-\frac {a}{b}} \arctan \left (\frac {2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {a}{b}} \cos \left (f x + e\right )}{{\left (a \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + b \sqrt {-\frac {a}{b}} \log \left (-\frac {a \cos \left (f x + e\right )^{3} + 4 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {-\frac {a}{b}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 5 \, a \cos \left (f x + e\right )^{2} - 5 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )^{3} + 3 \, \cos \left (f x + e\right )^{2} + 3 \, \cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 8 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{4 \, b^{2} f \sin \left (f x + e\right )}, \frac {2 \, b \sqrt {\frac {a}{b}} \arctan \left (\frac {2 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {a}{b}} \cos \left (f x + e\right )}{{\left (a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + b \sqrt {\frac {a}{b}} \log \left (\frac {4 \, {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sqrt {\frac {a}{b}} - {\left (a \cos \left (f x + e\right )^{2} + 6 \, a \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 8 \, \sqrt {a \sin \left (f x + e\right )} \sqrt {\frac {b \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{4 \, b^{2} f \sin \left (f x + e\right )}\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 {a \sin {\left (e + f x \right )}}}{\left (b \tan {\left (e + f x \right )}\right )^{\frac {3}{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 {a\,\sin \left (e+f\,x\right )}}{{\left (b\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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