Optimal. Leaf size=141 \[ -\frac{a \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{b^2 f \sqrt{a \sin (e+f x)}}-\frac{a \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{b^2 f \sqrt{a \sin (e+f x)}}+\frac{2 \sqrt{a \sin (e+f x)}}{b f \sqrt{b \tan (e+f x)}} \]
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Rubi [A] time = 0.143315, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {2595, 2601, 12, 2565, 329, 212, 206, 203} \[ -\frac{a \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)} \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{b^2 f \sqrt{a \sin (e+f x)}}-\frac{a \sqrt{\cos (e+f x)} \sqrt{b \tan (e+f x)} \tanh ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{b^2 f \sqrt{a \sin (e+f x)}}+\frac{2 \sqrt{a \sin (e+f x)}}{b f \sqrt{b \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2595
Rule 2601
Rule 12
Rule 2565
Rule 329
Rule 212
Rule 206
Rule 203
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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}
Mathematica [A] time = 0.326026, size = 88, normalized size = 0.62 \[ \frac{\sqrt{a \sin (e+f x)} \left (2 \sqrt [4]{\cos ^2(e+f x)}-\tan ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )-\tanh ^{-1}\left (\sqrt [4]{\cos ^2(e+f x)}\right )\right )}{b f \sqrt [4]{\cos ^2(e+f x)} \sqrt{b \tan (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.151, size = 237, normalized size = 1.7 \begin{align*} -{\frac{\cos \left ( fx+e \right ) -1}{2\,f\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) } \left ( 4\,\cos \left ( fx+e \right ) \sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}-\ln \left ( -{\frac{1}{ \left ( \sin \left ( fx+e \right ) \right ) ^{2}} \left ( 2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}} \left ( \cos \left ( fx+e \right ) \right ) ^{2}- \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}+2\,\cos \left ( fx+e \right ) -1 \right ) } \right ) +\arctan \left ({\frac{1}{2}{\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}}}} \right ) +4\,\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}} \right ) \sqrt{a\sin \left ( fx+e \right ) } \left ({\frac{b\sin \left ( fx+e \right ) }{\cos \left ( fx+e \right ) }} \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{-{\frac{\cos \left ( fx+e \right ) }{ \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}}}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \sin \left (f x + e\right )}}{\left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 3.89498, size = 1351, normalized size = 9.58 \begin{align*} \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{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a \sin \left (f x + e\right )}}{\left (b \tan \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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