Optimal. Leaf size=42 \[ \frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (x)}{\sqrt{2} \sqrt{\sin (x)} \sqrt{a-a \sin (x)}}\right )}{\sqrt{a}} \]
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Rubi [A] time = 0.064463, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2782, 208} \[ \frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (x)}{\sqrt{2} \sqrt{\sin (x)} \sqrt{a-a \sin (x)}}\right )}{\sqrt{a}} \]
Antiderivative was successfully verified.
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Rule 2782
Rule 208
Rubi steps
\begin{align*} \int \frac{1}{\sqrt{\sin (x)} \sqrt{a-a \sin (x)}} \, dx &=-\left ((2 a) \operatorname{Subst}\left (\int \frac{1}{2 a^2-a x^2} \, dx,x,-\frac{a \cos (x)}{\sqrt{\sin (x)} \sqrt{a-a \sin (x)}}\right )\right )\\ &=\frac{\sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (x)}{\sqrt{2} \sqrt{\sin (x)} \sqrt{a-a \sin (x)}}\right )}{\sqrt{a}}\\ \end{align*}
Mathematica [C] time = 0.10248, size = 128, normalized size = 3.05 \[ \frac{2 \sqrt{\sin (x)} \sec ^2\left (\frac{x}{4}\right ) \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right ) \left (F\left (\left .\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{4}\right )}}\right )\right |-1\right )+\Pi \left (-1-\sqrt{2};\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{4}\right )}}\right )\right |-1\right )+\Pi \left (-1+\sqrt{2};\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{4}\right )}}\right )\right |-1\right )\right )}{\tan ^{\frac{3}{2}}\left (\frac{x}{4}\right ) \sqrt{1-\cot ^2\left (\frac{x}{4}\right )} \sqrt{a-a \sin (x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.093, size = 53, normalized size = 1.3 \begin{align*} -2\,{\frac{ \left ( -1+\cos \left ( x \right ) +\sin \left ( x \right ) \right ) \sqrt{\sin \left ( x \right ) }}{\sqrt{-a \left ( -1+\sin \left ( x \right ) \right ) } \left ( -1+\cos \left ( x \right ) \right ) }\sqrt{-{\frac{-1+\cos \left ( x \right ) }{\sin \left ( x \right ) }}}{\it Artanh} \left ( \sqrt{-{\frac{-1+\cos \left ( x \right ) }{\sin \left ( x \right ) }}} \right ) } \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{1}{\sqrt{-a \sin \left (x\right ) + a} \sqrt{\sin \left (x\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.09568, size = 543, normalized size = 12.93 \begin{align*} \left [\frac{\sqrt{2} \log \left (\frac{17 \, \cos \left (x\right )^{3} + 3 \, \cos \left (x\right )^{2} + \frac{4 \, \sqrt{2}{\left (3 \, \cos \left (x\right )^{2} -{\left (3 \, \cos \left (x\right ) + 4\right )} \sin \left (x\right ) - \cos \left (x\right ) - 4\right )} \sqrt{-a \sin \left (x\right ) + a} \sqrt{\sin \left (x\right )}}{\sqrt{a}} -{\left (17 \, \cos \left (x\right )^{2} + 14 \, \cos \left (x\right ) - 4\right )} \sin \left (x\right ) - 18 \, \cos \left (x\right ) - 4}{\cos \left (x\right )^{3} + 3 \, \cos \left (x\right )^{2} -{\left (\cos \left (x\right )^{2} - 2 \, \cos \left (x\right ) - 4\right )} \sin \left (x\right ) - 2 \, \cos \left (x\right ) - 4}\right )}{4 \, \sqrt{a}}, -\frac{1}{2} \, \sqrt{2} \sqrt{-\frac{1}{a}} \arctan \left (\frac{\sqrt{2} \sqrt{-a \sin \left (x\right ) + a} \sqrt{-\frac{1}{a}}{\left (3 \, \sin \left (x\right ) + 1\right )}}{4 \, \cos \left (x\right ) \sqrt{\sin \left (x\right )}}\right )\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{- a \left (\sin{\left (x \right )} - 1\right )} \sqrt{\sin{\left (x \right )}}}\, dx \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{1}{\sqrt{-a \sin \left (x\right ) + a} \sqrt{\sin \left (x\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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