Optimal. Leaf size=132 \[ \frac{2 \sqrt{a+b \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{b f \sqrt{\frac{a+b \sin (e+f x)}{a+b}}}-\frac{2 a \sqrt{\frac{a+b \sin (e+f x)}{a+b}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{b f \sqrt{a+b \sin (e+f x)}} \]
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Rubi [A] time = 0.1069, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2752, 2663, 2661, 2655, 2653} \[ \frac{2 \sqrt{a+b \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{b f \sqrt{\frac{a+b \sin (e+f x)}{a+b}}}-\frac{2 a \sqrt{\frac{a+b \sin (e+f x)}{a+b}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{b f \sqrt{a+b \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 2752
Rule 2663
Rule 2661
Rule 2655
Rule 2653
Rubi steps
\begin{align*} \int \frac{\sin (e+f x)}{\sqrt{a+b \sin (e+f x)}} \, dx &=\frac{\int \sqrt{a+b \sin (e+f x)} \, dx}{b}-\frac{a \int \frac{1}{\sqrt{a+b \sin (e+f x)}} \, dx}{b}\\ &=\frac{\sqrt{a+b \sin (e+f x)} \int \sqrt{\frac{a}{a+b}+\frac{b \sin (e+f x)}{a+b}} \, dx}{b \sqrt{\frac{a+b \sin (e+f x)}{a+b}}}-\frac{\left (a \sqrt{\frac{a+b \sin (e+f x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \sin (e+f x)}{a+b}}} \, dx}{b \sqrt{a+b \sin (e+f x)}}\\ &=\frac{2 E\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 b}{a+b}\right ) \sqrt{a+b \sin (e+f x)}}{b f \sqrt{\frac{a+b \sin (e+f x)}{a+b}}}-\frac{2 a F\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 b}{a+b}\right ) \sqrt{\frac{a+b \sin (e+f x)}{a+b}}}{b f \sqrt{a+b \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 2.40032, size = 94, normalized size = 0.71 \[ -\frac{2 \sqrt{\frac{a+b \sin (e+f x)}{a+b}} \left ((a+b) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 b}{a+b}\right )\right )}{b f \sqrt{a+b \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.819, size = 202, normalized size = 1.5 \begin{align*} -2\,{\frac{a-b}{{b}^{2}\cos \left ( fx+e \right ) \sqrt{a+b\sin \left ( fx+e \right ) }f}\sqrt{{\frac{a+b\sin \left ( fx+e \right ) }{a-b}}}\sqrt{-{\frac{ \left ( -1+\sin \left ( fx+e \right ) \right ) b}{a+b}}}\sqrt{-{\frac{ \left ( 1+\sin \left ( fx+e \right ) \right ) b}{a-b}}} \left ({\it EllipticE} \left ( \sqrt{{\frac{a+b\sin \left ( fx+e \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) a+{\it EllipticE} \left ( \sqrt{{\frac{a+b\sin \left ( fx+e \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) b-{\it EllipticF} \left ( \sqrt{{\frac{a+b\sin \left ( fx+e \right ) }{a-b}}},\sqrt{{\frac{a-b}{a+b}}} \right ) b \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{\sin \left (f x + e\right )}{\sqrt{b \sin \left (f x + e\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sin \left (f x + e\right )}{\sqrt{b \sin \left (f x + e\right ) + a}}, x\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{\sin{\left (e + f x \right )}}{\sqrt{a + b \sin{\left (e + f 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{\sin \left (f x + e\right )}{\sqrt{b \sin \left (f x + e\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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