Optimal. Leaf size=229 \[ \frac{2 d (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b^2 f \sqrt{c+d \sin (e+f x)}}+\frac{2 (b c-a d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b^2 f (a+b) \sqrt{c+d \sin (e+f x)}}+\frac{2 d \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]
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Rubi [A] time = 0.491392, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2804, 2655, 2653, 2803, 2663, 2661, 2807, 2805} \[ \frac{2 d (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b^2 f \sqrt{c+d \sin (e+f x)}}+\frac{2 (b c-a d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b^2 f (a+b) \sqrt{c+d \sin (e+f x)}}+\frac{2 d \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{b f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]
Antiderivative was successfully verified.
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Rule 2804
Rule 2655
Rule 2653
Rule 2803
Rule 2663
Rule 2661
Rule 2807
Rule 2805
Rubi steps
\begin{align*} \int \frac{(c+d \sin (e+f x))^{3/2}}{a+b \sin (e+f x)} \, dx &=\frac{d \int \sqrt{c+d \sin (e+f x)} \, dx}{b}-\frac{(-b c+a d) \int \frac{\sqrt{c+d \sin (e+f x)}}{a+b \sin (e+f x)} \, dx}{b}\\ &=\frac{(d (b c-a d)) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{b^2}+\frac{(b c-a d)^2 \int \frac{1}{(a+b \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{b^2}+\frac{\left (d \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{b \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}\\ &=\frac{2 d E\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{c+d \sin (e+f x)}}{b f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (d (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}\right ) \int \frac{1}{\sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}}} \, dx}{b^2 \sqrt{c+d \sin (e+f x)}}+\frac{\left ((b c-a d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}\right ) \int \frac{1}{(a+b \sin (e+f x)) \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}}} \, dx}{b^2 \sqrt{c+d \sin (e+f x)}}\\ &=\frac{2 d E\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{c+d \sin (e+f x)}}{b f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{2 d (b c-a d) F\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}{b^2 f \sqrt{c+d \sin (e+f x)}}+\frac{2 (b c-a d)^2 \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}{b^2 (a+b) f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [C] time = 3.92112, size = 242, normalized size = 1.06 \[ \frac{2 i \sec (e+f x) \sqrt{-\frac{d (\sin (e+f x)-1)}{c+d}} \sqrt{-\frac{d (\sin (e+f x)+1)}{c-d}} \left ((a d+b (d-2 c)) F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )+(b c-a d) \Pi \left (\frac{b (c+d)}{b c-a d};i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )+b (c-d) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )\right )}{b^2 f \sqrt{-\frac{1}{c+d}}} \]
Antiderivative was successfully verified.
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Maple [A] time = 1.343, size = 391, normalized size = 1.7 \begin{align*} -2\,{\frac{c-d}{{b}^{2}\cos \left ( fx+e \right ) \sqrt{c+d\sin \left ( fx+e \right ) }f} \left ({\it EllipticE} \left ( \sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}},\sqrt{{\frac{c-d}{c+d}}} \right ) bc+{\it EllipticE} \left ( \sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}},\sqrt{{\frac{c-d}{c+d}}} \right ) bd+a{\it EllipticF} \left ( \sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}},\sqrt{{\frac{c-d}{c+d}}} \right ) d-2\,cb{\it EllipticF} \left ( \sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}},\sqrt{{\frac{c-d}{c+d}}} \right ) -{\it EllipticF} \left ( \sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}},\sqrt{{\frac{c-d}{c+d}}} \right ) bd-{\it EllipticPi} \left ( \sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}},-{\frac{b \left ( c-d \right ) }{da-cb}},\sqrt{{\frac{c-d}{c+d}}} \right ) ad+{\it EllipticPi} \left ( \sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}},-{\frac{b \left ( c-d \right ) }{da-cb}},\sqrt{{\frac{c-d}{c+d}}} \right ) bc \right ) \sqrt{-{\frac{d \left ( 1+\sin \left ( fx+e \right ) \right ) }{c-d}}}\sqrt{-{\frac{ \left ( -1+\sin \left ( fx+e \right ) \right ) d}{c+d}}}\sqrt{{\frac{c+d\sin \left ( fx+e \right ) }{c-d}}}} \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{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{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{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}{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(-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(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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