Optimal. Leaf size=244 \[ -\frac{d (c+3 d) \cos (e+f x)}{a f (c-d)^2 (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f (c-d) \sqrt{c+d \sin (e+f x)}}-\frac{(c+3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f (c-d)^2 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]
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Rubi [A] time = 0.326915, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2768, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac{d (c+3 d) \cos (e+f x)}{a f (c-d)^2 (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{f (c-d) (a \sin (e+f x)+a) \sqrt{c+d \sin (e+f x)}}+\frac{\sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f (c-d) \sqrt{c+d \sin (e+f x)}}-\frac{(c+3 d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{a f (c-d)^2 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]
Antiderivative was successfully verified.
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Rule 2768
Rule 2754
Rule 2752
Rule 2663
Rule 2661
Rule 2655
Rule 2653
Rubi steps
\begin{align*} \int \frac{1}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx &=-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}+\frac{d \int \frac{-\frac{3 a}{2}+\frac{1}{2} a \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{a^2 (c-d)}\\ &=-\frac{d (c+3 d) \cos (e+f x)}{a (c-d)^2 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}-\frac{(2 d) \int \frac{\frac{1}{4} a (3 c+d)+\frac{1}{4} a (c+3 d) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{a^2 (c-d)^2 (c+d)}\\ &=-\frac{d (c+3 d) \cos (e+f x)}{a (c-d)^2 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}+\frac{\int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{2 a (c-d)}-\frac{(c+3 d) \int \sqrt{c+d \sin (e+f x)} \, dx}{2 a (c-d)^2 (c+d)}\\ &=-\frac{d (c+3 d) \cos (e+f x)}{a (c-d)^2 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}-\frac{\left ((c+3 d) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{2 a (c-d)^2 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\sqrt{\frac{c+d \sin (e+f x)}{c+d}} \int \frac{1}{\sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}}} \, dx}{2 a (c-d) \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{d (c+3 d) \cos (e+f x)}{a (c-d)^2 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{(c-d) f (a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}}-\frac{(c+3 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)}}{a (c-d)^2 (c+d) f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{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}}}{a (c-d) f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 1.99742, size = 264, normalized size = 1.08 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (-\left (c^2-d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )+\left (c^2+4 c d+3 d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )+(c+3 d) (c+d \sin (e+f x))-\frac{2 \left ((c+d)^2 \cos \left (\frac{1}{2} (e+f x)\right )+d \sin \left (\frac{1}{2} (e+f x)\right ) ((c+3 d) \cos (e+f x)+2 (c+d))\right )}{\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )}\right )}{a f (c-d)^2 (c+d) (\sin (e+f x)+1) \sqrt{c+d \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 1.509, size = 925, normalized size = 3.8 \begin{align*} \text{result too large to display} \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}{{\left (a \sin \left (f x + e\right ) + a\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{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{\sqrt{d \sin \left (f x + e\right ) + c}}{a c^{2} + 2 \, a c d + a d^{2} -{\left (2 \, a c d + a d^{2}\right )} \cos \left (f x + e\right )^{2} -{\left (a d^{2} \cos \left (f x + e\right )^{2} - a c^{2} - 2 \, a c d - a d^{2}\right )} \sin \left (f x + e\right )}, 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*} \frac{\int \frac{1}{c \sqrt{c + d \sin{\left (e + f x \right )}} \sin{\left (e + f x \right )} + c \sqrt{c + d \sin{\left (e + f x \right )}} + d \sqrt{c + d \sin{\left (e + f x \right )}} \sin ^{2}{\left (e + f x \right )} + d \sqrt{c + d \sin{\left (e + f x \right )}} \sin{\left (e + f x \right )}}\, dx}{a} \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}{{\left (a \sin \left (f x + e\right ) + a\right )}{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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