Optimal. Leaf size=336 \[ -\frac{4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 f (c+d)^3 \sqrt{c+d \sin (e+f x)}}+\frac{4 a^3 \left (4 c^2+11 c d+15 d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^3 f (c+d)^2 \sqrt{c+d \sin (e+f x)}}-\frac{4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^3 f (c+d)^3 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 f (c+d)^2 (c+d \sin (e+f x))^{3/2}}+\frac{2 (c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}} \]
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Rubi [A] time = 0.720415, antiderivative size = 336, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2762, 2968, 3021, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac{4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 f (c+d)^3 \sqrt{c+d \sin (e+f x)}}+\frac{4 a^3 \left (4 c^2+11 c d+15 d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^3 f (c+d)^2 \sqrt{c+d \sin (e+f x)}}-\frac{4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^3 f (c+d)^3 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 f (c+d)^2 (c+d \sin (e+f x))^{3/2}}+\frac{2 (c-d) \cos (e+f x) \left (a^3 \sin (e+f x)+a^3\right )}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Rule 2762
Rule 2968
Rule 3021
Rule 2754
Rule 2752
Rule 2663
Rule 2661
Rule 2655
Rule 2653
Rubi steps
\begin{align*} \int \frac{(a+a \sin (e+f x))^3}{(c+d \sin (e+f x))^{7/2}} \, dx &=\frac{2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac{(2 a) \int \frac{(a+a \sin (e+f x)) (a (c-6 d)-a (2 c+3 d) \sin (e+f x))}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 d (c+d)}\\ &=\frac{2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac{(2 a) \int \frac{a^2 (c-6 d)+\left (a^2 (c-6 d)-a^2 (2 c+3 d)\right ) \sin (e+f x)-a^2 (2 c+3 d) \sin ^2(e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 d (c+d)}\\ &=\frac{2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac{8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 (c+d)^2 f (c+d \sin (e+f x))^{3/2}}+\frac{(4 a) \int \frac{\frac{3}{2} a^2 (c-d) d (c+9 d)+\frac{1}{2} a^2 (c-d) \left (4 c^2+11 c d+15 d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 (c-d) d^2 (c+d)^2}\\ &=\frac{2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac{8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac{4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 (c+d)^3 f \sqrt{c+d \sin (e+f x)}}-\frac{(8 a) \int \frac{\frac{1}{4} a^2 (c-15 d) (c-d)^2 d+\frac{1}{4} a^2 (c-d)^2 \left (4 c^2+15 c d+27 d^2\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{15 (c-d)^2 d^2 (c+d)^3}\\ &=\frac{2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac{8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac{4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 (c+d)^3 f \sqrt{c+d \sin (e+f x)}}+\frac{\left (2 a^3 \left (4 c^2+11 c d+15 d^2\right )\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{15 d^3 (c+d)^2}-\frac{\left (2 a^3 \left (4 c^2+15 c d+27 d^2\right )\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{15 d^3 (c+d)^3}\\ &=\frac{2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac{8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac{4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 (c+d)^3 f \sqrt{c+d \sin (e+f x)}}-\frac{\left (2 a^3 \left (4 c^2+15 c d+27 d^2\right ) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{15 d^3 (c+d)^3 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (2 a^3 \left (4 c^2+11 c d+15 d^2\right ) \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}{15 d^3 (c+d)^2 \sqrt{c+d \sin (e+f x)}}\\ &=\frac{2 (c-d) \cos (e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}+\frac{8 a^3 (c-d) (c+3 d) \cos (e+f x)}{15 d^2 (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac{4 a^3 \left (4 c^2+15 c d+27 d^2\right ) \cos (e+f x)}{15 d^2 (c+d)^3 f \sqrt{c+d \sin (e+f x)}}-\frac{4 a^3 \left (4 c^2+15 c d+27 d^2\right ) 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)}}{15 d^3 (c+d)^3 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{4 a^3 \left (4 c^2+11 c d+15 d^2\right ) 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}}}{15 d^3 (c+d)^2 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 2.10717, size = 298, normalized size = 0.89 \[ -\frac{2 a^3 (\sin (e+f x)+1)^3 \left (d \cos (e+f x) \left (2 d^2 \left (4 c^2+15 c d+27 d^2\right ) \sin ^2(e+f x)+d \left (45 c^2 d+9 c^3+115 c d^2+15 d^3\right ) \sin (e+f x)+55 c^2 d^2+15 c^3 d+4 c^4+15 c d^3+3 d^4\right )-2 (c+d \sin (e+f x))^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left (\left (4 c^2+15 c d+27 d^2\right ) \left ((c+d) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-c F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )+d^2 (c-15 d) F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )\right )}{15 d^3 f (c+d)^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6 (c+d \sin (e+f x))^{5/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 6.027, size = 1589, normalized size = 4.7 \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{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{7}{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{{\left (3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} +{\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{d \sin \left (f x + e\right ) + c}}{d^{4} \cos \left (f x + e\right )^{4} + c^{4} + 6 \, c^{2} d^{2} + d^{4} - 2 \,{\left (3 \, c^{2} d^{2} + d^{4}\right )} \cos \left (f x + e\right )^{2} - 4 \,{\left (c d^{3} \cos \left (f x + e\right )^{2} - c^{3} d - c d^{3}\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(-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{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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