Optimal. Leaf size=329 \[ \frac{2 \left (-a^2 d^2+2 a b c d+b^2 \left (2 c^2-3 d^2\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f \left (c^2-d^2\right ) \sqrt{c+d \sin (e+f x)}}+\frac{2 (b c-a d)^2 \cos (e+f x)}{3 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{3/2}}-\frac{4 \left (2 a c d+b \left (c^2-3 d^2\right )\right ) (b c-a d) \cos (e+f x)}{3 d f \left (c^2-d^2\right )^2 \sqrt{c+d \sin (e+f x)}}-\frac{4 \left (2 a c d+b \left (c^2-3 d^2\right )\right ) (b c-a d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f \left (c^2-d^2\right )^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]
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Rubi [A] time = 0.499152, antiderivative size = 329, 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 = {2790, 2754, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (-a^2 d^2+2 a b c d+b^2 \left (2 c^2-3 d^2\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f \left (c^2-d^2\right ) \sqrt{c+d \sin (e+f x)}}+\frac{2 (b c-a d)^2 \cos (e+f x)}{3 d f \left (c^2-d^2\right ) (c+d \sin (e+f x))^{3/2}}-\frac{4 \left (2 a c d+b \left (c^2-3 d^2\right )\right ) (b c-a d) \cos (e+f x)}{3 d f \left (c^2-d^2\right )^2 \sqrt{c+d \sin (e+f x)}}-\frac{4 \left (2 a c d+b \left (c^2-3 d^2\right )\right ) (b c-a d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{3 d^2 f \left (c^2-d^2\right )^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]
Antiderivative was successfully verified.
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Rule 2790
Rule 2754
Rule 2752
Rule 2663
Rule 2661
Rule 2655
Rule 2653
Rubi steps
\begin{align*} \int \frac{(a+b \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx &=\frac{2 (b c-a d)^2 \cos (e+f x)}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}+\frac{2 \int \frac{\frac{3}{2} d \left (a^2 c+b^2 c-2 a b d\right )+\frac{1}{2} \left (2 b^2 c^2+2 a b c d-\left (a^2+3 b^2\right ) d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 d \left (c^2-d^2\right )}\\ &=\frac{2 (b c-a d)^2 \cos (e+f x)}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac{4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \cos (e+f x)}{3 d \left (c^2-d^2\right )^2 f \sqrt{c+d \sin (e+f x)}}-\frac{4 \int \frac{\frac{1}{4} d \left (8 a b c d-a^2 \left (3 c^2+d^2\right )-b^2 \left (c^2+3 d^2\right )\right )+\frac{1}{2} (b c-a d) \left (b c^2+2 a c d-3 b d^2\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{3 d \left (c^2-d^2\right )^2}\\ &=\frac{2 (b c-a d)^2 \cos (e+f x)}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac{4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \cos (e+f x)}{3 d \left (c^2-d^2\right )^2 f \sqrt{c+d \sin (e+f x)}}-\frac{\left (2 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right )\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{3 d^2 \left (c^2-d^2\right )^2}+\frac{\left (2 a b c d-a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{3 d^2 \left (c^2-d^2\right )}\\ &=\frac{2 (b c-a d)^2 \cos (e+f x)}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac{4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \cos (e+f x)}{3 d \left (c^2-d^2\right )^2 f \sqrt{c+d \sin (e+f x)}}-\frac{\left (2 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{3 d^2 \left (c^2-d^2\right )^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (\left (2 a b c d-a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\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}{3 d^2 \left (c^2-d^2\right ) \sqrt{c+d \sin (e+f x)}}\\ &=\frac{2 (b c-a d)^2 \cos (e+f x)}{3 d \left (c^2-d^2\right ) f (c+d \sin (e+f x))^{3/2}}-\frac{4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \cos (e+f x)}{3 d \left (c^2-d^2\right )^2 f \sqrt{c+d \sin (e+f x)}}-\frac{4 (b c-a d) \left (2 a c d+b \left (c^2-3 d^2\right )\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)}}{3 d^2 \left (c^2-d^2\right )^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{2 \left (2 a b c d-a^2 d^2+b^2 \left (2 c^2-3 d^2\right )\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}}}{3 d^2 \left (c^2-d^2\right ) f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 2.40562, size = 302, normalized size = 0.92 \[ \frac{2 \left (\frac{(-c-d \sin (e+f x)) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left (d^2 \left (a^2 \left (3 c^2+d^2\right )-8 a b c d+b^2 \left (c^2+3 d^2\right )\right ) F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-2 \left (-2 a^2 c d^2+a b d \left (c^2+3 d^2\right )+b^2 \left (c^3-3 c d^2\right )\right ) \left ((c+d) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-c F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )\right )}{(c-d)^2 (c+d)^2}-\frac{d (a d-b c) \cos (e+f x) \left (-2 d \left (2 a c d+b \left (c^2-3 d^2\right )\right ) \sin (e+f x)-5 a c^2 d+a d^3-b c^3+5 b c d^2\right )}{\left (c^2-d^2\right )^2}\right )}{3 d^2 f (c+d \sin (e+f x))^{3/2}} \]
Antiderivative was successfully verified.
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Maple [B] time = 4.746, size = 1043, normalized size = 3.2 \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 (b \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{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 (b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}\right )} \sqrt{d \sin \left (f x + e\right ) + c}}{3 \, c d^{2} \cos \left (f x + e\right )^{2} - c^{3} - 3 \, c d^{2} +{\left (d^{3} \cos \left (f x + e\right )^{2} - 3 \, c^{2} d - 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 (b \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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