Optimal. Leaf size=254 \[ \frac{2 \left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{4 b \left (c^2-d^2\right ) (b c-5 a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{c+d \sin (e+f x)}}+\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f} \]
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Rubi [A] time = 0.411492, antiderivative size = 254, 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 = {2791, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (3 d^2 \left (5 a^2+3 b^2\right )-2 b c (b c-5 a d)\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{4 b \left (c^2-d^2\right ) (b c-5 a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{15 d^2 f \sqrt{c+d \sin (e+f x)}}+\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f} \]
Antiderivative was successfully verified.
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Rule 2791
Rule 2753
Rule 2752
Rule 2663
Rule 2661
Rule 2655
Rule 2653
Rubi steps
\begin{align*} \int (a+b \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)} \, dx &=-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{2 \int \sqrt{c+d \sin (e+f x)} \left (\frac{1}{2} \left (5 a^2+3 b^2\right ) d-b (b c-5 a d) \sin (e+f x)\right ) \, dx}{5 d}\\ &=\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{4 \int \frac{\frac{1}{4} d \left (15 a^2 c+7 b^2 c+10 a b d\right )+\frac{1}{4} \left (3 \left (5 a^2+3 b^2\right ) d^2-2 b c (b c-5 a d)\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{15 d}\\ &=\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{\left (2 b (b c-5 a d) \left (c^2-d^2\right )\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{15 d^2}+\frac{1}{15} \left (15 a^2+9 b^2-\frac{2 b c (b c-5 a d)}{d^2}\right ) \int \sqrt{c+d \sin (e+f x)} \, dx\\ &=\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{\left (\left (15 a^2+9 b^2-\frac{2 b c (b c-5 a d)}{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 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (2 b (b c-5 a d) \left (c^2-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^2 \sqrt{c+d \sin (e+f x)}}\\ &=\frac{4 b (b c-5 a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 d f}-\frac{2 b^2 \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{5 d f}+\frac{2 \left (15 a^2+9 b^2-\frac{2 b c (b c-5 a d)}{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 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{4 b (b c-5 a d) \left (c^2-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^2 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.895404, size = 214, normalized size = 0.84 \[ \frac{2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left (\left (-15 a^2 d^2-10 a b c d+b^2 \left (2 c^2-9 d^2\right )\right ) \left ((c+d) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-c F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )-d^2 \left (15 a^2 c+10 a b d+7 b^2 c\right ) F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )-2 b d \cos (e+f x) (c+d \sin (e+f x)) (10 a d+b c+3 b d \sin (e+f x))}{15 d^2 f \sqrt{c+d \sin (e+f x)}} \]
Antiderivative was successfully verified.
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Maple [B] time = 3.353, size = 1100, normalized size = 4.3 \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{\left (b \sin \left (f x + e\right ) + a\right )}^{2} \sqrt{d \sin \left (f x + e\right ) + c}\,{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 (-{\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}, 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 \left (a + b \sin{\left (e + f x \right )}\right )^{2} \sqrt{c + d \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(-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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