Optimal. Leaf size=265 \[ -\frac {c \sin (e+f x) \tan (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{4 f^3}-\frac {2 c \tan (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^3}+\frac {2 c x \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^2}+\frac {c x \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{2 f^2}+\frac {x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c \sin (e+f x)+c)^{5/2}}{2 c f}-\frac {3 c x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{4 f} \]
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Rubi [A] time = 0.27, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4604, 4422, 3317, 3296, 2637, 3310, 30} \[ \frac {2 c x \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^2}+\frac {c x \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{2 f^2}-\frac {c \sin (e+f x) \tan (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{4 f^3}-\frac {2 c \tan (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{f^3}+\frac {x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c \sin (e+f x)+c)^{5/2}}{2 c f}-\frac {3 c x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{4 f} \]
Antiderivative was successfully verified.
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Rule 30
Rule 2637
Rule 3296
Rule 3310
Rule 3317
Rule 4422
Rule 4604
Rubi steps
\begin {align*} \int x^2 \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2} \, dx &=\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int x^2 \cos (e+f x) (c+c \sin (e+f x)) \, dx\\ &=\frac {x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac {\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int x (c+c \sin (e+f x))^2 \, dx}{c f}\\ &=\frac {x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac {\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int \left (c^2 x+2 c^2 x \sin (e+f x)+c^2 x \sin ^2(e+f x)\right ) \, dx}{c f}\\ &=-\frac {c x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{2 f}+\frac {x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac {\left (c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int x \sin ^2(e+f x) \, dx}{f}-\frac {\left (2 c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int x \sin (e+f x) \, dx}{f}\\ &=\frac {2 c x \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^2}-\frac {c x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{2 f}+\frac {c x \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{2 f^2}+\frac {x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac {c \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{4 f^3}-\frac {\left (2 c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int \cos (e+f x) \, dx}{f^2}-\frac {\left (c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}\right ) \int x \, dx}{2 f}\\ &=\frac {2 c x \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{f^2}-\frac {3 c x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{4 f}+\frac {c x \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{2 f^2}+\frac {x^2 \sec (e+f x) \sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{5/2}}{2 c f}-\frac {2 c \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{f^3}-\frac {c \sin (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{4 f^3}\\ \end {align*}
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Mathematica [A] time = 0.81, size = 95, normalized size = 0.36 \[ \frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c (\sin (e+f x)+1)} \left (8 \left (f^2 x^2-2\right ) \tan (e+f x)-\left (2 f^2 x^2-1\right ) \cos (2 (e+f x)) \sec (e+f x)+4 f x \sin (e+f x)+16 f x\right )}{8 f^3} \]
Antiderivative was successfully verified.
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fricas [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.17, size = 0, normalized size = 0.00 \[ \int x^{2} \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-a \sin \left (f x + e\right ) + a} {\left (c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} x^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.77, size = 159, normalized size = 0.60 \[ \frac {c\,\sqrt {-a\,\left (\sin \left (e+f\,x\right )-1\right )}\,\sqrt {c\,\left (\sin \left (e+f\,x\right )+1\right )}\,\left (\cos \left (e+f\,x\right )+\cos \left (3\,e+3\,f\,x\right )-16\,\sin \left (2\,e+2\,f\,x\right )+16\,f\,x+16\,f\,x\,\cos \left (2\,e+2\,f\,x\right )+2\,f\,x\,\sin \left (3\,e+3\,f\,x\right )-2\,f^2\,x^2\,\cos \left (e+f\,x\right )+2\,f\,x\,\sin \left (e+f\,x\right )-2\,f^2\,x^2\,\cos \left (3\,e+3\,f\,x\right )+8\,f^2\,x^2\,\sin \left (2\,e+2\,f\,x\right )\right )}{8\,f^3\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (c \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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