Integrand size = 36, antiderivative size = 73 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {2 a (5 A+B) c^2 \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}-\frac {2 a B c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}} \]
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Time = 0.17 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3046, 2935, 2752} \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\frac {2 a c^2 (5 A+B) \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}-\frac {2 a B c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}} \]
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Rule 2752
Rule 2935
Rule 3046
Rubi steps \begin{align*} \text {integral}& = (a c) \int \frac {\cos ^2(e+f x) (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx \\ & = -\frac {2 a B c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}}+\frac {1}{5} (a (5 A+B) c) \int \frac {\cos ^2(e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx \\ & = \frac {2 a (5 A+B) c^2 \cos ^3(e+f x)}{15 f (c-c \sin (e+f x))^{3/2}}-\frac {2 a B c \cos ^3(e+f x)}{5 f \sqrt {c-c \sin (e+f x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(191\) vs. \(2(73)=146\).
Time = 1.15 (sec) , antiderivative size = 191, normalized size of antiderivative = 2.62 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right ) \left (32 B-30 \sqrt {2} A \sqrt {1+\cos (e+f x)}\right )+\sqrt {2} \sqrt {1+\cos (e+f x)} \left (5 (2 A+B) \cos \left (\frac {3}{2} (e+f x)\right )+3 B \cos \left (\frac {5}{2} (e+f x)\right )-2 (20 A+B+2 (5 A+B) \cos (e+f x)-3 B \cos (2 (e+f x))) \sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sqrt {c-c \sin (e+f x)}}{30 \sqrt {2} f \sqrt {1+\cos (e+f x)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]
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Time = 1.70 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.86
method | result | size |
default | \(-\frac {2 \left (\sin \left (f x +e \right )-1\right ) c \left (1+\sin \left (f x +e \right )\right )^{2} a \left (3 B \sin \left (f x +e \right )+5 A -2 B \right )}{15 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(63\) |
parts | \(-\frac {2 a A \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right ) c}{\cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 B a \left (\sin \left (f x +e \right )-1\right ) c \left (1+\sin \left (f x +e \right )\right ) \left (3 \left (\sin ^{2}\left (f x +e \right )\right )-4 \sin \left (f x +e \right )+8\right )}{15 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}-\frac {2 a \left (A +B \right ) \left (\sin \left (f x +e \right )-1\right ) c \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-2\right )}{3 \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\) | \(167\) |
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Time = 0.25 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.78 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {2 \, {\left (3 \, B a \cos \left (f x + e\right )^{3} + {\left (5 \, A + 4 \, B\right )} a \cos \left (f x + e\right )^{2} - {\left (5 \, A + B\right )} a \cos \left (f x + e\right ) - 2 \, {\left (5 \, A + B\right )} a + {\left (3 \, B a \cos \left (f x + e\right )^{2} - {\left (5 \, A + B\right )} a \cos \left (f x + e\right ) - 2 \, {\left (5 \, A + B\right )} a\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \]
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\[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=a \left (\int A \sqrt {- c \sin {\left (e + f x \right )} + c}\, dx + \int A \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int B \sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )}\, dx + \int B \sqrt {- c \sin {\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\, dx\right ) \]
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\[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )} \sqrt {-c \sin \left (f x + e\right ) + c} \,d x } \]
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Time = 0.38 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.62 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=-\frac {\sqrt {2} {\left (30 \, A a \cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 3 \, B a \cos \left (-\frac {5}{4} \, \pi + \frac {5}{2} \, f x + \frac {5}{2} \, e\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + 5 \, {\left (2 \, A a \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + B a \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \cos \left (-\frac {3}{4} \, \pi + \frac {3}{2} \, f x + \frac {3}{2} \, e\right )\right )} \sqrt {c}}{30 \, f} \]
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Timed out. \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) \sqrt {c-c \sin (e+f x)} \, dx=\int \left (A+B\,\sin \left (e+f\,x\right )\right )\,\left (a+a\,\sin \left (e+f\,x\right )\right )\,\sqrt {c-c\,\sin \left (e+f\,x\right )} \,d x \]
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