Integrand size = 12, antiderivative size = 75 \[ \int (c \sin (a+b x))^{5/2} \, dx=\frac {6 c^2 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b \sqrt {\sin (a+b x)}}-\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b} \]
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Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2715, 2721, 2719} \[ \int (c \sin (a+b x))^{5/2} \, dx=\frac {6 c^2 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b \sqrt {\sin (a+b x)}}-\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b} \]
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Rule 2715
Rule 2719
Rule 2721
Rubi steps \begin{align*} \text {integral}& = -\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b}+\frac {1}{5} \left (3 c^2\right ) \int \sqrt {c \sin (a+b x)} \, dx \\ & = -\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b}+\frac {\left (3 c^2 \sqrt {c \sin (a+b x)}\right ) \int \sqrt {\sin (a+b x)} \, dx}{5 \sqrt {\sin (a+b x)}} \\ & = \frac {6 c^2 E\left (\left .\frac {1}{2} \left (a-\frac {\pi }{2}+b x\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b \sqrt {\sin (a+b x)}}-\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b} \\ \end{align*}
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88 \[ \int (c \sin (a+b x))^{5/2} \, dx=-\frac {(c \sin (a+b x))^{5/2} \left (6 E\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right )+\sqrt {\sin (a+b x)} \sin (2 (a+b x))\right )}{5 b \sin ^{\frac {5}{2}}(a+b x)} \]
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Time = 0.14 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.03
method | result | size |
default | \(-\frac {c^{3} \left (6 \sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sqrt {\sin }\left (b x +a \right )\right ) E\left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sqrt {\sin }\left (b x +a \right )\right ) F\left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{4}\left (b x +a \right )\right )+2 \left (\sin ^{2}\left (b x +a \right )\right )\right )}{5 \cos \left (b x +a \right ) \sqrt {c \sin \left (b x +a \right )}\, b}\) | \(152\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.35 \[ \int (c \sin (a+b x))^{5/2} \, dx=-\frac {2 \, \sqrt {c \sin \left (b x + a\right )} c^{2} \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 3 i \, \sqrt {2} \sqrt {-i \, c} c^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right )\right ) + 3 i \, \sqrt {2} \sqrt {i \, c} c^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )\right )}{5 \, b} \]
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\[ \int (c \sin (a+b x))^{5/2} \, dx=\int \left (c \sin {\left (a + b x \right )}\right )^{\frac {5}{2}}\, dx \]
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\[ \int (c \sin (a+b x))^{5/2} \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \]
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\[ \int (c \sin (a+b x))^{5/2} \, dx=\int { \left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}} \,d x } \]
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Timed out. \[ \int (c \sin (a+b x))^{5/2} \, dx=\int {\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2} \,d x \]
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