Integrand size = 25, antiderivative size = 156 \[ \int (e \cos (c+d x))^p \sqrt {a+b \sin (c+d x)} \, dx=\frac {2 e \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {5}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{-1+p} (a+b \sin (c+d x))^{3/2} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}}}{3 b d} \]
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Time = 0.07 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2783, 143} \[ \int (e \cos (c+d x))^p \sqrt {a+b \sin (c+d x)} \, dx=\frac {2 e (a+b \sin (c+d x))^{3/2} (e \cos (c+d x))^{p-1} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {5}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{3 b d} \]
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Rule 143
Rule 2783
Rubi steps \begin{align*} \text {integral}& = \frac {\left (e (e \cos (c+d x))^{-1+p} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}}\right ) \text {Subst}\left (\int \sqrt {a+b x} \left (-\frac {b}{a-b}-\frac {b x}{a-b}\right )^{\frac {1}{2} (-1+p)} \left (\frac {b}{a+b}-\frac {b x}{a+b}\right )^{\frac {1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {2 e \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {5}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{-1+p} (a+b \sin (c+d x))^{3/2} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}}}{3 b d} \\ \end{align*}
Time = 1.01 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.20 \[ \int (e \cos (c+d x))^p \sqrt {a+b \sin (c+d x)} \, dx=\frac {2 e \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {5}{2},\frac {a+b \sin (c+d x)}{a-\sqrt {b^2}},\frac {a+b \sin (c+d x)}{a+\sqrt {b^2}}\right ) (e \cos (c+d x))^{-1+p} \left (\frac {\sqrt {b^2}-b \sin (c+d x)}{a+\sqrt {b^2}}\right )^{\frac {1-p}{2}} (a+b \sin (c+d x))^{3/2} \left (\frac {\sqrt {b^2}+b \sin (c+d x)}{-a+\sqrt {b^2}}\right )^{\frac {1-p}{2}}}{3 b d} \]
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\[\int \left (e \cos \left (d x +c \right )\right )^{p} \sqrt {a +b \sin \left (d x +c \right )}d x\]
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\[ \int (e \cos (c+d x))^p \sqrt {a+b \sin (c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \]
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\[ \int (e \cos (c+d x))^p \sqrt {a+b \sin (c+d x)} \, dx=\int \left (e \cos {\left (c + d x \right )}\right )^{p} \sqrt {a + b \sin {\left (c + d x \right )}}\, dx \]
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\[ \int (e \cos (c+d x))^p \sqrt {a+b \sin (c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \]
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\[ \int (e \cos (c+d x))^p \sqrt {a+b \sin (c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \]
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Timed out. \[ \int (e \cos (c+d x))^p \sqrt {a+b \sin (c+d x)} \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,\sqrt {a+b\,\sin \left (c+d\,x\right )} \,d x \]
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