Integrand size = 25, antiderivative size = 154 \[ \int \frac {(e \cos (c+d x))^p}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 e \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {3}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{-1+p} \sqrt {a+b \sin (c+d x)} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}}}{b d} \]
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Time = 0.07 (sec) , antiderivative size = 154, 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} \[ \int \frac {(e \cos (c+d x))^p}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 e \sqrt {a+b \sin (c+d x)} (e \cos (c+d x))^{p-1} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {3}{2},\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{b d} \]
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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 \frac {\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)}}{\sqrt {a+b x}} \, dx,x,\sin (c+d x)\right )}{d} \\ & = \frac {2 e \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {3}{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} \sqrt {a+b \sin (c+d x)} \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}}}{b d} \\ \end{align*}
Time = 3.46 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.20 \[ \int \frac {(e \cos (c+d x))^p}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 e \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-p}{2},\frac {1-p}{2},\frac {3}{2},\frac {a+b \sin (c+d x)}{a-\sqrt {b^2}},\frac {a+b \sin (c+d x)}{a+\sqrt {b^2}}\right ) (e \cos (c+d x))^{-1+p} \left (\frac {\sqrt {b^2}-b \sin (c+d x)}{a+\sqrt {b^2}}\right )^{\frac {1-p}{2}} \sqrt {a+b \sin (c+d x)} \left (\frac {\sqrt {b^2}+b \sin (c+d x)}{-a+\sqrt {b^2}}\right )^{\frac {1-p}{2}}}{b d} \]
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\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\sqrt {a +b \sin \left (d x +c \right )}}d x\]
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\[ \int \frac {(e \cos (c+d x))^p}{\sqrt {a+b \sin (c+d x)}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{\sqrt {b \sin \left (d x + c\right ) + a}} \,d x } \]
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\[ \int \frac {(e \cos (c+d x))^p}{\sqrt {a+b \sin (c+d x)}} \, dx=\int \frac {\left (e \cos {\left (c + d x \right )}\right )^{p}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \]
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\[ \int \frac {(e \cos (c+d x))^p}{\sqrt {a+b \sin (c+d x)}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{\sqrt {b \sin \left (d x + c\right ) + a}} \,d x } \]
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\[ \int \frac {(e \cos (c+d x))^p}{\sqrt {a+b \sin (c+d x)}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{\sqrt {b \sin \left (d x + c\right ) + a}} \,d x } \]
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Timed out. \[ \int \frac {(e \cos (c+d x))^p}{\sqrt {a+b \sin (c+d x)}} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \]
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