Integrand size = 23, antiderivative size = 158 \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^m \, dx=\frac {e \operatorname {AppellF1}\left (1+m,\frac {1-p}{2},\frac {1-p}{2},2+m,\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{-1+p} (a+b \sin (c+d x))^{1+m} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}}}{b d (1+m)} \]
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Time = 0.07 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2783, 143} \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^m \, dx=\frac {e (e \cos (c+d x))^{p-1} (a+b \sin (c+d x))^{m+1} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (m+1,\frac {1-p}{2},\frac {1-p}{2},m+2,\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{b d (m+1)} \]
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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 (a+b x)^m \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 {e \operatorname {AppellF1}\left (1+m,\frac {1-p}{2},\frac {1-p}{2},2+m,\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))^{1+m} \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 (1+m)} \\ \end{align*}
\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^m \, dx=\int (e \cos (c+d x))^p (a+b \sin (c+d x))^m \, dx \]
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\[\int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +b \sin \left (d x +c \right )\right )^{m}d x\]
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\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{p} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^m \, dx=\int \left (e \cos {\left (c + d x \right )}\right )^{p} \left (a + b \sin {\left (c + d x \right )}\right )^{m}\, dx \]
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\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{p} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^m \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{p} {\left (b \sin \left (d x + c\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^m \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m \,d x \]
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