Optimal. Leaf size=114 \[ -\frac{a 2^{m+\frac{p}{2}+\frac{1}{2}} (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{p+1} (\sin (c+d x)+1)^{\frac{1}{2} (-2 m-p+1)} \, _2F_1\left (\frac{1}{2} (-2 m-p+1),\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (p+1)} \]
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Rubi [A] time = 0.115, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2689, 70, 69} \[ -\frac{a 2^{m+\frac{p}{2}+\frac{1}{2}} (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{p+1} (\sin (c+d x)+1)^{\frac{1}{2} (-2 m-p+1)} \, _2F_1\left (\frac{1}{2} (-2 m-p+1),\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (p+1)} \]
Antiderivative was successfully verified.
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Rule 2689
Rule 70
Rule 69
Rubi steps
\begin{align*} \int (e \cos (c+d x))^p (a+a \sin (c+d x))^m \, dx &=\frac{\left (a^2 (e \cos (c+d x))^{1+p} (a-a \sin (c+d x))^{\frac{1}{2} (-1-p)} (a+a \sin (c+d x))^{\frac{1}{2} (-1-p)}\right ) \operatorname{Subst}\left (\int (a-a x)^{\frac{1}{2} (-1+p)} (a+a x)^{m+\frac{1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac{\left (2^{-\frac{1}{2}+m+\frac{p}{2}} a^2 (e \cos (c+d x))^{1+p} (a-a \sin (c+d x))^{\frac{1}{2} (-1-p)} (a+a \sin (c+d x))^{-\frac{1}{2}+m+\frac{1}{2} (-1-p)+\frac{p}{2}} \left (\frac{a+a \sin (c+d x)}{a}\right )^{\frac{1}{2}-m-\frac{p}{2}}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{x}{2}\right )^{m+\frac{1}{2} (-1+p)} (a-a x)^{\frac{1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=-\frac{2^{\frac{1}{2}+m+\frac{p}{2}} a (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac{1}{2} (1-2 m-p),\frac{1+p}{2};\frac{3+p}{2};\frac{1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac{1}{2} (1-2 m-p)} (a+a \sin (c+d x))^{-1+m}}{d e (1+p)}\\ \end{align*}
Mathematica [A] time = 0.18975, size = 112, normalized size = 0.98 \[ -\frac{2^{\frac{1}{2} (2 m+p+1)} \cos (c+d x) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^p (\sin (c+d x)+1)^{\frac{1}{2} (-2 m-p-1)} \, _2F_1\left (\frac{1}{2} (-2 m-p+1),\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d (p+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.882, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{p} \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{p}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (e \cos \left (d x + c\right )\right )^{p}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sin{\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos{\left (c + d x \right )}\right )^{p}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (e \cos \left (d x + c\right )\right )^{p}{\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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