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 = 0.11, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 69
Rule 70
Rule 2689
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*}
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Mathematica [A] time = 0.20, 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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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\left (e \cos \left (d x + c\right )\right )^{p} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{p} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.34, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x + c\right )\right )^{p} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^m \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{m} \left (e \cos {\left (c + d x \right )}\right )^{p}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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