Optimal. Leaf size=115 \[ -\frac{a 2^{\frac{m}{2}+\frac{1}{2}} (\sin (c+d x)+1)^{\frac{1-m}{2}} (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{1-m} \, _2F_1\left (\frac{1-m}{2},\frac{1-m}{2};\frac{3-m}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (1-m)} \]
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Rubi [A] time = 0.108866, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2689, 70, 69} \[ -\frac{a 2^{\frac{m}{2}+\frac{1}{2}} (\sin (c+d x)+1)^{\frac{1-m}{2}} (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{1-m} \, _2F_1\left (\frac{1-m}{2},\frac{1-m}{2};\frac{3-m}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d e (1-m)} \]
Antiderivative was successfully verified.
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Rule 2689
Rule 70
Rule 69
Rubi steps
\begin{align*} \int (e \cos (c+d x))^{-m} (a+a \sin (c+d x))^m \, dx &=\frac{\left (a^2 (e \cos (c+d x))^{1-m} (a-a \sin (c+d x))^{\frac{1}{2} (-1+m)} (a+a \sin (c+d x))^{\frac{1}{2} (-1+m)}\right ) \operatorname{Subst}\left (\int (a-a x)^{\frac{1}{2} (-1-m)} (a+a x)^{\frac{1}{2} (-1-m)+m} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac{\left (2^{-\frac{1}{2}+\frac{m}{2}} a^2 (e \cos (c+d x))^{1-m} (a-a \sin (c+d x))^{\frac{1}{2} (-1+m)} (a+a \sin (c+d x))^{-\frac{1}{2}+\frac{1}{2} (-1+m)+\frac{m}{2}} \left (\frac{a+a \sin (c+d x)}{a}\right )^{\frac{1}{2}-\frac{m}{2}}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{x}{2}\right )^{\frac{1}{2} (-1-m)+m} (a-a x)^{\frac{1}{2} (-1-m)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=-\frac{2^{\frac{1}{2}+\frac{m}{2}} a (e \cos (c+d x))^{1-m} \, _2F_1\left (\frac{1-m}{2},\frac{1-m}{2};\frac{3-m}{2};\frac{1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac{1-m}{2}} (a+a \sin (c+d x))^{-1+m}}{d e (1-m)}\\ \end{align*}
Mathematica [A] time = 0.128852, size = 108, normalized size = 0.94 \[ \frac{2^{\frac{m+1}{2}} \cos (c+d x) (\sin (c+d x)+1)^{\frac{1}{2} (-m-1)} (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-m} \, _2F_1\left (\frac{1-m}{2},\frac{1-m}{2};\frac{3-m}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d (m-1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.355, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m}}{ \left ( e\cos \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 \frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{\left (e \cos \left (d x + c\right )\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 (\frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{\left (e \cos \left (d x + c\right )\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 )^{- m}\, 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 \frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{\left (e \cos \left (d x + c\right )\right )^{m}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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