Optimal. Leaf size=89 \[ \frac{2^{\frac{3}{2}-m} (1-\sin (c+d x))^{m-\frac{3}{2}} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{3-2 m} \, _2F_1\left (\frac{3}{2},\frac{1}{2} (2 m-1);\frac{5}{2};\frac{1}{2} (\sin (c+d x)+1)\right )}{3 d e} \]
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Rubi [A] time = 0.0992501, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2689, 7, 70, 69} \[ \frac{2^{\frac{3}{2}-m} (1-\sin (c+d x))^{m-\frac{3}{2}} (a \sin (c+d x)+a)^m (e \cos (c+d x))^{3-2 m} \, _2F_1\left (\frac{3}{2},\frac{1}{2} (2 m-1);\frac{5}{2};\frac{1}{2} (\sin (c+d x)+1)\right )}{3 d e} \]
Antiderivative was successfully verified.
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Rule 2689
Rule 7
Rule 70
Rule 69
Rubi steps
\begin{align*} \int (e \cos (c+d x))^{2-2 m} (a+a \sin (c+d x))^m \, dx &=\frac{\left (a^2 (e \cos (c+d x))^{3-2 m} (a-a \sin (c+d x))^{\frac{1}{2} (-3+2 m)} (a+a \sin (c+d x))^{\frac{1}{2} (-3+2 m)}\right ) \operatorname{Subst}\left (\int (a-a x)^{\frac{1}{2} (1-2 m)} (a+a x)^{\frac{1}{2} (1-2 m)+m} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac{\left (a^2 (e \cos (c+d x))^{3-2 m} (a-a \sin (c+d x))^{\frac{1}{2} (-3+2 m)} (a+a \sin (c+d x))^{\frac{1}{2} (-3+2 m)}\right ) \operatorname{Subst}\left (\int (a-a x)^{\frac{1}{2} (1-2 m)} \sqrt{a+a x} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac{\left (2^{\frac{1}{2}-m} a^2 (e \cos (c+d x))^{3-2 m} (a-a \sin (c+d x))^{\frac{1}{2}-m+\frac{1}{2} (-3+2 m)} \left (\frac{a-a \sin (c+d x)}{a}\right )^{-\frac{1}{2}+m} (a+a \sin (c+d x))^{\frac{1}{2} (-3+2 m)}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{\frac{1}{2} (1-2 m)} \sqrt{a+a x} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac{2^{\frac{3}{2}-m} (e \cos (c+d x))^{3-2 m} \, _2F_1\left (\frac{3}{2},\frac{1}{2} (-1+2 m);\frac{5}{2};\frac{1}{2} (1+\sin (c+d x))\right ) (1-\sin (c+d x))^{-\frac{3}{2}+m} (a+a \sin (c+d x))^m}{3 d e}\\ \end{align*}
Mathematica [A] time = 0.181125, size = 96, normalized size = 1.08 \[ \frac{2 \sqrt{2} e^2 \cos ^3(c+d x) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-2 m} \, _2F_1\left (-\frac{1}{2},\frac{3}{2}-m;\frac{5}{2}-m;\frac{1}{2} (1-\sin (c+d x))\right )}{d (2 m-3) (\sin (c+d x)+1)^{3/2}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.927, size = 0, normalized size = 0. \begin{align*} \int \left ( e\cos \left ( dx+c \right ) \right ) ^{2-2\,m} \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 )^{-2 \, m + 2}{\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 )^{-2 \, m + 2}{\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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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 )^{-2 \, m + 2}{\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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