Optimal. Leaf size=93 \[ -\frac{2^{\frac{p-3}{2}} (\sin (c+d x)+1)^{\frac{1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{5-p}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{a^2 d e (p+1)} \]
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Rubi [A] time = 0.0901636, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2688, 69} \[ -\frac{2^{\frac{p-3}{2}} (\sin (c+d x)+1)^{\frac{1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{5-p}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{a^2 d e (p+1)} \]
Antiderivative was successfully verified.
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Rule 2688
Rule 69
Rubi steps
\begin{align*} \int \frac{(e \cos (c+d x))^p}{(a+a \sin (c+d x))^2} \, dx &=\frac{\left ((e \cos (c+d x))^{1+p} (1-\sin (c+d x))^{\frac{1}{2} (-1-p)} (1+\sin (c+d x))^{\frac{1}{2} (-1-p)}\right ) \operatorname{Subst}\left (\int (1-x)^{\frac{1}{2} (-1+p)} (1+x)^{-2+\frac{1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{a^2 d e}\\ &=-\frac{2^{\frac{1}{2} (-3+p)} (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac{5-p}{2},\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-p)}}{a^2 d e (1+p)}\\ \end{align*}
Mathematica [A] time = 0.159679, size = 94, normalized size = 1.01 \[ -\frac{2^{\frac{p-3}{2}} \cos (c+d x) (\sin (c+d x)+1)^{\frac{1}{2} (-p-1)} (e \cos (c+d x))^p \, _2F_1\left (\frac{5-p}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{a^2 d (p+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.26, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( e\cos \left ( dx+c \right ) \right ) ^{p}}{ \left ( a+a\sin \left ( dx+c \right ) \right ) ^{2}}}\, 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 (e \cos \left (d x + c\right )\right )^{p}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{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 (e \cos \left (d x + c\right )\right )^{p}}{a^{2} \cos \left (d x + c\right )^{2} - 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}}, 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 \frac{\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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