Optimal. Leaf size=105 \[ -\frac{2^{\frac{p}{2}-2} (\sin (c+d x)+1)^{1-\frac{p}{2}} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{6-p}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{a d e (p+1) (a \sin (c+d x)+a)^{3/2}} \]
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Rubi [A] time = 0.123095, antiderivative size = 105, 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} \[ -\frac{2^{\frac{p}{2}-2} (\sin (c+d x)+1)^{1-\frac{p}{2}} (e \cos (c+d x))^{p+1} \, _2F_1\left (\frac{6-p}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{a d e (p+1) (a \sin (c+d x)+a)^{3/2}} \]
Antiderivative was successfully verified.
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Rule 2689
Rule 70
Rule 69
Rubi steps
\begin{align*} \int \frac{(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, 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)^{-\frac{5}{2}+\frac{1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=\frac{\left (2^{-3+\frac{p}{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))^{-1+\frac{1}{2} (-1-p)+\frac{p}{2}} \left (\frac{a+a \sin (c+d x)}{a}\right )^{1-\frac{p}{2}}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}+\frac{x}{2}\right )^{-\frac{5}{2}+\frac{1}{2} (-1+p)} (a-a x)^{\frac{1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{d e}\\ &=-\frac{2^{-2+\frac{p}{2}} (e \cos (c+d x))^{1+p} \, _2F_1\left (\frac{6-p}{2},\frac{1+p}{2};\frac{3+p}{2};\frac{1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{1-\frac{p}{2}}}{a d e (1+p) (a+a \sin (c+d x))^{3/2}}\\ \end{align*}
Mathematica [A] time = 0.132011, size = 102, normalized size = 0.97 \[ -\frac{2^{\frac{p}{2}-2} \cos (c+d x) (\sin (c+d x)+1)^{-p/2} (e \cos (c+d x))^p \, _2F_1\left (3-\frac{p}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{a^2 d (p+1) \sqrt{a (\sin (c+d x)+1)}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.091, 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 ) ^{-{\frac{5}{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 )}^{\frac{5}{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{\sqrt{a \sin \left (d x + c\right ) + a} \left (e \cos \left (d x + c\right )\right )^{p}}{3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} +{\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )}, 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 )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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