Integrand size = 25, antiderivative size = 105 \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2^{-2+\frac {p}{2}} (e \cos (c+d x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {6-p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{1-\frac {p}{2}}}{a d e (1+p) (a+a \sin (c+d x))^{3/2}} \]
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Time = 0.10 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2768, 72, 71} \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2^{\frac {p}{2}-2} (\sin (c+d x)+1)^{1-\frac {p}{2}} (e \cos (c+d x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {6-p}{2},\frac {p+1}{2},\frac {p+3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{a d e (p+1) (a \sin (c+d x)+a)^{3/2}} \]
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Rule 71
Rule 72
Rule 2768
Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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} \operatorname {Hypergeometric2F1}\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*}
Time = 0.10 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.97 \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {2^{-2+\frac {p}{2}} \cos (c+d x) (e \cos (c+d x))^p \operatorname {Hypergeometric2F1}\left (3-\frac {p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{-p/2}}{a^2 d (1+p) \sqrt {a (1+\sin (c+d x))}} \]
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\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{\left (a +a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}d x\]
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\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {\left (e \cos {\left (c + d x \right )}\right )^{p}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]
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\[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(e \cos (c+d x))^p}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]
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