Integrand size = 23, antiderivative size = 95 \[ \int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx=-\frac {2^{-\frac {1}{2}+\frac {p}{2}} (e \cos (c+d x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {3-p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (-1-p)}}{a d e (1+p)} \]
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Time = 0.08 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2767, 71} \[ \int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx=-\frac {2^{\frac {p}{2}-\frac {1}{2}} (\sin (c+d x)+1)^{\frac {1}{2} (-p-1)} (e \cos (c+d x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {3-p}{2},\frac {p+1}{2},\frac {p+3}{2},\frac {1}{2} (1-\sin (c+d x))\right )}{a d e (p+1)} \]
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Rule 71
Rule 2767
Rubi steps \begin{align*} \text {integral}& = \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 ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-1+p)} (1+x)^{-1+\frac {1}{2} (-1+p)} \, dx,x,\sin (c+d x)\right )}{a d e} \\ & = -\frac {2^{-\frac {1}{2}+\frac {p}{2}} (e \cos (c+d x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {3-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 d e (1+p)} \\ \end{align*}
Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.99 \[ \int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx=-\frac {2^{\frac {1}{2} (-1+p)} \cos (c+d x) (e \cos (c+d x))^p \operatorname {Hypergeometric2F1}\left (\frac {3-p}{2},\frac {1+p}{2},\frac {3+p}{2},\frac {1}{2} (1-\sin (c+d x))\right ) (1+\sin (c+d x))^{\frac {1}{2} (-1-p)}}{a d (1+p)} \]
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\[\int \frac {\left (e \cos \left (d x +c \right )\right )^{p}}{a +a \sin \left (d x +c \right )}d x\]
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\[ \int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{a \sin \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {\left (e \cos {\left (c + d x \right )}\right )^{p}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]
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\[ \int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{a \sin \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx=\int { \frac {\left (e \cos \left (d x + c\right )\right )^{p}}{a \sin \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {(e \cos (c+d x))^p}{a+a \sin (c+d x)} \, dx=\int \frac {{\left (e\,\cos \left (c+d\,x\right )\right )}^p}{a+a\,\sin \left (c+d\,x\right )} \,d x \]
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