Integrand size = 25, antiderivative size = 115 \[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {2 e \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-m}{2},\frac {2-m}{2},\frac {5}{2},\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} \cos (c+d x) (1+\cos (c+d x))^{1-\frac {m}{2}} (e \sin (c+d x))^{-1+m}}{3 d \sqrt {a+a \sec (c+d x)}} \]
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Time = 0.43 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3961, 2965, 140, 138} \[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=-\frac {2 e \cos (c+d x) (1-\cos (c+d x))^{\frac {1-m}{2}} (\cos (c+d x)+1)^{1-\frac {m}{2}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-m}{2},\frac {2-m}{2},\frac {5}{2},\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{3 d \sqrt {a \sec (c+d x)+a}} \]
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Rule 138
Rule 140
Rule 2965
Rule 3961
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {-a-a \cos (c+d x)} \int \frac {\sqrt {-\cos (c+d x)} (e \sin (c+d x))^m}{\sqrt {-a-a \cos (c+d x)}} \, dx}{\sqrt {-\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \\ & = -\frac {\left (e (-a-a \cos (c+d x))^{\frac {1}{2}+\frac {1-m}{2}} (-a+a \cos (c+d x))^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}\right ) \text {Subst}\left (\int \sqrt {-x} (-a-a x)^{-\frac {1}{2}+\frac {1}{2} (-1+m)} (-a+a x)^{\frac {1}{2} (-1+m)} \, dx,x,\cos (c+d x)\right )}{d \sqrt {-\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \\ & = -\frac {\left (e (1+\cos (c+d x))^{1-\frac {m}{2}} (-a-a \cos (c+d x))^{-\frac {1}{2}+\frac {1-m}{2}+\frac {m}{2}} (-a+a \cos (c+d x))^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}\right ) \text {Subst}\left (\int \sqrt {-x} (1+x)^{-\frac {1}{2}+\frac {1}{2} (-1+m)} (-a+a x)^{\frac {1}{2} (-1+m)} \, dx,x,\cos (c+d x)\right )}{d \sqrt {-\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \\ & = -\frac {\left (e (1-\cos (c+d x))^{\frac {1}{2}-\frac {m}{2}} (1+\cos (c+d x))^{1-\frac {m}{2}} (-a-a \cos (c+d x))^{-\frac {1}{2}+\frac {1-m}{2}+\frac {m}{2}} (-a+a \cos (c+d x))^{-\frac {1}{2}+\frac {1-m}{2}+\frac {m}{2}} (e \sin (c+d x))^{-1+m}\right ) \text {Subst}\left (\int (1-x)^{\frac {1}{2} (-1+m)} \sqrt {-x} (1+x)^{-\frac {1}{2}+\frac {1}{2} (-1+m)} \, dx,x,\cos (c+d x)\right )}{d \sqrt {-\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \\ & = -\frac {2 e \operatorname {AppellF1}\left (\frac {3}{2},\frac {1-m}{2},\frac {2-m}{2},\frac {5}{2},\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} \cos (c+d x) (1+\cos (c+d x))^{1-\frac {m}{2}} (e \sin (c+d x))^{-1+m}}{3 d \sqrt {a+a \sec (c+d x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(277\) vs. \(2(115)=230\).
Time = 2.85 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.41 \[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\frac {4 (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right ) (e \sin (c+d x))^m}{d (1+m) \left (\left (2 (1+m) \operatorname {AppellF1}\left (\frac {3+m}{2},-\frac {1}{2},2+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+\operatorname {AppellF1}\left (\frac {3+m}{2},\frac {1}{2},1+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+(3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right ) \sqrt {a (1+\sec (c+d x))}} \]
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\[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\sqrt {a +a \sec \left (d x +c \right )}}d x\]
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\[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
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\[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]
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\[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
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\[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{\sqrt {a \sec \left (d x + c\right ) + a}} \,d x } \]
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Timed out. \[ \int \frac {(e \sin (c+d x))^m}{\sqrt {a+a \sec (c+d x)}} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]
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