Integrand size = 25, antiderivative size = 107 \[ \int \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^m \, dx=-\frac {2 e \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-\frac {m}{2},\frac {3}{2},\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} \cos (c+d x) (1+\cos (c+d x))^{-m/2} \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}}{d} \]
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Time = 0.39 (sec) , antiderivative size = 107, 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} \[ \int \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^m \, dx=-\frac {2 e \cos (c+d x) \sqrt {a \sec (c+d x)+a} (1-\cos (c+d x))^{\frac {1-m}{2}} (\cos (c+d x)+1)^{-m/2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-\frac {m}{2},\frac {3}{2},\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{d} \]
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Rule 138
Rule 140
Rule 2965
Rule 3961
Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {-\cos (c+d x)} \sqrt {a+a \sec (c+d x)}\right ) \int \frac {\sqrt {-a-a \cos (c+d x)} (e \sin (c+d x))^m}{\sqrt {-\cos (c+d x)}} \, dx}{\sqrt {-a-a \cos (c+d x)}} \\ & = -\frac {\left (e \sqrt {-\cos (c+d x)} (-a-a \cos (c+d x))^{-\frac {1}{2}+\frac {1-m}{2}} (-a+a \cos (c+d x))^{\frac {1-m}{2}} \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}\right ) \text {Subst}\left (\int \frac {(-a-a x)^{\frac {1}{2}+\frac {1}{2} (-1+m)} (-a+a x)^{\frac {1}{2} (-1+m)}}{\sqrt {-x}} \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {\left (e \sqrt {-\cos (c+d x)} (1+\cos (c+d x))^{-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}} \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}\right ) \text {Subst}\left (\int \frac {(1+x)^{\frac {1}{2}+\frac {1}{2} (-1+m)} (-a+a x)^{\frac {1}{2} (-1+m)}}{\sqrt {-x}} \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {\left (e (1-\cos (c+d x))^{\frac {1}{2}-\frac {m}{2}} \sqrt {-\cos (c+d x)} (1+\cos (c+d x))^{-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}} \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}\right ) \text {Subst}\left (\int \frac {(1-x)^{\frac {1}{2} (-1+m)} (1+x)^{\frac {1}{2}+\frac {1}{2} (-1+m)}}{\sqrt {-x}} \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {2 e \operatorname {AppellF1}\left (\frac {1}{2},\frac {1-m}{2},-\frac {m}{2},\frac {3}{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))^{-m/2} \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}}{d} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(433\) vs. \(2(107)=214\).
Time = 2.37 (sec) , antiderivative size = 433, normalized size of antiderivative = 4.05 \[ \int \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^m \, dx=\frac {4 (3+m) \left (\operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+\operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1}{2},m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \sin \left (\frac {1}{2} (c+d x)\right ) (e \sin (c+d x))^m}{d (1+m) \left (\left (2 (1+m) \operatorname {AppellF1}\left (\frac {3+m}{2},-\frac {1}{2},2+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+(1+2 m) \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {1}{2},1+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-\operatorname {AppellF1}\left (\frac {3+m}{2},\frac {3}{2},m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+(3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))+(3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},\frac {1}{2},m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right )} \]
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\[\int \sqrt {a +a \sec \left (d x +c \right )}\, \left (e \sin \left (d x +c \right )\right )^{m}d x\]
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\[ \int \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^m \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \]
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\[ \int \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^m \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (e \sin {\left (c + d x \right )}\right )^{m}\, dx \]
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\[ \int \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^m \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \]
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\[ \int \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^m \, dx=\int { \sqrt {a \sec \left (d x + c\right ) + a} \left (e \sin \left (d x + c\right )\right )^{m} \,d x } \]
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Timed out. \[ \int \sqrt {a+a \sec (c+d x)} (e \sin (c+d x))^m \, dx=\int {\left (e\,\sin \left (c+d\,x\right )\right )}^m\,\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}} \,d x \]
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