Integrand size = 25, antiderivative size = 120 \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {2 e \operatorname {AppellF1}\left (\frac {5}{2},\frac {1-m}{2},\frac {4-m}{2},\frac {7}{2},\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} \cos ^2(c+d x) (1+\cos (c+d x))^{1-\frac {m}{2}} (e \sin (c+d x))^{-1+m}}{5 a d \sqrt {a+a \sec (c+d x)}} \]
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Time = 0.46 (sec) , antiderivative size = 120, 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} \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {2 e \cos ^2(c+d x) (1-\cos (c+d x))^{\frac {1-m}{2}} (\cos (c+d x)+1)^{1-\frac {m}{2}} \operatorname {AppellF1}\left (\frac {5}{2},\frac {1-m}{2},\frac {4-m}{2},\frac {7}{2},\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{5 a d \sqrt {a \sec (c+d x)+a}} \]
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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 {(-\cos (c+d x))^{3/2} (e \sin (c+d x))^m}{(-a-a \cos (c+d x))^{3/2}} \, 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 (-x)^{3/2} (-a-a x)^{-\frac {3}{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 (-x)^{3/2} (1+x)^{-\frac {3}{2}+\frac {1}{2} (-1+m)} (-a+a x)^{\frac {1}{2} (-1+m)} \, dx,x,\cos (c+d x)\right )}{a 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)} (-x)^{3/2} (1+x)^{-\frac {3}{2}+\frac {1}{2} (-1+m)} \, dx,x,\cos (c+d x)\right )}{a d \sqrt {-\cos (c+d x)} \sqrt {a+a \sec (c+d x)}} \\ & = -\frac {2 e \operatorname {AppellF1}\left (\frac {5}{2},\frac {1-m}{2},\frac {4-m}{2},\frac {7}{2},\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac {1-m}{2}} \cos ^2(c+d x) (1+\cos (c+d x))^{1-\frac {m}{2}} (e \sin (c+d x))^{-1+m}}{5 a d \sqrt {a+a \sec (c+d x)}} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(484\) vs. \(2(120)=240\).
Time = 2.32 (sec) , antiderivative size = 484, normalized size of antiderivative = 4.03 \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {4 (3+m) \left (\operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {1}{2} (c+d x)\right ) (e \sin (c+d x))^m}{d (1+m) \left (-4 (3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},1+m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+\left (2 m \operatorname {AppellF1}\left (\frac {3+m}{2},-\frac {1}{2},1+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-4 (1+m) \operatorname {AppellF1}\left (\frac {3+m}{2},-\frac {1}{2},2+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )+\operatorname {AppellF1}\left (\frac {3+m}{2},\frac {1}{2},m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-2 \operatorname {AppellF1}\left (\frac {3+m}{2},\frac {1}{2},1+m,\frac {5+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right ) (-1+\cos (c+d x))+(3+m) \operatorname {AppellF1}\left (\frac {1+m}{2},-\frac {1}{2},m,\frac {3+m}{2},\tan ^2\left (\frac {1}{2} (c+d x)\right ),-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) (1+\cos (c+d x))\right ) (a (1+\sec (c+d x)))^{3/2}} \]
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\[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{\frac {3}{2}}}d x\]
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\[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
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\[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
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Exception generated. \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
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Timed out. \[ \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
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