Integrand size = 23, antiderivative size = 100 \[ \int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx=-\frac {e (e \sin (c+d x))^{-1+m}}{a d (1-m)}+\frac {e \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-1+m}}{a d (1-m) \sqrt {\cos ^2(c+d x)}} \]
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Time = 0.24 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3957, 2918, 2644, 30, 2657} \[ \int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx=\frac {e \cos (c+d x) (e \sin (c+d x))^{m-1} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {m-1}{2},\frac {m+1}{2},\sin ^2(c+d x)\right )}{a d (1-m) \sqrt {\cos ^2(c+d x)}}-\frac {e (e \sin (c+d x))^{m-1}}{a d (1-m)} \]
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Rule 30
Rule 2644
Rule 2657
Rule 2918
Rule 3957
Rubi steps \begin{align*} \text {integral}& = -\int \frac {\cos (c+d x) (e \sin (c+d x))^m}{-a-a \cos (c+d x)} \, dx \\ & = \frac {e^2 \int \cos (c+d x) (e \sin (c+d x))^{-2+m} \, dx}{a}-\frac {e^2 \int \cos ^2(c+d x) (e \sin (c+d x))^{-2+m} \, dx}{a} \\ & = \frac {e \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-1+m}}{a d (1-m) \sqrt {\cos ^2(c+d x)}}+\frac {e \text {Subst}\left (\int x^{-2+m} \, dx,x,e \sin (c+d x)\right )}{a d} \\ & = -\frac {e (e \sin (c+d x))^{-1+m}}{a d (1-m)}+\frac {e \cos (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-1+m}}{a d (1-m) \sqrt {\cos ^2(c+d x)}} \\ \end{align*}
Result contains complex when optimal does not.
Time = 16.38 (sec) , antiderivative size = 383, normalized size of antiderivative = 3.83 \[ \int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx=\frac {i 2^{1-m} \left (1-e^{2 i (c+d x)}\right )^{-m} \left (-i e^{-i (c+d x)} \left (-1+e^{2 i (c+d x)}\right )\right )^m \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (2 e^{i (c+d x)} (-2+m) m \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},1-m,\frac {3-m}{2},e^{2 i (c+d x)}\right )-4 e^{i (c+d x)} (-2+m) m \operatorname {Hypergeometric2F1}\left (\frac {1-m}{2},2-m,\frac {3-m}{2},e^{2 i (c+d x)}\right )-(-1+m) \left (2 (-2+m) \operatorname {Hypergeometric2F1}\left (1-m,-\frac {m}{2},1-\frac {m}{2},e^{2 i (c+d x)}\right )-2 e^{2 i (c+d x)} m \operatorname {Hypergeometric2F1}\left (2-m,1-\frac {m}{2},2-\frac {m}{2},e^{2 i (c+d x)}\right )-(-2+m) \left (2 \operatorname {Hypergeometric2F1}\left (2-m,-\frac {m}{2},1-\frac {m}{2},e^{2 i (c+d x)}\right )+\operatorname {Hypergeometric2F1}\left (-m,-\frac {m}{2},1-\frac {m}{2},e^{2 i (c+d x)}\right )\right )\right )\right ) \sec (c+d x) \sin ^{-m}(c+d x) (e \sin (c+d x))^m}{d (-2+m) (-1+m) m (a+a \sec (c+d x))} \]
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\[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{a +a \sec \left (d x +c \right )}d x\]
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\[ \int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{a \sec \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]
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\[ \int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{a \sec \left (d x + c\right ) + a} \,d x } \]
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\[ \int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx=\int { \frac {\left (e \sin \left (d x + c\right )\right )^{m}}{a \sec \left (d x + c\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx=\int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \]
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