Integrand size = 36, antiderivative size = 123 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {2^{\frac {1}{2}+m} (B+A m+B m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^m}{c f m}-\frac {B \sec (e+f x) (a+a \sin (e+f x))^{1+m}}{a c f m} \]
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Time = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3046, 2939, 2768, 72, 71} \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\frac {2^{m+\frac {1}{2}} (A m+B m+B) \sec (e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{c f m}-\frac {B \sec (e+f x) (a \sin (e+f x)+a)^{m+1}}{a c f m} \]
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Rule 71
Rule 72
Rule 2768
Rule 2939
Rule 3046
Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^2(e+f x) (a+a \sin (e+f x))^{1+m} (A+B \sin (e+f x)) \, dx}{a c} \\ & = -\frac {B \sec (e+f x) (a+a \sin (e+f x))^{1+m}}{a c f m}+\frac {(B+A m+B m) \int \sec ^2(e+f x) (a+a \sin (e+f x))^{1+m} \, dx}{a c m} \\ & = -\frac {B \sec (e+f x) (a+a \sin (e+f x))^{1+m}}{a c f m}+\frac {\left (a (B+A m+B m) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{c f m} \\ & = -\frac {B \sec (e+f x) (a+a \sin (e+f x))^{1+m}}{a c f m}+\frac {\left (2^{-\frac {1}{2}+m} a (B+A m+B m) \sec (e+f x) \sqrt {a-a \sin (e+f x)} (a+a \sin (e+f x))^m \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {1}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{c f m} \\ & = \frac {2^{\frac {1}{2}+m} (B+A m+B m) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {1}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^m}{c f m}-\frac {B \sec (e+f x) (a+a \sin (e+f x))^{1+m}}{a c f m} \\ \end{align*}
\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx \]
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\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{c -c \sin \left (f x +e \right )}d x\]
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\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\int { -\frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c \sin \left (f x + e\right ) - c} \,d x } \]
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\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=- \frac {\int \frac {A \left (a \sin {\left (e + f x \right )} + a\right )^{m}}{\sin {\left (e + f x \right )} - 1}\, dx + \int \frac {B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}}{\sin {\left (e + f x \right )} - 1}\, dx}{c} \]
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\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\int { -\frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c \sin \left (f x + e\right ) - c} \,d x } \]
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\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\int { -\frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{c \sin \left (f x + e\right ) - c} \,d x } \]
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Timed out. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{c-c \sin (e+f x)} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{c-c\,\sin \left (e+f\,x\right )} \,d x \]
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