Integrand size = 38, antiderivative size = 118 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=-\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}} \]
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Time = 0.20 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3052, 2824, 2746, 70} \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (1,m+\frac {1}{2},m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) \sqrt {c-c \sin (e+f x)}}-\frac {2 B \cos (e+f x) (a \sin (e+f x)+a)^m}{f (2 m+1) \sqrt {c-c \sin (e+f x)}} \]
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Rule 70
Rule 2746
Rule 2824
Rule 3052
Rubi steps \begin{align*} \text {integral}& = -\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+(A+B) \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c-c \sin (e+f x)}} \, dx \\ & = -\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {((A+B) \cos (e+f x)) \int \sec (e+f x) (a+a \sin (e+f x))^{\frac {1}{2}+m} \, dx}{\sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = -\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {(a (A+B) \cos (e+f x)) \text {Subst}\left (\int \frac {(a+x)^{-\frac {1}{2}+m}}{a-x} \, dx,x,a \sin (e+f x)\right )}{f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \\ & = -\frac {2 B \cos (e+f x) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (a+a \sin (e+f x))^m}{f (1+2 m) \sqrt {c-c \sin (e+f x)}} \\ \end{align*}
Time = 21.88 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.12 \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\frac {\cos (e+f x) (a (1+\sin (e+f x)))^m \left (2 A (3+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right )+B \left (-6-4 m+(1+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{2}+m,\frac {5}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (1+\sin (e+f x))\right )\right )}{2 f (1+2 m) (3+2 m) \sqrt {c-c \sin (e+f x)}} \]
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\[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right )}{\sqrt {c -c \sin \left (f x +e \right )}}d x\]
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\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]
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\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right )}{\sqrt {- c \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \]
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\[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{\sqrt {-c \sin \left (f x + e\right ) + c}} \,d x } \]
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Timed out. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(a+a \sin (e+f x))^m (A+B \sin (e+f x))}{\sqrt {c-c \sin (e+f x)}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]
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