Integrand size = 38, antiderivative size = 118 \[ \int \frac {(A+B \sin (e+f x)) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=-\frac {2 B \cos (e+f x) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \sin (e+f x)}}+\frac {(A+B) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \sin (e+f x)}} \]
[Out]
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} \[ \int \frac {(A+B \sin (e+f x)) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\frac {(A+B) \cos (e+f x) (c \sin (e+f x)+c)^m \operatorname {Hypergeometric2F1}\left (1,m+\frac {1}{2},m+\frac {3}{2},\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+1) \sqrt {a-a \sin (e+f x)}}-\frac {2 B \cos (e+f x) (c \sin (e+f x)+c)^m}{f (2 m+1) \sqrt {a-a \sin (e+f x)}} \]
[In]
[Out]
Rule 70
Rule 2746
Rule 2824
Rule 3052
Rubi steps \begin{align*} \text {integral}& = -\frac {2 B \cos (e+f x) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \sin (e+f x)}}-(-A-B) \int \frac {(c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx \\ & = -\frac {2 B \cos (e+f x) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \sin (e+f x)}}-\frac {((-A-B) \cos (e+f x)) \int \sec (e+f x) (c+c \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) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \sin (e+f x)}}-\frac {((-A-B) c \cos (e+f x)) \text {Subst}\left (\int \frac {(c+x)^{-\frac {1}{2}+m}}{c-x} \, dx,x,c \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) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \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 ) (c+c \sin (e+f x))^m}{f (1+2 m) \sqrt {a-a \sin (e+f x)}} \\ \end{align*}
Time = 21.97 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B \sin (e+f x)) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\frac {\cos (e+f x) (c (1+\sin (e+f x)))^m \left (2 A (3+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}+m,\frac {3}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right )+B \left (-6-4 m+(1+2 m) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{2}+m,\frac {5}{2}+m,\frac {1}{2} (1+\sin (e+f x))\right ) (1+\sin (e+f x))\right )\right )}{2 f (1+2 m) (3+2 m) \sqrt {a-a \sin (e+f x)}} \]
[In]
[Out]
\[\int \frac {\left (A +B \sin \left (f x +e \right )\right ) \left (c +c \sin \left (f x +e \right )\right )^{m}}{\sqrt {a -a \sin \left (f x +e \right )}}d x\]
[In]
[Out]
\[ \int \frac {(A+B \sin (e+f x)) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (c \sin \left (f x + e\right ) + c\right )}^{m}}{\sqrt {-a \sin \left (f x + e\right ) + a}} \,d x } \]
[In]
[Out]
\[ \int \frac {(A+B \sin (e+f x)) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\int \frac {\left (c \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + B \sin {\left (e + f x \right )}\right )}{\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )}}\, dx \]
[In]
[Out]
\[ \int \frac {(A+B \sin (e+f x)) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (c \sin \left (f x + e\right ) + c\right )}^{m}}{\sqrt {-a \sin \left (f x + e\right ) + a}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\text {Timed out} \]
[In]
[Out]
Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c+c \sin (e+f x))^m}{\sqrt {a-a \sin (e+f x)}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c+c\,\sin \left (e+f\,x\right )\right )}^m}{\sqrt {a-a\,\sin \left (e+f\,x\right )}} \,d x \]
[In]
[Out]