Integrand size = 27, antiderivative size = 116 \[ \int (-3+\sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=-\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,1-m,\frac {3-\sin (e+f x)}{1+\sin (e+f x)}\right ) (-3+\sin (e+f x))^{-m} \sqrt {-\frac {1-\sin (e+f x)}{1+\sin (e+f x)}} (3+3 \sin (e+f x))^m}{2 \sqrt {2} f m (1-\sin (e+f x))} \]
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Time = 0.07 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2867, 134} \[ \int (-3+\sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=-\frac {\sqrt {-\frac {1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (\sin (e+f x)-3)^{-m} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,1-m,\frac {3-\sin (e+f x)}{\sin (e+f x)+1}\right )}{2 \sqrt {2} f m (1-\sin (e+f x))} \]
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Rule 134
Rule 2867
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2 \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(-3+x)^{-1-m} (a+a x)^{-\frac {1}{2}+m}}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}} \\ & = -\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,1-m,\frac {3-\sin (e+f x)}{1+\sin (e+f x)}\right ) (-3+\sin (e+f x))^{-m} \sqrt {-\frac {1-\sin (e+f x)}{1+\sin (e+f x)}} (a+a \sin (e+f x))^m}{2 \sqrt {2} f m (1-\sin (e+f x))} \\ \end{align*}
Result contains complex when optimal does not.
Time = 10.61 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.33 \[ \int (-3+\sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\frac {\operatorname {Hypergeometric2F1}\left (1+m,1+2 m,2 (1+m),\frac {2 i \sqrt {2} (\cos (e+f x)+i (1+\sin (e+f x)))}{\left (-2+\sqrt {2}\right ) \left (3+2 \sqrt {2}+i \cos (e+f x)-\sin (e+f x)\right )}\right ) \left (\frac {9-6 \sqrt {2}+3 i \cos (e+f x)-3 \sin (e+f x)}{3+2 \sqrt {2}+i \cos (e+f x)-\sin (e+f x)}\right )^m \left (-\left (\left (-2+\sqrt {2}\right ) (-3+\sin (e+f x))\right )\right )^{-1-m} (1+\sin (e+f x))^m \left (-3+2 \sqrt {2}-i \cos (e+f x)+\sin (e+f x)\right ) (i \cos (e+f x)+\sin (e+f x)) (\cos (e+f x)+i (1+\sin (e+f x))) \left (\cosh \left (m \log \left (2+\sqrt {2}\right )\right )+\sinh \left (m \log \left (2+\sqrt {2}\right )\right )\right )}{2 f (1+2 m)} \]
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\[\int \left (-3+\sin \left (f x +e \right )\right )^{-1-m} \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]
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\[ \int (-3+\sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (\sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
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\[ \int (-3+\sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (\sin {\left (e + f x \right )} - 3\right )^{- m - 1}\, dx \]
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\[ \int (-3+\sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (\sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
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\[ \int (-3+\sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (\sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
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Timed out. \[ \int (-3+\sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (\sin \left (e+f\,x\right )-3\right )}^{m+1}} \,d x \]
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