Integrand size = 29, antiderivative size = 117 \[ \int (-3+2 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=-\frac {\cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,1-m,\frac {2 (3-2 \sin (e+f x))}{1+\sin (e+f x)}\right ) \sqrt {-\frac {1-\sin (e+f x)}{1+\sin (e+f x)}} (-3+2 \sin (e+f x))^{-m} (3+3 \sin (e+f x))^m}{\sqrt {5} f m (1-\sin (e+f x))} \]
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Time = 0.07 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2867, 134} \[ \int (-3+2 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=-\frac {\sqrt {-\frac {1-\sin (e+f x)}{\sin (e+f x)+1}} \cos (e+f x) (2 \sin (e+f x)-3)^{-m} (a \sin (e+f x)+a)^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,1-m,\frac {2 (3-2 \sin (e+f x))}{\sin (e+f x)+1}\right )}{\sqrt {5} f m (1-\sin (e+f x))} \]
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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+2 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 {2 (3-2 \sin (e+f x))}{1+\sin (e+f x)}\right ) \sqrt {-\frac {1-\sin (e+f x)}{1+\sin (e+f x)}} (-3+2 \sin (e+f x))^{-m} (a+a \sin (e+f x))^m}{\sqrt {5} f m (1-\sin (e+f x))} \\ \end{align*}
Result contains complex when optimal does not.
Time = 9.23 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.26 \[ \int (-3+2 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=-\frac {\operatorname {Hypergeometric2F1}\left (1+m,1+2 m,2 (1+m),\frac {4 i \sqrt {5} (\cos (e+f x)+i (1+\sin (e+f x)))}{\left (-5+\sqrt {5}\right ) \left (3+\sqrt {5}+2 i \cos (e+f x)-2 \sin (e+f x)\right )}\right ) \left (\frac {9-3 \sqrt {5}+6 i \cos (e+f x)-6 \sin (e+f x)}{3+\sqrt {5}+2 i \cos (e+f x)-2 \sin (e+f x)}\right )^m (\cos (e+f x)-i \sin (e+f x)) (1+\sin (e+f x))^m (1-i \cos (e+f x)+\sin (e+f x)) \left (-\left (\left (-5+\sqrt {5}\right ) (-3+2 \sin (e+f x))\right )\right )^{-1-m} \left (-3+\sqrt {5}-2 i \cos (e+f x)+2 \sin (e+f x)\right ) \left (\cosh \left (m \log \left (5+\sqrt {5}\right )\right )+\sinh \left (m \log \left (5+\sqrt {5}\right )\right )\right )}{f (1+2 m)} \]
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\[\int \left (-3+2 \sin \left (f x +e \right )\right )^{-1-m} \left (a +a \sin \left (f x +e \right )\right )^{m}d x\]
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\[ \int (-3+2 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (2 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
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\[ \int (-3+2 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (2 \sin {\left (e + f x \right )} - 3\right )^{- m - 1}\, dx \]
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\[ \int (-3+2 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (2 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
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\[ \int (-3+2 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int { {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (2 \, \sin \left (f x + e\right ) - 3\right )}^{-m - 1} \,d x } \]
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Timed out. \[ \int (-3+2 \sin (e+f x))^{-1-m} (3+3 \sin (e+f x))^m \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (2\,\sin \left (e+f\,x\right )-3\right )}^{m+1}} \,d x \]
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