Integrand size = 27, antiderivative size = 122 \[ \int (1+\sin (e+f x))^m (3+2 \sin (e+f x))^{-1-m} \, dx=-\frac {2^{\frac {1}{2}+m} 5^{-\frac {1}{2}-m} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1-\sin (e+f x)}{2 (3+2 \sin (e+f x))}\right ) (1+\sin (e+f x))^{-1+m} \left (\frac {1+\sin (e+f x)}{3+2 \sin (e+f x)}\right )^{\frac {1}{2}-m} (3+2 \sin (e+f x))^{-m}}{f} \]
[Out]
Time = 0.08 (sec) , antiderivative size = 122, 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} \[ \int (1+\sin (e+f x))^m (3+2 \sin (e+f x))^{-1-m} \, dx=-\frac {2^{m+\frac {1}{2}} 5^{-m-\frac {1}{2}} \cos (e+f x) (\sin (e+f x)+1)^{m-1} \left (\frac {\sin (e+f x)+1}{2 \sin (e+f x)+3}\right )^{\frac {1}{2}-m} (2 \sin (e+f x)+3)^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1-\sin (e+f x)}{2 (2 \sin (e+f x)+3)}\right )}{f} \]
[In]
[Out]
Rule 134
Rule 2867
Rubi steps \begin{align*} \text {integral}& = \frac {\cos (e+f x) \text {Subst}\left (\int \frac {(1+x)^{-\frac {1}{2}+m} (3+2 x)^{-1-m}}{\sqrt {1-x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}} \\ & = -\frac {2^{\frac {1}{2}+m} 5^{-\frac {1}{2}-m} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2}-m,\frac {3}{2},\frac {1-\sin (e+f x)}{2 (3+2 \sin (e+f x))}\right ) (1+\sin (e+f x))^{-1+m} \left (\frac {1+\sin (e+f x)}{3+2 \sin (e+f x)}\right )^{\frac {1}{2}-m} (3+2 \sin (e+f x))^{-m}}{f} \\ \end{align*}
Result contains complex when optimal does not.
Time = 7.88 (sec) , antiderivative size = 252, normalized size of antiderivative = 2.07 \[ \int (1+\sin (e+f x))^m (3+2 \sin (e+f x))^{-1-m} \, dx=\frac {2^{-m} \operatorname {Hypergeometric2F1}\left (1+m,1+2 m,2 (1+m),\frac {4 i \sqrt {5} (\cos (e+f x)+i (1+\sin (e+f x)))}{\left (1+\sqrt {5}\right ) \left (-3+\sqrt {5}+2 i \cos (e+f x)-2 \sin (e+f x)\right )}\right ) (\cos (e+f x)-i \sin (e+f x)) (1+\sin (e+f x))^m (1-i \cos (e+f x)+\sin (e+f x)) (3+2 \sin (e+f x))^{-1-m} \left (3+\sqrt {5}-2 i \cos (e+f x)+2 \sin (e+f x)\right ) \left (-\frac {\left (-3+\sqrt {5}\right ) \left (3+\sqrt {5}-2 i \cos (e+f x)+2 \sin (e+f x)\right )}{-3+\sqrt {5}+2 i \cos (e+f x)-2 \sin (e+f x)}\right )^m}{\left (1+\sqrt {5}\right ) f (1+2 m)} \]
[In]
[Out]
\[\int \left (\sin \left (f x +e \right )+1\right )^{m} \left (3+2 \sin \left (f x +e \right )\right )^{-1-m}d x\]
[In]
[Out]
\[ \int (1+\sin (e+f x))^m (3+2 \sin (e+f x))^{-1-m} \, dx=\int { {\left (2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} {\left (\sin \left (f x + e\right ) + 1\right )}^{m} \,d x } \]
[In]
[Out]
\[ \int (1+\sin (e+f x))^m (3+2 \sin (e+f x))^{-1-m} \, dx=\int \left (\sin {\left (e + f x \right )} + 1\right )^{m} \left (2 \sin {\left (e + f x \right )} + 3\right )^{- m - 1}\, dx \]
[In]
[Out]
\[ \int (1+\sin (e+f x))^m (3+2 \sin (e+f x))^{-1-m} \, dx=\int { {\left (2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} {\left (\sin \left (f x + e\right ) + 1\right )}^{m} \,d x } \]
[In]
[Out]
\[ \int (1+\sin (e+f x))^m (3+2 \sin (e+f x))^{-1-m} \, dx=\int { {\left (2 \, \sin \left (f x + e\right ) + 3\right )}^{-m - 1} {\left (\sin \left (f x + e\right ) + 1\right )}^{m} \,d x } \]
[In]
[Out]
Timed out. \[ \int (1+\sin (e+f x))^m (3+2 \sin (e+f x))^{-1-m} \, dx=\int \frac {{\left (\sin \left (e+f\,x\right )+1\right )}^m}{{\left (2\,\sin \left (e+f\,x\right )+3\right )}^{m+1}} \,d x \]
[In]
[Out]