Integrand size = 55, antiderivative size = 39 \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d-(c-d) m+(c+(c-d) m) \sin (e+f x)) \, dx=-\frac {\cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{f} \]
[Out]
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {3053} \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d-(c-d) m+(c+(c-d) m) \sin (e+f x)) \, dx=-\frac {\cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{f} \]
[In]
[Out]
Rule 3053
Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{f} \\ \end{align*}
Time = 2.55 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d-(c-d) m+(c+(c-d) m) \sin (e+f x)) \, dx=-\frac {\cos (e+f x) (a (1+\sin (e+f x)))^m (c+d \sin (e+f x))^{-1-m}}{f} \]
[In]
[Out]
\[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{-2-m} \left (d -\left (c -d \right ) m +\left (c +\left (c -d \right ) m \right ) \sin \left (f x +e \right )\right )d x\]
[In]
[Out]
none
Time = 0.32 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.44 \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d-(c-d) m+(c+(c-d) m) \sin (e+f x)) \, dx=-\frac {{\left (d \cos \left (f x + e\right ) \sin \left (f x + e\right ) + c \cos \left (f x + e\right )\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 2}}{f} \]
[In]
[Out]
Timed out. \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d-(c-d) m+(c+(c-d) m) \sin (e+f x)) \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d-(c-d) m+(c+(c-d) m) \sin (e+f x)) \, dx=\int { -{\left ({\left (c - d\right )} m - {\left ({\left (c - d\right )} m + c\right )} \sin \left (f x + e\right ) - d\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 2} \,d x } \]
[In]
[Out]
Timed out. \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d-(c-d) m+(c+(c-d) m) \sin (e+f x)) \, dx=\text {Timed out} \]
[In]
[Out]
Time = 16.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.51 \[ \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} (d-(c-d) m+(c+(c-d) m) \sin (e+f x)) \, dx=-\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,\left (d\,\sin \left (2\,e+2\,f\,x\right )-2\,c\,\left (2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1\right )\right )}{f\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^m\,\left (d^2\,\left (2\,{\sin \left (e+f\,x\right )}^2-1\right )+2\,c^2+d^2+4\,c\,d\,\sin \left (e+f\,x\right )\right )} \]
[In]
[Out]