Integrand size = 38, antiderivative size = 93 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=-\frac {2^{\frac {9}{4}+m} a^3 c^2 (g \cos (e+f x))^{13/2} \operatorname {Hypergeometric2F1}\left (\frac {13}{4},-\frac {1}{4}-m,\frac {17}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{4}-m} (a+a \sin (e+f x))^{-3+m}}{13 f g^5} \]
[Out]
Time = 0.18 (sec) , antiderivative size = 93, 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 = {2919, 2768, 72, 71} \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=-\frac {a^3 c^2 2^{m+\frac {9}{4}} (g \cos (e+f x))^{13/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-3} \operatorname {Hypergeometric2F1}\left (\frac {13}{4},-m-\frac {1}{4},\frac {17}{4},\frac {1}{2} (1-\sin (e+f x))\right )}{13 f g^5} \]
[In]
[Out]
Rule 71
Rule 72
Rule 2768
Rule 2919
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2 c^2\right ) \int (g \cos (e+f x))^{11/2} (a+a \sin (e+f x))^{-2+m} \, dx}{g^4} \\ & = \frac {\left (a^4 c^2 (g \cos (e+f x))^{13/2}\right ) \text {Subst}\left (\int (a-a x)^{9/4} (a+a x)^{\frac {1}{4}+m} \, dx,x,\sin (e+f x)\right )}{f g^5 (a-a \sin (e+f x))^{13/4} (a+a \sin (e+f x))^{13/4}} \\ & = \frac {\left (2^{\frac {1}{4}+m} a^4 c^2 (g \cos (e+f x))^{13/2} (a+a \sin (e+f x))^{-3+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{-\frac {1}{4}-m}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {x}{2}\right )^{\frac {1}{4}+m} (a-a x)^{9/4} \, dx,x,\sin (e+f x)\right )}{f g^5 (a-a \sin (e+f x))^{13/4}} \\ & = -\frac {2^{\frac {9}{4}+m} a^3 c^2 (g \cos (e+f x))^{13/2} \operatorname {Hypergeometric2F1}\left (\frac {13}{4},-\frac {1}{4}-m,\frac {17}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{4}-m} (a+a \sin (e+f x))^{-3+m}}{13 f g^5} \\ \end{align*}
Time = 0.52 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\frac {2^{\frac {9}{4}+m} c^2 g \sqrt {g \cos (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {13}{4},-\frac {1}{4}-m,\frac {17}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (-1+\sin (e+f x))^3 (1+\sin (e+f x))^{-\frac {1}{4}-m} (a (1+\sin (e+f x)))^m}{13 f} \]
[In]
[Out]
\[\int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{2}d x\]
[In]
[Out]
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (c \sin \left (f x + e\right ) - c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
[In]
[Out]
Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (c \sin \left (f x + e\right ) - c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
[In]
[Out]
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (c \sin \left (f x + e\right ) - c\right )}^{2} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
[In]
[Out]
Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^2 \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2 \,d x \]
[In]
[Out]