Integrand size = 40, antiderivative size = 114 \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=-\frac {2^{\frac {9}{4}+m} a^3 c^2 (g \cos (e+f x))^{15/2} \operatorname {Hypergeometric2F1}\left (\frac {15}{4},-\frac {1}{4}-m,\frac {19}{4},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{-\frac {1}{4}-m} (a+a \sin (e+f x))^{-3+m} \sqrt {c-c \sin (e+f x)}}{15 f g^6} \]
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Time = 0.26 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2932, 2768, 72, 71} \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=-\frac {a^3 c^2 2^{m+\frac {9}{4}} \sec (e+f x) \sqrt {c-c \sin (e+f x)} (g \cos (e+f x))^{15/2} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m-3} \operatorname {Hypergeometric2F1}\left (\frac {15}{4},-m-\frac {1}{4},\frac {19}{4},\frac {1}{2} (1-\sin (e+f x))\right )}{15 f g^6} \]
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Rule 71
Rule 72
Rule 2768
Rule 2932
Rubi steps \begin{align*} \text {integral}& = \frac {\left (a^2 c^2 \sec (e+f x) \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}\right ) \int (g \cos (e+f x))^{13/2} (a+a \sin (e+f x))^{-\frac {5}{2}+m} \, dx}{g^5} \\ & = \frac {\left (a^4 c^2 (g \cos (e+f x))^{15/2} \sec (e+f x) \sqrt {c-c \sin (e+f x)}\right ) \text {Subst}\left (\int (a-a x)^{11/4} (a+a x)^{\frac {1}{4}+m} \, dx,x,\sin (e+f x)\right )}{f g^6 (a-a \sin (e+f x))^{15/4} (a+a \sin (e+f x))^{13/4}} \\ & = \frac {\left (2^{\frac {1}{4}+m} a^4 c^2 (g \cos (e+f x))^{15/2} \sec (e+f x) (a+a \sin (e+f x))^{-3+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{-\frac {1}{4}-m} \sqrt {c-c \sin (e+f x)}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {x}{2}\right )^{\frac {1}{4}+m} (a-a x)^{11/4} \, dx,x,\sin (e+f x)\right )}{f g^6 (a-a \sin (e+f x))^{15/4}} \\ & = -\frac {2^{\frac {9}{4}+m} a^3 c^2 (g \cos (e+f x))^{15/2} \operatorname {Hypergeometric2F1}\left (\frac {15}{4},-\frac {1}{4}-m,\frac {19}{4},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{-\frac {1}{4}-m} (a+a \sin (e+f x))^{-3+m} \sqrt {c-c \sin (e+f x)}}{15 f g^6} \\ \end{align*}
\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx \]
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\[\int \left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{\frac {5}{2}}d x\]
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\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\text {Timed out} \]
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\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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\[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \,d x } \]
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Timed out. \[ \int (g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{5/2} \, dx=\int {\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]
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