Integrand size = 38, antiderivative size = 93 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {2^{\frac {9}{4}+m} g^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{4}-m,\frac {1}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{4}-m} (a+a \sin (e+f x))^{1+m}}{3 a c^2 f (g \cos (e+f x))^{3/2}} \]
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Time = 0.20 (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} \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {g^3 2^{m+\frac {9}{4}} (\sin (e+f x)+1)^{-m-\frac {1}{4}} (a \sin (e+f x)+a)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-m-\frac {1}{4},\frac {1}{4},\frac {1}{2} (1-\sin (e+f x))\right )}{3 a c^2 f (g \cos (e+f x))^{3/2}} \]
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Rule 71
Rule 72
Rule 2768
Rule 2919
Rubi steps \begin{align*} \text {integral}& = \frac {g^4 \int \frac {(a+a \sin (e+f x))^{2+m}}{(g \cos (e+f x))^{5/2}} \, dx}{a^2 c^2} \\ & = \frac {\left (g^3 (a-a \sin (e+f x))^{3/4} (a+a \sin (e+f x))^{3/4}\right ) \text {Subst}\left (\int \frac {(a+a x)^{\frac {1}{4}+m}}{(a-a x)^{7/4}} \, dx,x,\sin (e+f x)\right )}{c^2 f (g \cos (e+f x))^{3/2}} \\ & = \frac {\left (2^{\frac {1}{4}+m} g^3 (a-a \sin (e+f x))^{3/4} (a+a \sin (e+f x))^{1+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{-\frac {1}{4}-m}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{\frac {1}{4}+m}}{(a-a x)^{7/4}} \, dx,x,\sin (e+f x)\right )}{c^2 f (g \cos (e+f x))^{3/2}} \\ & = \frac {2^{\frac {9}{4}+m} g^3 \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{4}-m,\frac {1}{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))^{1+m}}{3 a c^2 f (g \cos (e+f x))^{3/2}} \\ \end{align*}
Time = 0.16 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03 \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=-\frac {2^{\frac {9}{4}+m} g \sqrt {g \cos (e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{4}-m,\frac {1}{4},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{4}-m} (a (1+\sin (e+f x)))^m}{3 c^2 f (-1+\sin (e+f x))} \]
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\[\int \frac {\left (g \cos \left (f x +e \right )\right )^{\frac {3}{2}} \left (a +a \sin \left (f x +e \right )\right )^{m}}{\left (c -c \sin \left (f x +e \right )\right )^{2}}d x\]
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\[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\text {Timed out} \]
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\[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
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\[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2} (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]
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