Integrand size = 34, antiderivative size = 81 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {2^{\frac {3}{2}+m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^{1+m}}{a c^2 f} \]
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Time = 0.15 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2919, 2768, 72, 71} \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {2^{m+\frac {3}{2}} \sec (e+f x) (\sin (e+f x)+1)^{-m-\frac {1}{2}} (a \sin (e+f x)+a)^{m+1} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-m-\frac {1}{2},\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right )}{a c^2 f} \]
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Rule 71
Rule 72
Rule 2768
Rule 2919
Rubi steps \begin{align*} \text {integral}& = \frac {\int \sec ^2(e+f x) (a+a \sin (e+f x))^{2+m} \, dx}{a^2 c^2} \\ & = \frac {\left (\sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}\right ) \text {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{c^2 f} \\ & = \frac {\left (2^{\frac {1}{2}+m} \sec (e+f x) \sqrt {a-a \sin (e+f x)} (a+a \sin (e+f x))^{1+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{-\frac {1}{2}-m}\right ) \text {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {x}{2}\right )^{\frac {1}{2}+m}}{(a-a x)^{3/2}} \, dx,x,\sin (e+f x)\right )}{c^2 f} \\ & = \frac {2^{\frac {3}{2}+m} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) \sec (e+f x) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a+a \sin (e+f x))^{1+m}}{a c^2 f} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {2^{\frac {3}{2}+m} \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{2}-m,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{-\frac {1}{2}-m} (a (1+\sin (e+f x)))^m}{c^2 f (1-\sin (e+f x))} \]
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\[\int \frac {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m}}{\left (c -c \sin \left (f x +e \right )\right )^{2}}d x\]
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\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
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\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\frac {\int \frac {\left (a \sin {\left (e + f x \right )} + a\right )^{m} \cos ^{2}{\left (e + f x \right )}}{\sin ^{2}{\left (e + f x \right )} - 2 \sin {\left (e + f x \right )} + 1}\, dx}{c^{2}} \]
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\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
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\[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}}{{\left (c \sin \left (f x + e\right ) - c\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {\cos ^2(e+f x) (a+a \sin (e+f x))^m}{(c-c \sin (e+f x))^2} \, dx=\int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]
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