Integrand size = 27, antiderivative size = 195 \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{3+b \sin (e+f x)} \, dx=\frac {b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {n p}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n}{\left (9-b^2\right ) f}-\frac {3 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (1-n p),1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{9-b^2}\right ) \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (9-b^2\right ) f} \]
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Time = 0.24 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2905, 2902, 3268, 440} \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{3+b \sin (e+f x)} \, dx=\frac {b \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n \operatorname {AppellF1}\left (\frac {1}{2},-\frac {n p}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )}-\frac {a \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (1-n p),1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )} \]
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Rule 440
Rule 2902
Rule 2905
Rule 3268
Rubi steps \begin{align*} \text {integral}& = \left ((d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int \frac {(d \sin (e+f x))^{n p}}{a+b \sin (e+f x)} \, dx \\ & = \left (a (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int \frac {(d \sin (e+f x))^{n p}}{a^2-b^2 \sin ^2(e+f x)} \, dx-\frac {\left (b (d \sin (e+f x))^{-n p} \left (c (d \sin (e+f x))^p\right )^n\right ) \int \frac {(d \sin (e+f x))^{1+n p}}{a^2-b^2 \sin ^2(e+f x)} \, dx}{d} \\ & = \frac {\left (b \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {n p}{2}}}{a^2-b^2+b^2 x^2} \, dx,x,\cos (e+f x)\right )}{f}-\frac {\left (a d (d \sin (e+f x))^{-n p+2 \left (-\frac {1}{2}+\frac {n p}{2}\right )} \sin ^2(e+f x)^{\frac {1}{2}-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n\right ) \text {Subst}\left (\int \frac {\left (1-x^2\right )^{\frac {1}{2} (-1+n p)}}{a^2-b^2+b^2 x^2} \, dx,x,\cos (e+f x)\right )}{f} \\ & = \frac {b \operatorname {AppellF1}\left (\frac {1}{2},-\frac {n p}{2},1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) \sin ^2(e+f x)^{-\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right ) f}-\frac {a \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{2} (1-n p),1,\frac {3}{2},\cos ^2(e+f x),-\frac {b^2 \cos ^2(e+f x)}{a^2-b^2}\right ) \cot (e+f x) \sin ^2(e+f x)^{\frac {1}{2} (1-n p)} \left (c (d \sin (e+f x))^p\right )^n}{\left (a^2-b^2\right ) f} \\ \end{align*}
Leaf count is larger than twice the leaf count of optimal. \(1730\) vs. \(2(195)=390\).
Time = 16.81 (sec) , antiderivative size = 1730, normalized size of antiderivative = 8.87 \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{3+b \sin (e+f x)} \, dx=-\frac {\sec ^2(e+f x)^{\frac {n p}{2}} \left (c (d \sin (e+f x))^p\right )^n \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (-3 b (2+n p) \operatorname {AppellF1}\left (\frac {1}{2} (1+n p),\frac {n p}{2},1,\frac {1}{2} (3+n p),-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right )+\left (-9+b^2\right ) (1+n p) \operatorname {AppellF1}\left (1+\frac {n p}{2},\frac {1}{2} (-1+n p),1,2+\frac {n p}{2},-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right ) \tan (e+f x)+9 (1+n p) \operatorname {Hypergeometric2F1}\left (1+\frac {n p}{2},\frac {1}{2} (1+n p),2+\frac {n p}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)\right )}{9 b f (1+n p) (2+n p) (3+b \sin (e+f x)) \left (-\frac {\sec ^2(e+f x)^{1+\frac {n p}{2}} \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (-3 b (2+n p) \operatorname {AppellF1}\left (\frac {1}{2} (1+n p),\frac {n p}{2},1,\frac {1}{2} (3+n p),-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right )+\left (-9+b^2\right ) (1+n p) \operatorname {AppellF1}\left (1+\frac {n p}{2},\frac {1}{2} (-1+n p),1,2+\frac {n p}{2},-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right ) \tan (e+f x)+9 (1+n p) \operatorname {Hypergeometric2F1}\left (1+\frac {n p}{2},\frac {1}{2} (1+n p),2+\frac {n p}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)\right )}{9 b (1+n p) (2+n p)}-\frac {n p \sec ^2(e+f x)^{\frac {n p}{2}} \tan ^2(e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (-3 b (2+n p) \operatorname {AppellF1}\left (\frac {1}{2} (1+n p),\frac {n p}{2},1,\frac {1}{2} (3+n p),-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right )+\left (-9+b^2\right ) (1+n p) \operatorname {AppellF1}\left (1+\frac {n p}{2},\frac {1}{2} (-1+n p),1,2+\frac {n p}{2},-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right ) \tan (e+f x)+9 (1+n p) \operatorname {Hypergeometric2F1}\left (1+\frac {n p}{2},\frac {1}{2} (1+n p),2+\frac {n p}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)\right )}{9 b (1+n p) (2+n p)}-\frac {n p \sec ^2(e+f x)^{\frac {n p}{2}} \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{-1+n p} \left (-3 b (2+n p) \operatorname {AppellF1}\left (\frac {1}{2} (1+n p),\frac {n p}{2},1,\frac {1}{2} (3+n p),-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right )+\left (-9+b^2\right ) (1+n p) \operatorname {AppellF1}\left (1+\frac {n p}{2},\frac {1}{2} (-1+n p),1,2+\frac {n p}{2},-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right ) \tan (e+f x)+9 (1+n p) \operatorname {Hypergeometric2F1}\left (1+\frac {n p}{2},\frac {1}{2} (1+n p),2+\frac {n p}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)\right ) \left (\sqrt {\sec ^2(e+f x)}-\frac {\tan ^2(e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )}{9 b (1+n p) (2+n p)}-\frac {\sec ^2(e+f x)^{\frac {n p}{2}} \tan (e+f x) \left (\frac {\tan (e+f x)}{\sqrt {\sec ^2(e+f x)}}\right )^{n p} \left (\left (-9+b^2\right ) (1+n p) \operatorname {AppellF1}\left (1+\frac {n p}{2},\frac {1}{2} (-1+n p),1,2+\frac {n p}{2},-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x)+9 (1+n p) \operatorname {Hypergeometric2F1}\left (1+\frac {n p}{2},\frac {1}{2} (1+n p),2+\frac {n p}{2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)+\left (-9+b^2\right ) (1+n p) \tan (e+f x) \left (\frac {2 \left (-9+b^2\right ) \left (1+\frac {n p}{2}\right ) \operatorname {AppellF1}\left (2+\frac {n p}{2},\frac {1}{2} (-1+n p),2,3+\frac {n p}{2},-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{9 \left (2+\frac {n p}{2}\right )}-\frac {\left (1+\frac {n p}{2}\right ) (-1+n p) \operatorname {AppellF1}\left (2+\frac {n p}{2},1+\frac {1}{2} (-1+n p),1,3+\frac {n p}{2},-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{2+\frac {n p}{2}}\right )-3 b (2+n p) \left (\frac {2 \left (-9+b^2\right ) (1+n p) \operatorname {AppellF1}\left (1+\frac {1}{2} (1+n p),\frac {n p}{2},2,1+\frac {1}{2} (3+n p),-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{9 (3+n p)}-\frac {n p (1+n p) \operatorname {AppellF1}\left (1+\frac {1}{2} (1+n p),1+\frac {n p}{2},1,1+\frac {1}{2} (3+n p),-\tan ^2(e+f x),\frac {1}{9} \left (-9+b^2\right ) \tan ^2(e+f x)\right ) \sec ^2(e+f x) \tan (e+f x)}{3+n p}\right )+18 \left (1+\frac {n p}{2}\right ) (1+n p) \sec ^2(e+f x) \left (-\operatorname {Hypergeometric2F1}\left (1+\frac {n p}{2},\frac {1}{2} (1+n p),2+\frac {n p}{2},-\tan ^2(e+f x)\right )+\left (1+\tan ^2(e+f x)\right )^{\frac {1}{2} (-1-n p)}\right )\right )}{9 b (1+n p) (2+n p)}\right )} \]
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\[\int \frac {\left (c \left (d \sin \left (f x +e \right )\right )^{p}\right )^{n}}{a +b \sin \left (f x +e \right )}d x\]
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\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{3+b \sin (e+f x)} \, dx=\int { \frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{b \sin \left (f x + e\right ) + a} \,d x } \]
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\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{3+b \sin (e+f x)} \, dx=\int \frac {\left (c \left (d \sin {\left (e + f x \right )}\right )^{p}\right )^{n}}{a + b \sin {\left (e + f x \right )}}\, dx \]
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\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{3+b \sin (e+f x)} \, dx=\int { \frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{b \sin \left (f x + e\right ) + a} \,d x } \]
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\[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{3+b \sin (e+f x)} \, dx=\int { \frac {\left (\left (d \sin \left (f x + e\right )\right )^{p} c\right )^{n}}{b \sin \left (f x + e\right ) + a} \,d x } \]
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Timed out. \[ \int \frac {\left (c (d \sin (e+f x))^p\right )^n}{3+b \sin (e+f x)} \, dx=\int \frac {{\left (c\,{\left (d\,\sin \left (e+f\,x\right )\right )}^p\right )}^n}{a+b\,\sin \left (e+f\,x\right )} \,d x \]
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