Integrand size = 23, antiderivative size = 131 \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}+\frac {(a-2 b (1+p)) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b \cos ^2(e+f x)}{a+b}\right )}{b f (3+2 p)} \]
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Time = 0.08 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3265, 396, 252, 251} \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {(a-2 b (p+1)) \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b \cos ^2(e+f x)}{a+b}\right )}{b f (2 p+3)}-\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{p+1}}{b f (2 p+3)} \]
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Rule 251
Rule 252
Rule 396
Rule 3265
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \left (1-x^2\right ) \left (a+b-b x^2\right )^p \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}+\frac {(a-2 b (1+p)) \text {Subst}\left (\int \left (a+b-b x^2\right )^p \, dx,x,\cos (e+f x)\right )}{b f (3+2 p)} \\ & = -\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}+\frac {\left ((a-2 b (1+p)) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p}\right ) \text {Subst}\left (\int \left (1-\frac {b x^2}{a+b}\right )^p \, dx,x,\cos (e+f x)\right )}{b f (3+2 p)} \\ & = -\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{1+p}}{b f (3+2 p)}+\frac {(a-2 b (1+p)) \cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^p \left (1-\frac {b \cos ^2(e+f x)}{a+b}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},\frac {b \cos ^2(e+f x)}{a+b}\right )}{b f (3+2 p)} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.41 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (2,\frac {1}{2},-p,3,\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) \sqrt {\cos ^2(e+f x)} \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {a+b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{4 f} \]
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\[\int \left (\sin ^{3}\left (f x +e \right )\right ) {\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}^{p}d x\]
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\[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]
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Timed out. \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
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\[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]
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\[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]
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Timed out. \[ \int \sin ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^p \, dx=\int {\sin \left (e+f\,x\right )}^3\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]
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