Integrand size = 23, antiderivative size = 117 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{3 a f}-\frac {(3 a+b-2 b p) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}}{3 a f} \]
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Time = 0.11 (sec) , antiderivative size = 117, 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 = {4219, 464, 372, 371} \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{3 a f}-\frac {(3 a-2 b p+b) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right )}{3 a f} \]
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Rule 371
Rule 372
Rule 464
Rule 4219
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right ) \left (a+b x^2\right )^p}{x^4} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{3 a f}+\frac {(3 a+b-2 b p) \text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{3 a f} \\ & = \frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{3 a f}+\frac {\left ((3 a+b-2 b p) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {b x^2}{a}\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{3 a f} \\ & = \frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{3 a f}-\frac {(3 a+b-2 b p) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}}{3 a f} \\ \end{align*}
Time = 2.40 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.52 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=-\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \sin ^2(e+f x) \left (-2 (3 a+b-2 b p) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right )+(a+2 b+a \cos (2 (e+f x))) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p\right )}{3 a f \left (-2 \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^p+\left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p+\cos (2 (e+f x)) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p\right )} \]
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\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \sin \left (f x +e \right )^{3}d x\]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]
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Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\text {Timed out} \]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{3} \,d x } \]
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Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^3(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^3\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]
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