Integrand size = 23, antiderivative size = 182 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}-\frac {\left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 p^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a}\right )^{-p}}{15 a^2 f} \]
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Time = 0.21 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4219, 473, 464, 372, 371} \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=-\frac {\left (15 a^2+10 a b (1-2 p)+b^2 \left (4 p^2-8 p+3\right )\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right )}{15 a^2 f}+\frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{5 a f} \]
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Rule 371
Rule 372
Rule 464
Rule 473
Rule 4219
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^2 \left (a+b x^2\right )^p}{x^6} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}+\frac {\text {Subst}\left (\int \frac {\left (-10 a-b (3-2 p)+5 a x^2\right ) \left (a+b x^2\right )^p}{x^4} \, dx,x,\sec (e+f x)\right )}{5 a f} \\ & = \frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}+\frac {\left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 p^2\right )\right ) \text {Subst}\left (\int \frac {\left (a+b x^2\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 a^2 f} \\ & = \frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}+\frac {\left (\left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 p^2\right )\right ) \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 )}{15 a^2 f} \\ & = \frac {(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}-\frac {\left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 p^2\right )\right ) \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}}{15 a^2 f} \\ \end{align*}
Time = 3.66 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.39 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\frac {2 \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \sin ^4(e+f x) \left (4 \left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 p^2\right )\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a}\right )+(a+2 b+a \cos (2 (e+f x))) (-17 a-6 b+4 b p+3 a \cos (2 (e+f x))) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p\right )}{15 a^2 f \left (4 \cos (2 (e+f x)) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p-2^{-p} \left (3 \left (\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x)}{a}\right )^p+2^p \cos (4 (e+f x)) \left (\frac {a+b+b \tan ^2(e+f x)}{a}\right )^p\right )\right )} \]
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\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \sin \left (f x +e \right )^{5}d x\]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
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Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\text {Timed out} \]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
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Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx=\int {\sin \left (e+f\,x\right )}^5\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]
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