Integrand size = 23, antiderivative size = 208 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\frac {(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}-\frac {\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cos (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a-b}\right ) \left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a-b}\right )^{-p}}{15 (a-b)^2 f} \]
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Time = 0.26 (sec) , antiderivative size = 208, 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 = {3745, 473, 464, 372, 371} \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=-\frac {\left (15 a^2-20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac {b \sec ^2(e+f x)}{a-b}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a-b}\right )}{15 f (a-b)^2}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{5 f (a-b)}+\frac {(10 a-2 b p-7 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{15 f (a-b)^2} \]
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Rule 371
Rule 372
Rule 464
Rule 473
Rule 3745
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^p}{x^6} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac {\text {Subst}\left (\int \frac {\left (-10 a+b (7+2 p)+5 (a-b) x^2\right ) \left (a-b+b x^2\right )^p}{x^4} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f} \\ & = \frac {(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac {\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \text {Subst}\left (\int \frac {\left (a-b+b x^2\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^2 f} \\ & = \frac {(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac {\left (\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a-b}\right )^{-p}\right ) \text {Subst}\left (\int \frac {\left (1+\frac {b x^2}{a-b}\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^2 f} \\ & = \frac {(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}-\frac {\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+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-b}\right ) \left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a-b}\right )^{-p}}{15 (a-b)^2 f} \\ \end{align*}
Time = 9.00 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.36 \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=-\frac {2^{3+p} \cos (e+f x) \sin ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-p,\frac {1}{2},-\frac {b \sec ^2(e+f x)}{a-b}\right )+\frac {1}{4} (a+b+(a-b) \cos (2 (e+f x))) (-17 a+b (11+4 p)+3 (a-b) \cos (2 (e+f x))) \left (\frac {a+b \tan ^2(e+f x)}{a-b}\right )^p\right )}{15 (a-b)^2 f \left (3 \left (\frac {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}{a-b}\right )^p-2^{2+p} \cos (2 (e+f x)) \left (\frac {a+b \tan ^2(e+f x)}{a-b}\right )^p+2^p \cos (4 (e+f x)) \left (\frac {a+b \tan ^2(e+f x)}{a-b}\right )^p\right )} \]
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\[\int \sin \left (f x +e \right )^{5} \left (a +b \tan \left (f x +e \right )^{2}\right )^{p}d x\]
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\[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
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Timed out. \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
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\[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
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\[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int { {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5} \,d x } \]
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Timed out. \[ \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx=\int {\sin \left (e+f\,x\right )}^5\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p \,d x \]
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