Integrand size = 23, antiderivative size = 122 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=-\frac {(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 b^2 f (1+p)}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 a f (1+p)}+\frac {\left (a+b \sec ^2(e+f x)\right )^{2+p}}{2 b^2 f (2+p)} \]
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Time = 0.18 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {4224, 457, 90, 67} \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=-\frac {(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{2 b^2 f (p+1)}+\frac {\left (a+b \sec ^2(e+f x)\right )^{p+2}}{2 b^2 f (p+2)}-\frac {\left (a+b \sec ^2(e+f x)\right )^{p+1} \operatorname {Hypergeometric2F1}\left (1,p+1,p+2,\frac {b \sec ^2(e+f x)}{a}+1\right )}{2 a f (p+1)} \]
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Rule 67
Rule 90
Rule 457
Rule 4224
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} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {(-1+x)^2 (a+b x)^p}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = \frac {\text {Subst}\left (\int \left (\frac {(-a-2 b) (a+b x)^p}{b}+\frac {(a+b x)^p}{x}+\frac {(a+b x)^{1+p}}{b}\right ) \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = -\frac {(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 b^2 f (1+p)}+\frac {\left (a+b \sec ^2(e+f x)\right )^{2+p}}{2 b^2 f (2+p)}+\frac {\text {Subst}\left (\int \frac {(a+b x)^p}{x} \, dx,x,\sec ^2(e+f x)\right )}{2 f} \\ & = -\frac {(a+2 b) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 b^2 f (1+p)}-\frac {\operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sec ^2(e+f x)}{a}\right ) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{2 a f (1+p)}+\frac {\left (a+b \sec ^2(e+f x)\right )^{2+p}}{2 b^2 f (2+p)} \\ \end{align*}
Time = 0.90 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.77 \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=-\frac {\left (a+b \sec ^2(e+f x)\right )^{1+p} \left (b^2 (2+p) \operatorname {Hypergeometric2F1}\left (1,1+p,2+p,1+\frac {b \sec ^2(e+f x)}{a}\right )+a \left (a+2 b (2+p)-b (1+p) \sec ^2(e+f x)\right )\right )}{2 a b^2 f (1+p) (2+p)} \]
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\[\int \left (a +b \sec \left (f x +e \right )^{2}\right )^{p} \tan \left (f x +e \right )^{5}d x\]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{5} \,d x } \]
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Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=\text {Timed out} \]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{5} \,d x } \]
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\[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \tan \left (f x + e\right )^{5} \,d x } \]
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Timed out. \[ \int \left (a+b \sec ^2(e+f x)\right )^p \tan ^5(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^5\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p \,d x \]
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