Integrand size = 23, antiderivative size = 216 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=-\frac {(3 a-2 b (2+p)) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{b^2 f (3+2 p) (5+2 p)} \]
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Time = 0.25 (sec) , antiderivative size = 216, 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 = {4231, 427, 396, 252, 251} \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\left (3 a^2-4 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^p \left (\frac {b \tan ^2(e+f x)}{a+b}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )}{b^2 f (2 p+3) (2 p+5)}-\frac {(3 a-2 b (p+2)) \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b^2 f (2 p+3) (2 p+5)}+\frac {\tan (e+f x) \sec ^2(e+f x) \left (a+b \tan ^2(e+f x)+b\right )^{p+1}}{b f (2 p+5)} \]
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Rule 251
Rule 252
Rule 396
Rule 427
Rule 4231
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (1+x^2\right )^2 \left (a+b+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\text {Subst}\left (\int \left (a+b+b x^2\right )^p \left (-a+2 b (2+p)-(3 a-2 b (2+p)) x^2\right ) \, dx,x,\tan (e+f x)\right )}{b f (5+2 p)} \\ & = -\frac {(3 a-2 b (2+p)) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \text {Subst}\left (\int \left (a+b+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)} \\ & = -\frac {(3 a-2 b (2+p)) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^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,\tan (e+f x)\right )}{b^2 f (3+2 p) (5+2 p)} \\ & = -\frac {(3 a-2 b (2+p)) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b^2 f (3+2 p) (5+2 p)}+\frac {\sec ^2(e+f x) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^{1+p}}{b f (5+2 p)}+\frac {\left (3 a^2-4 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan (e+f x) \left (a+b+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p}}{b^2 f (3+2 p) (5+2 p)} \\ \end{align*}
Time = 2.16 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.69 \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\frac {\left (a+b \sec ^2(e+f x)\right )^p \tan (e+f x) \left (1+\frac {b \tan ^2(e+f x)}{a+b}\right )^{-p} \left (15 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right )+10 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},-p,\frac {5}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan ^2(e+f x)+3 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},-p,\frac {7}{2},-\frac {b \tan ^2(e+f x)}{a+b}\right ) \tan ^4(e+f x)\right )}{15 f} \]
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\[\int \sec \left (f x +e \right )^{6} \left (a +b \sec \left (f x +e \right )^{2}\right )^{p}d x\]
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\[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{6} \,d x } \]
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Timed out. \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\text {Timed out} \]
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\[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{6} \,d x } \]
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\[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int { {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \sec \left (f x + e\right )^{6} \,d x } \]
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Timed out. \[ \int \sec ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^p \, dx=\int \frac {{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p}{{\cos \left (e+f\,x\right )}^6} \,d x \]
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