Integrand size = 21, antiderivative size = 172 \[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^2 \, dx=\frac {a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+n)}-\frac {a^2 (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) \sec ^{-1+n}(e+f x) \sin (e+f x)}{f \left (1-n^2\right ) \sqrt {\sin ^2(e+f x)}}+\frac {2 a^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right ) \sec ^n(e+f x) \sin (e+f x)}{f n \sqrt {\sin ^2(e+f x)}} \]
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Time = 0.16 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3873, 3857, 2722, 4131} \[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^2 \, dx=-\frac {a^2 (2 n+1) \sin (e+f x) \sec ^{n-1}(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right )}{f \left (1-n^2\right ) \sqrt {\sin ^2(e+f x)}}+\frac {2 a^2 \sin (e+f x) \sec ^n(e+f x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right )}{f n \sqrt {\sin ^2(e+f x)}}+\frac {a^2 \sin (e+f x) \sec ^{n+1}(e+f x)}{f (n+1)} \]
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Rule 2722
Rule 3857
Rule 3873
Rule 4131
Rubi steps \begin{align*} \text {integral}& = \left (2 a^2\right ) \int \sec ^{1+n}(e+f x) \, dx+\int \sec ^n(e+f x) \left (a^2+a^2 \sec ^2(e+f x)\right ) \, dx \\ & = \frac {a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+n)}+\frac {\left (a^2 (1+2 n)\right ) \int \sec ^n(e+f x) \, dx}{1+n}+\left (2 a^2 \cos ^n(e+f x) \sec ^n(e+f x)\right ) \int \cos ^{-1-n}(e+f x) \, dx \\ & = \frac {a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+n)}+\frac {2 a^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right ) \sec ^n(e+f x) \sin (e+f x)}{f n \sqrt {\sin ^2(e+f x)}}+\frac {\left (a^2 (1+2 n) \cos ^n(e+f x) \sec ^n(e+f x)\right ) \int \cos ^{-n}(e+f x) \, dx}{1+n} \\ & = \frac {a^2 \sec ^{1+n}(e+f x) \sin (e+f x)}{f (1+n)}-\frac {a^2 (1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1-n}{2},\frac {3-n}{2},\cos ^2(e+f x)\right ) \sec ^{-1+n}(e+f x) \sin (e+f x)}{f \left (1-n^2\right ) \sqrt {\sin ^2(e+f x)}}+\frac {2 a^2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-\frac {n}{2},\frac {2-n}{2},\cos ^2(e+f x)\right ) \sec ^n(e+f x) \sin (e+f x)}{f n \sqrt {\sin ^2(e+f x)}} \\ \end{align*}
Time = 0.74 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.78 \[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^2 \, dx=\frac {a^2 \csc (e+f x) \sec ^{-1+n}(e+f x) \left ((1+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {n}{2},\frac {2+n}{2},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}+n \left (\tan ^2(e+f x)+2 \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+n}{2},\frac {3+n}{2},\sec ^2(e+f x)\right ) \sec (e+f x) \sqrt {-\tan ^2(e+f x)}\right )\right )}{f n (1+n)} \]
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\[\int \sec \left (f x +e \right )^{n} \left (a +a \sec \left (f x +e \right )\right )^{2}d x\]
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\[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^2 \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{2} \sec \left (f x + e\right )^{n} \,d x } \]
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\[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^2 \, dx=a^{2} \left (\int 2 \sec {\left (e + f x \right )} \sec ^{n}{\left (e + f x \right )}\, dx + \int \sec ^{2}{\left (e + f x \right )} \sec ^{n}{\left (e + f x \right )}\, dx + \int \sec ^{n}{\left (e + f x \right )}\, dx\right ) \]
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\[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^2 \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{2} \sec \left (f x + e\right )^{n} \,d x } \]
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\[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^2 \, dx=\int { {\left (a \sec \left (f x + e\right ) + a\right )}^{2} \sec \left (f x + e\right )^{n} \,d x } \]
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Timed out. \[ \int \sec ^n(e+f x) (a+a \sec (e+f x))^2 \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]
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