Integrand size = 23, antiderivative size = 95 \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {e^{-2 i a} (2-p) x \left (c x^n\right )^{-\frac {2}{n (2-p)}} \left (1+e^{2 i a} \left (c x^n\right )^{\frac {2}{n (2-p)}}\right ) \sec ^p\left (a-\frac {i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)} \]
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Time = 0.11 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4599, 4603, 267} \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {e^{-2 i a} (2-p) x \left (c x^n\right )^{-\frac {2}{n (2-p)}} \left (1+e^{2 i a} \left (c x^n\right )^{\frac {2}{n (2-p)}}\right ) \sec ^p\left (a-\frac {i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)} \]
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Rule 267
Rule 4599
Rule 4603
Rubi steps \begin{align*} \text {integral}& = \frac {\left (x \left (c x^n\right )^{-1/n}\right ) \text {Subst}\left (\int x^{-1+\frac {1}{n}} \sec ^p\left (a+\frac {i \log (x)}{n (-2+p)}\right ) \, dx,x,c x^n\right )}{n} \\ & = \frac {\left (x \left (c x^n\right )^{-\frac {1}{n}+\frac {p}{n (-2+p)}} \left (1+e^{2 i a} \left (c x^n\right )^{-\frac {2}{n (-2+p)}}\right )^p \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right )\right ) \text {Subst}\left (\int x^{-1+\frac {1}{n}-\frac {p}{n (-2+p)}} \left (1+e^{2 i a} x^{-\frac {2}{n (-2+p)}}\right )^{-p} \, dx,x,c x^n\right )}{n} \\ & = \frac {e^{-2 i a} (2-p) x \left (c x^n\right )^{-\frac {2}{n (2-p)}} \left (1+e^{2 i a} \left (c x^n\right )^{\frac {2}{n (2-p)}}\right ) \sec ^p\left (a-\frac {i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)} \\ \end{align*}
Time = 1.22 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.23 \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {2^{-1+p} e^{-i a} (-2+p) x \left (c x^n\right )^{\frac {1}{n (-2+p)}} \left (\frac {e^{\frac {i a (2+p)}{-2+p}} \left (c x^n\right )^{\frac {1}{n (-2+p)}}}{e^{\frac {2 i a p}{-2+p}}+e^{\frac {4 i a}{-2+p}} \left (c x^n\right )^{\frac {2}{n (-2+p)}}}\right )^{-1+p}}{-1+p} \]
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\[\int {\sec \left (a +\frac {i \ln \left (c \,x^{n}\right )}{n \left (-2+p \right )}\right )}^{p}d x\]
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none
Time = 0.25 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.57 \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {{\left ({\left (p - 2\right )} x e^{\left (\frac {2 \, {\left (i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )} + {\left (p - 2\right )} x\right )} \left (\frac {2 \, e^{\left (\frac {i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )}{n p - 2 \, n}\right )}}{e^{\left (\frac {2 \, {\left (i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )} + 1}\right )^{p} e^{\left (-\frac {2 \, {\left (i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )}}{2 \, {\left (p - 1\right )}} \]
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\[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int \sec ^{p}{\left (a + \frac {i \log {\left (c x^{n} \right )}}{n \left (p - 2\right )} \right )}\, dx \]
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\[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int { \sec \left (a + \frac {i \, \log \left (c x^{n}\right )}{n {\left (p - 2\right )}}\right )^{p} \,d x } \]
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\[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int { \sec \left (a + \frac {i \, \log \left (c x^{n}\right )}{n {\left (p - 2\right )}}\right )^{p} \,d x } \]
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Timed out. \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int {\left (\frac {1}{\cos \left (a+\frac {\ln \left (c\,x^n\right )\,1{}\mathrm {i}}{n\,\left (p-2\right )}\right )}\right )}^p \,d x \]
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