Integrand size = 23, antiderivative size = 95 \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {e^{-2 i a} (2-p) x \left (c x^n\right )^{-\frac {2}{n (2-p)}} \left (1+e^{2 i a} \left (c x^n\right )^{\frac {2}{n (2-p)}}\right ) \sec ^p\left (a-\frac {i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)} \]
1/2*(2-p)*x*(1+exp(2*I*a)*(c*x^n)^(2/n/(2-p)))*sec(a-I*ln(c*x^n)/n/(2-p))^ p/exp(2*I*a)/(1-p)/((c*x^n)^(2/n/(2-p)))
Time = 1.22 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.23 \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {2^{-1+p} e^{-i a} (-2+p) x \left (c x^n\right )^{\frac {1}{n (-2+p)}} \left (\frac {e^{\frac {i a (2+p)}{-2+p}} \left (c x^n\right )^{\frac {1}{n (-2+p)}}}{e^{\frac {2 i a p}{-2+p}}+e^{\frac {4 i a}{-2+p}} \left (c x^n\right )^{\frac {2}{n (-2+p)}}}\right )^{-1+p}}{-1+p} \]
(2^(-1 + p)*(-2 + p)*x*(c*x^n)^(1/(n*(-2 + p)))*((E^((I*a*(2 + p))/(-2 + p ))*(c*x^n)^(1/(n*(-2 + p))))/(E^(((2*I)*a*p)/(-2 + p)) + E^(((4*I)*a)/(-2 + p))*(c*x^n)^(2/(n*(-2 + p)))))^(-1 + p))/(E^(I*a)*(-1 + p))
Time = 0.33 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.07, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {5014, 5018, 793}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (p-2)}\right ) \, dx\) |
\(\Big \downarrow \) 5014 |
\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \left (c x^n\right )^{\frac {1}{n}-1} \sec ^p\left (a-\frac {i \log \left (c x^n\right )}{n (2-p)}\right )d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 5018 |
\(\displaystyle \frac {x \left (c x^n\right )^{-\frac {p}{n (2-p)}-\frac {1}{n}} \left (1+e^{2 i a} \left (c x^n\right )^{\frac {2}{n (2-p)}}\right )^p \sec ^p\left (a-\frac {i \log \left (c x^n\right )}{n (2-p)}\right ) \int \left (c x^n\right )^{\frac {p}{2 n-n p}+\frac {1}{n}-1} \left (e^{2 i a} \left (c x^n\right )^{\frac {2}{n (2-p)}}+1\right )^{-p}d\left (c x^n\right )}{n}\) |
\(\Big \downarrow \) 793 |
\(\displaystyle \frac {e^{-2 i a} (2-p) x \left (c x^n\right )^{-\frac {p}{n (2-p)}-\frac {1}{n}} \left (1+e^{2 i a} \left (c x^n\right )^{\frac {2}{n (2-p)}}\right ) \sec ^p\left (a-\frac {i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)}\) |
((2 - p)*x*(c*x^n)^(-n^(-1) - p/(n*(2 - p)))*(1 + E^((2*I)*a)*(c*x^n)^(2/( n*(2 - p))))*Sec[a - (I*Log[c*x^n])/(n*(2 - p))]^p)/(2*E^((2*I)*a)*(1 - p) )
3.3.64.3.1 Defintions of rubi rules used
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n) ^(p + 1)/(b*n*(p + 1)), x] /; FreeQ[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]
Int[Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Si mp[x/(n*(c*x^n)^(1/n)) Subst[Int[x^(1/n - 1)*Sec[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, c, d, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])
Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] :> Simp[Sec[d*(a + b*Log[x])]^p*((1 + E^(2*I*a*d)*x^(2*I*b*d))^p/x^(I*b*d*p )) Int[(e*x)^m*(x^(I*b*d*p)/(1 + E^(2*I*a*d)*x^(2*I*b*d))^p), x], x] /; F reeQ[{a, b, d, e, m, p}, x] && !IntegerQ[p]
\[\int {\sec \left (a +\frac {i \ln \left (c \,x^{n}\right )}{n \left (-2+p \right )}\right )}^{p}d x\]
Time = 0.25 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.57 \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {{\left ({\left (p - 2\right )} x e^{\left (\frac {2 \, {\left (i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )} + {\left (p - 2\right )} x\right )} \left (\frac {2 \, e^{\left (\frac {i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )}{n p - 2 \, n}\right )}}{e^{\left (\frac {2 \, {\left (i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )} + 1}\right )^{p} e^{\left (-\frac {2 \, {\left (i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )}}{2 \, {\left (p - 1\right )}} \]
1/2*((p - 2)*x*e^(2*(I*a*n*p - 2*I*a*n - n*log(x) - log(c))/(n*p - 2*n)) + (p - 2)*x)*(2*e^((I*a*n*p - 2*I*a*n - n*log(x) - log(c))/(n*p - 2*n))/(e^ (2*(I*a*n*p - 2*I*a*n - n*log(x) - log(c))/(n*p - 2*n)) + 1))^p*e^(-2*(I*a *n*p - 2*I*a*n - n*log(x) - log(c))/(n*p - 2*n))/(p - 1)
\[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int \sec ^{p}{\left (a + \frac {i \log {\left (c x^{n} \right )}}{n \left (p - 2\right )} \right )}\, dx \]
\[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int { \sec \left (a + \frac {i \, \log \left (c x^{n}\right )}{n {\left (p - 2\right )}}\right )^{p} \,d x } \]
\[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int { \sec \left (a + \frac {i \, \log \left (c x^{n}\right )}{n {\left (p - 2\right )}}\right )^{p} \,d x } \]
Timed out. \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int {\left (\frac {1}{\cos \left (a+\frac {\ln \left (c\,x^n\right )\,1{}\mathrm {i}}{n\,\left (p-2\right )}\right )}\right )}^p \,d x \]