\(\int \sec ^p(a+\frac {i \log (c x^n)}{n (-2+p)}) \, dx\) [264]

Optimal result

Integrand size = 23, antiderivative size = 95 \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {e^{-2 i a} (2-p) x \left (c x^n\right )^{-\frac {2}{n (2-p)}} \left (1+e^{2 i a} \left (c x^n\right )^{\frac {2}{n (2-p)}}\right ) \sec ^p\left (a-\frac {i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)} \] Output:

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)))
 

Mathematica [A] (warning: unable to verify)

Time = 1.18 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.23 \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {2^{-1+p} e^{-i a} (-2+p) x \left (c x^n\right )^{\frac {1}{n (-2+p)}} \left (\frac {e^{\frac {i a (2+p)}{-2+p}} \left (c x^n\right )^{\frac {1}{n (-2+p)}}}{e^{\frac {2 i a p}{-2+p}}+e^{\frac {4 i a}{-2+p}} \left (c x^n\right )^{\frac {2}{n (-2+p)}}}\right )^{-1+p}}{-1+p} \] Input:

Integrate[Sec[a + (I*Log[c*x^n])/(n*(-2 + p))]^p,x]
 

Output:

(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))
 

Rubi [A] (verified)

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

\(\displaystyle \frac {e^{-2 i a} (2-p) x \left (c x^n\right )^{-\frac {p}{n (2-p)}-\frac {1}{n}} \left (1+e^{2 i a} \left (c x^n\right )^{\frac {2}{n (2-p)}}\right ) \sec ^p\left (a-\frac {i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)}\)

Input:

Int[Sec[a + (I*Log[c*x^n])/(n*(-2 + p))]^p,x]
 

Output:

((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) 
)
 

Defintions of rubi rules used

rule 793

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]
 

rule 5014

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])
 

rule 5018

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]
 

Maple [F]

\[\int {\sec \left (a +\frac {i \ln \left (c \,x^{n}\right )}{n \left (-2+p \right )}\right )}^{p}d x\]

Input:

int(sec(a+I*ln(c*x^n)/n/(-2+p))^p,x)
 

Output:

int(sec(a+I*ln(c*x^n)/n/(-2+p))^p,x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.57 \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {{\left ({\left (p - 2\right )} x e^{\left (\frac {2 \, {\left (i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )} + {\left (p - 2\right )} x\right )} \left (\frac {2 \, e^{\left (\frac {i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )}{n p - 2 \, n}\right )}}{e^{\left (\frac {2 \, {\left (i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )} + 1}\right )^{p} e^{\left (-\frac {2 \, {\left (i \, a n p - 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )}}{2 \, {\left (p - 1\right )}} \] Input:

integrate(sec(a+I*log(c*x^n)/n/(-2+p))^p,x, algorithm="fricas")
 

Output:

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)
 

Sympy [F]

\[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int \sec ^{p}{\left (a + \frac {i \log {\left (c x^{n} \right )}}{n \left (p - 2\right )} \right )}\, dx \] Input:

integrate(sec(a+I*ln(c*x**n)/n/(-2+p))**p,x)
 

Output:

Integral(sec(a + I*log(c*x**n)/(n*(p - 2)))**p, x)
 

Maxima [F]

\[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int { \sec \left (a + \frac {i \, \log \left (c x^{n}\right )}{n {\left (p - 2\right )}}\right )^{p} \,d x } \] Input:

integrate(sec(a+I*log(c*x^n)/n/(-2+p))^p,x, algorithm="maxima")
 

Output:

integrate(sec(a + I*log(c*x^n)/(n*(p - 2)))^p, x)
 

Giac [F]

\[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int { \sec \left (a + \frac {i \, \log \left (c x^{n}\right )}{n {\left (p - 2\right )}}\right )^{p} \,d x } \] Input:

integrate(sec(a+I*log(c*x^n)/n/(-2+p))^p,x, algorithm="giac")
 

Output:

integrate(sec(a + I*log(c*x^n)/(n*(p - 2)))^p, x)
 

Mupad [F(-1)]

Timed out. \[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\int {\left (\frac {1}{\cos \left (a+\frac {\ln \left (c\,x^n\right )\,1{}\mathrm {i}}{n\,\left (p-2\right )}\right )}\right )}^p \,d x \] Input:

int((1/cos(a + (log(c*x^n)*1i)/(n*(p - 2))))^p,x)
 

Output:

int((1/cos(a + (log(c*x^n)*1i)/(n*(p - 2))))^p, x)
 

Reduce [F]

\[ \int \sec ^p\left (a+\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {{\sec \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +a n p -2 a n}{n p -2 n}\right )}^{p} p x -2 {\sec \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +a n p -2 a n}{n p -2 n}\right )}^{p} x -\left (\int {\sec \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +a n p -2 a n}{n p -2 n}\right )}^{p} \tan \left (\frac {\mathrm {log}\left (x^{n} c \right ) i +a n p -2 a n}{n p -2 n}\right )d x \right ) i p}{p -2} \] Input:

int(sec(a+I*log(c*x^n)/n/(-2+p))^p,x)
 

Output:

(sec((log(x**n*c)*i + a*n*p - 2*a*n)/(n*p - 2*n))**p*p*x - 2*sec((log(x**n 
*c)*i + a*n*p - 2*a*n)/(n*p - 2*n))**p*x - int(sec((log(x**n*c)*i + a*n*p 
- 2*a*n)/(n*p - 2*n))**p*tan((log(x**n*c)*i + a*n*p - 2*a*n)/(n*p - 2*n)), 
x)*i*p)/(p - 2)