\(\int \csc ^p(a-\frac {i \log (c x^n)}{n (-2+p)}) \, dx\) [307]

Optimal result

Integrand size = 23, antiderivative size = 71 \[ \int \csc ^p\left (a-\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {(2-p) x \left (1-e^{2 i a} \left (c x^n\right )^{-\frac {2}{n (2-p)}}\right ) \csc ^p\left (a+\frac {i \log \left (c x^n\right )}{n (2-p)}\right )}{2 (1-p)} \] Output:

(2-p)*x*(1-exp(2*I*a)/((c*x^n)^(2/n/(2-p))))*csc(a+I*ln(c*x^n)/n/(2-p))^p/ 
(2-2*p)
 

Mathematica [A] (verified)

Time = 1.96 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.80 \[ \int \csc ^p\left (a-\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {2^{-1+p} (-2+p) x \left (\frac {i e^{i a} \left (c x^n\right )^{\frac {1}{n (-2+p)}}}{-1+e^{2 i a} \left (c x^n\right )^{\frac {2}{n (-2+p)}}}\right )^p \left (1+e^{2 i a} \left (c x^n\right )^{\frac {2}{n (-2+p)}} \left (-1+\left (1-e^{-2 i a} \left (c x^n\right )^{-\frac {2}{n (-2+p)}}\right )^p\right )\right )}{-1+p} \] Input:

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

Output:

(2^(-1 + p)*(-2 + p)*x*((I*E^(I*a)*(c*x^n)^(1/(n*(-2 + p))))/(-1 + E^((2*I 
)*a)*(c*x^n)^(2/(n*(-2 + p)))))^p*(1 + E^((2*I)*a)*(c*x^n)^(2/(n*(-2 + p)) 
)*(-1 + (1 - 1/(E^((2*I)*a)*(c*x^n)^(2/(n*(-2 + p)))))^p)))/(-1 + p)
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.58, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {5015, 5019, 796}

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.

\(\displaystyle \int \csc ^p\left (a-\frac {i \log \left (c x^n\right )}{n (p-2)}\right ) \, dx\)

\(\Big \downarrow \) 5015

\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \left (c x^n\right )^{\frac {1}{n}-1} \csc ^p\left (a+\frac {i \log \left (c x^n\right )}{n (2-p)}\right )d\left (c x^n\right )}{n}\)

\(\Big \downarrow \) 5019

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

\(\Big \downarrow \) 796

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

Input:

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

Output:

((2 - p)*x*(c*x^n)^(-n^(-1) + (2*(1 - p))/(n*(2 - p)) + p/(n*(2 - p)))*(1 
- E^((2*I)*a)/(c*x^n)^(2/(n*(2 - p))))*Csc[a + (I*Log[c*x^n])/(n*(2 - p))] 
^p)/(2*(1 - p))
 

Defintions of rubi rules used

rule 796

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] /; FreeQ[{a, b, c, m, n, 
 p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]
 

rule 5015

Int[Csc[((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)*Csc[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 5019

Int[Csc[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] 
:> Simp[Csc[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]

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 150 vs. \(2 (55) = 110\).

Time = 0.07 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.11 \[ \int \csc ^p\left (a-\frac {i \log \left (c x^n\right )}{n (-2+p)}\right ) \, dx=\frac {{\left ({\left (p - 2\right )} x e^{\left (\frac {2 \, {\left (-i \, a n p + 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )} - {\left (p - 2\right )} x\right )} \left (-\frac {2 i \, e^{\left (\frac {-i \, a n p + 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )}{n p - 2 \, n}\right )}}{e^{\left (\frac {2 \, {\left (-i \, a n p + 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )} - 1}\right )^{p} e^{\left (-\frac {2 \, {\left (-i \, a n p + 2 i \, a n - n \log \left (x\right ) - \log \left (c\right )\right )}}{n p - 2 \, n}\right )}}{2 \, {\left (p - 1\right )}} \] Input:

integrate(csc(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*I*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]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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