\(\int x^m \sec (a+b \log (c x^n)) \, dx\) [280]

Optimal result

Integrand size = 15, antiderivative size = 103 \[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 e^{i a} x^{1+m} \left (c x^n\right )^{i b} \operatorname {Hypergeometric2F1}\left (1,-\frac {i+i m-b n}{2 b n},-\frac {i (1+m)-3 b n}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{1+m+i b n} \] Output:

2*exp(I*a)*x^(1+m)*(c*x^n)^(I*b)*hypergeom([1, -1/2*(I+I*m-b*n)/b/n],[-1/2 
*(I*(1+m)-3*b*n)/b/n],-exp(2*I*a)*(c*x^n)^(2*I*b))/(1+m+I*b*n)
 

Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.96 \[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 e^{i a} x^{1+m} \left (c x^n\right )^{i b} \operatorname {Hypergeometric2F1}\left (1,\frac {-i-i m+b n}{2 b n},-\frac {i (1+m+3 i b n)}{2 b n},-e^{2 i \left (a+b \log \left (c x^n\right )\right )}\right )}{1+m+i b n} \] Input:

Integrate[x^m*Sec[a + b*Log[c*x^n]],x]
 

Output:

(2*E^(I*a)*x^(1 + m)*(c*x^n)^(I*b)*Hypergeometric2F1[1, (-I - I*m + b*n)/( 
2*b*n), ((-1/2*I)*(1 + m + (3*I)*b*n))/(b*n), -E^((2*I)*(a + b*Log[c*x^n]) 
)])/(1 + m + I*b*n)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5020, 5016, 888}

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.

\(\displaystyle \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx\)

\(\Big \downarrow \) 5020

\(\displaystyle \frac {x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}} \int \left (c x^n\right )^{\frac {m+1}{n}-1} \sec \left (a+b \log \left (c x^n\right )\right )d\left (c x^n\right )}{n}\)

\(\Big \downarrow \) 5016

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

\(\Big \downarrow \) 888

\(\displaystyle \frac {2 e^{i a} x^{m+1} \left (c x^n\right )^{-\frac {m+1}{n}+\frac {i b n+m+1}{n}} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} \left (1-\frac {i (m+1)}{b n}\right ),-\frac {i (m+1)-3 b n}{2 b n},-e^{2 i a} \left (c x^n\right )^{2 i b}\right )}{i b n+m+1}\)

Input:

Int[x^m*Sec[a + b*Log[c*x^n]],x]
 

Output:

(2*E^(I*a)*x^(1 + m)*(c*x^n)^(-((1 + m)/n) + (1 + m + I*b*n)/n)*Hypergeome 
tric2F1[1, (1 - (I*(1 + m))/(b*n))/2, -1/2*(I*(1 + m) - 3*b*n)/(b*n), -(E^ 
((2*I)*a)*(c*x^n)^((2*I)*b))])/(1 + m + I*b*n)
 

Defintions of rubi rules used

rule 888

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p 
*((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 
, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] && (ILt 
Q[p, 0] || GtQ[a, 0])
 

rule 5016

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.), x_Symbol] 
:> Simp[2^p*E^(I*a*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] /; FreeQ[{a, b, d, e, m}, x] && IntegerQ[p]
 

rule 5020

Int[((e_.)*(x_))^(m_.)*Sec[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_ 
.), x_Symbol] :> Simp[(e*x)^(m + 1)/(e*n*(c*x^n)^((m + 1)/n))   Subst[Int[x 
^((m + 1)/n - 1)*Sec[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, b, 
 c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])
 

Maple [F]

\[\int x^{m} \sec \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]

Input:

int(x^m*sec(a+b*ln(c*x^n)),x)
 

Output:

int(x^m*sec(a+b*ln(c*x^n)),x)
 

Fricas [F]

\[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right ) \,d x } \] Input:

integrate(x^m*sec(a+b*log(c*x^n)),x, algorithm="fricas")
 

Output:

integral(x^m*sec(b*log(c*x^n) + a), x)
 

Sympy [F]

\[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^{m} \sec {\left (a + b \log {\left (c x^{n} \right )} \right )}\, dx \] Input:

integrate(x**m*sec(a+b*ln(c*x**n)),x)
 

Output:

Integral(x**m*sec(a + b*log(c*x**n)), x)
 

Maxima [F]

\[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right ) \,d x } \] Input:

integrate(x^m*sec(a+b*log(c*x^n)),x, algorithm="maxima")
 

Output:

integrate(x^m*sec(b*log(c*x^n) + a), x)
 

Giac [F]

\[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { x^{m} \sec \left (b \log \left (c x^{n}\right ) + a\right ) \,d x } \] Input:

integrate(x^m*sec(a+b*log(c*x^n)),x, algorithm="giac")
 

Output:

integrate(x^m*sec(b*log(c*x^n) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\int \frac {x^m}{\cos \left (a+b\,\ln \left (c\,x^n\right )\right )} \,d x \] Input:

int(x^m/cos(a + b*log(c*x^n)),x)
 

Output:

int(x^m/cos(a + b*log(c*x^n)), x)
 

Reduce [F]

\[ \int x^m \sec \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {too large to display} \] Input:

int(x^m*sec(a+b*log(c*x^n)),x)
 

Output:

(2*x**m*cos(log(x**n*c)*b + a)*sec(log(x**n*c)*b + a)*b**2*n**2*x - x**m*c 
os(log(x**n*c)*b + a)*sec(log(x**n*c)*b + a)*m**2*x - 2*x**m*cos(log(x**n* 
c)*b + a)*sec(log(x**n*c)*b + a)*m*x - x**m*cos(log(x**n*c)*b + a)*sec(log 
(x**n*c)*b + a)*x - x**m*cos(log(x**n*c)*b + a)*b**2*n**2*x + 8*cos(log(x* 
*n*c)*b + a)*int(x**m/(2*tan((log(x**n*c)*b + a)/2)**4*b**2*n**2 - tan((lo 
g(x**n*c)*b + a)/2)**4*m**2 - 2*tan((log(x**n*c)*b + a)/2)**4*m - tan((log 
(x**n*c)*b + a)/2)**4 - 4*tan((log(x**n*c)*b + a)/2)**2*b**2*n**2 + 2*tan( 
(log(x**n*c)*b + a)/2)**2*m**2 + 4*tan((log(x**n*c)*b + a)/2)**2*m + 2*tan 
((log(x**n*c)*b + a)/2)**2 + 2*b**2*n**2 - m**2 - 2*m - 1),x)*b**4*m*n**4 
+ 8*cos(log(x**n*c)*b + a)*int(x**m/(2*tan((log(x**n*c)*b + a)/2)**4*b**2* 
n**2 - tan((log(x**n*c)*b + a)/2)**4*m**2 - 2*tan((log(x**n*c)*b + a)/2)** 
4*m - tan((log(x**n*c)*b + a)/2)**4 - 4*tan((log(x**n*c)*b + a)/2)**2*b**2 
*n**2 + 2*tan((log(x**n*c)*b + a)/2)**2*m**2 + 4*tan((log(x**n*c)*b + a)/2 
)**2*m + 2*tan((log(x**n*c)*b + a)/2)**2 + 2*b**2*n**2 - m**2 - 2*m - 1),x 
)*b**4*n**4 - 4*cos(log(x**n*c)*b + a)*int(x**m/(2*tan((log(x**n*c)*b + a) 
/2)**4*b**2*n**2 - tan((log(x**n*c)*b + a)/2)**4*m**2 - 2*tan((log(x**n*c) 
*b + a)/2)**4*m - tan((log(x**n*c)*b + a)/2)**4 - 4*tan((log(x**n*c)*b + a 
)/2)**2*b**2*n**2 + 2*tan((log(x**n*c)*b + a)/2)**2*m**2 + 4*tan((log(x**n 
*c)*b + a)/2)**2*m + 2*tan((log(x**n*c)*b + a)/2)**2 + 2*b**2*n**2 - m**2 
- 2*m - 1),x)*b**2*m**3*n**2 - 12*cos(log(x**n*c)*b + a)*int(x**m/(2*ta...