\(\int \frac {1}{\cosh ^{\frac {7}{2}}(a+\frac {2 \log (c x^n)}{n})} \, dx\) [262]

Optimal result

Integrand size = 18, antiderivative size = 87 \[ \int \frac {1}{\cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=-\frac {x \left (1+e^{-2 a} \left (c x^n\right )^{-4/n}\right )}{10 \cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}-\frac {x \left (1+e^{-2 a} \left (c x^n\right )^{-4/n}\right )^2}{15 \cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \] Output:

-1/10*x*(1+1/exp(2*a)/((c*x^n)^(4/n)))/cosh(a+2*ln(c*x^n)/n)^(7/2)-1/15*x* 
(1+1/exp(2*a)/((c*x^n)^(4/n)))^2/cosh(a+2*ln(c*x^n)/n)^(7/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.39 \[ \int \frac {1}{\cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\frac {\left (\left (2+5 x^4\right ) \cosh \left (a-2 \log (x)+\frac {2 \log \left (c x^n\right )}{n}\right )+\left (-2+5 x^4\right ) \sinh \left (a-2 \log (x)+\frac {2 \log \left (c x^n\right )}{n}\right )\right ) \left (-\cosh \left (2 a-4 \log (x)+\frac {4 \log \left (c x^n\right )}{n}\right )+\sinh \left (2 a-4 \log (x)+\frac {4 \log \left (c x^n\right )}{n}\right )\right )}{15 x^5 \cosh ^{\frac {5}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \] Input:

Integrate[Cosh[a + (2*Log[c*x^n])/n]^(-7/2),x]
 

Output:

(((2 + 5*x^4)*Cosh[a - 2*Log[x] + (2*Log[c*x^n])/n] + (-2 + 5*x^4)*Sinh[a 
- 2*Log[x] + (2*Log[c*x^n])/n])*(-Cosh[2*a - 4*Log[x] + (4*Log[c*x^n])/n] 
+ Sinh[2*a - 4*Log[x] + (4*Log[c*x^n])/n]))/(15*x^5*Cosh[a + (2*Log[c*x^n] 
)/n]^(5/2))
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.62, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6052, 6060, 803, 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.

\(\displaystyle \int \frac {1}{\cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx\)

\(\Big \downarrow \) 6052

\(\displaystyle \frac {x \left (c x^n\right )^{-1/n} \int \frac {\left (c x^n\right )^{\frac {1}{n}-1}}{\cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}d\left (c x^n\right )}{n}\)

\(\Big \downarrow \) 6060

\(\displaystyle \frac {x \left (c x^n\right )^{6/n} \left (e^{-2 a} \left (c x^n\right )^{-4/n}+1\right )^{7/2} \int \frac {\left (c x^n\right )^{-1-\frac {6}{n}}}{\left (e^{-2 a} \left (c x^n\right )^{-4/n}+1\right )^{7/2}}d\left (c x^n\right )}{n \cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}\)

\(\Big \downarrow \) 803

\(\displaystyle \frac {x \left (c x^n\right )^{6/n} \left (e^{-2 a} \left (c x^n\right )^{-4/n}+1\right )^{7/2} \left (\frac {2}{3} e^{-2 a} \int \frac {\left (c x^n\right )^{-1-\frac {10}{n}}}{\left (e^{-2 a} \left (c x^n\right )^{-4/n}+1\right )^{7/2}}d\left (c x^n\right )-\frac {n \left (c x^n\right )^{-6/n}}{6 \left (e^{-2 a} \left (c x^n\right )^{-4/n}+1\right )^{5/2}}\right )}{n \cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}\)

\(\Big \downarrow \) 796

\(\displaystyle \frac {x \left (c x^n\right )^{6/n} \left (e^{-2 a} \left (c x^n\right )^{-4/n}+1\right )^{7/2} \left (-\frac {e^{-2 a} n \left (c x^n\right )^{-10/n}}{15 \left (e^{-2 a} \left (c x^n\right )^{-4/n}+1\right )^{5/2}}-\frac {n \left (c x^n\right )^{-6/n}}{6 \left (e^{-2 a} \left (c x^n\right )^{-4/n}+1\right )^{5/2}}\right )}{n \cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )}\)

Input:

Int[Cosh[a + (2*Log[c*x^n])/n]^(-7/2),x]
 

Output:

(x*(c*x^n)^(6/n)*(1 + 1/(E^(2*a)*(c*x^n)^(4/n)))^(7/2)*(-1/15*n/(E^(2*a)*( 
c*x^n)^(10/n)*(1 + 1/(E^(2*a)*(c*x^n)^(4/n)))^(5/2)) - n/(6*(c*x^n)^(6/n)* 
(1 + 1/(E^(2*a)*(c*x^n)^(4/n)))^(5/2))))/(n*Cosh[a + (2*Log[c*x^n])/n]^(7/ 
2))
 

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 803

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

rule 6052

Int[Cosh[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.), x_Symbol] :> S 
imp[x/(n*(c*x^n)^(1/n))   Subst[Int[x^(1/n - 1)*Cosh[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 6060

Int[Cosh[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_)*((e_.)*(x_))^(m_.), x_Symbol] 
:> Simp[Cosh[d*(a + b*Log[x])]^p/(x^(b*d*p)*(1 + 1/(E^(2*a*d)*x^(2*b*d)))^p 
)   Int[(e*x)^m*x^(b*d*p)*(1 + 1/(E^(2*a*d)*x^(2*b*d)))^p, x], x] /; FreeQ[ 
{a, b, d, e, m, p}, x] &&  !IntegerQ[p]
 

Maple [F]

\[\int \frac {1}{{\cosh \left (a +\frac {2 \ln \left (c \,x^{n}\right )}{n}\right )}^{\frac {7}{2}}}d x\]

Input:

int(1/cosh(a+2*ln(c*x^n)/n)^(7/2),x)
 

Output:

int(1/cosh(a+2*ln(c*x^n)/n)^(7/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=-\frac {8 \, \sqrt {\frac {1}{2}} {\left (5 \, x^{5} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )} + 2 \, x\right )} \sqrt {\frac {x^{4} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )} + 1}{x^{2}}} e^{\left (-\frac {a n + 2 \, \log \left (c\right )}{2 \, n}\right )}}{15 \, {\left (x^{12} e^{\left (\frac {6 \, {\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )} + 3 \, x^{8} e^{\left (\frac {4 \, {\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )} + 3 \, x^{4} e^{\left (\frac {2 \, {\left (a n + 2 \, \log \left (c\right )\right )}}{n}\right )} + 1\right )}} \] Input:

integrate(1/cosh(a+2*log(c*x^n)/n)^(7/2),x, algorithm="fricas")
 

Output:

-8/15*sqrt(1/2)*(5*x^5*e^(2*(a*n + 2*log(c))/n) + 2*x)*sqrt((x^4*e^(2*(a*n 
 + 2*log(c))/n) + 1)/x^2)*e^(-1/2*(a*n + 2*log(c))/n)/(x^12*e^(6*(a*n + 2* 
log(c))/n) + 3*x^8*e^(4*(a*n + 2*log(c))/n) + 3*x^4*e^(2*(a*n + 2*log(c))/ 
n) + 1)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\text {Timed out} \] Input:

integrate(1/cosh(a+2*ln(c*x**n)/n)**(7/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{\cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\int { \frac {1}{\cosh \left (a + \frac {2 \, \log \left (c x^{n}\right )}{n}\right )^{\frac {7}{2}}} \,d x } \] Input:

integrate(1/cosh(a+2*log(c*x^n)/n)^(7/2),x, algorithm="maxima")
 

Output:

integrate(cosh(a + 2*log(c*x^n)/n)^(-7/2), x)
 

Giac [A] (verification not implemented)

Time = 3.30 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=-\frac {4 \, \sqrt {2} c^{\frac {7}{n}} {\left (\frac {5 \, e^{a}}{c^{\frac {4}{n}} \mathrm {sgn}\left (x\right )} + \frac {2 \, e^{\left (-a\right )}}{c^{\frac {8}{n}} x^{4} \mathrm {sgn}\left (x\right )}\right )} e^{\left (3 \, a\right )}}{15 \, {\left (c^{\frac {4}{n}} e^{\left (3 \, a\right )} + \frac {e^{a}}{x^{4}}\right )}^{\frac {5}{2}} x^{6}} \] Input:

integrate(1/cosh(a+2*log(c*x^n)/n)^(7/2),x, algorithm="giac")
 

Output:

-4/15*sqrt(2)*c^(7/n)*(5*e^a/(c^(4/n)*sgn(x)) + 2*e^(-a)/(c^(8/n)*x^4*sgn( 
x)))*e^(3*a)/((c^(4/n)*e^(3*a) + e^a/x^4)^(5/2)*x^6)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\int \frac {1}{{\mathrm {cosh}\left (a+\frac {2\,\ln \left (c\,x^n\right )}{n}\right )}^{7/2}} \,d x \] Input:

int(1/cosh(a + (2*log(c*x^n))/n)^(7/2),x)
 

Output:

int(1/cosh(a + (2*log(c*x^n))/n)^(7/2), x)
 

Reduce [F]

\[ \int \frac {1}{\cosh ^{\frac {7}{2}}\left (a+\frac {2 \log \left (c x^n\right )}{n}\right )} \, dx=\int \frac {\sqrt {\cosh \left (\frac {2 \,\mathrm {log}\left (x^{n} c \right )+a n}{n}\right )}}{{\cosh \left (\frac {2 \,\mathrm {log}\left (x^{n} c \right )+a n}{n}\right )}^{4}}d x \] Input:

int(1/cosh(a+2*log(c*x^n)/n)^(7/2),x)
 

Output:

int(sqrt(cosh((2*log(x**n*c) + a*n)/n))/cosh((2*log(x**n*c) + a*n)/n)**4,x 
)