\(\int \frac {1}{x \cosh ^{\frac {5}{2}}(a+b \log (c x^n))} \, dx\) [257]

Optimal result

Integrand size = 19, antiderivative size = 67 \[ \int \frac {1}{x \cosh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=-\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} i \left (a+b \log \left (c x^n\right )\right ),2\right )}{3 b n}+\frac {2 \sinh \left (a+b \log \left (c x^n\right )\right )}{3 b n \cosh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \] Output:

-2/3*I*InverseJacobiAM(1/2*I*(a+b*ln(c*x^n)),2^(1/2))/b/n+2/3*sinh(a+b*ln( 
c*x^n))/b/n/cosh(a+b*ln(c*x^n))^(3/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.05 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.82 \[ \int \frac {1}{x \cosh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {2 \left (\sinh \left (a+b \log \left (c x^n\right )\right )+\cosh \left (a+b \log \left (c x^n\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-\sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right ) \sqrt {1+\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )}\right )}{3 b n \cosh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \] Input:

Integrate[1/(x*Cosh[a + b*Log[c*x^n]]^(5/2)),x]
 

Output:

(2*(Sinh[a + b*Log[c*x^n]] + Cosh[a + b*Log[c*x^n]]*Hypergeometric2F1[1/4, 
 1/2, 5/4, -Cosh[2*(a + b*Log[c*x^n])] - Sinh[2*(a + b*Log[c*x^n])]]*Sqrt[ 
1 + Cosh[2*(a + b*Log[c*x^n])] + Sinh[2*(a + b*Log[c*x^n])]]))/(3*b*n*Cosh 
[a + b*Log[c*x^n]]^(3/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3039, 3042, 3116, 3042, 3120}

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.

\(\displaystyle \int \frac {1}{x \cosh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx\)

\(\Big \downarrow \) 3039

\(\displaystyle \frac {\int \frac {1}{\cosh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )}d\log \left (c x^n\right )}{n}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{\sin \left (i a+i b \log \left (c x^n\right )+\frac {\pi }{2}\right )^{5/2}}d\log \left (c x^n\right )}{n}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\frac {1}{3} \int \frac {1}{\sqrt {\cosh \left (a+b \log \left (c x^n\right )\right )}}d\log \left (c x^n\right )+\frac {2 \sinh \left (a+b \log \left (c x^n\right )\right )}{3 b \cosh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}}{n}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 \sinh \left (a+b \log \left (c x^n\right )\right )}{3 b \cosh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (i a+i b \log \left (c x^n\right )+\frac {\pi }{2}\right )}}d\log \left (c x^n\right )}{n}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {2 \sinh \left (a+b \log \left (c x^n\right )\right )}{3 b \cosh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}-\frac {2 i \operatorname {EllipticF}\left (\frac {1}{2} i \left (a+b \log \left (c x^n\right )\right ),2\right )}{3 b}}{n}\)

Input:

Int[1/(x*Cosh[a + b*Log[c*x^n]]^(5/2)),x]
 

Output:

((((-2*I)/3)*EllipticF[(I/2)*(a + b*Log[c*x^n]), 2])/b + (2*Sinh[a + b*Log 
[c*x^n]])/(3*b*Cosh[a + b*Log[c*x^n]]^(3/2)))/n
 

Defintions of rubi rules used

rule 3039

Int[u_, x_Symbol] :> With[{lst = FunctionOfLog[Cancel[x*u], x]}, Simp[1/lst 
[[3]]   Subst[Int[lst[[1]], x], x, Log[lst[[2]]]], x] /;  !FalseQ[lst]] /; 
NonsumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3116

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(257\) vs. \(2(59)=118\).

Time = 1.60 (sec) , antiderivative size = 258, normalized size of antiderivative = 3.85

method	result	size
derivativedivides	\(\frac {\sqrt {\left (2 {\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}-1\right ) {\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}\, \left (\frac {\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) \sqrt {2 {\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{4}+{\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}}{3 {\left ({\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}-\frac {1}{2}\right )}^{2}}+\frac {2 \sqrt {-{\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}\, \sqrt {-2 {\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}+1}\, \operatorname {EllipticF}\left (\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ), \sqrt {2}\right )}{3 \sqrt {2 {\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{4}+{\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}}\right )}{n \sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) \sqrt {2 {\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}-1}\, b}\)	\(258\)
default	\(\frac {\sqrt {\left (2 {\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}-1\right ) {\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}\, \left (\frac {\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) \sqrt {2 {\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{4}+{\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}}{3 {\left ({\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}-\frac {1}{2}\right )}^{2}}+\frac {2 \sqrt {-{\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}\, \sqrt {-2 {\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}+1}\, \operatorname {EllipticF}\left (\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ), \sqrt {2}\right )}{3 \sqrt {2 {\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{4}+{\sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}}}\right )}{n \sinh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right ) \sqrt {2 {\cosh \left (\frac {a}{2}+\frac {b \ln \left (c \,x^{n}\right )}{2}\right )}^{2}-1}\, b}\)	\(258\)

Input:

int(1/x/cosh(a+b*ln(c*x^n))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/n*((2*cosh(1/2*a+1/2*b*ln(c*x^n))^2-1)*sinh(1/2*a+1/2*b*ln(c*x^n))^2)^(1 
/2)*(1/3*cosh(1/2*a+1/2*b*ln(c*x^n))*(2*sinh(1/2*a+1/2*b*ln(c*x^n))^4+sinh 
(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)/(cosh(1/2*a+1/2*b*ln(c*x^n))^2-1/2)^2+2/3 
*(-sinh(1/2*a+1/2*b*ln(c*x^n))^2)^(1/2)*(-2*cosh(1/2*a+1/2*b*ln(c*x^n))^2+ 
1)^(1/2)/(2*sinh(1/2*a+1/2*b*ln(c*x^n))^4+sinh(1/2*a+1/2*b*ln(c*x^n))^2)^( 
1/2)*EllipticF(cosh(1/2*a+1/2*b*ln(c*x^n)),2^(1/2)))/sinh(1/2*a+1/2*b*ln(c 
*x^n))/(2*cosh(1/2*a+1/2*b*ln(c*x^n))^2-1)^(1/2)/b
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 501 vs. \(2 (58) = 116\).

Time = 0.12 (sec) , antiderivative size = 501, normalized size of antiderivative = 7.48 \[ \int \frac {1}{x \cosh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx =\text {Too large to display} \] Input:

integrate(1/x/cosh(a+b*log(c*x^n))^(5/2),x, algorithm="fricas")
 

Output:

2/3*((sqrt(2)*cosh(b*n*log(x) + b*log(c) + a)^4 + 4*sqrt(2)*cosh(b*n*log(x 
) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a)^3 + sqrt(2)*sinh(b*n*log 
(x) + b*log(c) + a)^4 + 2*(3*sqrt(2)*cosh(b*n*log(x) + b*log(c) + a)^2 + s 
qrt(2))*sinh(b*n*log(x) + b*log(c) + a)^2 + 2*sqrt(2)*cosh(b*n*log(x) + b* 
log(c) + a)^2 + 4*(sqrt(2)*cosh(b*n*log(x) + b*log(c) + a)^3 + sqrt(2)*cos 
h(b*n*log(x) + b*log(c) + a))*sinh(b*n*log(x) + b*log(c) + a) + sqrt(2))*w 
eierstrassPInverse(-4, 0, cosh(b*n*log(x) + b*log(c) + a) + sinh(b*n*log(x 
) + b*log(c) + a)) + 2*(cosh(b*n*log(x) + b*log(c) + a)^3 + 3*cosh(b*n*log 
(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a)^2 + sinh(b*n*log(x) + 
b*log(c) + a)^3 + (3*cosh(b*n*log(x) + b*log(c) + a)^2 - 1)*sinh(b*n*log(x 
) + b*log(c) + a) - cosh(b*n*log(x) + b*log(c) + a))*sqrt(cosh(b*n*log(x) 
+ b*log(c) + a)))/(b*n*cosh(b*n*log(x) + b*log(c) + a)^4 + 4*b*n*cosh(b*n* 
log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a)^3 + b*n*sinh(b*n*lo 
g(x) + b*log(c) + a)^4 + 2*b*n*cosh(b*n*log(x) + b*log(c) + a)^2 + 2*(3*b* 
n*cosh(b*n*log(x) + b*log(c) + a)^2 + b*n)*sinh(b*n*log(x) + b*log(c) + a) 
^2 + b*n + 4*(b*n*cosh(b*n*log(x) + b*log(c) + a)^3 + b*n*cosh(b*n*log(x) 
+ b*log(c) + a))*sinh(b*n*log(x) + b*log(c) + a))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x \cosh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\text {Timed out} \] Input:

integrate(1/x/cosh(a+b*ln(c*x**n))**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{x \cosh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{x \cosh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/x/cosh(a+b*log(c*x^n))^(5/2),x, algorithm="maxima")
 

Output:

integrate(1/(x*cosh(b*log(c*x^n) + a)^(5/2)), x)
 

Giac [F]

\[ \int \frac {1}{x \cosh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{x \cosh \left (b \log \left (c x^{n}\right ) + a\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate(1/x/cosh(a+b*log(c*x^n))^(5/2),x, algorithm="giac")
 

Output:

integrate(1/(x*cosh(b*log(c*x^n) + a)^(5/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \cosh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{x\,{\mathrm {cosh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^{5/2}} \,d x \] Input:

int(1/(x*cosh(a + b*log(c*x^n))^(5/2)),x)
 

Output:

int(1/(x*cosh(a + b*log(c*x^n))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x \cosh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {\sqrt {\cosh \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}}{{\cosh \left (\mathrm {log}\left (x^{n} c \right ) b +a \right )}^{3} x}d x \] Input:

int(1/x/cosh(a+b*log(c*x^n))^(5/2),x)
                                                                                    
                                                                                    
 

Output:

int(sqrt(cosh(log(x**n*c)*b + a))/(cosh(log(x**n*c)*b + a)**3*x),x)