\(\int \frac {1}{x \sqrt {\sinh (a+b \log (c x^n))}} \, dx\) [282]

Optimal result

Integrand size = 19, antiderivative size = 70 \[ \int \frac {1}{x \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}} \, dx=-\frac {2 i \operatorname {EllipticF}\left (\frac {1}{4} \left (2 i a-\pi +2 i b \log \left (c x^n\right )\right ),2\right ) \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )}}{b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}} \] Output:

-2*I*InverseJacobiAM(1/2*I*a-1/4*Pi+1/2*I*b*ln(c*x^n),2^(1/2))*(I*sinh(a+b 
*ln(c*x^n)))^(1/2)/b/n/sinh(a+b*ln(c*x^n))^(1/2)
 

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}} \, dx=-\frac {2 \operatorname {EllipticF}\left (\frac {1}{4} \left (-2 i a+\pi -2 i b \log \left (c x^n\right )\right ),2\right ) \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}{b n \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )}} \] Input:

Integrate[1/(x*Sqrt[Sinh[a + b*Log[c*x^n]]]),x]
 

Output:

(-2*EllipticF[((-2*I)*a + Pi - (2*I)*b*Log[c*x^n])/4, 2]*Sqrt[Sinh[a + b*L 
og[c*x^n]]])/(b*n*Sqrt[I*Sinh[a + b*Log[c*x^n]]])
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3039, 3042, 3121, 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.

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

\(\Big \downarrow \) 3039

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

Int[1/(x*Sqrt[Sinh[a + b*Log[c*x^n]]]),x]
 

Output:

((-2*I)*EllipticF[(I*a - Pi/2 + I*b*Log[c*x^n])/2, 2]*Sqrt[I*Sinh[a + b*Lo 
g[c*x^n]]])/(b*n*Sqrt[Sinh[a + b*Log[c*x^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 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]
 

rule 3121

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

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.71

method	result	size
derivativedivides	\(\frac {i \sqrt {-i \left (\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (-\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+i\right )}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-i \left (\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+i\right )}, \frac {\sqrt {2}}{2}\right )}{n \cosh \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, b}\)	\(120\)
default	\(\frac {i \sqrt {-i \left (\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+i\right )}\, \sqrt {2}\, \sqrt {-i \left (-\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+i\right )}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {-i \left (\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+i\right )}, \frac {\sqrt {2}}{2}\right )}{n \cosh \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, b}\)	\(120\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}} \, dx=\frac {2 \, \sqrt {2} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )}{b n} \] Input:

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

Output:

2*sqrt(2)*weierstrassPInverse(4, 0, cosh(b*n*log(x) + b*log(c) + a) + sinh 
(b*n*log(x) + b*log(c) + a))/(b*n)
 

Sympy [F]

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

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

Output:

Integral(1/(x*sqrt(sinh(a + b*log(c*x**n)))), x)
 

Maxima [F]

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

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

Output:

integrate(1/(x*sqrt(sinh(b*log(c*x^n) + a))), x)
 

Giac [F]

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

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

Output:

integrate(1/(x*sqrt(sinh(b*log(c*x^n) + a))), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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