\(\int \frac {1}{x \sinh ^{\frac {5}{2}}(a+b \log (c x^n))} \, dx\) [284]

Optimal result

Integrand size = 19, antiderivative size = 109 \[ \int \frac {1}{x \sinh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=-\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{3 b n \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}+\frac {2 i \operatorname {EllipticF}\left (\frac {1}{4} \left (2 i a-\pi +2 i b \log \left (c x^n\right )\right ),2\right ) \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )}}{3 b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}} \] Output:

-2/3*cosh(a+b*ln(c*x^n))/b/n/sinh(a+b*ln(c*x^n))^(3/2)+2/3*I*InverseJacobi 
AM(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 [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x \sinh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=-\frac {2 \left (\cosh \left (a+b \log \left (c x^n\right )\right )+\operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right ) \sinh \left (a+b \log \left (c x^n\right )\right ) \sqrt {1-\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-\sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )}\right )}{3 b n \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3039, 3042, 3116, 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.

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

\(\Big \downarrow \) 3039

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

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

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 = 0.19 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(-\frac {i \sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}, \frac {\sqrt {2}}{2}\right ) \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+2 {\cosh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{3 n {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}} \cosh \left (a +b \ln \left (c \,x^{n}\right )\right ) b}\)	\(144\)
default	\(-\frac {i \sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}, \frac {\sqrt {2}}{2}\right ) \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )+2 {\cosh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{3 n {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{\frac {3}{2}} \cosh \left (a +b \ln \left (c \,x^{n}\right )\right ) b}\)	\(144\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 504 vs. \(2 (89) = 178\).

Time = 0.13 (sec) , antiderivative size = 504, normalized size of antiderivative = 4.62 \[ \int \frac {1}{x \sinh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx =\text {Too large to display} \] Input:

integrate(1/x/sinh(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*lo 
g(x) + b*log(c) + a)^4 + 2*(3*sqrt(2)*cosh(b*n*log(x) + b*log(c) + a)^2 - 
sqrt(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)*co 
sh(b*n*log(x) + b*log(c) + a))*sinh(b*n*log(x) + b*log(c) + a) + sqrt(2))* 
weierstrassPInverse(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(sinh(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. \[ \int \frac {1}{x \sinh ^{\frac {5}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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