3.3.83 \(\int \frac {1}{x \sinh ^{\frac {3}{2}}(a+b \log (c x^n))} \, dx\) [283]

3.3.83.1 Optimal result
3.3.83.2 Mathematica [A] (verified)
3.3.83.3 Rubi [A] (verified)
3.3.83.4 Maple [A] (verified)
3.3.83.5 Fricas [C] (verification not implemented)
3.3.83.6 Sympy [F]
3.3.83.7 Maxima [F]
3.3.83.8 Giac [F]
3.3.83.9 Mupad [F(-1)]

3.3.83.1 Optimal result

Integrand size = 19, antiderivative size = 107 \[ \int \frac {1}{x \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=-\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}-\frac {2 i E\left (\left .\frac {1}{2} \left (i a-\frac {\pi }{2}+i b \log \left (c x^n\right )\right )\right |2\right ) \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}{b n \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )}} \]

output
-2*cosh(a+b*ln(c*x^n))/b/n/sinh(a+b*ln(c*x^n))^(1/2)+2*I*(sin(1/2*I*a+1/4* 
Pi+1/2*I*b*ln(c*x^n))^2)^(1/2)/sin(1/2*I*a+1/4*Pi+1/2*I*b*ln(c*x^n))*Ellip 
ticE(cos(1/2*I*a+1/4*Pi+1/2*I*b*ln(c*x^n)),2^(1/2))*sinh(a+b*ln(c*x^n))^(1 
/2)/b/n/(I*sinh(a+b*ln(c*x^n)))^(1/2)
 
3.3.83.2 Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=-\frac {2 \left (\cosh \left (a+b \log \left (c x^n\right )\right )-E\left (\left .\frac {1}{4} \left (-2 i a+\pi -2 i b \log \left (c x^n\right )\right )\right |2\right ) \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )}\right )}{b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}} \]

input
Integrate[1/(x*Sinh[a + b*Log[c*x^n]]^(3/2)),x]
 
output
(-2*(Cosh[a + b*Log[c*x^n]] - EllipticE[((-2*I)*a + Pi - (2*I)*b*Log[c*x^n 
])/4, 2]*Sqrt[I*Sinh[a + b*Log[c*x^n]]]))/(b*n*Sqrt[Sinh[a + b*Log[c*x^n]] 
])
 
3.3.83.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.98, 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, 3119}

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.

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

\(\Big \downarrow \) 3039

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {-\frac {2 \cosh \left (a+b \log \left (c x^n\right )\right )}{b \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}}-\frac {2 i \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )} E\left (\left .\frac {1}{2} \left (i a+i b \log \left (c x^n\right )-\frac {\pi }{2}\right )\right |2\right )}{b \sqrt {i \sinh \left (a+b \log \left (c x^n\right )\right )}}}{n}\)

input
Int[1/(x*Sinh[a + b*Log[c*x^n]]^(3/2)),x]
 
output
((-2*Cosh[a + b*Log[c*x^n]])/(b*Sqrt[Sinh[a + b*Log[c*x^n]]]) - ((2*I)*Ell 
ipticE[(I*a - Pi/2 + I*b*Log[c*x^n])/2, 2]*Sqrt[Sinh[a + b*Log[c*x^n]]])/( 
b*Sqrt[I*Sinh[a + b*Log[c*x^n]]]))/n
 

3.3.83.3.1 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 3119
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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]
 
3.3.83.4 Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.98

method result size
derivativedivides \(\frac {2 \sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticE}\left (\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}, \frac {\sqrt {2}}{2}\right )-2 {\cosh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{n \cosh \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, b}\) \(212\)
default \(\frac {2 \sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticE}\left (\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}, \frac {\sqrt {2}}{2}\right )-\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}, \frac {\sqrt {2}}{2}\right )-2 {\cosh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2}}{n \cosh \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, b}\) \(212\)

input
int(1/x/sinh(a+b*ln(c*x^n))^(3/2),x,method=_RETURNVERBOSE)
 
output
1/n*(2*(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)*EllipticE((1-I*sinh(a+b*ln(c*x^n)))^(1/ 
2),1/2*2^(1/2))-(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))-2*cosh(a+b*ln(c*x^n))^2)/cosh(a+b*ln(c*x^n))/sinh 
(a+b*ln(c*x^n))^(1/2)/b
 
3.3.83.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.10 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.30 \[ \int \frac {1}{x \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=-\frac {2 \, {\left ({\left (\sqrt {2} \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 2 \, \sqrt {2} \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + \sqrt {2} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - \sqrt {2}\right )} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )\right ) + 2 \, {\left (\cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 2 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2}\right )} \sqrt {\sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}\right )}}{b n \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 2 \, b n \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + b n \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} - b n} \]

input
integrate(1/x/sinh(a+b*log(c*x^n))^(3/2),x, algorithm="fricas")
 
output
-2*((sqrt(2)*cosh(b*n*log(x) + b*log(c) + a)^2 + 2*sqrt(2)*cosh(b*n*log(x) 
 + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a) + sqrt(2)*sinh(b*n*log(x) 
 + b*log(c) + a)^2 - sqrt(2))*weierstrassZeta(4, 0, 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)^2 + 2*cosh(b*n*log(x) + b*log(c) + a)*s 
inh(b*n*log(x) + b*log(c) + a) + sinh(b*n*log(x) + b*log(c) + a)^2)*sqrt(s 
inh(b*n*log(x) + b*log(c) + a)))/(b*n*cosh(b*n*log(x) + b*log(c) + a)^2 + 
2*b*n*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a) + b* 
n*sinh(b*n*log(x) + b*log(c) + a)^2 - b*n)
 
3.3.83.6 Sympy [F]

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

input
integrate(1/x/sinh(a+b*ln(c*x**n))**(3/2),x)
 
output
Integral(1/(x*sinh(a + b*log(c*x**n))**(3/2)), x)
 
3.3.83.7 Maxima [F]

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

input
integrate(1/x/sinh(a+b*log(c*x^n))^(3/2),x, algorithm="maxima")
 
output
integrate(1/(x*sinh(b*log(c*x^n) + a)^(3/2)), x)
 
3.3.83.8 Giac [F]

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

input
integrate(1/x/sinh(a+b*log(c*x^n))^(3/2),x, algorithm="giac")
 
output
integrate(1/(x*sinh(b*log(c*x^n) + a)^(3/2)), x)
 
3.3.83.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x \sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{x\,{\mathrm {sinh}\left (a+b\,\ln \left (c\,x^n\right )\right )}^{3/2}} \,d x \]

input
int(1/(x*sinh(a + b*log(c*x^n))^(3/2)),x)
 
output
int(1/(x*sinh(a + b*log(c*x^n))^(3/2)), x)