\(\int \frac {\sinh ^{\frac {3}{2}}(a+b \log (c x^n))}{x} \, dx\) [280]

Optimal result

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

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

Mathematica [C] (verified)

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

Time = 0.08 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.05 \[ \int \frac {\sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {-2 \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )+\sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )\right ) \sqrt {1-\cosh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )-\sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )}+\sinh \left (2 \left (a+b \log \left (c x^n\right )\right )\right )}{3 b n \sqrt {\sinh \left (a+b \log \left (c x^n\right )\right )}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.38 (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, 3115, 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.

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

\(\Big \downarrow \) 3039

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

((((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]]]) + (2*Cosh[a + b*Log[c*x^n 
]]*Sqrt[Sinh[a + b*Log[c*x^n]]])/(3*b))/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 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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 = 0.20 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(\frac {-\frac {i \sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}, \frac {\sqrt {2}}{2}\right )}{3}+\frac {2 {\cosh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2} \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}{3}}{n \cosh \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, b}\)	\(143\)
default	\(\frac {-\frac {i \sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {2}\, \sqrt {1+i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \sqrt {i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, \operatorname {EllipticF}\left (\sqrt {1-i \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}, \frac {\sqrt {2}}{2}\right )}{3}+\frac {2 {\cosh \left (a +b \ln \left (c \,x^{n}\right )\right )}^{2} \sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}{3}}{n \cosh \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {\sinh \left (a +b \ln \left (c \,x^{n}\right )\right )}\, b}\)	\(143\)

Input:

int(sinh(a+b*ln(c*x^n))^(3/2)/x,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.57 \[ \int \frac {\sinh ^{\frac {3}{2}}\left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-\frac {2 \, {\left (\sqrt {2} \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + \sqrt {2} \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )} {\rm weierstrassPInverse}\left (4, 0, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right ) - {\left (\cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 2 \, \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )^{2} + 1\right )} \sqrt {\sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )}}{3 \, {\left (b n \cosh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right ) + b n \sinh \left (b n \log \left (x\right ) + b \log \left (c\right ) + a\right )\right )}} \] Input:

integrate(sinh(a+b*log(c*x^n))^(3/2)/x,x, algorithm="fricas")
 

Output:

-1/3*(2*(sqrt(2)*cosh(b*n*log(x) + b*log(c) + a) + sqrt(2)*sinh(b*n*log(x) 
 + b*log(c) + a))*weierstrassPInverse(4, 0, cosh(b*n*log(x) + b*log(c) + a 
) + sinh(b*n*log(x) + b*log(c) + a)) - (cosh(b*n*log(x) + b*log(c) + a)^2 
+ 2*cosh(b*n*log(x) + b*log(c) + a)*sinh(b*n*log(x) + b*log(c) + a) + sinh 
(b*n*log(x) + b*log(c) + a)^2 + 1)*sqrt(sinh(b*n*log(x) + b*log(c) + a)))/ 
(b*n*cosh(b*n*log(x) + b*log(c) + a) + b*n*sinh(b*n*log(x) + b*log(c) + a) 
)
 

Sympy [F]

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

integrate(sinh(a+b*ln(c*x**n))**(3/2)/x,x)
 

Output:

Integral(sinh(a + b*log(c*x**n))**(3/2)/x, x)
 

Maxima [F]

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

integrate(sinh(a+b*log(c*x^n))^(3/2)/x,x, algorithm="maxima")
 

Output:

integrate(sinh(b*log(c*x^n) + a)^(3/2)/x, x)
 

Giac [F]

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

integrate(sinh(a+b*log(c*x^n))^(3/2)/x,x, algorithm="giac")
 

Output:

integrate(sinh(b*log(c*x^n) + a)^(3/2)/x, x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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