\(\int \frac {x}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx\) [150]

Optimal result

Integrand size = 13, antiderivative size = 130 \[ \int \frac {x}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=-\frac {6}{5 \left (c^4-\frac {1}{x^4}\right ) x^2 \text {csch}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^2}{5 \text {csch}^{\frac {3}{2}}(2 \log (c x))}-\frac {12 E\left (\left .\csc ^{-1}(c x)\right |-1\right )}{5 c^5 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {csch}^{\frac {3}{2}}(2 \log (c x))}+\frac {12 \operatorname {EllipticF}\left (\csc ^{-1}(c x),-1\right )}{5 c^5 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {csch}^{\frac {3}{2}}(2 \log (c x))} \] Output:

-6/5/(c^4-1/x^4)/x^2/csch(2*ln(c*x))^(3/2)+1/5*x^2/csch(2*ln(c*x))^(3/2)-1 
2/5*EllipticE(1/c/x,I)/c^5/(1-1/c^4/x^4)^(3/2)/x^3/csch(2*ln(c*x))^(3/2)+1 
2/5*InverseJacobiAM(arccsc(c*x),I)/c^5/(1-1/c^4/x^4)^(3/2)/x^3/csch(2*ln(c 
*x))^(3/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 0.12 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.46 \[ \int \frac {x}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {1}{4},\frac {3}{4},c^4 x^4\right )}{2 c^2 \sqrt {2-2 c^4 x^4} \sqrt {\frac {c^2 x^2}{-1+c^4 x^4}}} \] Input:

Integrate[x/Csch[2*Log[c*x]]^(3/2),x]
 

Output:

Hypergeometric2F1[-3/2, -1/4, 3/4, c^4*x^4]/(2*c^2*Sqrt[2 - 2*c^4*x^4]*Sqr 
t[(c^2*x^2)/(-1 + c^4*x^4)])
 

Rubi [A] (warning: unable to verify)

Time = 0.37 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.85, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {6086, 6084, 858, 809, 809, 836, 762, 1388, 327}

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.

\(\displaystyle \int \frac {x}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx\)

\(\Big \downarrow \) 6086

\(\displaystyle \frac {\int \frac {c x}{\text {csch}^{\frac {3}{2}}(2 \log (c x))}d(c x)}{c^2}\)

\(\Big \downarrow \) 6084

\(\displaystyle \frac {\int c^4 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} x^4d(c x)}{c^5 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {\int \frac {\left (1-c^4 x^4\right )^{3/2}}{c^6 x^6}d\frac {1}{c x}}{c^5 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 809

\(\displaystyle -\frac {-\frac {6}{5} \int \frac {\sqrt {1-c^4 x^4}}{c^2 x^2}d\frac {1}{c x}-\frac {\left (1-c^4 x^4\right )^{3/2}}{5 c^5 x^5}}{c^5 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 809

\(\displaystyle -\frac {-\frac {6}{5} \left (-2 \int \frac {c^2 x^2}{\sqrt {1-c^4 x^4}}d\frac {1}{c x}-\frac {\sqrt {1-c^4 x^4}}{c x}\right )-\frac {\left (1-c^4 x^4\right )^{3/2}}{5 c^5 x^5}}{c^5 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 836

\(\displaystyle -\frac {-\frac {6}{5} \left (-2 \left (\int \frac {c^2 x^2+1}{\sqrt {1-c^4 x^4}}d\frac {1}{c x}-\int \frac {1}{\sqrt {1-c^4 x^4}}d\frac {1}{c x}\right )-\frac {\sqrt {1-c^4 x^4}}{c x}\right )-\frac {\left (1-c^4 x^4\right )^{3/2}}{5 c^5 x^5}}{c^5 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 762

\(\displaystyle -\frac {-\frac {6}{5} \left (-2 \left (\int \frac {c^2 x^2+1}{\sqrt {1-c^4 x^4}}d\frac {1}{c x}-\operatorname {EllipticF}\left (\arcsin \left (\frac {1}{c x}\right ),-1\right )\right )-\frac {\sqrt {1-c^4 x^4}}{c x}\right )-\frac {\left (1-c^4 x^4\right )^{3/2}}{5 c^5 x^5}}{c^5 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 1388

\(\displaystyle -\frac {-\frac {6}{5} \left (-2 \left (\int \frac {\sqrt {c^2 x^2+1}}{\sqrt {1-c^2 x^2}}d\frac {1}{c x}-\operatorname {EllipticF}\left (\arcsin \left (\frac {1}{c x}\right ),-1\right )\right )-\frac {\sqrt {1-c^4 x^4}}{c x}\right )-\frac {\left (1-c^4 x^4\right )^{3/2}}{5 c^5 x^5}}{c^5 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 327

\(\displaystyle -\frac {-\frac {6}{5} \left (-2 \left (E\left (\left .\arcsin \left (\frac {1}{c x}\right )\right |-1\right )-\operatorname {EllipticF}\left (\arcsin \left (\frac {1}{c x}\right ),-1\right )\right )-\frac {\sqrt {1-c^4 x^4}}{c x}\right )-\frac {\left (1-c^4 x^4\right )^{3/2}}{5 c^5 x^5}}{c^5 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

Input:

Int[x/Csch[2*Log[c*x]]^(3/2),x]
 

Output:

-((-1/5*(1 - c^4*x^4)^(3/2)/(c^5*x^5) - (6*(-(Sqrt[1 - c^4*x^4]/(c*x)) - 2 
*(EllipticE[ArcSin[1/(c*x)], -1] - EllipticF[ArcSin[1/(c*x)], -1])))/5)/(c 
^5*(1 - 1/(c^4*x^4))^(3/2)*x^3*Csch[2*Log[c*x]]^(3/2)))
 

Defintions of rubi rules used

rule 327

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ 
(Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) 
)], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
 

rule 762

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) 
)*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] 
 && GtQ[a, 0]
 

rule 809

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1))), x] - Simp[b*n*(p/(c^n*(m + 1)))   I 
nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ 
[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntB 
inomialQ[a, b, c, n, m, p, x]
 

rule 836

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[-b/a, 2]}, 
Simp[-q^(-1)   Int[1/Sqrt[a + b*x^4], x], x] + Simp[1/q   Int[(1 + q*x^2)/S 
qrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && NegQ[b/a]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

rule 1388

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), 
x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, 
 c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer 
Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
 

rule 6084

Int[Csch[((a_.) + Log[x_]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] 
 :> Simp[Csch[d*(a + b*Log[x])]^p*((1 - 1/(E^(2*a*d)*x^(2*b*d)))^p/x^((-b)* 
d*p))   Int[(e*x)^m*(1/(x^(b*d*p)*(1 - 1/(E^(2*a*d)*x^(2*b*d)))^p)), x], x] 
 /; FreeQ[{a, b, d, e, m, p}, x] &&  !IntegerQ[p]
 

rule 6086

Int[Csch[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(d_.)]^(p_.)*((e_.)*(x_))^(m 
_.), x_Symbol] :> Simp[(e*x)^(m + 1)/(e*n*(c*x^n)^((m + 1)/n))   Subst[Int[ 
x^((m + 1)/n - 1)*Csch[d*(a + b*Log[x])]^p, x], x, c*x^n], x] /; FreeQ[{a, 
b, c, d, e, m, n, p}, x] && (NeQ[c, 1] || NeQ[n, 1])
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.17

method	result	size
risch	\(\frac {\left (c^{8} x^{8}+4 c^{4} x^{4}-5\right ) \sqrt {2}}{20 \left (c^{4} x^{4}-1\right ) c^{2} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}-1}}}-\frac {3 \sqrt {c^{2} x^{2}+1}\, \sqrt {-c^{2} x^{2}+1}\, \left (\operatorname {EllipticF}\left (x \sqrt {-c^{2}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {-c^{2}}, i\right )\right ) \sqrt {2}\, x}{5 \sqrt {-c^{2}}\, \left (c^{4} x^{4}-1\right ) \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}-1}}}\)	\(152\)

Input:

int(x/csch(2*ln(x*c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/20*(c^8*x^8+4*c^4*x^4-5)/(c^4*x^4-1)*2^(1/2)/c^2/(c^2*x^2/(c^4*x^4-1))^( 
1/2)-3/5/(-c^2)^(1/2)*(c^2*x^2+1)^(1/2)*(-c^2*x^2+1)^(1/2)/(c^4*x^4-1)*(El 
lipticF(x*(-c^2)^(1/2),I)-EllipticE(x*(-c^2)^(1/2),I))*2^(1/2)*x/(c^2*x^2/ 
(c^4*x^4-1))^(1/2)
 

Fricas [F]

\[ \int \frac {x}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {x}{\operatorname {csch}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(x/csch(2*log(c*x))^(3/2),x, algorithm="fricas")
 

Output:

integral(x/csch(2*log(c*x))^(3/2), x)
 

Sympy [F]

\[ \int \frac {x}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x}{\operatorname {csch}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \] Input:

integrate(x/csch(2*ln(c*x))**(3/2),x)
 

Output:

Integral(x/csch(2*log(c*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {x}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {x}{\operatorname {csch}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(x/csch(2*log(c*x))^(3/2),x, algorithm="maxima")
 

Output:

integrate(x/csch(2*log(c*x))^(3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {x}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \] Input:

integrate(x/csch(2*log(c*x))^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x}{{\left (\frac {1}{\mathrm {sinh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \] Input:

int(x/(1/sinh(2*log(c*x)))^(3/2),x)
                                                                                    
                                                                                    
 

Output:

int(x/(1/sinh(2*log(c*x)))^(3/2), x)
 

Reduce [F]

\[ \int \frac {x}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {\sqrt {\mathrm {csch}\left (2 \,\mathrm {log}\left (c x \right )\right )}\, x}{\mathrm {csch}\left (2 \,\mathrm {log}\left (c x \right )\right )^{2}}d x \] Input:

int(x/csch(2*log(c*x))^(3/2),x)
 

Output:

int((sqrt(csch(2*log(c*x)))*x)/csch(2*log(c*x))**2,x)