\(\int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx\) [149]

Optimal result

Integrand size = 15, antiderivative size = 91 \[ \int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=-\frac {1}{2 \left (c^4-\frac {1}{x^4}\right ) x \text {csch}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^3}{6 \text {csch}^{\frac {3}{2}}(2 \log (c x))}-\frac {\csc ^{-1}\left (c^2 x^2\right )}{2 c^6 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {csch}^{\frac {3}{2}}(2 \log (c x))} \] Output:

-1/2/(c^4-1/x^4)/x/csch(2*ln(c*x))^(3/2)+1/6*x^3/csch(2*ln(c*x))^(3/2)-1/2 
*arccsc(c^2*x^2)/c^6/(1-1/c^4/x^4)^(3/2)/x^3/csch(2*ln(c*x))^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97 \[ \int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {x \left (\left (-4+c^4 x^4\right ) \sqrt {-1+c^4 x^4}+3 \arctan \left (\sqrt {-1+c^4 x^4}\right )\right )}{12 \sqrt {2} c^2 \sqrt {\frac {c^2 x^2}{-1+c^4 x^4}} \sqrt {-1+c^4 x^4}} \] Input:

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

Output:

(x*((-4 + c^4*x^4)*Sqrt[-1 + c^4*x^4] + 3*ArcTan[Sqrt[-1 + c^4*x^4]]))/(12 
*Sqrt[2]*c^2*Sqrt[(c^2*x^2)/(-1 + c^4*x^4)]*Sqrt[-1 + c^4*x^4])
 

Rubi [A] (warning: unable to verify)

Time = 0.34 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {6086, 6084, 858, 807, 247, 247, 223}

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.

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

\(\Big \downarrow \) 6086

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

\(\Big \downarrow \) 6084

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

\(\Big \downarrow \) 858

\(\displaystyle -\frac {\int \frac {\left (1-c^4 x^4\right )^{3/2}}{c^7 x^7}d\frac {1}{c x}}{c^6 x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 223

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

Input:

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

Output:

-1/2*(c^2*x^2*Sqrt[1 - c^2*x^2] - (1 - c^2*x^2)^(3/2)/(3*c^3*x^3) + ArcSin 
[c^2*x^2])/(c^6*(1 - 1/(c^4*x^4))^(3/2)*x^3*Csch[2*Log[c*x]]^(3/2))
 

Defintions of rubi rules used

rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 247

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, 
m, p, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

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 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 [F]

\[\int \frac {x^{2}}{\operatorname {csch}\left (2 \ln \left (x c \right )\right )^{\frac {3}{2}}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.03 \[ \int \frac {x^2}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {3 \, \sqrt {2} c x \arctan \left (\frac {{\left (c^{4} x^{4} - 1\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}}}{c x}\right ) + \sqrt {2} {\left (c^{8} x^{8} - 5 \, c^{4} x^{4} + 4\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}}}{24 \, c^{4} x} \] Input:

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

Output:

1/24*(3*sqrt(2)*c*x*arctan((c^4*x^4 - 1)*sqrt(c^2*x^2/(c^4*x^4 - 1))/(c*x) 
) + sqrt(2)*(c^8*x^8 - 5*c^4*x^4 + 4)*sqrt(c^2*x^2/(c^4*x^4 - 1)))/(c^4*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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