\(\int \frac {x^8}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx\) [143]

Optimal result

Integrand size = 15, antiderivative size = 128 \[ \int \frac {x^8}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {x}{32 c^4 \left (c^4-\frac {1}{x^4}\right ) \text {csch}^{\frac {3}{2}}(2 \log (c x))}-\frac {x^5}{16 \left (c^4-\frac {1}{x^4}\right ) \text {csch}^{\frac {3}{2}}(2 \log (c x))}+\frac {x^9}{12 \text {csch}^{\frac {3}{2}}(2 \log (c x))}+\frac {\text {arctanh}\left (\sqrt {1-\frac {1}{c^4 x^4}}\right )}{32 c^{12} \left (1-\frac {1}{c^4 x^4}\right )^{3/2} x^3 \text {csch}^{\frac {3}{2}}(2 \log (c x))} \] Output:

1/32*x/c^4/(c^4-1/x^4)/csch(2*ln(c*x))^(3/2)-1/16*x^5/(c^4-1/x^4)/csch(2*l 
n(c*x))^(3/2)+1/12*x^9/csch(2*ln(c*x))^(3/2)+1/32*arctanh((1-1/c^4/x^4)^(1 
/2))/c^12/(1-1/c^4/x^4)^(3/2)/x^3/csch(2*ln(c*x))^(3/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.74 \[ \int \frac {x^8}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {c^3 x^3 \sqrt {1-c^4 x^4} \left (3-14 c^4 x^4+8 c^8 x^8\right )-3 c x \arcsin \left (c^2 x^2\right )}{192 c^9 \sqrt {2-2 c^4 x^4} \sqrt {\frac {c^2 x^2}{-1+c^4 x^4}}} \] Input:

Integrate[x^8/Csch[2*Log[c*x]]^(3/2),x]
 

Output:

(c^3*x^3*Sqrt[1 - c^4*x^4]*(3 - 14*c^4*x^4 + 8*c^8*x^8) - 3*c*x*ArcSin[c^2 
*x^2])/(192*c^9*Sqrt[2 - 2*c^4*x^4]*Sqrt[(c^2*x^2)/(-1 + c^4*x^4)])
 

Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {6086, 6084, 798, 51, 51, 52, 73, 219}

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.

\(\displaystyle \int \frac {x^8}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx\)

\(\Big \downarrow \) 6086

\(\displaystyle \frac {\int \frac {c^8 x^8}{\text {csch}^{\frac {3}{2}}(2 \log (c x))}d(c x)}{c^9}\)

\(\Big \downarrow \) 6084

\(\displaystyle \frac {\int c^{11} \left (1-\frac {1}{c^4 x^4}\right )^{3/2} x^{11}d(c x)}{c^{12} x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 798

\(\displaystyle -\frac {\int \frac {\left (1-\frac {1}{c^4 x^4}\right )^{3/2}}{c^4 x^4}d\frac {1}{c^4 x^4}}{4 c^{12} x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 51

\(\displaystyle -\frac {-\frac {1}{2} \int \frac {\sqrt {1-\frac {1}{c^4 x^4}}}{c^3 x^3}d\frac {1}{c^4 x^4}-\frac {\left (1-\frac {1}{c^4 x^4}\right )^{3/2}}{3 c^3 x^3}}{4 c^{12} x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 51

\(\displaystyle -\frac {\frac {1}{2} \left (\frac {1}{4} \int \frac {1}{c^2 \sqrt {1-\frac {1}{c^4 x^4}} x^2}d\frac {1}{c^4 x^4}+\frac {\sqrt {1-\frac {1}{c^4 x^4}}}{2 c^2 x^2}\right )-\frac {\left (1-\frac {1}{c^4 x^4}\right )^{3/2}}{3 c^3 x^3}}{4 c^{12} x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 52

\(\displaystyle -\frac {\frac {1}{2} \left (\frac {1}{4} \left (\frac {1}{2} \int \frac {1}{c \sqrt {1-\frac {1}{c^4 x^4}} x}d\frac {1}{c^4 x^4}-\frac {\sqrt {1-\frac {1}{c^4 x^4}}}{c x}\right )+\frac {\sqrt {1-\frac {1}{c^4 x^4}}}{2 c^2 x^2}\right )-\frac {\left (1-\frac {1}{c^4 x^4}\right )^{3/2}}{3 c^3 x^3}}{4 c^{12} x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\frac {1}{2} \left (\frac {1}{4} \left (-\int \frac {1}{1-c^2 x^2}d\sqrt {1-\frac {1}{c^4 x^4}}-\frac {\sqrt {1-\frac {1}{c^4 x^4}}}{c x}\right )+\frac {\sqrt {1-\frac {1}{c^4 x^4}}}{2 c^2 x^2}\right )-\frac {\left (1-\frac {1}{c^4 x^4}\right )^{3/2}}{3 c^3 x^3}}{4 c^{12} x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {\frac {1}{2} \left (\frac {1}{4} \left (-\text {arctanh}\left (\sqrt {1-\frac {1}{c^4 x^4}}\right )-\frac {\sqrt {1-\frac {1}{c^4 x^4}}}{c x}\right )+\frac {\sqrt {1-\frac {1}{c^4 x^4}}}{2 c^2 x^2}\right )-\frac {\left (1-\frac {1}{c^4 x^4}\right )^{3/2}}{3 c^3 x^3}}{4 c^{12} x^3 \left (1-\frac {1}{c^4 x^4}\right )^{3/2} \text {csch}^{\frac {3}{2}}(2 \log (c x))}\)

Input:

Int[x^8/Csch[2*Log[c*x]]^(3/2),x]
 

Output:

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

Defintions of rubi rules used

rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

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

rule 798

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

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 = 0.14 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.95

method	result	size
risch	\(\frac {x^{3} \left (8 c^{8} x^{8}-14 c^{4} x^{4}+3\right ) \sqrt {2}}{384 c^{6} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}-1}}}+\frac {\ln \left (\frac {c^{4} x^{2}}{\sqrt {c^{4}}}+\sqrt {c^{4} x^{4}-1}\right ) \sqrt {2}\, x}{128 c^{6} \sqrt {c^{4}}\, \sqrt {c^{4} x^{4}-1}\, \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4}-1}}}\)	\(121\)

Input:

int(x^8/csch(2*ln(x*c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/384*x^3*(8*c^8*x^8-14*c^4*x^4+3)/c^6*2^(1/2)/(c^2*x^2/(c^4*x^4-1))^(1/2) 
+1/128/c^6*ln(c^4*x^2/(c^4)^(1/2)+(c^4*x^4-1)^(1/2))/(c^4)^(1/2)*2^(1/2)*x 
/(c^4*x^4-1)^(1/2)/(c^2*x^2/(c^4*x^4-1))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.86 \[ \int \frac {x^8}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\frac {2 \, \sqrt {2} {\left (8 \, c^{13} x^{13} - 22 \, c^{9} x^{9} + 17 \, c^{5} x^{5} - 3 \, c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}} + 3 \, \sqrt {2} \log \left (2 \, c^{4} x^{4} + 2 \, {\left (c^{5} x^{5} - c x\right )} \sqrt {\frac {c^{2} x^{2}}{c^{4} x^{4} - 1}} - 1\right )}{768 \, c^{9}} \] Input:

integrate(x^8/csch(2*log(c*x))^(3/2),x, algorithm="fricas")
 

Output:

1/768*(2*sqrt(2)*(8*c^13*x^13 - 22*c^9*x^9 + 17*c^5*x^5 - 3*c*x)*sqrt(c^2* 
x^2/(c^4*x^4 - 1)) + 3*sqrt(2)*log(2*c^4*x^4 + 2*(c^5*x^5 - c*x)*sqrt(c^2* 
x^2/(c^4*x^4 - 1)) - 1))/c^9
 

Sympy [F]

\[ \int \frac {x^8}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^{8}}{\operatorname {csch}^{\frac {3}{2}}{\left (2 \log {\left (c x \right )} \right )}}\, dx \] Input:

integrate(x**8/csch(2*ln(c*x))**(3/2),x)
 

Output:

Integral(x**8/csch(2*log(c*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {x^8}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int { \frac {x^{8}}{\operatorname {csch}\left (2 \, \log \left (c x\right )\right )^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^8/csch(2*log(c*x))^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^8/csch(2*log(c*x))^(3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {x^8}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\text {Timed out} \] Input:

integrate(x^8/csch(2*log(c*x))^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^8}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {x^8}{{\left (\frac {1}{\mathrm {sinh}\left (2\,\ln \left (c\,x\right )\right )}\right )}^{3/2}} \,d x \] Input:

int(x^8/(1/sinh(2*log(c*x)))^(3/2),x)
 

Output:

int(x^8/(1/sinh(2*log(c*x)))^(3/2), x)
 

Reduce [F]

\[ \int \frac {x^8}{\text {csch}^{\frac {3}{2}}(2 \log (c x))} \, dx=\int \frac {\sqrt {\mathrm {csch}\left (2 \,\mathrm {log}\left (c x \right )\right )}\, x^{8}}{\mathrm {csch}\left (2 \,\mathrm {log}\left (c x \right )\right )^{2}}d x \] Input:

int(x^8/csch(2*log(c*x))^(3/2),x)
 

Output:

int((sqrt(csch(2*log(c*x)))*x**8)/csch(2*log(c*x))**2,x)