3.50 \(\int \frac {1}{1+\coth ^3(x)} \, dx\)

Optimal. Leaf size=38 \[ \frac {x}{2}-\frac {1}{6 (\coth (x)+1)}-\frac {2 \tan ^{-1}\left (\frac {1-2 \coth (x)}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

[Out]

1/2*x-1/6/(1+coth(x))-2/9*arctan(1/3*(1-2*coth(x))*3^(1/2))*3^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {3661, 2074, 207, 618, 204} \[ \frac {x}{2}-\frac {1}{6 (\coth (x)+1)}-\frac {2 \tan ^{-1}\left (\frac {1-2 \coth (x)}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Coth[x]^3)^(-1),x]

[Out]

x/2 - (2*ArcTan[(1 - 2*Coth[x])/Sqrt[3]])/(3*Sqrt[3]) - 1/(6*(1 + Coth[x]))

Rule 204

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

Rule 207

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \frac {1}{1+\coth ^3(x)} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \left (1+x^3\right )} \, dx,x,\coth (x)\right )\\ &=\operatorname {Subst}\left (\int \left (\frac {1}{6 (1+x)^2}-\frac {1}{2 \left (-1+x^2\right )}+\frac {1}{3 \left (1-x+x^2\right )}\right ) \, dx,x,\coth (x)\right )\\ &=-\frac {1}{6 (1+\coth (x))}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,\coth (x)\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\coth (x)\right )\\ &=\frac {x}{2}-\frac {1}{6 (1+\coth (x))}-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 \coth (x)\right )\\ &=\frac {x}{2}-\frac {2 \tan ^{-1}\left (\frac {1-2 \coth (x)}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {1}{6 (1+\coth (x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 40, normalized size = 1.05 \[ \frac {1}{2} \tanh ^{-1}(\tanh (x))+\frac {1}{6 (\tanh (x)+1)}+\frac {2 \tan ^{-1}\left (\frac {1-2 \tanh (x)}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Coth[x]^3)^(-1),x]

[Out]

(2*ArcTan[(1 - 2*Tanh[x])/Sqrt[3]])/(3*Sqrt[3]) + ArcTanh[Tanh[x]]/2 + 1/(6*(1 + Tanh[x]))

________________________________________________________________________________________

fricas [B]  time = 0.45, size = 95, normalized size = 2.50 \[ \frac {18 \, x \cosh \relax (x)^{2} + 36 \, x \cosh \relax (x) \sinh \relax (x) + 18 \, x \sinh \relax (x)^{2} + 8 \, {\left (\sqrt {3} \cosh \relax (x)^{2} + 2 \, \sqrt {3} \cosh \relax (x) \sinh \relax (x) + \sqrt {3} \sinh \relax (x)^{2}\right )} \arctan \left (-\frac {\sqrt {3} \cosh \relax (x) + \sqrt {3} \sinh \relax (x)}{3 \, {\left (\cosh \relax (x) - \sinh \relax (x)\right )}}\right ) + 3}{36 \, {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+coth(x)^3),x, algorithm="fricas")

[Out]

1/36*(18*x*cosh(x)^2 + 36*x*cosh(x)*sinh(x) + 18*x*sinh(x)^2 + 8*(sqrt(3)*cosh(x)^2 + 2*sqrt(3)*cosh(x)*sinh(x
) + sqrt(3)*sinh(x)^2)*arctan(-1/3*(sqrt(3)*cosh(x) + sqrt(3)*sinh(x))/(cosh(x) - sinh(x))) + 3)/(cosh(x)^2 +
2*cosh(x)*sinh(x) + sinh(x)^2)

________________________________________________________________________________________

giac [A]  time = 0.13, size = 25, normalized size = 0.66 \[ -\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} e^{\left (2 \, x\right )}\right ) + \frac {1}{2} \, x + \frac {1}{12} \, e^{\left (-2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+coth(x)^3),x, algorithm="giac")

[Out]

-2/9*sqrt(3)*arctan(1/3*sqrt(3)*e^(2*x)) + 1/2*x + 1/12*e^(-2*x)

________________________________________________________________________________________

maple [A]  time = 0.08, size = 41, normalized size = 1.08 \[ -\frac {\ln \left (\coth \relax (x )-1\right )}{4}-\frac {1}{6 \left (1+\coth \relax (x )\right )}+\frac {\ln \left (1+\coth \relax (x )\right )}{4}+\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 \coth \relax (x )-1\right ) \sqrt {3}}{3}\right )}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1+coth(x)^3),x)

[Out]

-1/4*ln(coth(x)-1)-1/6/(1+coth(x))+1/4*ln(1+coth(x))+2/9*3^(1/2)*arctan(1/3*(2*coth(x)-1)*3^(1/2))

________________________________________________________________________________________

maxima [B]  time = 0.41, size = 73, normalized size = 1.92 \[ -\frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (2 \, \sqrt {3} e^{\left (-x\right )} + 3^{\frac {1}{4}} \sqrt {2}\right )}\right ) + \frac {2}{9} \, \sqrt {3} \arctan \left (\frac {1}{6} \cdot 3^{\frac {3}{4}} \sqrt {2} {\left (2 \, \sqrt {3} e^{\left (-x\right )} - 3^{\frac {1}{4}} \sqrt {2}\right )}\right ) + \frac {1}{2} \, x + \frac {1}{12} \, e^{\left (-2 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+coth(x)^3),x, algorithm="maxima")

[Out]

-2/9*sqrt(3)*arctan(1/6*3^(3/4)*sqrt(2)*(2*sqrt(3)*e^(-x) + 3^(1/4)*sqrt(2))) + 2/9*sqrt(3)*arctan(1/6*3^(3/4)
*sqrt(2)*(2*sqrt(3)*e^(-x) - 3^(1/4)*sqrt(2))) + 1/2*x + 1/12*e^(-2*x)

________________________________________________________________________________________

mupad [B]  time = 0.09, size = 38, normalized size = 1.00 \[ \frac {\frac {x}{2}+\frac {\mathrm {coth}\relax (x)}{6}+\frac {x\,\mathrm {coth}\relax (x)}{2}}{\mathrm {coth}\relax (x)+1}+\frac {2\,\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}\,\left (2\,\mathrm {coth}\relax (x)-1\right )}{3}\right )}{9} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(coth(x)^3 + 1),x)

[Out]

(x/2 + coth(x)/6 + (x*coth(x))/2)/(coth(x) + 1) + (2*3^(1/2)*atan((3^(1/2)*(2*coth(x) - 1))/3))/9

________________________________________________________________________________________

sympy [B]  time = 0.98, size = 102, normalized size = 2.68 \[ \frac {9 x \tanh {\relax (x )}}{18 \tanh {\relax (x )} + 18} + \frac {9 x}{18 \tanh {\relax (x )} + 18} - \frac {4 \sqrt {3} \tanh {\relax (x )} \operatorname {atan}{\left (\frac {2 \sqrt {3} \tanh {\relax (x )}}{3} - \frac {\sqrt {3}}{3} \right )}}{18 \tanh {\relax (x )} + 18} - \frac {4 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} \tanh {\relax (x )}}{3} - \frac {\sqrt {3}}{3} \right )}}{18 \tanh {\relax (x )} + 18} + \frac {3}{18 \tanh {\relax (x )} + 18} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+coth(x)**3),x)

[Out]

9*x*tanh(x)/(18*tanh(x) + 18) + 9*x/(18*tanh(x) + 18) - 4*sqrt(3)*tanh(x)*atan(2*sqrt(3)*tanh(x)/3 - sqrt(3)/3
)/(18*tanh(x) + 18) - 4*sqrt(3)*atan(2*sqrt(3)*tanh(x)/3 - sqrt(3)/3)/(18*tanh(x) + 18) + 3/(18*tanh(x) + 18)

________________________________________________________________________________________