### 3.258 $$\int \frac{1}{1+\tanh ^3(x)} \, dx$$

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

[Out]

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

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Rubi [A]  time = 0.0662628, antiderivative size = 38, normalized size of antiderivative = 1., 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} $\frac{x}{2}-\frac{1}{6 (\tanh (x)+1)}-\frac{2 \tan ^{-1}\left (\frac{1-2 \tanh (x)}{\sqrt{3}}\right )}{3 \sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

Int[(1 + Tanh[x]^3)^(-1),x]

[Out]

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

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 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 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])

Rubi steps

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

Mathematica [A]  time = 0.0652679, 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 + Tanh[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]))

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Maple [A]  time = 0.024, size = 41, normalized size = 1.1 \begin{align*} -{\frac{1}{6+6\,\tanh \left ( x \right ) }}+{\frac{\ln \left ( 1+\tanh \left ( x \right ) \right ) }{4}}+{\frac{2\,\sqrt{3}}{9}\arctan \left ({\frac{ \left ( 2\,\tanh \left ( x \right ) -1 \right ) \sqrt{3}}{3}} \right ) }-{\frac{\ln \left ( \tanh \left ( x \right ) -1 \right ) }{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.50688, size = 99, normalized size = 2.61 \begin{align*} \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 )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(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)

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Fricas [B]  time = 2.46051, size = 340, normalized size = 8.95 \begin{align*} \frac{18 \, x \cosh \left (x\right )^{2} + 36 \, x \cosh \left (x\right ) \sinh \left (x\right ) + 18 \, x \sinh \left (x\right )^{2} - 8 \,{\left (\sqrt{3} \cosh \left (x\right )^{2} + 2 \, \sqrt{3} \cosh \left (x\right ) \sinh \left (x\right ) + \sqrt{3} \sinh \left (x\right )^{2}\right )} \arctan \left (-\frac{\sqrt{3} \cosh \left (x\right ) + \sqrt{3} \sinh \left (x\right )}{3 \,{\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}}\right ) - 3}{36 \,{\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(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)

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Sympy [B]  time = 0.822486, size = 102, normalized size = 2.68 \begin{align*} \frac{9 x \tanh{\left (x \right )}}{18 \tanh{\left (x \right )} + 18} + \frac{9 x}{18 \tanh{\left (x \right )} + 18} + \frac{4 \sqrt{3} \tanh{\left (x \right )} \operatorname{atan}{\left (\frac{2 \sqrt{3} \tanh{\left (x \right )}}{3} - \frac{\sqrt{3}}{3} \right )}}{18 \tanh{\left (x \right )} + 18} + \frac{4 \sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} \tanh{\left (x \right )}}{3} - \frac{\sqrt{3}}{3} \right )}}{18 \tanh{\left (x \right )} + 18} - \frac{3}{18 \tanh{\left (x \right )} + 18} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(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)

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Giac [A]  time = 1.29784, size = 34, normalized size = 0.89 \begin{align*} \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 )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1+tanh(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)