3.111 \(\int \frac{\text{sech}^3(x)}{1+\coth (x)} \, dx\)

Optimal. Leaf size=20 \[ -\text{sech}(x)-\frac{1}{2} \tan ^{-1}(\sinh (x))+\frac{1}{2} \tanh (x) \text{sech}(x) \]

[Out]

-ArcTan[Sinh[x]]/2 - Sech[x] + (Sech[x]*Tanh[x])/2

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Rubi [A]  time = 0.167309, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {3518, 3108, 3107, 2606, 8, 2611, 3770} \[ -\text{sech}(x)-\frac{1}{2} \tan ^{-1}(\sinh (x))+\frac{1}{2} \tanh (x) \text{sech}(x) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(1 + Coth[x]),x]

[Out]

-ArcTan[Sinh[x]]/2 - Sech[x] + (Sech[x]*Tanh[x])/2

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3108

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[a^p*b^p, Int[(Cos[c + d*x]^m*Sin[c + d*x]^n)/(b*Cos[c + d*x] + a*Sin
[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[a^2 + b^2, 0] && ILtQ[p, 0]

Rule 3107

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_
.) + (d_.)*(x_)])^(p_.), x_Symbol] :> Int[ExpandTrig[cos[c + d*x]^m*sin[c + d*x]^n*(a*cos[c + d*x] + b*sin[c +
d*x])^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IGtQ[p, 0]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\text{sech}^3(x)}{1+\coth (x)} \, dx &=-\left (i \int \frac{\text{sech}^2(x) \tanh (x)}{-i \cosh (x)-i \sinh (x)} \, dx\right )\\ &=-\int \text{sech}^2(x) (-\cosh (x)+\sinh (x)) \tanh (x) \, dx\\ &=i \int \left (-i \text{sech}(x) \tanh (x)+i \text{sech}(x) \tanh ^2(x)\right ) \, dx\\ &=\int \text{sech}(x) \tanh (x) \, dx-\int \text{sech}(x) \tanh ^2(x) \, dx\\ &=\frac{1}{2} \text{sech}(x) \tanh (x)-\frac{1}{2} \int \text{sech}(x) \, dx-\operatorname{Subst}(\int 1 \, dx,x,\text{sech}(x))\\ &=-\frac{1}{2} \tan ^{-1}(\sinh (x))-\text{sech}(x)+\frac{1}{2} \text{sech}(x) \tanh (x)\\ \end{align*}

Mathematica [A]  time = 0.0391703, size = 20, normalized size = 1. \[ \frac{1}{2} (\tanh (x)-2) \text{sech}(x)-\tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^3/(1 + Coth[x]),x]

[Out]

-ArcTan[Tanh[x/2]] + (Sech[x]*(-2 + Tanh[x]))/2

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Maple [B]  time = 0.029, size = 45, normalized size = 2.3 \begin{align*} 4\,{\frac{-1/4\, \left ( \tanh \left ( x/2 \right ) \right ) ^{3}-1/2\, \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1/4\,\tanh \left ( x/2 \right ) -1/2}{ \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.52306, size = 45, normalized size = 2.25 \begin{align*} -\frac{e^{\left (-x\right )} + 3 \, e^{\left (-3 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} + \arctan \left (e^{\left (-x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(e^(-x) + 3*e^(-3*x))/(2*e^(-2*x) + e^(-4*x) + 1) + arctan(e^(-x))

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Fricas [B]  time = 2.5068, size = 508, normalized size = 25.4 \begin{align*} -\frac{\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3} +{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + 3 \,{\left (\cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right ) + 3 \, \cosh \left (x\right )}{\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cosh(x)^3 + 3*cosh(x)*sinh(x)^2 + sinh(x)^3 + (cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2
+ 1)*sinh(x)^2 + 2*cosh(x)^2 + 4*(cosh(x)^3 + cosh(x))*sinh(x) + 1)*arctan(cosh(x) + sinh(x)) + 3*(cosh(x)^2 +
1)*sinh(x) + 3*cosh(x))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 + 1)*sinh(x)^2 + 2*cosh
(x)^2 + 4*(cosh(x)^3 + cosh(x))*sinh(x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{3}{\left (x \right )}}{\coth{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(x)**3/(coth(x) + 1), x)

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Giac [A]  time = 1.14922, size = 34, normalized size = 1.7 \begin{align*} -\frac{e^{\left (3 \, x\right )} + 3 \, e^{x}}{{\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} - \arctan \left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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