3.135 \(\int \frac{\coth (x)}{(1+\coth (x))^{3/2}} \, dx\)

Optimal. Leaf size=49 \[ -\frac{1}{2 \sqrt{\coth (x)+1}}+\frac{1}{3 (\coth (x)+1)^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{\coth (x)+1}}{\sqrt{2}}\right )}{2 \sqrt{2}} \]

[Out]

ArcTanh[Sqrt[1 + Coth[x]]/Sqrt[2]]/(2*Sqrt[2]) + 1/(3*(1 + Coth[x])^(3/2)) - 1/(2*Sqrt[1 + Coth[x]])

________________________________________________________________________________________

Rubi [A]  time = 0.0508973, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {3526, 3479, 3480, 206} \[ -\frac{1}{2 \sqrt{\coth (x)+1}}+\frac{1}{3 (\coth (x)+1)^{3/2}}+\frac{\tanh ^{-1}\left (\frac{\sqrt{\coth (x)+1}}{\sqrt{2}}\right )}{2 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]/(1 + Coth[x])^(3/2),x]

[Out]

ArcTanh[Sqrt[1 + Coth[x]]/Sqrt[2]]/(2*Sqrt[2]) + 1/(3*(1 + Coth[x])^(3/2)) - 1/(2*Sqrt[1 + Coth[x]])

Rule 3526

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^m)/(2*a*f*m), x] + Dist[(b*c + a*d)/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0]

Rule 3479

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

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 206

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

Rubi steps

\begin{align*} \int \frac{\coth (x)}{(1+\coth (x))^{3/2}} \, dx &=\frac{1}{3 (1+\coth (x))^{3/2}}+\frac{1}{2} \int \frac{1}{\sqrt{1+\coth (x)}} \, dx\\ &=\frac{1}{3 (1+\coth (x))^{3/2}}-\frac{1}{2 \sqrt{1+\coth (x)}}+\frac{1}{4} \int \sqrt{1+\coth (x)} \, dx\\ &=\frac{1}{3 (1+\coth (x))^{3/2}}-\frac{1}{2 \sqrt{1+\coth (x)}}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{2-x^2} \, dx,x,\sqrt{1+\coth (x)}\right )\\ &=\frac{\tanh ^{-1}\left (\frac{\sqrt{1+\coth (x)}}{\sqrt{2}}\right )}{2 \sqrt{2}}+\frac{1}{3 (1+\coth (x))^{3/2}}-\frac{1}{2 \sqrt{1+\coth (x)}}\\ \end{align*}

Mathematica [C]  time = 0.33222, size = 84, normalized size = 1.71 \[ \left (\frac{1}{4}+\frac{i}{4}\right ) \sqrt{\coth (x)+1} \left (\left (\frac{1}{6}-\frac{i}{6}\right ) (-\sinh (2 x)-\sinh (4 x)+\cosh (2 x)+\cosh (4 x)-2)-\frac{i \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt{i (\coth (x)+1)}\right )}{\sqrt{i (\coth (x)+1)}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]/(1 + Coth[x])^(3/2),x]

[Out]

(1/4 + I/4)*Sqrt[1 + Coth[x]]*(((-I)*ArcTan[(1/2 + I/2)*Sqrt[I*(1 + Coth[x])]])/Sqrt[I*(1 + Coth[x])] + (1/6 -
I/6)*(-2 + Cosh[2*x] + Cosh[4*x] - Sinh[2*x] - Sinh[4*x]))

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 35, normalized size = 0.7 \begin{align*}{\frac{1}{3} \left ( 1+{\rm coth} \left (x\right ) \right ) ^{-{\frac{3}{2}}}}+{\frac{\sqrt{2}}{4}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{1+{\rm coth} \left (x\right )}} \right ) }-{\frac{1}{2}{\frac{1}{\sqrt{1+{\rm coth} \left (x\right )}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)/(1+coth(x))^(3/2),x)

[Out]

1/3/(1+coth(x))^(3/2)+1/4*arctanh(1/2*(1+coth(x))^(1/2)*2^(1/2))*2^(1/2)-1/2/(1+coth(x))^(1/2)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth \left (x\right )}{{\left (\coth \left (x\right ) + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(1+coth(x))^(3/2),x, algorithm="maxima")

[Out]

integrate(coth(x)/(coth(x) + 1)^(3/2), x)

________________________________________________________________________________________

Fricas [B]  time = 2.95701, size = 578, normalized size = 11.8 \begin{align*} -\frac{2 \, \sqrt{2}{\left (2 \, \sqrt{2} \cosh \left (x\right )^{2} + 4 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sqrt{2} \sinh \left (x\right )^{2} + \sqrt{2}\right )} \sqrt{\frac{\sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}} - 3 \,{\left (\sqrt{2} \cosh \left (x\right )^{3} + 3 \, \sqrt{2} \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, \sqrt{2} \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sqrt{2} \sinh \left (x\right )^{3}\right )} \log \left (2 \, \sqrt{2} \sqrt{\frac{\sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}}{\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} + 2 \, \cosh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sinh \left (x\right )^{2} - 1\right )}{24 \,{\left (\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right )^{2} \sinh \left (x\right ) + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(1+coth(x))^(3/2),x, algorithm="fricas")

[Out]

-1/24*(2*sqrt(2)*(2*sqrt(2)*cosh(x)^2 + 4*sqrt(2)*cosh(x)*sinh(x) + 2*sqrt(2)*sinh(x)^2 + sqrt(2))*sqrt(sinh(x
)/(cosh(x) - sinh(x))) - 3*(sqrt(2)*cosh(x)^3 + 3*sqrt(2)*cosh(x)^2*sinh(x) + 3*sqrt(2)*cosh(x)*sinh(x)^2 + sq
rt(2)*sinh(x)^3)*log(2*sqrt(2)*sqrt(sinh(x)/(cosh(x) - sinh(x)))*(cosh(x) + sinh(x)) + 2*cosh(x)^2 + 4*cosh(x)
*sinh(x) + 2*sinh(x)^2 - 1))/(cosh(x)^3 + 3*cosh(x)^2*sinh(x) + 3*cosh(x)*sinh(x)^2 + sinh(x)^3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth{\left (x \right )}}{\left (\coth{\left (x \right )} + 1\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(1+coth(x))**(3/2),x)

[Out]

Integral(coth(x)/(coth(x) + 1)**(3/2), x)

________________________________________________________________________________________

Giac [B]  time = 1.18417, size = 144, normalized size = 2.94 \begin{align*} -\frac{1}{24} \, \sqrt{2}{\left (\frac{3 \, \log \left ({\left | 2 \, \sqrt{e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - 2 \, e^{\left (2 \, x\right )} + 1 \right |}\right )}{\mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right )} + \frac{2 \,{\left (3 \, \sqrt{e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - 3 \, e^{\left (2 \, x\right )} + 1\right )}}{{\left (\sqrt{e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )}} - e^{\left (2 \, x\right )}\right )}^{3} \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right )} - 4 \, \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(1+coth(x))^(3/2),x, algorithm="giac")

[Out]

-1/24*sqrt(2)*(3*log(abs(2*sqrt(e^(4*x) - e^(2*x)) - 2*e^(2*x) + 1))/sgn(e^(2*x) - 1) + 2*(3*sqrt(e^(4*x) - e^
(2*x)) - 3*e^(2*x) + 1)/((sqrt(e^(4*x) - e^(2*x)) - e^(2*x))^3*sgn(e^(2*x) - 1)) - 4*sgn(e^(2*x) - 1))