\(\int \frac {\coth (x)}{1+\tanh (x)} \, dx\) [121]

Optimal result

Integrand size = 9, antiderivative size = 19 \[ \int \frac {\coth (x)}{1+\tanh (x)} \, dx=-\frac {x}{2}+\log (\sinh (x))+\frac {1}{2 (1+\tanh (x))} \]

[Out]

-1/2*x+ln(sinh(x))+1/2/(1+tanh(x))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3632, 3560, 8, 3556} \[ \int \frac {\coth (x)}{1+\tanh (x)} \, dx=-\frac {x}{2}+\frac {1}{2 (\tanh (x)+1)}+\log (\sinh (x)) \]

[In]

Int[Coth[x]/(1 + Tanh[x]),x]

[Out]

-1/2*x + Log[Sinh[x]] + 1/(2*(1 + Tanh[x]))

Rule 8

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

Rule 3556

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

Rule 3560

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 3632

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

Rubi steps \begin{align*} \text {integral}& = \int \coth (x) \, dx-\int \frac {1}{1+\tanh (x)} \, dx \\ & = \log (\sinh (x))+\frac {1}{2 (1+\tanh (x))}-\frac {\int 1 \, dx}{2} \\ & = -\frac {x}{2}+\log (\sinh (x))+\frac {1}{2 (1+\tanh (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.79 \[ \int \frac {\coth (x)}{1+\tanh (x)} \, dx=-\frac {1}{4} \log (1-\tanh (x))+\log (\tanh (x))-\frac {3}{4} \log (1+\tanh (x))+\frac {1}{2 (1+\tanh (x))} \]

[In]

Integrate[Coth[x]/(1 + Tanh[x]),x]

[Out]

-1/4*Log[1 - Tanh[x]] + Log[Tanh[x]] - (3*Log[1 + Tanh[x]])/4 + 1/(2*(1 + Tanh[x]))

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95

method	result	size
risch	\(-\frac {3 x}{2}+\frac {{\mathrm e}^{-2 x}}{4}+\ln \left ({\mathrm e}^{2 x}-1\right )\)	\(18\)
default	\(\ln \left (\tanh \left (\frac {x}{2}\right )\right )-\frac {1}{\tanh \left (\frac {x}{2}\right )+1}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {3 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}\)	\(43\)
parallelrisch	\(\frac {\left (-2-2 \tanh \left (x \right )\right ) \ln \left (1-\tanh \left (x \right )\right )+\left (2+2 \tanh \left (x \right )\right ) \ln \left (\tanh \left (x \right )\right )-3 \tanh \left (x \right ) x -3 x +1}{2+2 \tanh \left (x \right )}\)	\(44\)

[In]

int(coth(x)/(1+tanh(x)),x,method=_RETURNVERBOSE)

[Out]

-3/2*x+1/4*exp(-2*x)+ln(exp(2*x)-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (15) = 30\).

Time = 0.26 (sec) , antiderivative size = 73, normalized size of antiderivative = 3.84 \[ \int \frac {\coth (x)}{1+\tanh (x)} \, dx=-\frac {6 \, x \cosh \left (x\right )^{2} + 12 \, x \cosh \left (x\right ) \sinh \left (x\right ) + 6 \, x \sinh \left (x\right )^{2} - 4 \, {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )} \log \left (\frac {2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - 1}{4 \, {\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2}\right )}} \]

[In]

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

[Out]

-1/4*(6*x*cosh(x)^2 + 12*x*cosh(x)*sinh(x) + 6*x*sinh(x)^2 - 4*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)*log
(2*sinh(x)/(cosh(x) - sinh(x))) - 1)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)

Sympy [F]

\[ \int \frac {\coth (x)}{1+\tanh (x)} \, dx=\int \frac {\coth {\left (x \right )}}{\tanh {\left (x \right )} + 1}\, dx \]

[In]

integrate(coth(x)/(1+tanh(x)),x)

[Out]

Integral(coth(x)/(tanh(x) + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.26 \[ \int \frac {\coth (x)}{1+\tanh (x)} \, dx=\frac {1}{2} \, x + \frac {1}{4} \, e^{\left (-2 \, x\right )} + \log \left (e^{\left (-x\right )} + 1\right ) + \log \left (e^{\left (-x\right )} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {\coth (x)}{1+\tanh (x)} \, dx=-\frac {3}{2} \, x + \frac {1}{4} \, e^{\left (-2 \, x\right )} + \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {\coth (x)}{1+\tanh (x)} \, dx=\ln \left ({\mathrm {e}}^{2\,x}-1\right )-\frac {3\,x}{2}+\frac {{\mathrm {e}}^{-2\,x}}{4} \]

[In]

int(coth(x)/(tanh(x) + 1),x)

[Out]

log(exp(2*x) - 1) - (3*x)/2 + exp(-2*x)/4