\(\int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx\) [122]

Optimal result

Integrand size = 11, antiderivative size = 29 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {3 x}{2}-\frac {3 \coth (x)}{2}-\log (\sinh (x))+\frac {\coth (x)}{2 (1+\tanh (x))} \]

[Out]

3/2*x-3/2*coth(x)-ln(sinh(x))+1/2*coth(x)/(1+tanh(x))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3633, 3610, 3612, 3556} \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {3 x}{2}-\frac {3 \coth (x)}{2}-\log (\sinh (x))+\frac {\coth (x)}{2 (\tanh (x)+1)} \]

[In]

Int[Coth[x]^2/(1 + Tanh[x]),x]

[Out]

(3*x)/2 - (3*Coth[x])/2 - Log[Sinh[x]] + Coth[x]/(2*(1 + Tanh[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 3610

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 + 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3612

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

Rule 3633

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

Rubi steps \begin{align*} \text {integral}& = \frac {\coth (x)}{2 (1+\tanh (x))}-\frac {1}{2} \int \coth ^2(x) (-3+2 \tanh (x)) \, dx \\ & = -\frac {3 \coth (x)}{2}+\frac {\coth (x)}{2 (1+\tanh (x))}-\frac {1}{2} i \int \coth (x) (-2 i+3 i \tanh (x)) \, dx \\ & = \frac {3 x}{2}-\frac {3 \coth (x)}{2}+\frac {\coth (x)}{2 (1+\tanh (x))}-\int \coth (x) \, dx \\ & = \frac {3 x}{2}-\frac {3 \coth (x)}{2}-\log (\sinh (x))+\frac {\coth (x)}{2 (1+\tanh (x))} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.62 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \left (\coth ^2(x)+\frac {\coth ^4(x)}{1+\coth (x)}-\coth ^3(x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\tanh ^2(x)\right )-2 (\log (\cosh (x))+\log (\tanh (x)))\right ) \]

[In]

Integrate[Coth[x]^2/(1 + Tanh[x]),x]

[Out]

(Coth[x]^2 + Coth[x]^4/(1 + Coth[x]) - Coth[x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Tanh[x]^2] - 2*(Log[Cosh[x]]
+ Log[Tanh[x]]))/2

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03

method	result	size
risch	\(\frac {5 x}{2}-\frac {{\mathrm e}^{-2 x}}{4}-\frac {2}{{\mathrm e}^{2 x}-1}-\ln \left ({\mathrm e}^{2 x}-1\right )\)	\(30\)
parallelrisch	\(\frac {\left (2+2 \tanh \left (x \right )\right ) \ln \left (1-\tanh \left (x \right )\right )+\left (-2-2 \tanh \left (x \right )\right ) \ln \left (\tanh \left (x \right )\right )+5 \tanh \left (x \right ) x +5 x -2 \coth \left (x \right )-3}{2+2 \tanh \left (x \right )}\)	\(48\)
default	\(-\frac {\tanh \left (\frac {x}{2}\right )}{2}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}-\frac {1}{2 \tanh \left (\frac {x}{2}\right )}-\ln \left (\tanh \left (\frac {x}{2}\right )\right )-\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{\tanh \left (\frac {x}{2}\right )+1}+\frac {5 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}\)	\(59\)

[In]

int(coth(x)^2/(1+tanh(x)),x,method=_RETURNVERBOSE)

[Out]

5/2*x-1/4*exp(-2*x)-2/(exp(2*x)-1)-ln(exp(2*x)-1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 196 vs. \(2 (23) = 46\).

Time = 0.26 (sec) , antiderivative size = 196, normalized size of antiderivative = 6.76 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {10 \, x \cosh \left (x\right )^{4} + 40 \, x \cosh \left (x\right ) \sinh \left (x\right )^{3} + 10 \, x \sinh \left (x\right )^{4} - {\left (10 \, x + 9\right )} \cosh \left (x\right )^{2} + {\left (60 \, x \cosh \left (x\right )^{2} - 10 \, x - 9\right )} \sinh \left (x\right )^{2} - 4 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + {\left (6 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - \cosh \left (x\right )^{2} + 2 \, {\left (2 \, \cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right )\right )} \log \left (\frac {2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + 2 \, {\left (20 \, x \cosh \left (x\right )^{3} - {\left (10 \, x + 9\right )} \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1}{4 \, {\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + {\left (6 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right )^{2} - \cosh \left (x\right )^{2} + 2 \, {\left (2 \, \cosh \left (x\right )^{3} - \cosh \left (x\right )\right )} \sinh \left (x\right )\right )}} \]

[In]

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

[Out]

1/4*(10*x*cosh(x)^4 + 40*x*cosh(x)*sinh(x)^3 + 10*x*sinh(x)^4 - (10*x + 9)*cosh(x)^2 + (60*x*cosh(x)^2 - 10*x
- 9)*sinh(x)^2 - 4*(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 - 1)*sinh(x)^2 - cosh(x)^2 + 2*
(2*cosh(x)^3 - cosh(x))*sinh(x))*log(2*sinh(x)/(cosh(x) - sinh(x))) + 2*(20*x*cosh(x)^3 - (10*x + 9)*cosh(x))*
sinh(x) + 1)/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + (6*cosh(x)^2 - 1)*sinh(x)^2 - cosh(x)^2 + 2*(2*cos
h(x)^3 - cosh(x))*sinh(x))

Sympy [F]

\[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\int \frac {\coth ^{2}{\left (x \right )}}{\tanh {\left (x \right )} + 1}\, dx \]

[In]

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

[Out]

Integral(coth(x)**2/(tanh(x) + 1), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {1}{2} \, x + \frac {2}{e^{\left (-2 \, x\right )} - 1} - \frac {1}{4} \, e^{\left (-2 \, x\right )} - \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {5}{2} \, x - \frac {{\left (9 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-2 \, x\right )}}{4 \, {\left (e^{\left (2 \, x\right )} - 1\right )}} - \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \]

[In]

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

[Out]

5/2*x - 1/4*(9*e^(2*x) - 1)*e^(-2*x)/(e^(2*x) - 1) - log(abs(e^(2*x) - 1))

Mupad [B] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx=\frac {5\,x}{2}-\ln \left ({\mathrm {e}}^{2\,x}-1\right )-\frac {{\mathrm {e}}^{-2\,x}}{4}-\frac {2}{{\mathrm {e}}^{2\,x}-1} \]

[In]

int(coth(x)^2/(tanh(x) + 1),x)

[Out]

(5*x)/2 - log(exp(2*x) - 1) - exp(-2*x)/4 - 2/(exp(2*x) - 1)