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

Optimal. Leaf size=29 \[ \frac {3 x}{2}-\frac {3 \coth (x)}{2}-\log (\sinh (x))+\frac {\coth (x)}{2 (\tanh (x)+1)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 29, normalized size of antiderivative = 1.00, 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 = {3552, 3529, 3531, 3475} \[ \frac {3 x}{2}-\frac {3 \coth (x)}{2}-\log (\sinh (x))+\frac {\coth (x)}{2 (\tanh (x)+1)} \]

Antiderivative was successfully verified.

[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 3475

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

Rule 3529

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 3531

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 3552

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*} \int \frac {\coth ^2(x)}{1+\tanh (x)} \, dx &=\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 [A]  time = 0.05, size = 27, normalized size = 0.93 \[ \frac {1}{4} (6 x+\sinh (2 x)-\cosh (2 x)-4 \coth (x)-4 \log (\sinh (x))) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^2/(1 + Tanh[x]),x]

[Out]

(6*x - Cosh[2*x] - 4*Coth[x] - 4*Log[Sinh[x]] + Sinh[2*x])/4

________________________________________________________________________________________

fricas [B]  time = 0.58, size = 196, normalized size = 6.76 \[ \frac {10 \, x \cosh \relax (x)^{4} + 40 \, x \cosh \relax (x) \sinh \relax (x)^{3} + 10 \, x \sinh \relax (x)^{4} - {\left (10 \, x + 9\right )} \cosh \relax (x)^{2} + {\left (60 \, x \cosh \relax (x)^{2} - 10 \, x - 9\right )} \sinh \relax (x)^{2} - 4 \, {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4} + {\left (6 \, \cosh \relax (x)^{2} - 1\right )} \sinh \relax (x)^{2} - \cosh \relax (x)^{2} + 2 \, {\left (2 \, \cosh \relax (x)^{3} - \cosh \relax (x)\right )} \sinh \relax (x)\right )} \log \left (\frac {2 \, \sinh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 2 \, {\left (20 \, x \cosh \relax (x)^{3} - {\left (10 \, x + 9\right )} \cosh \relax (x)\right )} \sinh \relax (x) + 1}{4 \, {\left (\cosh \relax (x)^{4} + 4 \, \cosh \relax (x) \sinh \relax (x)^{3} + \sinh \relax (x)^{4} + {\left (6 \, \cosh \relax (x)^{2} - 1\right )} \sinh \relax (x)^{2} - \cosh \relax (x)^{2} + 2 \, {\left (2 \, \cosh \relax (x)^{3} - \cosh \relax (x)\right )} \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

giac [A]  time = 0.14, size = 36, normalized size = 1.24 \[ \frac {5}{2} \, x - \frac {{\left (9 \, e^{\left (2 \, x\right )} - 1\right )} e^{\left (-2 \, x\right )}}{4 \, {\left (e^{\left (2 \, x\right )} - 1\right )}} - \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

maple [B]  time = 0.09, size = 59, normalized size = 2.03 \[ -\frac {\tanh \left (\frac {x}{2}\right )}{2}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}-\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{\tanh \left (\frac {x}{2}\right )+1}+\frac {5 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}-\frac {1}{2 \tanh \left (\frac {x}{2}\right )}-\ln \left (\tanh \left (\frac {x}{2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*tanh(1/2*x)-1/2*ln(tanh(1/2*x)-1)-1/(tanh(1/2*x)+1)^2+1/(tanh(1/2*x)+1)+5/2*ln(tanh(1/2*x)+1)-1/2/tanh(1/
2*x)-ln(tanh(1/2*x))

________________________________________________________________________________________

maxima [A]  time = 0.31, size = 38, normalized size = 1.31 \[ \frac {1}{2} \, x + \frac {2}{e^{\left (-2 \, x\right )} - 1} - \frac {1}{4} \, e^{\left (-2 \, x\right )} - \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [B]  time = 1.03, size = 29, normalized size = 1.00 \[ \frac {5\,x}{2}-\ln \left ({\mathrm {e}}^{2\,x}-1\right )-\frac {{\mathrm {e}}^{-2\,x}}{4}-\frac {2}{{\mathrm {e}}^{2\,x}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{2}{\relax (x )}}{\tanh {\relax (x )} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(coth(x)**2/(tanh(x) + 1), x)

________________________________________________________________________________________