3.123 \(\int \frac {\coth ^3(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=37 \[ -\frac {3 x}{2}-\coth ^2(x)+\frac {3 \coth (x)}{2}+2 \log (\sinh (x))+\frac {\coth ^2(x)}{2 (\tanh (x)+1)} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3552, 3529, 3531, 3475} \[ -\frac {3 x}{2}-\coth ^2(x)+\frac {3 \coth (x)}{2}+2 \log (\sinh (x))+\frac {\coth ^2(x)}{2 (\tanh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^3/(1 + Tanh[x]),x]

[Out]

(-3*x)/2 + (3*Coth[x])/2 - Coth[x]^2 + 2*Log[Sinh[x]] + Coth[x]^2/(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 ^3(x)}{1+\tanh (x)} \, dx &=\frac {\coth ^2(x)}{2 (1+\tanh (x))}-\frac {1}{2} \int \coth ^3(x) (-4+3 \tanh (x)) \, dx\\ &=-\coth ^2(x)+\frac {\coth ^2(x)}{2 (1+\tanh (x))}-\frac {1}{2} i \int \coth ^2(x) (-3 i+4 i \tanh (x)) \, dx\\ &=\frac {3 \coth (x)}{2}-\coth ^2(x)+\frac {\coth ^2(x)}{2 (1+\tanh (x))}+\frac {1}{2} \int \coth (x) (4-3 \tanh (x)) \, dx\\ &=-\frac {3 x}{2}+\frac {3 \coth (x)}{2}-\coth ^2(x)+\frac {\coth ^2(x)}{2 (1+\tanh (x))}+2 \int \coth (x) \, dx\\ &=-\frac {3 x}{2}+\frac {3 \coth (x)}{2}-\coth ^2(x)+2 \log (\sinh (x))+\frac {\coth ^2(x)}{2 (1+\tanh (x))}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 33, normalized size = 0.89 \[ \frac {1}{4} \left (-6 x-\sinh (2 x)+\cosh (2 x)+4 \coth (x)-2 \text {csch}^2(x)+8 \log (\sinh (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^3/(1 + Tanh[x]),x]

[Out]

(-6*x + Cosh[2*x] + 4*Coth[x] - 2*Csch[x]^2 + 8*Log[Sinh[x]] - Sinh[2*x])/4

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fricas [B]  time = 0.54, size = 357, normalized size = 9.65 \[ -\frac {14 \, x \cosh \relax (x)^{6} + 84 \, x \cosh \relax (x) \sinh \relax (x)^{5} + 14 \, x \sinh \relax (x)^{6} - {\left (28 \, x + 1\right )} \cosh \relax (x)^{4} + {\left (210 \, x \cosh \relax (x)^{2} - 28 \, x - 1\right )} \sinh \relax (x)^{4} + 4 \, {\left (70 \, x \cosh \relax (x)^{3} - {\left (28 \, x + 1\right )} \cosh \relax (x)\right )} \sinh \relax (x)^{3} + 2 \, {\left (7 \, x + 5\right )} \cosh \relax (x)^{2} + 2 \, {\left (105 \, x \cosh \relax (x)^{4} - 3 \, {\left (28 \, x + 1\right )} \cosh \relax (x)^{2} + 7 \, x + 5\right )} \sinh \relax (x)^{2} - 8 \, {\left (\cosh \relax (x)^{6} + 6 \, \cosh \relax (x) \sinh \relax (x)^{5} + \sinh \relax (x)^{6} + {\left (15 \, \cosh \relax (x)^{2} - 2\right )} \sinh \relax (x)^{4} - 2 \, \cosh \relax (x)^{4} + 4 \, {\left (5 \, \cosh \relax (x)^{3} - 2 \, \cosh \relax (x)\right )} \sinh \relax (x)^{3} + {\left (15 \, \cosh \relax (x)^{4} - 12 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{2} + \cosh \relax (x)^{2} + 2 \, {\left (3 \, \cosh \relax (x)^{5} - 4 \, \cosh \relax (x)^{3} + \cosh \relax (x)\right )} \sinh \relax (x)\right )} \log \left (\frac {2 \, \sinh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) + 4 \, {\left (21 \, x \cosh \relax (x)^{5} - {\left (28 \, x + 1\right )} \cosh \relax (x)^{3} + {\left (7 \, x + 5\right )} \cosh \relax (x)\right )} \sinh \relax (x) - 1}{4 \, {\left (\cosh \relax (x)^{6} + 6 \, \cosh \relax (x) \sinh \relax (x)^{5} + \sinh \relax (x)^{6} + {\left (15 \, \cosh \relax (x)^{2} - 2\right )} \sinh \relax (x)^{4} - 2 \, \cosh \relax (x)^{4} + 4 \, {\left (5 \, \cosh \relax (x)^{3} - 2 \, \cosh \relax (x)\right )} \sinh \relax (x)^{3} + {\left (15 \, \cosh \relax (x)^{4} - 12 \, \cosh \relax (x)^{2} + 1\right )} \sinh \relax (x)^{2} + \cosh \relax (x)^{2} + 2 \, {\left (3 \, \cosh \relax (x)^{5} - 4 \, \cosh \relax (x)^{3} + \cosh \relax (x)\right )} \sinh \relax (x)\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(14*x*cosh(x)^6 + 84*x*cosh(x)*sinh(x)^5 + 14*x*sinh(x)^6 - (28*x + 1)*cosh(x)^4 + (210*x*cosh(x)^2 - 28*
x - 1)*sinh(x)^4 + 4*(70*x*cosh(x)^3 - (28*x + 1)*cosh(x))*sinh(x)^3 + 2*(7*x + 5)*cosh(x)^2 + 2*(105*x*cosh(x
)^4 - 3*(28*x + 1)*cosh(x)^2 + 7*x + 5)*sinh(x)^2 - 8*(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 + sinh(x)^6 + (15*cosh(
x)^2 - 2)*sinh(x)^4 - 2*cosh(x)^4 + 4*(5*cosh(x)^3 - 2*cosh(x))*sinh(x)^3 + (15*cosh(x)^4 - 12*cosh(x)^2 + 1)*
sinh(x)^2 + cosh(x)^2 + 2*(3*cosh(x)^5 - 4*cosh(x)^3 + cosh(x))*sinh(x))*log(2*sinh(x)/(cosh(x) - sinh(x))) +
4*(21*x*cosh(x)^5 - (28*x + 1)*cosh(x)^3 + (7*x + 5)*cosh(x))*sinh(x) - 1)/(cosh(x)^6 + 6*cosh(x)*sinh(x)^5 +
sinh(x)^6 + (15*cosh(x)^2 - 2)*sinh(x)^4 - 2*cosh(x)^4 + 4*(5*cosh(x)^3 - 2*cosh(x))*sinh(x)^3 + (15*cosh(x)^4
- 12*cosh(x)^2 + 1)*sinh(x)^2 + cosh(x)^2 + 2*(3*cosh(x)^5 - 4*cosh(x)^3 + cosh(x))*sinh(x))

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giac [A]  time = 0.14, size = 40, normalized size = 1.08 \[ -\frac {7}{2} \, x + \frac {{\left (e^{\left (4 \, x\right )} - 10 \, e^{\left (2 \, x\right )} + 1\right )} e^{\left (-2 \, x\right )}}{4 \, {\left (e^{\left (2 \, x\right )} - 1\right )}^{2}} + 2 \, \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/2*x + 1/4*(e^(4*x) - 10*e^(2*x) + 1)*e^(-2*x)/(e^(2*x) - 1)^2 + 2*log(abs(e^(2*x) - 1))

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maple [B]  time = 0.10, size = 75, normalized size = 2.03 \[ -\frac {\left (\tanh ^{2}\left (\frac {x}{2}\right )\right )}{8}+\frac {\tanh \left (\frac {x}{2}\right )}{2}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2}+\frac {1}{\left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {1}{\tanh \left (\frac {x}{2}\right )+1}-\frac {7 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2}-\frac {1}{8 \tanh \left (\frac {x}{2}\right )^{2}}+\frac {1}{2 \tanh \left (\frac {x}{2}\right )}+2 \ln \left (\tanh \left (\frac {x}{2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*tanh(1/2*x)^2+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)-7/2*ln(tanh(1/2
*x)+1)-1/8/tanh(1/2*x)^2+1/2/tanh(1/2*x)+2*ln(tanh(1/2*x))

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maxima [A]  time = 0.30, size = 54, normalized size = 1.46 \[ \frac {1}{2} \, x + \frac {2 \, {\left (2 \, e^{\left (-2 \, x\right )} - 1\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} + \frac {1}{4} \, e^{\left (-2 \, x\right )} + 2 \, \log \left (e^{\left (-x\right )} + 1\right ) + 2 \, \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 35, normalized size = 0.95 \[ 2\,\ln \left ({\mathrm {e}}^{2\,x}-1\right )-\frac {7\,x}{2}+\frac {{\mathrm {e}}^{-2\,x}}{4}-\frac {2}{{\mathrm {e}}^{4\,x}-2\,{\mathrm {e}}^{2\,x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^3/(tanh(x) + 1),x)

[Out]

2*log(exp(2*x) - 1) - (7*x)/2 + exp(-2*x)/4 - 2/(exp(4*x) - 2*exp(2*x) + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{3}{\relax (x )}}{\tanh {\relax (x )} + 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(coth(x)**3/(tanh(x) + 1), x)

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