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

Optimal. Leaf size=36 \[ \frac {x}{8}-\frac {1}{8 (\coth (x)+1)}-\frac {1}{8 (\coth (x)+1)^2}-\frac {1}{6 (\coth (x)+1)^3} \]

[Out]

1/8*x-1/6/(1+coth(x))^3-1/8/(1+coth(x))^2-1/8/(1+coth(x))

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Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3479, 8} \[ \frac {x}{8}-\frac {1}{8 (\coth (x)+1)}-\frac {1}{8 (\coth (x)+1)^2}-\frac {1}{6 (\coth (x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Int[(1 + Coth[x])^(-3),x]

[Out]

x/8 - 1/(6*(1 + Coth[x])^3) - 1/(8*(1 + Coth[x])^2) - 1/(8*(1 + Coth[x]))

Rule 8

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

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]

Rubi steps

\begin {align*} \int \frac {1}{(1+\coth (x))^3} \, dx &=-\frac {1}{6 (1+\coth (x))^3}+\frac {1}{2} \int \frac {1}{(1+\coth (x))^2} \, dx\\ &=-\frac {1}{6 (1+\coth (x))^3}-\frac {1}{8 (1+\coth (x))^2}+\frac {1}{4} \int \frac {1}{1+\coth (x)} \, dx\\ &=-\frac {1}{6 (1+\coth (x))^3}-\frac {1}{8 (1+\coth (x))^2}-\frac {1}{8 (1+\coth (x))}+\frac {\int 1 \, dx}{8}\\ &=\frac {x}{8}-\frac {1}{6 (1+\coth (x))^3}-\frac {1}{8 (1+\coth (x))^2}-\frac {1}{8 (1+\coth (x))}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 44, normalized size = 1.22 \[ \frac {1}{96} (12 x-18 \sinh (2 x)+9 \sinh (4 x)-2 \sinh (6 x)+18 \cosh (2 x)-9 \cosh (4 x)+2 \cosh (6 x)) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Coth[x])^(-3),x]

[Out]

(12*x + 18*Cosh[2*x] - 9*Cosh[4*x] + 2*Cosh[6*x] - 18*Sinh[2*x] + 9*Sinh[4*x] - 2*Sinh[6*x])/96

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fricas [B]  time = 0.38, size = 86, normalized size = 2.39 \[ \frac {2 \, {\left (6 \, x + 1\right )} \cosh \relax (x)^{3} + 6 \, {\left (6 \, x + 1\right )} \cosh \relax (x) \sinh \relax (x)^{2} + 2 \, {\left (6 \, x - 1\right )} \sinh \relax (x)^{3} + 3 \, {\left (2 \, {\left (6 \, x - 1\right )} \cosh \relax (x)^{2} + 9\right )} \sinh \relax (x) + 9 \, \cosh \relax (x)}{96 \, {\left (\cosh \relax (x)^{3} + 3 \, \cosh \relax (x)^{2} \sinh \relax (x) + 3 \, \cosh \relax (x) \sinh \relax (x)^{2} + \sinh \relax (x)^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(2*(6*x + 1)*cosh(x)^3 + 6*(6*x + 1)*cosh(x)*sinh(x)^2 + 2*(6*x - 1)*sinh(x)^3 + 3*(2*(6*x - 1)*cosh(x)^2
+ 9)*sinh(x) + 9*cosh(x))/(cosh(x)^3 + 3*cosh(x)^2*sinh(x) + 3*cosh(x)*sinh(x)^2 + sinh(x)^3)

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giac [A]  time = 0.13, size = 24, normalized size = 0.67 \[ \frac {1}{96} \, {\left (18 \, e^{\left (4 \, x\right )} - 9 \, e^{\left (2 \, x\right )} + 2\right )} e^{\left (-6 \, x\right )} + \frac {1}{8} \, x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(18*e^(4*x) - 9*e^(2*x) + 2)*e^(-6*x) + 1/8*x

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maple [A]  time = 0.05, size = 40, normalized size = 1.11 \[ -\frac {\ln \left (\coth \relax (x )-1\right )}{16}-\frac {1}{6 \left (1+\coth \relax (x )\right )^{3}}-\frac {1}{8 \left (1+\coth \relax (x )\right )^{2}}-\frac {1}{8 \left (1+\coth \relax (x )\right )}+\frac {\ln \left (1+\coth \relax (x )\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*ln(coth(x)-1)-1/6/(1+coth(x))^3-1/8/(1+coth(x))^2-1/8/(1+coth(x))+1/16*ln(1+coth(x))

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maxima [A]  time = 0.31, size = 22, normalized size = 0.61 \[ \frac {1}{8} \, x + \frac {3}{16} \, e^{\left (-2 \, x\right )} - \frac {3}{32} \, e^{\left (-4 \, x\right )} + \frac {1}{48} \, e^{\left (-6 \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x + 3/16*e^(-2*x) - 3/32*e^(-4*x) + 1/48*e^(-6*x)

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mupad [B]  time = 0.06, size = 22, normalized size = 0.61 \[ \frac {x}{8}+\frac {3\,{\mathrm {e}}^{-2\,x}}{16}-\frac {3\,{\mathrm {e}}^{-4\,x}}{32}+\frac {{\mathrm {e}}^{-6\,x}}{48} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(coth(x) + 1)^3,x)

[Out]

x/8 + (3*exp(-2*x))/16 - (3*exp(-4*x))/32 + exp(-6*x)/48

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sympy [B]  time = 1.05, size = 182, normalized size = 5.06 \[ \frac {3 x \tanh ^{3}{\relax (x )}}{24 \tanh ^{3}{\relax (x )} + 72 \tanh ^{2}{\relax (x )} + 72 \tanh {\relax (x )} + 24} + \frac {9 x \tanh ^{2}{\relax (x )}}{24 \tanh ^{3}{\relax (x )} + 72 \tanh ^{2}{\relax (x )} + 72 \tanh {\relax (x )} + 24} + \frac {9 x \tanh {\relax (x )}}{24 \tanh ^{3}{\relax (x )} + 72 \tanh ^{2}{\relax (x )} + 72 \tanh {\relax (x )} + 24} + \frac {3 x}{24 \tanh ^{3}{\relax (x )} + 72 \tanh ^{2}{\relax (x )} + 72 \tanh {\relax (x )} + 24} + \frac {21 \tanh ^{2}{\relax (x )}}{24 \tanh ^{3}{\relax (x )} + 72 \tanh ^{2}{\relax (x )} + 72 \tanh {\relax (x )} + 24} + \frac {27 \tanh {\relax (x )}}{24 \tanh ^{3}{\relax (x )} + 72 \tanh ^{2}{\relax (x )} + 72 \tanh {\relax (x )} + 24} + \frac {10}{24 \tanh ^{3}{\relax (x )} + 72 \tanh ^{2}{\relax (x )} + 72 \tanh {\relax (x )} + 24} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*x*tanh(x)**3/(24*tanh(x)**3 + 72*tanh(x)**2 + 72*tanh(x) + 24) + 9*x*tanh(x)**2/(24*tanh(x)**3 + 72*tanh(x)*
*2 + 72*tanh(x) + 24) + 9*x*tanh(x)/(24*tanh(x)**3 + 72*tanh(x)**2 + 72*tanh(x) + 24) + 3*x/(24*tanh(x)**3 + 7
2*tanh(x)**2 + 72*tanh(x) + 24) + 21*tanh(x)**2/(24*tanh(x)**3 + 72*tanh(x)**2 + 72*tanh(x) + 24) + 27*tanh(x)
/(24*tanh(x)**3 + 72*tanh(x)**2 + 72*tanh(x) + 24) + 10/(24*tanh(x)**3 + 72*tanh(x)**2 + 72*tanh(x) + 24)

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