3.64 \(\int (1+\coth (x))^2 \, dx\)

Optimal. Leaf size=13 \[ 2 x-\coth (x)+2 \log (\sinh (x)) \]

[Out]

2*x - Coth[x] + 2*Log[Sinh[x]]

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Rubi [A]  time = 0.0133208, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3477, 3475} \[ 2 x-\coth (x)+2 \log (\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[(1 + Coth[x])^2,x]

[Out]

2*x - Coth[x] + 2*Log[Sinh[x]]

Rule 3477

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

Rule 3475

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

Rubi steps

\begin{align*} \int (1+\coth (x))^2 \, dx &=2 x-\coth (x)+2 \int \coth (x) \, dx\\ &=2 x-\coth (x)+2 \log (\sinh (x))\\ \end{align*}

Mathematica [A]  time = 0.0043895, size = 13, normalized size = 1. \[ 2 x-\coth (x)+2 \log (\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + Coth[x])^2,x]

[Out]

2*x - Coth[x] + 2*Log[Sinh[x]]

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Maple [A]  time = 0.003, size = 13, normalized size = 1. \begin{align*} -{\rm coth} \left (x\right )-2\,\ln \left ({\rm coth} \left (x\right )-1 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-coth(x)-2*ln(coth(x)-1)

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Maxima [A]  time = 1.13003, size = 26, normalized size = 2. \begin{align*} 2 \, x + \frac{2}{e^{\left (-2 \, x\right )} - 1} + 2 \, \log \left (\sinh \left (x\right )\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.95735, size = 189, normalized size = 14.54 \begin{align*} \frac{2 \,{\left ({\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} - 1\right )} \log \left (\frac{2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - 1\right )}}{\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.745568, size = 22, normalized size = 1.69 \begin{align*} 4 x - 2 \log{\left (\tanh{\left (x \right )} + 1 \right )} + 2 \log{\left (\tanh{\left (x \right )} \right )} - \frac{1}{\tanh{\left (x \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*x - 2*log(tanh(x) + 1) + 2*log(tanh(x)) - 1/tanh(x)

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Giac [A]  time = 1.18667, size = 28, normalized size = 2.15 \begin{align*} -\frac{2}{e^{\left (2 \, x\right )} - 1} + 2 \, \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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