3.94 \(\int \coth ^{-1}(\frac{1}{x}) \, dx\)

Optimal. Leaf size=19 \[ \frac{1}{2} \log \left (1-x^2\right )+x \coth ^{-1}\left (\frac{1}{x}\right ) \]

[Out]

x*ArcCoth[x^(-1)] + Log[1 - x^2]/2

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Rubi [A]  time = 0.0060927, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 4, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {6092, 263, 260} \[ \frac{1}{2} \log \left (1-x^2\right )+x \coth ^{-1}\left (\frac{1}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcCoth[x^(-1)],x]

[Out]

x*ArcCoth[x^(-1)] + Log[1 - x^2]/2

Rule 6092

Int[ArcCoth[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[x*ArcCoth[c*x^n], x] - Dist[c*n, Int[x^n/(1 - c^2*x^(2*n)), x]
, x] /; FreeQ[{c, n}, x]

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [A]  time = 0.0018144, size = 19, normalized size = 1. \[ \frac{1}{2} \log \left (1-x^2\right )+x \coth ^{-1}\left (\frac{1}{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCoth[x^(-1)],x]

[Out]

x*ArcCoth[x^(-1)] + Log[1 - x^2]/2

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Maple [A]  time = 0.056, size = 30, normalized size = 1.6 \begin{align*} x{\rm arccoth} \left ({x}^{-1}\right )+{\frac{\ln \left ({x}^{-1}-1 \right ) }{2}}-\ln \left ({x}^{-1} \right ) +{\frac{\ln \left ({x}^{-1}+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(1/x),x)

[Out]

x*arccoth(1/x)+1/2*ln(1/x-1)-ln(1/x)+1/2*ln(1/x+1)

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Maxima [A]  time = 0.976526, size = 20, normalized size = 1.05 \begin{align*} x \operatorname{arcoth}\left (\frac{1}{x}\right ) + \frac{1}{2} \, \log \left (x^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1/x),x, algorithm="maxima")

[Out]

x*arccoth(1/x) + 1/2*log(x^2 - 1)

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Fricas [A]  time = 1.85874, size = 65, normalized size = 3.42 \begin{align*} \frac{1}{2} \, x \log \left (-\frac{x + 1}{x - 1}\right ) + \frac{1}{2} \, \log \left (x^{2} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1/x),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.254185, size = 15, normalized size = 0.79 \begin{align*} x \operatorname{acoth}{\left (\frac{1}{x} \right )} + \log{\left (x + 1 \right )} - \operatorname{acoth}{\left (\frac{1}{x} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acoth(1/x),x)

[Out]

x*acoth(1/x) + log(x + 1) - acoth(1/x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{arcoth}\left (\frac{1}{x}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccoth(1/x),x, algorithm="giac")

[Out]

integrate(arccoth(1/x), x)