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

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.00, size = 19, normalized size = 1.00 \[ \frac {1}{2} \log \left (1-x^2\right )+x \coth ^{-1}\left (\frac {1}{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*ArcCoth[x^(-1)] + Log[1 - x^2]/2

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fricas [A]  time = 0.99, size = 23, normalized size = 1.21 \[ \frac {1}{2} \, x \log \left (-\frac {x + 1}{x - 1}\right ) + \frac {1}{2} \, \log \left (x^{2} - 1\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \operatorname {arcoth}\left (\frac {1}{x}\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(1/x), x)

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maple [A]  time = 0.07, size = 30, normalized size = 1.58 \[ x \,\mathrm {arccoth}\left (\frac {1}{x}\right )-\ln \left (\frac {1}{x}\right )+\frac {\ln \left (\frac {1}{x}-1\right )}{2}+\frac {\ln \left (\frac {1}{x}+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccoth(1/x),x)

[Out]

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

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maxima [A]  time = 0.30, size = 15, normalized size = 0.79 \[ x \operatorname {arcoth}\left (\frac {1}{x}\right ) + \frac {1}{2} \, \log \left (x^{2} - 1\right ) \]

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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mupad [B]  time = 1.13, size = 26, normalized size = 1.37 \[ \frac {\ln \left (x^2-1\right )}{2}+x\,\left (\frac {\ln \left (x+1\right )}{2}-\frac {\ln \left (1-x\right )}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(1/x),x)

[Out]

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

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sympy [A]  time = 0.20, size = 15, normalized size = 0.79 \[ x \operatorname {acoth}{\left (\frac {1}{x} \right )} + \log {\left (x + 1 \right )} - \operatorname {acoth}{\left (\frac {1}{x} \right )} \]

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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