3.82 \(\int \cot ^{-1}(x) \, dx\)

Optimal. Leaf size=15 \[ \frac {1}{2} \log \left (x^2+1\right )+x \cot ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.00, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {4847, 260} \[ \frac {1}{2} \log \left (x^2+1\right )+x \cot ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[x],x]

[Out]

x*ArcCot[x] + 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 4847

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcCot[c*x])^p, x] + Dist[b*c*p, Int[
(x*(a + b*ArcCot[c*x])^(p - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \[ \frac {1}{2} \log \left (x^2+1\right )+x \cot ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[x],x]

[Out]

x*ArcCot[x] + Log[1 + x^2]/2

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fricas [A]  time = 0.42, size = 13, normalized size = 0.87 \[ x \operatorname {arccot}\relax (x) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x),x, algorithm="fricas")

[Out]

x*arccot(x) + 1/2*log(x^2 + 1)

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giac [A]  time = 0.01, size = 21, normalized size = 1.40 \[ x \arctan \left (\frac {1}{x}\right ) + \frac {1}{2} \, \log \left (\frac {1}{x^{2}} + 1\right ) - \frac {1}{2} \, \log \left (\frac {1}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 14, normalized size = 0.93 \[ x \,\mathrm {arccot}\relax (x )+\frac {\ln \left (x^{2}+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(x),x)

[Out]

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

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maxima [A]  time = 0.54, size = 13, normalized size = 0.87 \[ x \operatorname {arccot}\relax (x) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x),x, algorithm="maxima")

[Out]

x*arccot(x) + 1/2*log(x^2 + 1)

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mupad [B]  time = 0.10, size = 13, normalized size = 0.87 \[ \frac {\ln \left (x^2+1\right )}{2}+x\,\mathrm {acot}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(x),x)

[Out]

log(x^2 + 1)/2 + x*acot(x)

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sympy [A]  time = 0.21, size = 12, normalized size = 0.80 \[ x \operatorname {acot}{\relax (x )} + \frac {\log {\left (x^{2} + 1 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(x),x)

[Out]

x*acot(x) + log(x**2 + 1)/2

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