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

Optimal. Leaf size=17 \[ x \cot ^{-1}\left (\frac {1}{x}\right )-\frac {1}{2} \log \left (x^2+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 17, 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 = {5028, 263, 260} \[ x \cot ^{-1}\left (\frac {1}{x}\right )-\frac {1}{2} \log \left (x^2+1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*ArcCot[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 5028

Int[ArcCot[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[x*ArcCot[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 \cot ^{-1}\left (\frac {1}{x}\right ) \, dx &=x \cot ^{-1}\left (\frac {1}{x}\right )-\int \frac {1}{\left (1+\frac {1}{x^2}\right ) x} \, dx\\ &=x \cot ^{-1}\left (\frac {1}{x}\right )-\int \frac {x}{1+x^2} \, dx\\ &=x \cot ^{-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 = 17, normalized size = 1.00 \[ x \cot ^{-1}\left (\frac {1}{x}\right )-\frac {1}{2} \log \left (x^2+1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*ArcCot[x^(-1)] - Log[1 + x^2]/2

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fricas [A]  time = 0.48, size = 15, normalized size = 0.88 \[ x \operatorname {arccot}\left (\frac {1}{x}\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arccot(1/x) - 1/2*log(x^2 + 1)

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giac [A]  time = 0.12, size = 13, normalized size = 0.76 \[ x \arctan \relax (x) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan(x) - 1/2*log(x^2 + 1)

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maple [A]  time = 0.07, size = 20, normalized size = 1.18 \[ x \,\mathrm {arccot}\left (\frac {1}{x}\right )+\ln \left (\frac {1}{x}\right )-\frac {\ln \left (\frac {1}{x^{2}}+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(1/x),x)

[Out]

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

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maxima [A]  time = 0.33, size = 15, normalized size = 0.88 \[ x \operatorname {arccot}\left (\frac {1}{x}\right ) - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arccot(1/x) - 1/2*log(x^2 + 1)

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mupad [B]  time = 0.06, size = 15, normalized size = 0.88 \[ x\,\mathrm {acot}\left (\frac {1}{x}\right )-\frac {\ln \left (x^2+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(1/x),x)

[Out]

x*acot(1/x) - log(x^2 + 1)/2

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sympy [A]  time = 0.15, size = 14, normalized size = 0.82 \[ x \operatorname {acot}{\left (\frac {1}{x} \right )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(1/x),x)

[Out]

x*acot(1/x) - log(x**2 + 1)/2

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