3.1.82 \(\int \cot ^{-1}(x) \, dx\) [82]

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

[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 = {4931, 266} \begin {gather*} \frac {1}{2} \log \left (x^2+1\right )+x \cot ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCot[x],x]

[Out]

x*ArcCot[x] + Log[1 + x^2]/2

Rule 266

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 4931

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

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 \begin {gather*} x \cot ^{-1}(x)+\frac {1}{2} \log \left (1+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[x],x]

[Out]

x*ArcCot[x] + Log[1 + x^2]/2

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Maple [A]
time = 0.02, size = 14, normalized size = 0.93

method	result	size
lookup	\(x \,\mathrm {arccot}\left (x \right )+\frac {\ln \left (x^{2}+1\right )}{2}\)	\(14\)
default	\(x \,\mathrm {arccot}\left (x \right )+\frac {\ln \left (x^{2}+1\right )}{2}\)	\(14\)
risch	\(\frac {i x \ln \left (i x +1\right )}{2}-\frac {i x \ln \left (-i x +1\right )}{2}+\frac {\pi x}{2}+\frac {\ln \left (x^{2}+1\right )}{2}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 3.45, size = 13, normalized size = 0.87 \begin {gather*} x \operatorname {arccot}\left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \end {gather*}

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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Fricas [A]
time = 1.00, size = 13, normalized size = 0.87 \begin {gather*} x \operatorname {arccot}\left (x\right ) + \frac {1}{2} \, \log \left (x^{2} + 1\right ) \end {gather*}

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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Sympy [A]
time = 0.06, size = 12, normalized size = 0.80 \begin {gather*} x \operatorname {acot}{\left (x \right )} + \frac {\log {\left (x^{2} + 1 \right )}}{2} \end {gather*}

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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Giac [A]
time = 0.52, size = 21, normalized size = 1.40 \begin {gather*} 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 ) \end {gather*}

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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Mupad [B]
time = 0.10, size = 13, normalized size = 0.87 \begin {gather*} \frac {\ln \left (x^2+1\right )}{2}+x\,\mathrm {acot}\left (x\right ) \end {gather*}

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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