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

Optimal. Leaf size=24 \[ \frac {\log \left (a^2 x^2+1\right )}{2 a}+x \cot ^{-1}(a x) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4847, 260} \[ \frac {\log \left (a^2 x^2+1\right )}{2 a}+x \cot ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[a*x],x]

[Out]

x*ArcCot[a*x] + Log[1 + a^2*x^2]/(2*a)

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

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Mathematica [A]  time = 0.00, size = 24, normalized size = 1.00 \[ \frac {\log \left (a^2 x^2+1\right )}{2 a}+x \cot ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[a*x],x]

[Out]

x*ArcCot[a*x] + Log[1 + a^2*x^2]/(2*a)

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fricas [A]  time = 0.75, size = 24, normalized size = 1.00 \[ \frac {2 \, a x \operatorname {arccot}\left (a x\right ) + \log \left (a^{2} x^{2} + 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*x*arccot(a*x) + log(a^2*x^2 + 1))/a

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giac [B]  time = 0.11, size = 45, normalized size = 1.88 \[ \frac {1}{2} \, a {\left (\frac {2 \, x \arctan \left (\frac {1}{a x}\right )}{a} + \frac {\log \left (\frac {1}{a^{2} x^{2}} + 1\right )}{a^{2}} - \frac {\log \left (\frac {1}{a^{2} x^{2}}\right )}{a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 23, normalized size = 0.96 \[ x \,\mathrm {arccot}\left (a x \right )+\frac {\ln \left (a^{2} x^{2}+1\right )}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(a*x),x)

[Out]

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

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maxima [A]  time = 0.32, size = 24, normalized size = 1.00 \[ \frac {2 \, a x \operatorname {arccot}\left (a x\right ) + \log \left (a^{2} x^{2} + 1\right )}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*a*x*arccot(a*x) + log(a^2*x^2 + 1))/a

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mupad [B]  time = 0.14, size = 22, normalized size = 0.92 \[ x\,\mathrm {acot}\left (a\,x\right )+\frac {\ln \left (a^2\,x^2+1\right )}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(a*x),x)

[Out]

x*acot(a*x) + log(a^2*x^2 + 1)/(2*a)

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sympy [A]  time = 0.21, size = 24, normalized size = 1.00 \[ \begin {cases} x \operatorname {acot}{\left (a x \right )} + \frac {\log {\left (a^{2} x^{2} + 1 \right )}}{2 a} & \text {for}\: a \neq 0 \\\frac {\pi x}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(a*x),x)

[Out]

Piecewise((x*acot(a*x) + log(a**2*x**2 + 1)/(2*a), Ne(a, 0)), (pi*x/2, True))

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