3.70 \(\int \frac {x \cot ^{-1}(x)}{(1+x^2)^2} \, dx\)

Optimal. Leaf size=32 \[ -\frac {x}{4 \left (x^2+1\right )}-\frac {\cot ^{-1}(x)}{2 \left (x^2+1\right )}-\frac {1}{4} \tan ^{-1}(x) \]

[Out]

-1/4*x/(x^2+1)-1/2*arccot(x)/(x^2+1)-1/4*arctan(x)

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Rubi [A]  time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4931, 199, 203} \[ -\frac {x}{4 \left (x^2+1\right )}-\frac {\cot ^{-1}(x)}{2 \left (x^2+1\right )}-\frac {1}{4} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcCot[x])/(1 + x^2)^2,x]

[Out]

-x/(4*(1 + x^2)) - ArcCot[x]/(2*(1 + x^2)) - ArcTan[x]/4

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 4931

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*(x_)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[((d + e*x^2)^
(q + 1)*(a + b*ArcCot[c*x])^p)/(2*e*(q + 1)), x] + Dist[(b*p)/(2*c*(q + 1)), Int[(d + e*x^2)^q*(a + b*ArcCot[c
*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, q}, x] && EqQ[e, c^2*d] && GtQ[p, 0] && NeQ[q, -1]

Rubi steps

\begin {align*} \int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx &=-\frac {\cot ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {1}{2} \int \frac {1}{\left (1+x^2\right )^2} \, dx\\ &=-\frac {x}{4 \left (1+x^2\right )}-\frac {\cot ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {1}{4} \int \frac {1}{1+x^2} \, dx\\ &=-\frac {x}{4 \left (1+x^2\right )}-\frac {\cot ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {1}{4} \tan ^{-1}(x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 0.78 \[ -\frac {x^2 \tan ^{-1}(x)+x+\tan ^{-1}(x)+2 \cot ^{-1}(x)}{4 x^2+4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcCot[x])/(1 + x^2)^2,x]

[Out]

-((x + 2*ArcCot[x] + ArcTan[x] + x^2*ArcTan[x])/(4 + 4*x^2))

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fricas [A]  time = 0.67, size = 21, normalized size = 0.66 \[ \frac {{\left (x^{2} - 1\right )} \operatorname {arccot}\relax (x) - x}{4 \, {\left (x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(x)/(x^2+1)^2,x, algorithm="fricas")

[Out]

1/4*((x^2 - 1)*arccot(x) - x)/(x^2 + 1)

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giac [A]  time = 0.13, size = 32, normalized size = 1.00 \[ -\frac {\arctan \left (\frac {1}{x}\right )}{2 \, {\left (x^{2} + 1\right )}} - \frac {1}{4 \, x {\left (\frac {1}{x^{2}} + 1\right )}} + \frac {1}{4} \, \arctan \left (\frac {1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(x)/(x^2+1)^2,x, algorithm="giac")

[Out]

-1/2*arctan(1/x)/(x^2 + 1) - 1/4/(x*(1/x^2 + 1)) + 1/4*arctan(1/x)

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maple [A]  time = 0.04, size = 27, normalized size = 0.84 \[ -\frac {x}{4 \left (x^{2}+1\right )}-\frac {\mathrm {arccot}\relax (x )}{2 \left (x^{2}+1\right )}-\frac {\arctan \relax (x )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccot(x)/(x^2+1)^2,x)

[Out]

-1/4*x/(x^2+1)-1/2*arccot(x)/(x^2+1)-1/4*arctan(x)

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maxima [A]  time = 0.45, size = 26, normalized size = 0.81 \[ -\frac {x}{4 \, {\left (x^{2} + 1\right )}} - \frac {\operatorname {arccot}\relax (x)}{2 \, {\left (x^{2} + 1\right )}} - \frac {1}{4} \, \arctan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(x)/(x^2+1)^2,x, algorithm="maxima")

[Out]

-1/4*x/(x^2 + 1) - 1/2*arccot(x)/(x^2 + 1) - 1/4*arctan(x)

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mupad [B]  time = 0.07, size = 22, normalized size = 0.69 \[ \frac {\mathrm {acot}\relax (x)}{4}-\frac {\frac {x}{4}+\frac {\mathrm {acot}\relax (x)}{2}}{x^2+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*acot(x))/(x^2 + 1)^2,x)

[Out]

acot(x)/4 - (x/4 + acot(x)/2)/(x^2 + 1)

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sympy [A]  time = 0.50, size = 31, normalized size = 0.97 \[ \frac {x^{2} \operatorname {acot}{\relax (x )}}{4 x^{2} + 4} - \frac {x}{4 x^{2} + 4} - \frac {\operatorname {acot}{\relax (x )}}{4 x^{2} + 4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acot(x)/(x**2+1)**2,x)

[Out]

x**2*acot(x)/(4*x**2 + 4) - x/(4*x**2 + 4) - acot(x)/(4*x**2 + 4)

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