3.71 \(\int \frac {x \cot ^{-1}(x)}{(1+x^2)^3} \, dx\)

Optimal. Leaf size=44 \[ -\frac {3 x}{32 \left (x^2+1\right )}-\frac {x}{16 \left (x^2+1\right )^2}-\frac {\cot ^{-1}(x)}{4 \left (x^2+1\right )^2}-\frac {3}{32} \tan ^{-1}(x) \]

[Out]

-1/16*x/(x^2+1)^2-3/32*x/(x^2+1)-1/4*arccot(x)/(x^2+1)^2-3/32*arctan(x)

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Rubi [A]  time = 0.03, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {4931, 199, 203} \[ -\frac {3 x}{32 \left (x^2+1\right )}-\frac {x}{16 \left (x^2+1\right )^2}-\frac {\cot ^{-1}(x)}{4 \left (x^2+1\right )^2}-\frac {3}{32} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x*ArcCot[x])/(1 + x^2)^3,x]

[Out]

-x/(16*(1 + x^2)^2) - (3*x)/(32*(1 + x^2)) - ArcCot[x]/(4*(1 + x^2)^2) - (3*ArcTan[x])/32

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 )^3} \, dx &=-\frac {\cot ^{-1}(x)}{4 \left (1+x^2\right )^2}-\frac {1}{4} \int \frac {1}{\left (1+x^2\right )^3} \, dx\\ &=-\frac {x}{16 \left (1+x^2\right )^2}-\frac {\cot ^{-1}(x)}{4 \left (1+x^2\right )^2}-\frac {3}{16} \int \frac {1}{\left (1+x^2\right )^2} \, dx\\ &=-\frac {x}{16 \left (1+x^2\right )^2}-\frac {3 x}{32 \left (1+x^2\right )}-\frac {\cot ^{-1}(x)}{4 \left (1+x^2\right )^2}-\frac {3}{32} \int \frac {1}{1+x^2} \, dx\\ &=-\frac {x}{16 \left (1+x^2\right )^2}-\frac {3 x}{32 \left (1+x^2\right )}-\frac {\cot ^{-1}(x)}{4 \left (1+x^2\right )^2}-\frac {3}{32} \tan ^{-1}(x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 36, normalized size = 0.82 \[ -\frac {x \left (3 x^2+5\right )+3 \left (x^2+1\right )^2 \tan ^{-1}(x)+8 \cot ^{-1}(x)}{32 \left (x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*ArcCot[x])/(1 + x^2)^3,x]

[Out]

-1/32*(x*(5 + 3*x^2) + 8*ArcCot[x] + 3*(1 + x^2)^2*ArcTan[x])/(1 + x^2)^2

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fricas [A]  time = 1.97, size = 39, normalized size = 0.89 \[ -\frac {3 \, x^{3} - {\left (3 \, x^{4} + 6 \, x^{2} - 5\right )} \operatorname {arccot}\relax (x) + 5 \, x}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(x)/(x^2+1)^3,x, algorithm="fricas")

[Out]

-1/32*(3*x^3 - (3*x^4 + 6*x^2 - 5)*arccot(x) + 5*x)/(x^4 + 2*x^2 + 1)

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giac [A]  time = 0.13, size = 40, normalized size = 0.91 \[ -\frac {\frac {3}{x} + \frac {5}{x^{3}}}{32 \, {\left (\frac {1}{x^{2}} + 1\right )}^{2}} - \frac {\arctan \left (\frac {1}{x}\right )}{4 \, {\left (x^{2} + 1\right )}^{2}} + \frac {3}{32} \, \arctan \left (\frac {1}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(x)/(x^2+1)^3,x, algorithm="giac")

[Out]

-1/32*(3/x + 5/x^3)/(1/x^2 + 1)^2 - 1/4*arctan(1/x)/(x^2 + 1)^2 + 3/32*arctan(1/x)

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maple [A]  time = 0.04, size = 37, normalized size = 0.84 \[ -\frac {x}{16 \left (x^{2}+1\right )^{2}}-\frac {3 x}{32 \left (x^{2}+1\right )}-\frac {\mathrm {arccot}\relax (x )}{4 \left (x^{2}+1\right )^{2}}-\frac {3 \arctan \relax (x )}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*arccot(x)/(x^2+1)^3,x)

[Out]

-1/16*x/(x^2+1)^2-3/32*x/(x^2+1)-1/4*arccot(x)/(x^2+1)^2-3/32*arctan(x)

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maxima [A]  time = 0.44, size = 39, normalized size = 0.89 \[ -\frac {3 \, x^{3} + 5 \, x}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} - \frac {\operatorname {arccot}\relax (x)}{4 \, {\left (x^{2} + 1\right )}^{2}} - \frac {3}{32} \, \arctan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*arccot(x)/(x^2+1)^3,x, algorithm="maxima")

[Out]

-1/32*(3*x^3 + 5*x)/(x^4 + 2*x^2 + 1) - 1/4*arccot(x)/(x^2 + 1)^2 - 3/32*arctan(x)

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mupad [B]  time = 0.60, size = 27, normalized size = 0.61 \[ -\frac {3\,\mathrm {atan}\relax (x)}{32}-\frac {\frac {5\,x}{32}+\frac {\mathrm {acot}\relax (x)}{4}+\frac {3\,x^3}{32}}{{\left (x^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*acot(x))/(x^2 + 1)^3,x)

[Out]

- (3*atan(x))/32 - ((5*x)/32 + acot(x)/4 + (3*x^3)/32)/(x^2 + 1)^2

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sympy [B]  time = 0.80, size = 88, normalized size = 2.00 \[ \frac {3 x^{4} \operatorname {acot}{\relax (x )}}{32 x^{4} + 64 x^{2} + 32} - \frac {3 x^{3}}{32 x^{4} + 64 x^{2} + 32} + \frac {6 x^{2} \operatorname {acot}{\relax (x )}}{32 x^{4} + 64 x^{2} + 32} - \frac {5 x}{32 x^{4} + 64 x^{2} + 32} - \frac {5 \operatorname {acot}{\relax (x )}}{32 x^{4} + 64 x^{2} + 32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*acot(x)/(x**2+1)**3,x)

[Out]

3*x**4*acot(x)/(32*x**4 + 64*x**2 + 32) - 3*x**3/(32*x**4 + 64*x**2 + 32) + 6*x**2*acot(x)/(32*x**4 + 64*x**2
+ 32) - 5*x/(32*x**4 + 64*x**2 + 32) - 5*acot(x)/(32*x**4 + 64*x**2 + 32)

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