3.1.71 \(\int \frac {x \cot ^{-1}(x)}{(1+x^2)^3} \, dx\) [71]

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

[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 = 0.02, 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 = {5051, 205, 209} \begin {gather*} -\frac {3 \text {ArcTan}(x)}{32}-\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*ArcCot[x])/(1 + x^2)^3,x]

[Out]

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

Rule 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 209

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

Rule 5051

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 = 0.02, size = 36, normalized size = 0.82 \begin {gather*} -\frac {x \left (5+3 x^2\right )+8 \cot ^{-1}(x)+3 \left (1+x^2\right )^2 \text {ArcTan}(x)}{32 \left (1+x^2\right )^2} \end {gather*}

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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Maple [A]
time = 0.14, size = 37, normalized size = 0.84

method result size
default \(-\frac {x}{16 \left (x^{2}+1\right )^{2}}-\frac {3 x}{32 \left (x^{2}+1\right )}-\frac {\mathrm {arccot}\left (x \right )}{4 \left (x^{2}+1\right )^{2}}-\frac {3 \arctan \left (x \right )}{32}\) \(37\)
risch \(-\frac {i \ln \left (i x +1\right )}{8 \left (x^{2}+1\right )^{2}}-\frac {-8 i \ln \left (-i x +1\right )-3 i \ln \left (x -i\right ) x^{4}-6 i \ln \left (x -i\right ) x^{2}-3 i \ln \left (x -i\right )+3 i \ln \left (i+x \right ) x^{4}+6 i \ln \left (i+x \right ) x^{2}+3 i \ln \left (i+x \right )+6 x^{3}+8 \pi +10 x}{64 \left (i+x \right ) \left (x^{2}+1\right ) \left (x -i\right )}\) \(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.46, size = 39, normalized size = 0.89 \begin {gather*} -\frac {3 \, x^{3} + 5 \, x}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} - \frac {\operatorname {arccot}\left (x\right )}{4 \, {\left (x^{2} + 1\right )}^{2}} - \frac {3}{32} \, \arctan \left (x\right ) \end {gather*}

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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Fricas [A]
time = 1.85, size = 39, normalized size = 0.89 \begin {gather*} -\frac {3 \, x^{3} - {\left (3 \, x^{4} + 6 \, x^{2} - 5\right )} \operatorname {arccot}\left (x\right ) + 5 \, x}{32 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \end {gather*}

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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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (39) = 78\).
time = 0.30, size = 88, normalized size = 2.00 \begin {gather*} \frac {3 x^{4} \operatorname {acot}{\left (x \right )}}{32 x^{4} + 64 x^{2} + 32} - \frac {3 x^{3}}{32 x^{4} + 64 x^{2} + 32} + \frac {6 x^{2} \operatorname {acot}{\left (x \right )}}{32 x^{4} + 64 x^{2} + 32} - \frac {5 x}{32 x^{4} + 64 x^{2} + 32} - \frac {5 \operatorname {acot}{\left (x \right )}}{32 x^{4} + 64 x^{2} + 32} \end {gather*}

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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Giac [A]
time = 0.42, size = 40, normalized size = 0.91 \begin {gather*} -\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 ) \end {gather*}

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

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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