3.1.73 \(\int \frac {\cot ^{-1}(x)^2}{(1+x^2)^2} \, dx\) [73]

Optimal. Leaf size=56 \[ -\frac {x}{4 \left (1+x^2\right )}-\frac {\cot ^{-1}(x)}{2 \left (1+x^2\right )}+\frac {x \cot ^{-1}(x)^2}{2 \left (1+x^2\right )}-\frac {1}{6} \cot ^{-1}(x)^3-\frac {\text {ArcTan}(x)}{4} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5013, 5051, 205, 209} \begin {gather*} -\frac {\text {ArcTan}(x)}{4}-\frac {x}{4 \left (x^2+1\right )}+\frac {x \cot ^{-1}(x)^2}{2 \left (x^2+1\right )}-\frac {\cot ^{-1}(x)}{2 \left (x^2+1\right )}-\frac {1}{6} \cot ^{-1}(x)^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCot[x]^2/(1 + x^2)^2,x]

[Out]

-1/4*x/(1 + x^2) - ArcCot[x]/(2*(1 + x^2)) + (x*ArcCot[x]^2)/(2*(1 + x^2)) - ArcCot[x]^3/6 - ArcTan[x]/4

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 5013

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2)^2, x_Symbol] :> Simp[x*((a + b*ArcCot[c*x])
^p/(2*d*(d + e*x^2))), x] + (Dist[b*c*(p/2), Int[x*((a + b*ArcCot[c*x])^(p - 1)/(d + e*x^2)^2), x], x] - Simp[
(a + b*ArcCot[c*x])^(p + 1)/(2*b*c*d^2*(p + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[p, 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 {\cot ^{-1}(x)^2}{\left (1+x^2\right )^2} \, dx &=\frac {x \cot ^{-1}(x)^2}{2 \left (1+x^2\right )}-\frac {1}{6} \cot ^{-1}(x)^3+\int \frac {x \cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx\\ &=-\frac {\cot ^{-1}(x)}{2 \left (1+x^2\right )}+\frac {x \cot ^{-1}(x)^2}{2 \left (1+x^2\right )}-\frac {1}{6} \cot ^{-1}(x)^3-\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 {x \cot ^{-1}(x)^2}{2 \left (1+x^2\right )}-\frac {1}{6} \cot ^{-1}(x)^3-\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 {x \cot ^{-1}(x)^2}{2 \left (1+x^2\right )}-\frac {1}{6} \cot ^{-1}(x)^3-\frac {1}{4} \tan ^{-1}(x)\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 46, normalized size = 0.82 \begin {gather*} -\frac {6 \cot ^{-1}(x)-6 x \cot ^{-1}(x)^2+2 \left (1+x^2\right ) \cot ^{-1}(x)^3+3 \left (x+\left (1+x^2\right ) \text {ArcTan}(x)\right )}{12 \left (1+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[x]^2/(1 + x^2)^2,x]

[Out]

-1/12*(6*ArcCot[x] - 6*x*ArcCot[x]^2 + 2*(1 + x^2)*ArcCot[x]^3 + 3*(x + (1 + x^2)*ArcTan[x]))/(1 + x^2)

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Maple [A]
time = 0.27, size = 65, normalized size = 1.16

method	result	size
default	\(\frac {x \mathrm {arccot}\left (x \right )^{2}}{2 x^{2}+2}+\frac {\mathrm {arccot}\left (x \right )^{2} \arctan \left (x \right )}{2}-\frac {\mathrm {arccot}\left (x \right )^{2} \pi }{4}+\frac {\mathrm {arccot}\left (x \right )^{3}}{3}+\frac {x^{2} \mathrm {arccot}\left (x \right )}{2 x^{2}+2}-\frac {x}{4 \left (x^{2}+1\right )}-\frac {\mathrm {arccot}\left (x \right )}{4}\)	\(65\)
risch	\(\frac {i \ln \left (i x +1\right )^{3}}{48}+\frac {\left (-i \ln \left (-i x +1\right ) x^{2}+\pi \,x^{2}-i \ln \left (-i x +1\right )+\pi -2 x \right ) \ln \left (i x +1\right )^{2}}{16 x^{2}+16}-\frac {\left (-i x^{2} \ln \left (-i x +1\right )^{2}-i \ln \left (-i x +1\right )^{2}-4 \ln \left (-i x +1\right ) x +2 \pi \ln \left (-i x +1\right ) x^{2}+2 \pi \ln \left (-i x +1\right )-4 i \pi x +4 i\right ) \ln \left (i x +1\right )}{16 \left (i+x \right ) \left (x -i\right )}-\frac {i \left (3 \ln \left (x -i\right ) \pi ^{2} x^{2}-3 \ln \left (i+x \right ) \pi ^{2} x^{2}+x^{2} \ln \left (-i x +1\right )^{3}+3 i \pi \,x^{2} \ln \left (-i x +1\right )^{2}+3 \ln \left (x -i\right ) \pi ^{2}-3 \ln \left (i+x \right ) \pi ^{2}+12 \pi \ln \left (-i x +1\right ) x -6 \ln \left (x -i\right ) x^{2}+6 \ln \left (i+x \right ) x^{2}+\ln \left (-i x +1\right )^{3}-6 i \ln \left (-i x +1\right )^{2} x +3 i \pi \ln \left (-i x +1\right )^{2}+6 i \pi ^{2} x -6 \ln \left (x -i\right )+6 \ln \left (i+x \right )-12 \ln \left (-i x +1\right )-12 i x -12 i \pi \right )}{48 \left (i+x \right ) \left (x -i\right )}\)	\(349\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 75, normalized size = 1.34 \begin {gather*} \frac {1}{2} \, {\left (\frac {x}{x^{2} + 1} + \arctan \left (x\right )\right )} \operatorname {arccot}\left (x\right )^{2} + \frac {{\left ({\left (x^{2} + 1\right )} \arctan \left (x\right )^{2} - 1\right )} \operatorname {arccot}\left (x\right )}{2 \, {\left (x^{2} + 1\right )}} + \frac {2 \, {\left (x^{2} + 1\right )} \arctan \left (x\right )^{3} - 3 \, {\left (x^{2} + 1\right )} \arctan \left (x\right ) - 3 \, x}{12 \, {\left (x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x/(x^2 + 1) + arctan(x))*arccot(x)^2 + 1/2*((x^2 + 1)*arctan(x)^2 - 1)*arccot(x)/(x^2 + 1) + 1/12*(2*(x^2
+ 1)*arctan(x)^3 - 3*(x^2 + 1)*arctan(x) - 3*x)/(x^2 + 1)

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Fricas [A]
time = 1.87, size = 40, normalized size = 0.71 \begin {gather*} -\frac {2 \, {\left (x^{2} + 1\right )} \operatorname {arccot}\left (x\right )^{3} - 6 \, x \operatorname {arccot}\left (x\right )^{2} - 3 \, {\left (x^{2} - 1\right )} \operatorname {arccot}\left (x\right ) + 3 \, x}{12 \, {\left (x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(2*(x^2 + 1)*arccot(x)^3 - 6*x*arccot(x)^2 - 3*(x^2 - 1)*arccot(x) + 3*x)/(x^2 + 1)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {acot}^{2}{\left (x \right )}}{\left (x^{2} + 1\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acot(x)**2/(x**2 + 1)**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccot(x)^2/(x^2 + 1)^2, x)

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Mupad [B]
time = 0.06, size = 51, normalized size = 0.91 \begin {gather*} \frac {x\,{\mathrm {acot}\left (x\right )}^2}{2\,\left (x^2+1\right )}-\frac {{\mathrm {acot}\left (x\right )}^3}{6}-\frac {x}{4\,\left (x^2+1\right )}-\frac {\mathrm {acot}\left (x\right )}{2\,\left (x^2+1\right )}-\frac {\mathrm {atan}\left (x\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(x)^2/(x^2 + 1)^2,x)

[Out]

(x*acot(x)^2)/(2*(x^2 + 1)) - acot(x)^3/6 - x/(4*(x^2 + 1)) - acot(x)/(2*(x^2 + 1)) - atan(x)/4

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