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

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

[Out]

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

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Rubi [A]  time = 0.0438201, antiderivative size = 56, normalized size of antiderivative = 1., 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 = {4893, 4931, 199, 203} \[ -\frac{x}{4 \left (x^2+1\right )}+\frac{x \cot ^{-1}(x)^2}{2 \left (x^2+1\right )}-\frac{\cot ^{-1}(x)}{2 \left (x^2+1\right )}-\frac{1}{4} \tan ^{-1}(x)-\frac{1}{6} \cot ^{-1}(x)^3 \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[x]^2/(1 + x^2)^2,x]

[Out]

-x/(4*(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 4893

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 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]

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])

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*}

Mathematica [A]  time = 0.0235425, size = 46, normalized size = 0.82 \[ -\frac{3 \left (\left (x^2+1\right ) \tan ^{-1}(x)+x\right )+2 \left (x^2+1\right ) \cot ^{-1}(x)^3-6 x \cot ^{-1}(x)^2+6 \cot ^{-1}(x)}{12 \left (x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[x]^2/(1 + x^2)^2,x]

[Out]

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

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Maple [A]  time = 0.169, size = 61, normalized size = 1.1 \begin{align*} -{\frac{ \left ({\rm arccot} \left (x\right ) \right ) ^{2} \left ({x}^{2}{\rm arccot} \left (x\right )+{\rm arccot} \left (x\right )-x \right ) }{2\,{x}^{2}+2}}+{\frac{{x}^{2}{\rm arccot} \left (x\right )}{2\,{x}^{2}+2}}-{\frac{x}{4\,{x}^{2}+4}}-{\frac{{\rm arccot} \left (x\right )}{4}}+{\frac{ \left ({\rm arccot} \left (x\right ) \right ) ^{3}}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.52471, size = 101, normalized size = 1.8 \begin{align*} \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{align*}

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 = 2.08247, size = 123, normalized size = 2.2 \begin{align*} -\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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acot}^{2}{\left (x \right )}}{\left (x^{2} + 1\right )^{2}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arccot}\left (x\right )^{2}}{{\left (x^{2} + 1\right )}^{2}}\,{d x} \end{align*}

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)