\(\int \frac {\cot ^{-1}(x)}{(1+x^2)^2} \, dx\) [72]

Optimal result

Integrand size = 10, antiderivative size = 34 \[ \int \frac {\cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=-\frac {1}{4 \left (1+x^2\right )}+\frac {x \cot ^{-1}(x)}{2 \left (1+x^2\right )}-\frac {1}{4} \cot ^{-1}(x)^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5013, 267} \[ \int \frac {\cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=-\frac {1}{4 \left (x^2+1\right )}+\frac {x \cot ^{-1}(x)}{2 \left (x^2+1\right )}-\frac {1}{4} \cot ^{-1}(x)^2 \]

[In]

Int[ArcCot[x]/(1 + x^2)^2,x]

[Out]

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

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

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
]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {\cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=-\frac {1-2 x \cot ^{-1}(x)+\left (1+x^2\right ) \cot ^{-1}(x)^2}{4 \left (1+x^2\right )} \]

[In]

Integrate[ArcCot[x]/(1 + x^2)^2,x]

[Out]

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

Maple [A] (verified)

Time = 0.75 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {x \,\operatorname {arccot}\left (x \right )}{2 x^{2}+2}+\frac {\operatorname {arccot}\left (x \right ) \arctan \left (x \right )}{2}-\frac {1}{4 \left (x^{2}+1\right )}+\frac {\arctan \left (x \right )^{2}}{4}\)	\(35\)
parts	\(\frac {x \,\operatorname {arccot}\left (x \right )}{2 x^{2}+2}+\frac {\operatorname {arccot}\left (x \right ) \arctan \left (x \right )}{2}-\frac {1}{4 \left (x^{2}+1\right )}+\frac {\arctan \left (x \right )^{2}}{4}\)	\(35\)
risch	\(\frac {\ln \left (i x +1\right )^{2}}{16}-\frac {\left (x^{2} \ln \left (-i x +1\right )-2 i x +\ln \left (-i x +1\right )\right ) \ln \left (i x +1\right )}{8 \left (x^{2}+1\right )}+\frac {x^{2} \ln \left (-i x +1\right )^{2}+\ln \left (-i x +1\right )^{2}+2 i \pi \ln \left (i+x \right )+2 i \pi \ln \left (i+x \right ) x^{2}-2 i \pi \ln \left (x -i\right )-2 i \pi \ln \left (x -i\right ) x^{2}+4 \pi x -4-4 i x \ln \left (-i x +1\right )}{16 \left (i+x \right ) \left (x -i\right )}\)	\(147\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {\cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=-\frac {{\left (x^{2} + 1\right )} \operatorname {arccot}\left (x\right )^{2} - 2 \, x \operatorname {arccot}\left (x\right ) + 1}{4 \, {\left (x^{2} + 1\right )}} \]

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {\cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=\text {Exception raised: RecursionError} \]

[In]

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

[Out]

Exception raised: RecursionError >> maximum recursion depth exceeded in comparison

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {\cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=\frac {1}{2} \, {\left (\frac {x}{x^{2} + 1} + \arctan \left (x\right )\right )} \operatorname {arccot}\left (x\right ) + \frac {{\left (x^{2} + 1\right )} \arctan \left (x\right )^{2} - 1}{4 \, {\left (x^{2} + 1\right )}} \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=\int { \frac {\operatorname {arccot}\left (x\right )}{{\left (x^{2} + 1\right )}^{2}} \,d x } \]

[In]

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

[Out]

integrate(arccot(x)/(x^2 + 1)^2, x)

Mupad [B] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {\cot ^{-1}(x)}{\left (1+x^2\right )^2} \, dx=\frac {\frac {x\,\mathrm {acot}\left (x\right )}{2}-\frac {1}{4}}{x^2+1}-\frac {{\mathrm {acot}\left (x\right )}^2}{4} \]

[In]

int(acot(x)/(x^2 + 1)^2,x)

[Out]

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