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

Optimal. Leaf size=30 \[ \frac {1}{2} \log \left (x^2+1\right )-\log (x)+\frac {1}{2} \cot ^{-1}(x)^2-\frac {\cot ^{-1}(x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {4919, 4853, 266, 36, 29, 31, 4885} \[ \frac {1}{2} \log \left (x^2+1\right )-\log (x)+\frac {1}{2} \cot ^{-1}(x)^2-\frac {\cot ^{-1}(x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[x]/(x^2*(1 + x^2)),x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 266

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

Rule 4853

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCo
t[c*x])^p)/(d*(m + 1)), x] + Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCot[c*x])^(p - 1))/(1 + c^
2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 4885

Int[((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(a + b*ArcCot[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[e, c^2*d] && NeQ[p, -1]

Rule 4919

Int[(((a_.) + ArcCot[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d,
Int[(f*x)^m*(a + b*ArcCot[c*x])^p, x], x] - Dist[e/(d*f^2), Int[((f*x)^(m + 2)*(a + b*ArcCot[c*x])^p)/(d + e*
x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\cot ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx &=\int \frac {\cot ^{-1}(x)}{x^2} \, dx-\int \frac {\cot ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac {\cot ^{-1}(x)}{x}+\frac {1}{2} \cot ^{-1}(x)^2-\int \frac {1}{x \left (1+x^2\right )} \, dx\\ &=-\frac {\cot ^{-1}(x)}{x}+\frac {1}{2} \cot ^{-1}(x)^2-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right )\\ &=-\frac {\cot ^{-1}(x)}{x}+\frac {1}{2} \cot ^{-1}(x)^2-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x} \, dx,x,x^2\right )\\ &=-\frac {\cot ^{-1}(x)}{x}+\frac {1}{2} \cot ^{-1}(x)^2-\log (x)+\frac {1}{2} \log \left (1+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 1.00 \[ \frac {1}{2} \log \left (x^2+1\right )-\log (x)+\frac {1}{2} \cot ^{-1}(x)^2-\frac {\cot ^{-1}(x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.02, size = 29, normalized size = 0.97 \[ \frac {x \operatorname {arccot}\relax (x)^{2} + x \log \left (x^{2} + 1\right ) - 2 \, x \log \relax (x) - 2 \, \operatorname {arccot}\relax (x)}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.11, size = 26, normalized size = 0.87 \[ \frac {1}{2} \, \arctan \left (\frac {1}{x}\right )^{2} - \frac {\arctan \left (\frac {1}{x}\right )}{x} + \frac {1}{2} \, \log \left (\frac {1}{x^{2}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(1/x)^2 - arctan(1/x)/x + 1/2*log(1/x^2 + 1)

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maple [A]  time = 0.05, size = 33, normalized size = 1.10 \[ -\frac {\mathrm {arccot}\relax (x )}{x}-\mathrm {arccot}\relax (x ) \arctan \relax (x )-\ln \relax (x )+\frac {\ln \left (x^{2}+1\right )}{2}-\frac {\arctan \relax (x )^{2}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 29, normalized size = 0.97 \[ -{\left (\frac {1}{x} + \arctan \relax (x)\right )} \operatorname {arccot}\relax (x) - \frac {1}{2} \, \arctan \relax (x)^{2} + \frac {1}{2} \, \log \left (x^{2} + 1\right ) - \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.09, size = 26, normalized size = 0.87 \[ \frac {\ln \left (x^2+1\right )}{2}-\ln \relax (x)-\frac {\mathrm {acot}\relax (x)}{x}+\frac {{\mathrm {acot}\relax (x)}^2}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(x)/(x^2*(x^2 + 1)),x)

[Out]

log(x^2 + 1)/2 - log(x) - acot(x)/x + acot(x)^2/2

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sympy [A]  time = 0.43, size = 22, normalized size = 0.73 \[ - \log {\relax (x )} + \frac {\log {\left (x^{2} + 1 \right )}}{2} + \frac {\operatorname {acot}^{2}{\relax (x )}}{2} - \frac {\operatorname {acot}{\relax (x )}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x) + log(x**2 + 1)/2 + acot(x)**2/2 - acot(x)/x

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