3.1.43 \(\int \frac {\cot ^{-1}(x)}{x^2 (1+x^2)} \, dx\) [43]

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

[Out]

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

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Rubi [A]
time = 0.04, 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 = {5039, 4947, 272, 36, 29, 31, 5005} \begin {gather*} \frac {1}{2} \log \left (x^2+1\right )-\log (x)+\frac {1}{2} \cot ^{-1}(x)^2-\frac {\cot ^{-1}(x)}{x} \end {gather*}

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 272

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 4947

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

Rule 5005

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 5039

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} \text {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} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \text {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 \begin {gather*} -\frac {\cot ^{-1}(x)}{x}+\frac {1}{2} \cot ^{-1}(x)^2-\log (x)+\frac {1}{2} \log \left (1+x^2\right ) \end {gather*}

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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Maple [A]
time = 0.16, size = 33, normalized size = 1.10

method	result	size
default	\(-\mathrm {arccot}\left (x \right ) \arctan \left (x \right )-\frac {\mathrm {arccot}\left (x \right )}{x}-\ln \left (x \right )+\frac {\ln \left (x^{2}+1\right )}{2}-\frac {\arctan \left (x \right )^{2}}{2}\)	\(33\)
risch	\(-\frac {\ln \left (i x +1\right )^{2}}{8}+\frac {\left (\ln \left (-i x +1\right ) x -2 i\right ) \ln \left (i x +1\right )}{4 x}-\frac {-2 i \ln \left (\left (-\pi +6 i\right ) x +6+i \pi \right ) \pi x +2 i \ln \left (\left (-\pi -6 i\right ) x +6-i \pi \right ) \pi x +\ln \left (-i x +1\right )^{2} x -4 i \ln \left (-i x +1\right )-4 \ln \left (\left (-\pi +6 i\right ) x +6+i \pi \right ) x -4 \ln \left (\left (-\pi -6 i\right ) x +6-i \pi \right ) x +8 \ln \left (-x \right ) x +4 \pi }{8 x}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.48, size = 29, normalized size = 0.97 \begin {gather*} -{\left (\frac {1}{x} + \arctan \left (x\right )\right )} \operatorname {arccot}\left (x\right ) - \frac {1}{2} \, \arctan \left (x\right )^{2} + \frac {1}{2} \, \log \left (x^{2} + 1\right ) - \log \left (x\right ) \end {gather*}

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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Fricas [A]
time = 4.09, size = 29, normalized size = 0.97 \begin {gather*} \frac {x \operatorname {arccot}\left (x\right )^{2} + x \log \left (x^{2} + 1\right ) - 2 \, x \log \left (x\right ) - 2 \, \operatorname {arccot}\left (x\right )}{2 \, x} \end {gather*}

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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Sympy [A]
time = 0.18, size = 22, normalized size = 0.73 \begin {gather*} - \log {\left (x \right )} + \frac {\log {\left (x^{2} + 1 \right )}}{2} + \frac {\operatorname {acot}^{2}{\left (x \right )}}{2} - \frac {\operatorname {acot}{\left (x \right )}}{x} \end {gather*}

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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Giac [A]
time = 0.41, size = 26, normalized size = 0.87 \begin {gather*} \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 ) \end {gather*}

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

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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