3.1.42 \(\int \frac {\cot ^{-1}(x)}{x (1+x^2)} \, dx\) [42]

Optimal. Leaf size=49 \[ \frac {1}{2} i \cot ^{-1}(x)^2+\cot ^{-1}(x) \log \left (2-\frac {2}{1-i x}\right )+\frac {1}{2} i \text {PolyLog}\left (2,-1+\frac {2}{1-i x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {5045, 4989, 2497} \begin {gather*} \frac {1}{2} i \text {Li}_2\left (\frac {2}{1-i x}-1\right )+\frac {1}{2} i \cot ^{-1}(x)^2+\log \left (2-\frac {2}{1-i x}\right ) \cot ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcCot[x]/(x*(1 + x^2)),x]

[Out]

(I/2)*ArcCot[x]^2 + ArcCot[x]*Log[2 - 2/(1 - I*x)] + (I/2)*PolyLog[2, -1 + 2/(1 - I*x)]

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 4989

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

Rule 5045

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

Rubi steps

\begin {align*} \int \frac {\cot ^{-1}(x)}{x \left (1+x^2\right )} \, dx &=\frac {1}{2} i \cot ^{-1}(x)^2+i \int \frac {\cot ^{-1}(x)}{x (i+x)} \, dx\\ &=\frac {1}{2} i \cot ^{-1}(x)^2+\cot ^{-1}(x) \log \left (2-\frac {2}{1-i x}\right )+\int \frac {\log \left (2-\frac {2}{1-i x}\right )}{1+x^2} \, dx\\ &=\frac {1}{2} i \cot ^{-1}(x)^2+\cot ^{-1}(x) \log \left (2-\frac {2}{1-i x}\right )+\frac {1}{2} i \text {Li}_2\left (-1+\frac {2}{1-i x}\right )\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(251\) vs. \(2(49)=98\).
time = 0.05, size = 251, normalized size = 5.12 \begin {gather*} \frac {1}{8} i \log ^2(i-x)-\frac {1}{4} i \log (i-x) \log \left (-\frac {i-x}{x}\right )-\frac {1}{4} i \log (i-x) \log \left (-\frac {1}{2} i (i+x)\right )+\frac {1}{4} i \log \left (-\frac {1}{2} i (i-x)\right ) \log (i+x)-\frac {1}{4} i \log \left (-\frac {i-x}{x}\right ) \log (i+x)-\frac {1}{8} i \log ^2(i+x)+\frac {1}{4} i \log (i-x) \log \left (\frac {i+x}{x}\right )+\frac {1}{4} i \log (i+x) \log \left (\frac {i+x}{x}\right )-\frac {1}{4} i \text {PolyLog}\left (2,-\frac {1}{2} i (i-x)\right )-\frac {1}{2} i \text {PolyLog}\left (2,-\frac {i}{x}\right )+\frac {1}{2} i \text {PolyLog}\left (2,\frac {i}{x}\right )+\frac {1}{4} i \text {PolyLog}\left (2,-\frac {1}{2} i (i+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcCot[x]/(x*(1 + x^2)),x]

[Out]

(I/8)*Log[I - x]^2 - (I/4)*Log[I - x]*Log[-((I - x)/x)] - (I/4)*Log[I - x]*Log[(-1/2*I)*(I + x)] + (I/4)*Log[(
-1/2*I)*(I - x)]*Log[I + x] - (I/4)*Log[-((I - x)/x)]*Log[I + x] - (I/8)*Log[I + x]^2 + (I/4)*Log[I - x]*Log[(
I + x)/x] + (I/4)*Log[I + x]*Log[(I + x)/x] - (I/4)*PolyLog[2, (-1/2*I)*(I - x)] - (I/2)*PolyLog[2, (-I)/x] +
(I/2)*PolyLog[2, I/x] + (I/4)*PolyLog[2, (-1/2*I)*(I + x)]

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (41 ) = 82\).
time = 0.12, size = 163, normalized size = 3.33

method	result	size
risch	\(-\frac {\pi \ln \left (x^{2}+1\right )}{4}+\frac {\pi \ln \left (-i x \right )}{2}+\frac {i \ln \left (-i x +1\right )^{2}}{8}+\frac {i \dilog \left (-i x +1\right )}{2}+\frac {i \ln \left (\frac {1}{2}+\frac {i x}{2}\right ) \ln \left (-i x +1\right )}{4}-\frac {i \dilog \left (\frac {1}{2}-\frac {i x}{2}\right )}{4}-\frac {i \ln \left (i x +1\right )^{2}}{8}-\frac {i \dilog \left (i x +1\right )}{2}-\frac {i \ln \left (\frac {1}{2}-\frac {i x}{2}\right ) \ln \left (i x +1\right )}{4}+\frac {i \dilog \left (\frac {1}{2}+\frac {i x}{2}\right )}{4}\)	\(117\)
default	\(\mathrm {arccot}\left (x \right ) \ln \left (x \right )-\frac {\mathrm {arccot}\left (x \right ) \ln \left (x^{2}+1\right )}{2}-\frac {i \ln \left (x \right ) \ln \left (i x +1\right )}{2}+\frac {i \ln \left (x \right ) \ln \left (-i x +1\right )}{2}-\frac {i \dilog \left (i x +1\right )}{2}+\frac {i \dilog \left (-i x +1\right )}{2}+\frac {i \ln \left (x -i\right ) \ln \left (x^{2}+1\right )}{4}-\frac {i \dilog \left (-\frac {i \left (i+x \right )}{2}\right )}{4}-\frac {i \ln \left (x -i\right ) \ln \left (-\frac {i \left (i+x \right )}{2}\right )}{4}-\frac {i \ln \left (x -i\right )^{2}}{8}-\frac {i \ln \left (i+x \right ) \ln \left (x^{2}+1\right )}{4}+\frac {i \dilog \left (\frac {i \left (x -i\right )}{2}\right )}{4}+\frac {i \ln \left (i+x \right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )}{4}+\frac {i \ln \left (i+x \right )^{2}}{8}\)	\(163\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arccot(x)*ln(x)-1/2*arccot(x)*ln(x^2+1)-1/2*I*ln(x)*ln(1+I*x)+1/2*I*ln(x)*ln(1-I*x)-1/2*I*dilog(1+I*x)+1/2*I*d
ilog(1-I*x)+1/4*I*ln(x-I)*ln(x^2+1)-1/4*I*dilog(-1/2*I*(I+x))-1/4*I*ln(x-I)*ln(-1/2*I*(I+x))-1/8*I*ln(x-I)^2-1
/4*I*ln(I+x)*ln(x^2+1)+1/4*I*dilog(1/2*I*(x-I))+1/4*I*ln(I+x)*ln(1/2*I*(x-I))+1/8*I*ln(I+x)^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccot(x)/((x^2 + 1)*x), x)

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Fricas [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)/x/(x^2+1),x, algorithm="fricas")

[Out]

integral(arccot(x)/(x^3 + x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acot(x)/(x*(x**2 + 1)), 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)/x/(x^2+1),x, algorithm="giac")

[Out]

integrate(arccot(x)/((x^2 + 1)*x), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\mathrm {acot}\left (x\right )}{x\,\left (x^2+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(x)/(x*(x^2 + 1)),x)

[Out]

int(acot(x)/(x*(x^2 + 1)), x)

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