3.44 \(\int \frac {\cot ^{-1}(x)}{x^3 (1+x^2)} \, dx\)

Optimal. Leaf size=72 \[ -\frac {1}{2} i \text {Li}_2\left (\frac {2}{1-i x}-1\right )-\frac {\cot ^{-1}(x)}{2 x^2}+\frac {1}{2 x}+\frac {1}{2} \tan ^{-1}(x)-\frac {1}{2} i \cot ^{-1}(x)^2-\log \left (2-\frac {2}{1-i x}\right ) \cot ^{-1}(x) \]

[Out]

1/2/x-1/2*arccot(x)/x^2-1/2*I*arccot(x)^2+1/2*arctan(x)-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 = 0.11, antiderivative size = 72, 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, 325, 203, 4925, 4869, 2447} \[ -\frac {1}{2} i \text {PolyLog}\left (2,-1+\frac {2}{1-i x}\right )-\frac {\cot ^{-1}(x)}{2 x^2}+\frac {1}{2 x}+\frac {1}{2} \tan ^{-1}(x)-\frac {1}{2} i \cot ^{-1}(x)^2-\log \left (2-\frac {2}{1-i x}\right ) \cot ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[x]/(x^3*(1 + x^2)),x]

[Out]

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

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])

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2447

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 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 4869

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 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]

Rule 4925

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^3 \left (1+x^2\right )} \, dx &=\int \frac {\cot ^{-1}(x)}{x^3} \, dx-\int \frac {\cot ^{-1}(x)}{x \left (1+x^2\right )} \, dx\\ &=-\frac {\cot ^{-1}(x)}{2 x^2}-\frac {1}{2} i \cot ^{-1}(x)^2-i \int \frac {\cot ^{-1}(x)}{x (i+x)} \, dx-\frac {1}{2} \int \frac {1}{x^2 \left (1+x^2\right )} \, dx\\ &=\frac {1}{2 x}-\frac {\cot ^{-1}(x)}{2 x^2}-\frac {1}{2} i \cot ^{-1}(x)^2-\cot ^{-1}(x) \log \left (2-\frac {2}{1-i x}\right )+\frac {1}{2} \int \frac {1}{1+x^2} \, dx-\int \frac {\log \left (2-\frac {2}{1-i x}\right )}{1+x^2} \, dx\\ &=\frac {1}{2 x}-\frac {\cot ^{-1}(x)}{2 x^2}-\frac {1}{2} i \cot ^{-1}(x)^2+\frac {1}{2} \tan ^{-1}(x)-\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 [C]  time = 0.07, size = 280, normalized size = 3.89 \[ \frac {\, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-x^2\right )}{2 x}+\frac {1}{4} i \text {Li}_2\left (-\frac {1}{2} i (i-x)\right )+\frac {1}{2} i \text {Li}_2\left (-\frac {i}{x}\right )-\frac {1}{2} i \text {Li}_2\left (\frac {i}{x}\right )-\frac {1}{4} i \text {Li}_2\left (-\frac {1}{2} i (x+i)\right )-\frac {\cot ^{-1}(x)}{2 x^2}-\frac {1}{8} i \log ^2(-x+i)+\frac {1}{8} i \log ^2(x+i)+\frac {1}{4} i \log \left (-\frac {-x+i}{x}\right ) \log (-x+i)+\frac {1}{4} i \log \left (-\frac {1}{2} i (x+i)\right ) \log (-x+i)-\frac {1}{4} i \log \left (\frac {x+i}{x}\right ) \log (-x+i)-\frac {1}{4} i \log \left (-\frac {1}{2} i (-x+i)\right ) \log (x+i)+\frac {1}{4} i \log \left (-\frac {-x+i}{x}\right ) \log (x+i)-\frac {1}{4} i \log (x+i) \log \left (\frac {x+i}{x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCot[x]/(x^3*(1 + x^2)),x]

[Out]

-1/2*ArcCot[x]/x^2 + Hypergeometric2F1[-1/2, 1, 1/2, -x^2]/(2*x) - (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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fricas [F]  time = 1.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arccot}\relax (x)}{x^{5} + x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/x^3/(x^2+1),x, algorithm="fricas")

[Out]

integral(arccot(x)/(x^5 + x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arccot}\relax (x)}{{\left (x^{2} + 1\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/x^3/(x^2+1),x, algorithm="giac")

[Out]

integrate(arccot(x)/((x^2 + 1)*x^3), x)

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maple [B]  time = 0.11, size = 180, normalized size = 2.50 \[ -\frac {\mathrm {arccot}\relax (x )}{2 x^{2}}-\mathrm {arccot}\relax (x ) \ln \relax (x )+\frac {\mathrm {arccot}\relax (x ) \ln \left (x^{2}+1\right )}{2}+\frac {i \dilog \left (-\frac {i \left (x +i\right )}{2}\right )}{4}-\frac {i \ln \left (x +i\right ) \ln \left (\frac {i \left (x -i\right )}{2}\right )}{4}-\frac {i \ln \relax (x ) \ln \left (-i x +1\right )}{2}+\frac {i \ln \relax (x ) \ln \left (i x +1\right )}{2}-\frac {i \ln \left (x -i\right ) \ln \left (x^{2}+1\right )}{4}-\frac {i \dilog \left (\frac {i \left (x -i\right )}{2}\right )}{4}-\frac {i \ln \left (x +i\right )^{2}}{8}+\frac {i \dilog \left (i x +1\right )}{2}+\frac {1}{2 x}+\frac {\arctan \relax (x )}{2}-\frac {i \dilog \left (-i x +1\right )}{2}+\frac {i \ln \left (x -i\right ) \ln \left (-\frac {i \left (x +i\right )}{2}\right )}{4}+\frac {i \ln \left (x -i\right )^{2}}{8}+\frac {i \ln \left (x +i\right ) \ln \left (x^{2}+1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(x)/x^3/(x^2+1),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arccot}\relax (x)}{{\left (x^{2} + 1\right )} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/x^3/(x^2+1),x, algorithm="maxima")

[Out]

integrate(arccot(x)/((x^2 + 1)*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acot}\relax (x)}{x^3\,\left (x^2+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(x)/(x^3*(x^2 + 1)),x)

[Out]

int(acot(x)/(x^3*(x^2 + 1)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acot}{\relax (x )}}{x^{3} \left (x^{2} + 1\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(x)/x**3/(x**2+1),x)

[Out]

Integral(acot(x)/(x**3*(x**2 + 1)), x)

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