3.45 \(\int \frac {\cot ^{-1}(x)}{x^4 (1+x^2)} \, dx\)

Optimal. Leaf size=47 \[ -\frac {\cot ^{-1}(x)}{3 x^3}+\frac {1}{6 x^2}-\frac {2}{3} \log \left (x^2+1\right )+\frac {4 \log (x)}{3}-\frac {1}{2} \cot ^{-1}(x)^2+\frac {\cot ^{-1}(x)}{x} \]

[Out]

1/6/x^2-1/3*arccot(x)/x^3+arccot(x)/x-1/2*arccot(x)^2+4/3*ln(x)-2/3*ln(x^2+1)

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Rubi [A]  time = 0.11, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {4919, 4853, 266, 44, 36, 29, 31, 4885} \[ \frac {1}{6 x^2}-\frac {2}{3} \log \left (x^2+1\right )-\frac {\cot ^{-1}(x)}{3 x^3}+\frac {4 \log (x)}{3}-\frac {1}{2} \cot ^{-1}(x)^2+\frac {\cot ^{-1}(x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCot[x]/(x^4*(1 + x^2)),x]

[Out]

1/(6*x^2) - ArcCot[x]/(3*x^3) + ArcCot[x]/x - ArcCot[x]^2/2 + (4*Log[x])/3 - (2*Log[1 + x^2])/3

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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 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^4 \left (1+x^2\right )} \, dx &=\int \frac {\cot ^{-1}(x)}{x^4} \, dx-\int \frac {\cot ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx\\ &=-\frac {\cot ^{-1}(x)}{3 x^3}-\frac {1}{3} \int \frac {1}{x^3 \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)}{3 x^3}+\frac {\cot ^{-1}(x)}{x}-\frac {1}{2} \cot ^{-1}(x)^2-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{x^2 (1+x)} \, dx,x,x^2\right )+\int \frac {1}{x \left (1+x^2\right )} \, dx\\ &=-\frac {\cot ^{-1}(x)}{3 x^3}+\frac {\cot ^{-1}(x)}{x}-\frac {1}{2} \cot ^{-1}(x)^2-\frac {1}{6} \operatorname {Subst}\left (\int \left (\frac {1}{x^2}-\frac {1}{x}+\frac {1}{1+x}\right ) \, dx,x,x^2\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right )\\ &=\frac {1}{6 x^2}-\frac {\cot ^{-1}(x)}{3 x^3}+\frac {\cot ^{-1}(x)}{x}-\frac {1}{2} \cot ^{-1}(x)^2+\frac {\log (x)}{3}-\frac {1}{6} \log \left (1+x^2\right )+\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 {1}{6 x^2}-\frac {\cot ^{-1}(x)}{3 x^3}+\frac {\cot ^{-1}(x)}{x}-\frac {1}{2} \cot ^{-1}(x)^2+\frac {4 \log (x)}{3}-\frac {2}{3} \log \left (1+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 47, normalized size = 1.00 \[ -\frac {\cot ^{-1}(x)}{3 x^3}+\frac {1}{6 x^2}-\frac {2}{3} \log \left (x^2+1\right )+\frac {4 \log (x)}{3}-\frac {1}{2} \cot ^{-1}(x)^2+\frac {\cot ^{-1}(x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(6*x^2) - ArcCot[x]/(3*x^3) + ArcCot[x]/x - ArcCot[x]^2/2 + (4*Log[x])/3 - (2*Log[1 + x^2])/3

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fricas [A]  time = 0.66, size = 47, normalized size = 1.00 \[ -\frac {3 \, x^{3} \operatorname {arccot}\relax (x)^{2} + 4 \, x^{3} \log \left (x^{2} + 1\right ) - 8 \, x^{3} \log \relax (x) - 2 \, {\left (3 \, x^{2} - 1\right )} \operatorname {arccot}\relax (x) - x}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/x^4/(x^2+1),x, algorithm="fricas")

[Out]

-1/6*(3*x^3*arccot(x)^2 + 4*x^3*log(x^2 + 1) - 8*x^3*log(x) - 2*(3*x^2 - 1)*arccot(x) - x)/x^3

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giac [A]  time = 0.12, size = 39, normalized size = 0.83 \[ -\frac {1}{2} \, \arctan \left (\frac {1}{x}\right )^{2} + \frac {\arctan \left (\frac {1}{x}\right )}{x} + \frac {1}{6 \, x^{2}} - \frac {\arctan \left (\frac {1}{x}\right )}{3 \, x^{3}} - \frac {2}{3} \, \log \left (\frac {1}{x^{2}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/x^4/(x^2+1),x, algorithm="giac")

[Out]

-1/2*arctan(1/x)^2 + arctan(1/x)/x + 1/6/x^2 - 1/3*arctan(1/x)/x^3 - 2/3*log(1/x^2 + 1)

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maple [A]  time = 0.06, size = 43, normalized size = 0.91 \[ -\frac {\mathrm {arccot}\relax (x )}{3 x^{3}}+\frac {\mathrm {arccot}\relax (x )}{x}+\mathrm {arccot}\relax (x ) \arctan \relax (x )+\frac {1}{6 x^{2}}+\frac {4 \ln \relax (x )}{3}-\frac {2 \ln \left (x^{2}+1\right )}{3}+\frac {\arctan \relax (x )^{2}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccot(x)/x^4/(x^2+1),x)

[Out]

-1/3*arccot(x)/x^3+arccot(x)/x+arccot(x)*arctan(x)+1/6/x^2+4/3*ln(x)-2/3*ln(x^2+1)+1/2*arctan(x)^2

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maxima [A]  time = 0.42, size = 55, normalized size = 1.17 \[ \frac {1}{3} \, {\left (\frac {3 \, x^{2} - 1}{x^{3}} + 3 \, \arctan \relax (x)\right )} \operatorname {arccot}\relax (x) + \frac {3 \, x^{2} \arctan \relax (x)^{2} - 4 \, x^{2} \log \left (x^{2} + 1\right ) + 8 \, x^{2} \log \relax (x) + 1}{6 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccot(x)/x^4/(x^2+1),x, algorithm="maxima")

[Out]

1/3*((3*x^2 - 1)/x^3 + 3*arctan(x))*arccot(x) + 1/6*(3*x^2*arctan(x)^2 - 4*x^2*log(x^2 + 1) + 8*x^2*log(x) + 1
)/x^2

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mupad [B]  time = 0.10, size = 35, normalized size = 0.74 \[ \frac {4\,\ln \relax (x)}{3}-\frac {2\,\ln \left (x^2+1\right )}{3}-\frac {{\mathrm {acot}\relax (x)}^2}{2}+\frac {1}{6\,x^2}+\frac {\mathrm {acot}\relax (x)\,\left (x^2-\frac {1}{3}\right )}{x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acot(x)/(x^4*(x^2 + 1)),x)

[Out]

(4*log(x))/3 - (2*log(x^2 + 1))/3 - acot(x)^2/2 + 1/(6*x^2) + (acot(x)*(x^2 - 1/3))/x^3

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sympy [A]  time = 0.85, size = 42, normalized size = 0.89 \[ \frac {4 \log {\relax (x )}}{3} - \frac {2 \log {\left (x^{2} + 1 \right )}}{3} - \frac {\operatorname {acot}^{2}{\relax (x )}}{2} + \frac {\operatorname {acot}{\relax (x )}}{x} + \frac {1}{6 x^{2}} - \frac {\operatorname {acot}{\relax (x )}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acot(x)/x**4/(x**2+1),x)

[Out]

4*log(x)/3 - 2*log(x**2 + 1)/3 - acot(x)**2/2 + acot(x)/x + 1/(6*x**2) - acot(x)/(3*x**3)

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