3.679 \(\int \frac{\tan ^{-1}(x)}{x^2 (1+x^2)} \, dx\)

Optimal. Leaf size=28 \[ -\frac{1}{2} \log \left (x^2+1\right )+\log (x)-\frac{1}{2} \tan ^{-1}(x)^2-\frac{\tan ^{-1}(x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.0574401, antiderivative size = 28, normalized size of antiderivative = 1., 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 = {4918, 4852, 266, 36, 29, 31, 4884} \[ -\frac{1}{2} \log \left (x^2+1\right )+\log (x)-\frac{1}{2} \tan ^{-1}(x)^2-\frac{\tan ^{-1}(x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[x]/(x^2*(1 + x^2)),x]

[Out]

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

Rule 4918

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

Rule 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[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 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 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 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 4884

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan[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]

Rubi steps

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

Mathematica [A]  time = 0.0064808, size = 28, normalized size = 1. \[ -\frac{1}{2} \log \left (x^2+1\right )+\log (x)-\frac{1}{2} \tan ^{-1}(x)^2-\frac{\tan ^{-1}(x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 25, normalized size = 0.9 \begin{align*} -{\frac{\arctan \left ( x \right ) }{x}}-{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{2}}+\ln \left ( x \right ) -{\frac{\ln \left ({x}^{2}+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.43856, size = 36, normalized size = 1.29 \begin{align*} -{\left (\frac{1}{x} + \arctan \left (x\right )\right )} \arctan \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{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.50508, size = 92, normalized size = 3.29 \begin{align*} -\frac{x \arctan \left (x\right )^{2} + x \log \left (x^{2} + 1\right ) - 2 \, x \log \left (x\right ) + 2 \, \arctan \left (x\right )}{2 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.595404, size = 22, normalized size = 0.79 \begin{align*} \log{\left (x \right )} - \frac{\log{\left (x^{2} + 1 \right )}}{2} - \frac{\operatorname{atan}^{2}{\left (x \right )}}{2} - \frac{\operatorname{atan}{\left (x \right )}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (x\right )}{{\left (x^{2} + 1\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arctan(x)/((x^2 + 1)*x^2), x)