3.680 \(\int \frac{\tan ^{-1}(x)^2}{x^3} \, dx\)

Optimal. Leaf size=39 \[ -\frac{1}{2} \log \left (x^2+1\right )-\frac{\tan ^{-1}(x)^2}{2 x^2}+\log (x)-\frac{1}{2} \tan ^{-1}(x)^2-\frac{\tan ^{-1}(x)}{x} \]

[Out]

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

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Rubi [A]  time = 0.069604, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {4852, 4918, 266, 36, 29, 31, 4884} \[ -\frac{1}{2} \log \left (x^2+1\right )-\frac{\tan ^{-1}(x)^2}{2 x^2}+\log (x)-\frac{1}{2} \tan ^{-1}(x)^2-\frac{\tan ^{-1}(x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[x]^2/x^3,x]

[Out]

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

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 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 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)^2}{x^3} \, dx &=-\frac{\tan ^{-1}(x)^2}{2 x^2}+\int \frac{\tan ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx\\ &=-\frac{\tan ^{-1}(x)^2}{2 x^2}+\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-\frac{\tan ^{-1}(x)^2}{2 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{\tan ^{-1}(x)^2}{2 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{\tan ^{-1}(x)^2}{2 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-\frac{\tan ^{-1}(x)^2}{2 x^2}+\log (x)-\frac{1}{2} \log \left (1+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0194049, size = 38, normalized size = 0.97 \[ -\frac{1}{2} \log \left (x^2+1\right )+\frac{\left (-x^2-1\right ) \tan ^{-1}(x)^2}{2 x^2}+\log (x)-\frac{\tan ^{-1}(x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[x]^2/x^3,x]

[Out]

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

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Maple [A]  time = 0.001, size = 34, normalized size = 0.9 \begin{align*} -{\frac{\arctan \left ( x \right ) }{x}}-{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{2}}-{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{2\,{x}^{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)^2/x^3,x)

[Out]

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

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Maxima [A]  time = 1.42613, size = 49, normalized size = 1.26 \begin{align*} -{\left (\frac{1}{x} + \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac{1}{2} \, \arctan \left (x\right )^{2} - \frac{\arctan \left (x\right )^{2}}{2 \, x^{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)^2/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.50901, size = 113, normalized size = 2.9 \begin{align*} -\frac{{\left (x^{2} + 1\right )} \arctan \left (x\right )^{2} + x^{2} \log \left (x^{2} + 1\right ) - 2 \, x^{2} \log \left (x\right ) + 2 \, x \arctan \left (x\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x)^2/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.547456, size = 32, normalized size = 0.82 \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} - \frac{\operatorname{atan}^{2}{\left (x \right )}}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(x)**2/x**3,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (x\right )^{2}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x)^2/x^3,x, algorithm="giac")

[Out]

integrate(arctan(x)^2/x^3, x)