3.648 \(\int \frac{\tan ^{-1}(x)^2}{x^5} \, dx\)

Optimal. Leaf size=61 \[ -\frac{1}{12 x^2}+\frac{1}{3} \log \left (x^2+1\right )-\frac{\tan ^{-1}(x)^2}{4 x^4}-\frac{\tan ^{-1}(x)}{6 x^3}-\frac{2 \log (x)}{3}+\frac{1}{4} \tan ^{-1}(x)^2+\frac{\tan ^{-1}(x)}{2 x} \]

[Out]

-1/(12*x^2) - ArcTan[x]/(6*x^3) + ArcTan[x]/(2*x) + ArcTan[x]^2/4 - ArcTan[x]^2/(4*x^4) - (2*Log[x])/3 + Log[1
+ x^2]/3

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Rubi [A]  time = 0.126128, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {4852, 4918, 266, 44, 36, 29, 31, 4884} \[ -\frac{1}{12 x^2}+\frac{1}{3} \log \left (x^2+1\right )-\frac{\tan ^{-1}(x)^2}{4 x^4}-\frac{\tan ^{-1}(x)}{6 x^3}-\frac{2 \log (x)}{3}+\frac{1}{4} \tan ^{-1}(x)^2+\frac{\tan ^{-1}(x)}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcTan[x]^2/x^5,x]

[Out]

-1/(12*x^2) - ArcTan[x]/(6*x^3) + ArcTan[x]/(2*x) + ArcTan[x]^2/4 - ArcTan[x]^2/(4*x^4) - (2*Log[x])/3 + Log[1
+ x^2]/3

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 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 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^5} \, dx &=-\frac{\tan ^{-1}(x)^2}{4 x^4}+\frac{1}{2} \int \frac{\tan ^{-1}(x)}{x^4 \left (1+x^2\right )} \, dx\\ &=-\frac{\tan ^{-1}(x)^2}{4 x^4}+\frac{1}{2} \int \frac{\tan ^{-1}(x)}{x^4} \, dx-\frac{1}{2} \int \frac{\tan ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx\\ &=-\frac{\tan ^{-1}(x)}{6 x^3}-\frac{\tan ^{-1}(x)^2}{4 x^4}+\frac{1}{6} \int \frac{1}{x^3 \left (1+x^2\right )} \, dx-\frac{1}{2} \int \frac{\tan ^{-1}(x)}{x^2} \, dx+\frac{1}{2} \int \frac{\tan ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac{\tan ^{-1}(x)}{6 x^3}+\frac{\tan ^{-1}(x)}{2 x}+\frac{1}{4} \tan ^{-1}(x)^2-\frac{\tan ^{-1}(x)^2}{4 x^4}+\frac{1}{12} \operatorname{Subst}\left (\int \frac{1}{x^2 (1+x)} \, dx,x,x^2\right )-\frac{1}{2} \int \frac{1}{x \left (1+x^2\right )} \, dx\\ &=-\frac{\tan ^{-1}(x)}{6 x^3}+\frac{\tan ^{-1}(x)}{2 x}+\frac{1}{4} \tan ^{-1}(x)^2-\frac{\tan ^{-1}(x)^2}{4 x^4}+\frac{1}{12} \operatorname{Subst}\left (\int \left (\frac{1}{x^2}-\frac{1}{x}+\frac{1}{1+x}\right ) \, dx,x,x^2\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x (1+x)} \, dx,x,x^2\right )\\ &=-\frac{1}{12 x^2}-\frac{\tan ^{-1}(x)}{6 x^3}+\frac{\tan ^{-1}(x)}{2 x}+\frac{1}{4} \tan ^{-1}(x)^2-\frac{\tan ^{-1}(x)^2}{4 x^4}-\frac{\log (x)}{6}+\frac{1}{12} \log \left (1+x^2\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,x^2\right )\\ &=-\frac{1}{12 x^2}-\frac{\tan ^{-1}(x)}{6 x^3}+\frac{\tan ^{-1}(x)}{2 x}+\frac{1}{4} \tan ^{-1}(x)^2-\frac{\tan ^{-1}(x)^2}{4 x^4}-\frac{2 \log (x)}{3}+\frac{1}{3} \log \left (1+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0205726, size = 56, normalized size = 0.92 \[ -\frac{1}{12 x^2}+\frac{1}{3} \log \left (x^2+1\right )+\frac{\left (x^4-1\right ) \tan ^{-1}(x)^2}{4 x^4}+\frac{\left (3 x^2-1\right ) \tan ^{-1}(x)}{6 x^3}-\frac{2 \log (x)}{3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcTan[x]^2/x^5,x]

[Out]

-1/(12*x^2) + ((-1 + 3*x^2)*ArcTan[x])/(6*x^3) + ((-1 + x^4)*ArcTan[x]^2)/(4*x^4) - (2*Log[x])/3 + Log[1 + x^2
]/3

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Maple [A]  time = 0.017, size = 48, normalized size = 0.8 \begin{align*} -{\frac{1}{12\,{x}^{2}}}-{\frac{\arctan \left ( x \right ) }{6\,{x}^{3}}}+{\frac{\arctan \left ( x \right ) }{2\,x}}+{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{4}}-{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{4\,{x}^{4}}}-{\frac{2\,\ln \left ( x \right ) }{3}}+{\frac{\ln \left ({x}^{2}+1 \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x)^2/x^5,x)

[Out]

-1/12/x^2-1/6*arctan(x)/x^3+1/2*arctan(x)/x+1/4*arctan(x)^2-1/4*arctan(x)^2/x^4-2/3*ln(x)+1/3*ln(x^2+1)

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Maxima [A]  time = 1.43238, size = 86, normalized size = 1.41 \begin{align*} \frac{1}{6} \,{\left (\frac{3 \, x^{2} - 1}{x^{3}} + 3 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) - \frac{3 \, x^{2} \arctan \left (x\right )^{2} - 4 \, x^{2} \log \left (x^{2} + 1\right ) + 8 \, x^{2} \log \left (x\right ) + 1}{12 \, x^{2}} - \frac{\arctan \left (x\right )^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x)^2/x^5,x, algorithm="maxima")

[Out]

1/6*((3*x^2 - 1)/x^3 + 3*arctan(x))*arctan(x) - 1/12*(3*x^2*arctan(x)^2 - 4*x^2*log(x^2 + 1) + 8*x^2*log(x) +
1)/x^2 - 1/4*arctan(x)^2/x^4

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Fricas [A]  time = 2.51856, size = 140, normalized size = 2.3 \begin{align*} \frac{4 \, x^{4} \log \left (x^{2} + 1\right ) - 8 \, x^{4} \log \left (x\right ) + 3 \,{\left (x^{4} - 1\right )} \arctan \left (x\right )^{2} - x^{2} + 2 \,{\left (3 \, x^{3} - x\right )} \arctan \left (x\right )}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x)^2/x^5,x, algorithm="fricas")

[Out]

1/12*(4*x^4*log(x^2 + 1) - 8*x^4*log(x) + 3*(x^4 - 1)*arctan(x)^2 - x^2 + 2*(3*x^3 - x)*arctan(x))/x^4

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Sympy [A]  time = 0.901114, size = 53, normalized size = 0.87 \begin{align*} - \frac{2 \log{\left (x \right )}}{3} + \frac{\log{\left (x^{2} + 1 \right )}}{3} + \frac{\operatorname{atan}^{2}{\left (x \right )}}{4} + \frac{\operatorname{atan}{\left (x \right )}}{2 x} - \frac{1}{12 x^{2}} - \frac{\operatorname{atan}{\left (x \right )}}{6 x^{3}} - \frac{\operatorname{atan}^{2}{\left (x \right )}}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(atan(x)**2/x**5,x)

[Out]

-2*log(x)/3 + log(x**2 + 1)/3 + atan(x)**2/4 + atan(x)/(2*x) - 1/(12*x**2) - atan(x)/(6*x**3) - atan(x)**2/(4*
x**4)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (x\right )^{2}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctan(x)^2/x^5,x, algorithm="giac")

[Out]

integrate(arctan(x)^2/x^5, x)