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

Optimal. Leaf size=60 \[ -\frac{1}{12 x^2}-\frac{1}{6} \log \left (x^2+1\right )-\frac{\left (x^2+1\right )^2 \tan ^{-1}(x)^2}{4 x^4}-\frac{\tan ^{-1}(x)}{6 x^3}+\frac{\log (x)}{3}-\frac{\tan ^{-1}(x)}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.0779685, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {4944, 4950, 4852, 266, 44, 36, 29, 31} \[ -\frac{1}{12 x^2}-\frac{1}{6} \log \left (x^2+1\right )-\frac{\left (x^2+1\right )^2 \tan ^{-1}(x)^2}{4 x^4}-\frac{\tan ^{-1}(x)}{6 x^3}+\frac{\log (x)}{3}-\frac{\tan ^{-1}(x)}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[((1 + x^2)*ArcTan[x]^2)/x^5,x]

[Out]

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

Rule 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 1)), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 4950

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0318336, size = 56, normalized size = 0.93 \[ \frac{x^2 \left (4 x^2 \log (x)-2 x^2 \log \left (x^2+1\right )-1\right )-3 \left (x^2+1\right )^2 \tan ^{-1}(x)^2-2 \left (3 x^3+x\right ) \tan ^{-1}(x)}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((1 + x^2)*ArcTan[x]^2)/x^5,x]

[Out]

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

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Maple [A]  time = 0.019, size = 57, normalized size = 1. \begin{align*} -{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{4\,{x}^{4}}}-{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{2\,{x}^{2}}}-{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{4}}-{\frac{\arctan \left ( x \right ) }{6\,{x}^{3}}}-{\frac{\arctan \left ( x \right ) }{2\,x}}-{\frac{\ln \left ({x}^{2}+1 \right ) }{6}}-{\frac{1}{12\,{x}^{2}}}+{\frac{\ln \left ( x \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+1)*arctan(x)^2/x^5,x)

[Out]

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

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Maxima [A]  time = 1.43972, size = 96, normalized size = 1.6 \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} - 2 \, x^{2} \log \left (x^{2} + 1\right ) + 4 \, x^{2} \log \left (x\right ) - 1}{12 \, x^{2}} - \frac{{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*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 - 2*x^2*log(x^2 + 1) + 4*x^2*log(x) -
1)/x^2 - 1/4*(2*x^2 + 1)*arctan(x)^2/x^4

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Fricas [A]  time = 2.42932, size = 153, normalized size = 2.55 \begin{align*} -\frac{2 \, x^{4} \log \left (x^{2} + 1\right ) - 4 \, x^{4} \log \left (x\right ) + 3 \,{\left (x^{4} + 2 \, x^{2} + 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((x^2+1)*arctan(x)^2/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.907907, size = 61, normalized size = 1.02 \begin{align*} \frac{\log{\left (x \right )}}{3} - \frac{\log{\left (x^{2} + 1 \right )}}{6} - \frac{\operatorname{atan}^{2}{\left (x \right )}}{4} - \frac{\operatorname{atan}{\left (x \right )}}{2 x} - \frac{\operatorname{atan}^{2}{\left (x \right )}}{2 x^{2}} - \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((x**2+1)*atan(x)**2/x**5,x)

[Out]

log(x)/3 - log(x**2 + 1)/6 - atan(x)**2/4 - atan(x)/(2*x) - atan(x)**2/(2*x**2) - 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{{\left (x^{2} + 1\right )} \arctan \left (x\right )^{2}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*arctan(x)^2/x^5,x, algorithm="giac")

[Out]

integrate((x^2 + 1)*arctan(x)^2/x^5, x)