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

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

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Rubi [A]  time = 0.08, antiderivative size = 60, normalized size of antiderivative = 1.00, 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 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 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 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]))

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*}

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Mathematica [A]  time = 0.03, size = 56, normalized size = 0.93 \[ \frac {-2 \left (3 x^3+x\right ) \tan ^{-1}(x)+x^2 \left (4 x^2 \log (x)-2 x^2 \log \left (x^2+1\right )-1\right )-3 \left (x^2+1\right )^2 \tan ^{-1}(x)^2}{12 x^4} \]

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (1+x^2\right ) \tan ^{-1}(x)^2}{x^5} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.00, size = 54, normalized size = 0.90 \[ -\frac {2 \, x^{4} \log \left (x^{2} + 1\right ) - 4 \, x^{4} \log \relax (x) + 3 \, {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \relax (x)^{2} + x^{2} + 2 \, {\left (3 \, x^{3} + x\right )} \arctan \relax (x)}{12 \, x^{4}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (x^{2} + 1\right )} \arctan \relax (x)^{2}}{x^{5}}\,{d x} \]

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)

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maple [A]  time = 0.15, size = 57, normalized size = 0.95

method	result	size
default	\(-\frac {\arctan \relax (x )^{2}}{2 x^{2}}-\frac {\arctan \relax (x )^{2}}{4 x^{4}}-\frac {\arctan \relax (x )}{6 x^{3}}-\frac {\arctan \relax (x )}{2 x}-\frac {\arctan \relax (x )^{2}}{4}-\frac {\ln \left (x^{2}+1\right )}{6}-\frac {1}{12 x^{2}}+\frac {\ln \relax (x )}{3}\)	\(57\)
risch	\(\frac {\left (x^{4}+2 x^{2}+1\right ) \ln \left (i x +1\right )^{2}}{16 x^{4}}-\frac {\left (3 x^{4} \ln \left (-i x +1\right )-6 i x^{3}+6 x^{2} \ln \left (-i x +1\right )-2 i x +3 \ln \left (-i x +1\right )\right ) \ln \left (i x +1\right )}{24 x^{4}}+\frac {3 x^{4} \ln \left (-i x +1\right )^{2}+16 x^{4} \ln \relax (x )-8 \ln \left (x^{2}+1\right ) x^{4}-12 i x^{3} \ln \left (-i x +1\right )+6 x^{2} \ln \left (-i x +1\right )^{2}-4 i \ln \left (-i x +1\right ) x -4 x^{2}+3 \ln \left (-i x +1\right )^{2}}{48 x^{4}}\)	\(174\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.97, size = 71, normalized size = 1.18 \[ -\frac {1}{6} \, {\left (\frac {3 \, x^{2} + 1}{x^{3}} + 3 \, \arctan \relax (x)\right )} \arctan \relax (x) + \frac {3 \, x^{2} \arctan \relax (x)^{2} - 2 \, x^{2} \log \left (x^{2} + 1\right ) + 4 \, x^{2} \log \relax (x) - 1}{12 \, x^{2}} - \frac {{\left (2 \, x^{2} + 1\right )} \arctan \relax (x)^{2}}{4 \, x^{4}} \]

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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mupad [B]  time = 0.12, size = 51, normalized size = 0.85 \[ \frac {\ln \relax (x)}{3}-\frac {\ln \left (x^2+1\right )}{6}-{\mathrm {atan}\relax (x)}^2\,\left (\frac {\frac {x^2}{2}+\frac {1}{4}}{x^4}+\frac {1}{4}\right )-\frac {1}{12\,x^2}-\frac {\mathrm {atan}\relax (x)\,\left (\frac {x^2}{2}+\frac {1}{6}\right )}{x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atan(x)^2*(x^2 + 1))/x^5,x)

[Out]

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

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sympy [A]  time = 0.90, size = 61, normalized size = 1.02 \[ \frac {\log {\relax (x )}}{3} - \frac {\log {\left (x^{2} + 1 \right )}}{6} - \frac {\operatorname {atan}^{2}{\relax (x )}}{4} - \frac {\operatorname {atan}{\relax (x )}}{2 x} - \frac {\operatorname {atan}^{2}{\relax (x )}}{2 x^{2}} - \frac {1}{12 x^{2}} - \frac {\operatorname {atan}{\relax (x )}}{6 x^{3}} - \frac {\operatorname {atan}^{2}{\relax (x )}}{4 x^{4}} \]

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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