3.7.48 \(\int \frac {\tan ^{-1}(x)^2}{x^5} \, dx\) [648]

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

[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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Rubi [A]
time = 0.09, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {4946, 5038, 272, 46, 36, 29, 31, 5004} \begin {gather*} -\frac {\text {ArcTan}(x)^2}{4 x^4}-\frac {\text {ArcTan}(x)}{6 x^3}+\frac {\text {ArcTan}(x)^2}{4}+\frac {\text {ArcTan}(x)}{2 x}-\frac {1}{12 x^2}+\frac {1}{3} \log \left (x^2+1\right )-\frac {2 \log (x)}{3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*1/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 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 46

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] && Lt
Q[m + n + 2, 0])

Rule 272

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 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 5004

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]

Rule 5038

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]

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} \text {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} \text {Subst}\left (\int \left (\frac {1}{x^2}-\frac {1}{x}+\frac {1}{1+x}\right ) \, dx,x,x^2\right )-\frac {1}{4} \text {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} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{4} \text {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*}

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Mathematica [A]
time = 0.01, size = 56, normalized size = 0.92 \begin {gather*} -\frac {1}{12 x^2}+\frac {\left (-1+3 x^2\right ) \tan ^{-1}(x)}{6 x^3}+\frac {\left (-1+x^4\right ) \tan ^{-1}(x)^2}{4 x^4}-\frac {2 \log (x)}{3}+\frac {1}{3} \log \left (1+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*1/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 = 0.08, size = 48, normalized size = 0.79

method	result	size
default	\(-\frac {1}{12 x^{2}}-\frac {\arctan \left (x \right )}{6 x^{3}}+\frac {\arctan \left (x \right )}{2 x}+\frac {\arctan \left (x \right )^{2}}{4}-\frac {\arctan \left (x \right )^{2}}{4 x^{4}}-\frac {2 \ln \left (x \right )}{3}+\frac {\ln \left (x^{2}+1\right )}{3}\)	\(48\)
risch	\(-\frac {\left (x^{4}-1\right ) \ln \left (i x +1\right )^{2}}{16 x^{4}}+\frac {\left (3 x^{4} \ln \left (-i x +1\right )-6 i x^{3}+2 i x -3 \ln \left (-i x +1\right )\right ) \ln \left (i x +1\right )}{24 x^{4}}-\frac {3 x^{4} \ln \left (-i x +1\right )^{2}-12 i x^{3} \ln \left (-i x +1\right )+32 x^{4} \ln \left (x \right )-16 x^{4} \ln \left (x^{2}+1\right )+4 i x \ln \left (-i x +1\right )+4 x^{2}-3 \ln \left (-i x +1\right )^{2}}{48 x^{4}}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/x^2-1/6/x^3*arctan(x)+1/2/x*arctan(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 = 3.29, size = 64, normalized size = 1.05 \begin {gather*} \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 {gather*}

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 = 0.42, size = 53, normalized size = 0.87 \begin {gather*} \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 {gather*}

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 = 0.26, size = 53, normalized size = 0.87 \begin {gather*} - \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 0.11, size = 44, normalized size = 0.72 \begin {gather*} \frac {\ln \left (x^2+1\right )}{3}-\frac {2\,\ln \left (x\right )}{3}-{\mathrm {atan}\left (x\right )}^2\,\left (\frac {1}{4\,x^4}-\frac {1}{4}\right )-\frac {1}{12\,x^2}+\frac {\mathrm {atan}\left (x\right )\,\left (\frac {x^2}{2}-\frac {1}{6}\right )}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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