3.7.80 \(\int \frac {\tan ^{-1}(x)^2}{x^3} \, dx\) [680]

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

[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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Rubi [A]
time = 0.05, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 = {4946, 5038, 272, 36, 29, 31, 5004} \begin {gather*} -\frac {\text {ArcTan}(x)^2}{2 x^2}-\frac {\text {ArcTan}(x)^2}{2}-\frac {\text {ArcTan}(x)}{x}-\frac {1}{2} \log \left (x^2+1\right )+\log (x) \end {gather*}

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

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Mathematica [A]
time = 0.01, size = 38, normalized size = 0.97 \begin {gather*} -\frac {\tan ^{-1}(x)}{x}+\frac {\left (-1-x^2\right ) \tan ^{-1}(x)^2}{2 x^2}+\log (x)-\frac {1}{2} \log \left (1+x^2\right ) \end {gather*}

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 = 0.07, size = 34, normalized size = 0.87

method	result	size
default	\(-\frac {\arctan \left (x \right )}{x}-\frac {\arctan \left (x \right )^{2}}{2}-\frac {\arctan \left (x \right )^{2}}{2 x^{2}}+\ln \left (x \right )-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(34\)
risch	\(\frac {\left (x^{2}+1\right ) \ln \left (i x +1\right )^{2}}{8 x^{2}}-\frac {\left (x^{2} \ln \left (-i x +1\right )-2 i x +\ln \left (-i x +1\right )\right ) \ln \left (i x +1\right )}{4 x^{2}}+\frac {x^{2} \ln \left (-i x +1\right )^{2}+8 x^{2} \ln \left (x \right )-4 x^{2} \ln \left (x^{2}+1\right )-4 i x \ln \left (-i x +1\right )+\ln \left (-i x +1\right )^{2}}{8 x^{2}}\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctan(x)^2/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/x*arctan(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 = 3.77, size = 36, normalized size = 0.92 \begin {gather*} -{\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 {gather*}

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 = 0.58, size = 38, normalized size = 0.97 \begin {gather*} -\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 {gather*}

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 = 0.19, size = 32, normalized size = 0.82 \begin {gather*} \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 {gather*}

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 = 0.00, 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^3,x, algorithm="giac")

[Out]

integrate(arctan(x)^2/x^3, x)

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Mupad [B]
time = 0.07, size = 31, normalized size = 0.79 \begin {gather*} \ln \left (x\right )-\frac {\ln \left (x^2+1\right )}{2}-\frac {\mathrm {atan}\left (x\right )}{x}-{\mathrm {atan}\left (x\right )}^2\,\left (\frac {1}{2\,x^2}+\frac {1}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(x)^2/x^3,x)

[Out]

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

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