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

Optimal. Leaf size=23 \[ -\frac {1}{2} \log \left (x^2+1\right )-\frac {1}{2} \tan ^{-1}(x)^2+x \tan ^{-1}(x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {4916, 4846, 260, 4884} \[ -\frac {1}{2} \log \left (x^2+1\right )-\frac {1}{2} \tan ^{-1}(x)^2+x \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^2*ArcTan[x])/(1 + x^2),x]

[Out]

x*ArcTan[x] - ArcTan[x]^2/2 - Log[1 + x^2]/2

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 4846

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

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]

Rule 4916

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2/
e, Int[(f*x)^(m - 2)*(a + b*ArcTan[c*x])^p, x], x] - Dist[(d*f^2)/e, 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] && GtQ[m, 1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 23, normalized size = 1.00 \[ -\frac {1}{2} \log \left (x^2+1\right )-\frac {1}{2} \tan ^{-1}(x)^2+x \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

x*ArcTan[x] - ArcTan[x]^2/2 - Log[1 + x^2]/2

________________________________________________________________________________________

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

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

________________________________________________________________________________________

fricas [A]  time = 0.87, size = 19, normalized size = 0.83 \[ x \arctan \relax (x) - \frac {1}{2} \, \arctan \relax (x)^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.99, size = 19, normalized size = 0.83 \[ x \arctan \relax (x) - \frac {1}{2} \, \arctan \relax (x)^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.30, size = 20, normalized size = 0.87

method	result	size
default	\(x \arctan \relax (x )-\frac {\arctan \relax (x )^{2}}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(20\)
risch	\(\frac {\ln \left (i x +1\right )^{2}}{8}+\frac {i \left (-x +\frac {i \ln \left (-i x +1\right )}{2}\right ) \ln \left (i x +1\right )}{2}+\frac {\ln \left (-i x +1\right )^{2}}{8}+\frac {i \ln \left (-i x +1\right ) x}{2}-\frac {\ln \left (x^{2}+1\right )}{2}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*arctan(x)-1/2*arctan(x)^2-1/2*ln(x^2+1)

________________________________________________________________________________________

maxima [A]  time = 0.99, size = 24, normalized size = 1.04 \[ {\left (x - \arctan \relax (x)\right )} \arctan \relax (x) + \frac {1}{2} \, \arctan \relax (x)^{2} - \frac {1}{2} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.22, size = 19, normalized size = 0.83 \[ -\frac {{\mathrm {atan}\relax (x)}^2}{2}+x\,\mathrm {atan}\relax (x)-\frac {\ln \left (x^2+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*atan(x))/(x^2 + 1),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.39, size = 19, normalized size = 0.83 \[ x \operatorname {atan}{\relax (x )} - \frac {\log {\left (x^{2} + 1 \right )}}{2} - \frac {\operatorname {atan}^{2}{\relax (x )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atan(x)/(x**2+1),x)

[Out]

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

________________________________________________________________________________________