3.366 \(\int x^2 \tan ^{-1}(x) \, dx\)

Optimal. Leaf size=27 \[ -\frac{x^2}{6}+\frac{1}{6} \log \left (x^2+1\right )+\frac{1}{3} x^3 \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0162793, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4852, 266, 43} \[ -\frac{x^2}{6}+\frac{1}{6} \log \left (x^2+1\right )+\frac{1}{3} x^3 \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTan[x],x]

[Out]

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

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0042958, size = 23, normalized size = 0.85 \[ \frac{1}{6} \left (-x^2+\log \left (x^2+1\right )+2 x^3 \tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTan[x],x]

[Out]

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

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Maple [A]  time = 0., size = 22, normalized size = 0.8 \begin{align*} -{\frac{{x}^{2}}{6}}+{\frac{{x}^{3}\arctan \left ( x \right ) }{3}}+{\frac{\ln \left ({x}^{2}+1 \right ) }{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctan(x),x)

[Out]

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

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Maxima [A]  time = 0.920129, size = 28, normalized size = 1.04 \begin{align*} \frac{1}{3} \, x^{3} \arctan \left (x\right ) - \frac{1}{6} \, x^{2} + \frac{1}{6} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctan(x) - 1/6*x^2 + 1/6*log(x^2 + 1)

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Fricas [A]  time = 2.03142, size = 65, normalized size = 2.41 \begin{align*} \frac{1}{3} \, x^{3} \arctan \left (x\right ) - \frac{1}{6} \, x^{2} + \frac{1}{6} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctan(x) - 1/6*x^2 + 1/6*log(x^2 + 1)

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Sympy [A]  time = 0.325198, size = 20, normalized size = 0.74 \begin{align*} \frac{x^{3} \operatorname{atan}{\left (x \right )}}{3} - \frac{x^{2}}{6} + \frac{\log{\left (x^{2} + 1 \right )}}{6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atan(x),x)

[Out]

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

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Giac [A]  time = 1.06491, size = 28, normalized size = 1.04 \begin{align*} \frac{1}{3} \, x^{3} \arctan \left (x\right ) - \frac{1}{6} \, x^{2} + \frac{1}{6} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*arctan(x) - 1/6*x^2 + 1/6*log(x^2 + 1)