3.4.66 \(\int x^2 \tan ^{-1}(x) \, dx\) [366]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4946, 272, 45} \begin {gather*} \frac {1}{3} x^3 \text {ArcTan}(x)-\frac {x^2}{6}+\frac {1}{6} \log \left (x^2+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 45

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])

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]

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

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Mathematica [A]
time = 0.00, size = 23, normalized size = 0.85 \begin {gather*} \frac {1}{6} \left (-x^2+2 x^3 \tan ^{-1}(x)+\log \left (1+x^2\right )\right ) \end {gather*}

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 = 0.00, size = 22, normalized size = 0.81

method	result	size
default	\(-\frac {x^{2}}{6}+\frac {x^{3} \arctan \left (x \right )}{3}+\frac {\ln \left (x^{2}+1\right )}{6}\)	\(22\)
meijerg	\(-\frac {x^{2}}{6}+\frac {x^{4} \arctan \left (\sqrt {x^{2}}\right )}{3 \sqrt {x^{2}}}+\frac {\ln \left (x^{2}+1\right )}{6}\)	\(31\)
risch	\(-\frac {i x^{3} \ln \left (i x +1\right )}{6}+\frac {i x^{3} \ln \left (-i x +1\right )}{6}-\frac {x^{2}}{6}+\frac {\ln \left (x^{2}+1\right )}{6}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 2.04, size = 21, normalized size = 0.78 \begin {gather*} \frac {1}{3} \, x^{3} \arctan \left (x\right ) - \frac {1}{6} \, x^{2} + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \end {gather*}

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 = 1.09, size = 21, normalized size = 0.78 \begin {gather*} \frac {1}{3} \, x^{3} \arctan \left (x\right ) - \frac {1}{6} \, x^{2} + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \end {gather*}

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 = 0.09, size = 20, normalized size = 0.74 \begin {gather*} \frac {x^{3} \operatorname {atan}{\left (x \right )}}{3} - \frac {x^{2}}{6} + \frac {\log {\left (x^{2} + 1 \right )}}{6} \end {gather*}

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 = 1.14, size = 21, normalized size = 0.78 \begin {gather*} \frac {1}{3} \, x^{3} \arctan \left (x\right ) - \frac {1}{6} \, x^{2} + \frac {1}{6} \, \log \left (x^{2} + 1\right ) \end {gather*}

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)

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Mupad [B]
time = 0.00, size = 21, normalized size = 0.78 \begin {gather*} \frac {\ln \left (x^2+1\right )}{6}+\frac {x^3\,\mathrm {atan}\left (x\right )}{3}-\frac {x^2}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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