3.647 \(\int x^3 \tan ^{-1}(x)^2 \, dx\)

Optimal. Leaf size=53 \[ \frac{x^2}{12}-\frac{1}{3} \log \left (x^2+1\right )+\frac{1}{4} x^4 \tan ^{-1}(x)^2-\frac{1}{6} x^3 \tan ^{-1}(x)+\frac{1}{2} x \tan ^{-1}(x)-\frac{1}{4} \tan ^{-1}(x)^2 \]

[Out]

x^2/12 + (x*ArcTan[x])/2 - (x^3*ArcTan[x])/6 - ArcTan[x]^2/4 + (x^4*ArcTan[x]^2)/4 - Log[1 + x^2]/3

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Rubi [A]  time = 0.116431, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.875, Rules used = {4852, 4916, 266, 43, 4846, 260, 4884} \[ \frac{x^2}{12}-\frac{1}{3} \log \left (x^2+1\right )+\frac{1}{4} x^4 \tan ^{-1}(x)^2-\frac{1}{6} x^3 \tan ^{-1}(x)+\frac{1}{2} x \tan ^{-1}(x)-\frac{1}{4} \tan ^{-1}(x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTan[x]^2,x]

[Out]

x^2/12 + (x*ArcTan[x])/2 - (x^3*ArcTan[x])/6 - ArcTan[x]^2/4 + (x^4*ArcTan[x]^2)/4 - Log[1 + x^2]/3

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 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]

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])

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 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 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]

Rubi steps

\begin{align*} \int x^3 \tan ^{-1}(x)^2 \, dx &=\frac{1}{4} x^4 \tan ^{-1}(x)^2-\frac{1}{2} \int \frac{x^4 \tan ^{-1}(x)}{1+x^2} \, dx\\ &=\frac{1}{4} x^4 \tan ^{-1}(x)^2-\frac{1}{2} \int x^2 \tan ^{-1}(x) \, dx+\frac{1}{2} \int \frac{x^2 \tan ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac{1}{6} x^3 \tan ^{-1}(x)+\frac{1}{4} x^4 \tan ^{-1}(x)^2+\frac{1}{6} \int \frac{x^3}{1+x^2} \, dx+\frac{1}{2} \int \tan ^{-1}(x) \, dx-\frac{1}{2} \int \frac{\tan ^{-1}(x)}{1+x^2} \, dx\\ &=\frac{1}{2} x \tan ^{-1}(x)-\frac{1}{6} x^3 \tan ^{-1}(x)-\frac{1}{4} \tan ^{-1}(x)^2+\frac{1}{4} x^4 \tan ^{-1}(x)^2+\frac{1}{12} \operatorname{Subst}\left (\int \frac{x}{1+x} \, dx,x,x^2\right )-\frac{1}{2} \int \frac{x}{1+x^2} \, dx\\ &=\frac{1}{2} x \tan ^{-1}(x)-\frac{1}{6} x^3 \tan ^{-1}(x)-\frac{1}{4} \tan ^{-1}(x)^2+\frac{1}{4} x^4 \tan ^{-1}(x)^2-\frac{1}{4} \log \left (1+x^2\right )+\frac{1}{12} \operatorname{Subst}\left (\int \left (1+\frac{1}{-1-x}\right ) \, dx,x,x^2\right )\\ &=\frac{x^2}{12}+\frac{1}{2} x \tan ^{-1}(x)-\frac{1}{6} x^3 \tan ^{-1}(x)-\frac{1}{4} \tan ^{-1}(x)^2+\frac{1}{4} x^4 \tan ^{-1}(x)^2-\frac{1}{3} \log \left (1+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0149791, size = 37, normalized size = 0.7 \[ \frac{1}{12} \left (x^2-4 \log \left (x^2+1\right )-2 \left (x^2-3\right ) x \tan ^{-1}(x)+3 \left (x^4-1\right ) \tan ^{-1}(x)^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcTan[x]^2,x]

[Out]

(x^2 - 2*x*(-3 + x^2)*ArcTan[x] + 3*(-1 + x^4)*ArcTan[x]^2 - 4*Log[1 + x^2])/12

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Maple [A]  time = 0.01, size = 42, normalized size = 0.8 \begin{align*}{\frac{{x}^{2}}{12}}+{\frac{x\arctan \left ( x \right ) }{2}}-{\frac{{x}^{3}\arctan \left ( x \right ) }{6}}-{\frac{ \left ( \arctan \left ( x \right ) \right ) ^{2}}{4}}+{\frac{{x}^{4} \left ( \arctan \left ( x \right ) \right ) ^{2}}{4}}-{\frac{\ln \left ({x}^{2}+1 \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctan(x)^2,x)

[Out]

1/12*x^2+1/2*x*arctan(x)-1/6*x^3*arctan(x)-1/4*arctan(x)^2+1/4*x^4*arctan(x)^2-1/3*ln(x^2+1)

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Maxima [A]  time = 1.44738, size = 59, normalized size = 1.11 \begin{align*} \frac{1}{4} \, x^{4} \arctan \left (x\right )^{2} + \frac{1}{12} \, x^{2} - \frac{1}{6} \,{\left (x^{3} - 3 \, x + 3 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac{1}{4} \, \arctan \left (x\right )^{2} - \frac{1}{3} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.54543, size = 115, normalized size = 2.17 \begin{align*} \frac{1}{4} \,{\left (x^{4} - 1\right )} \arctan \left (x\right )^{2} + \frac{1}{12} \, x^{2} - \frac{1}{6} \,{\left (x^{3} - 3 \, x\right )} \arctan \left (x\right ) - \frac{1}{3} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.61409, size = 44, normalized size = 0.83 \begin{align*} \frac{x^{4} \operatorname{atan}^{2}{\left (x \right )}}{4} - \frac{x^{3} \operatorname{atan}{\left (x \right )}}{6} + \frac{x^{2}}{12} + \frac{x \operatorname{atan}{\left (x \right )}}{2} - \frac{\log{\left (x^{2} + 1 \right )}}{3} - \frac{\operatorname{atan}^{2}{\left (x \right )}}{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atan(x)**2,x)

[Out]

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

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Giac [A]  time = 1.08217, size = 55, normalized size = 1.04 \begin{align*} \frac{1}{4} \, x^{4} \arctan \left (x\right )^{2} - \frac{1}{6} \, x^{3} \arctan \left (x\right ) + \frac{1}{12} \, x^{2} + \frac{1}{2} \, x \arctan \left (x\right ) - \frac{1}{4} \, \arctan \left (x\right )^{2} - \frac{1}{3} \, \log \left (x^{2} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(x)^2,x, algorithm="giac")

[Out]

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