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

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

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Rubi [A]  time = 0.12, antiderivative size = 53, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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 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*}

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Mathematica [A]  time = 0.02, size = 37, normalized size = 0.70 \[ \frac {1}{12} \left (3 \left (x^4-1\right ) \tan ^{-1}(x)^2+x^2-4 \log \left (x^2+1\right )-2 \left (x^2-3\right ) x \tan ^{-1}(x)\right ) \]

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^3 \tan ^{-1}(x)^2 \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.01, size = 36, normalized size = 0.68 \[ \frac {1}{4} \, {\left (x^{4} - 1\right )} \arctan \relax (x)^{2} + \frac {1}{12} \, x^{2} - \frac {1}{6} \, {\left (x^{3} - 3 \, x\right )} \arctan \relax (x) - \frac {1}{3} \, \log \left (x^{2} + 1\right ) \]

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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giac [A]  time = 1.09, size = 41, normalized size = 0.77 \[ \frac {1}{4} \, x^{4} \arctan \relax (x)^{2} - \frac {1}{6} \, x^{3} \arctan \relax (x) + \frac {1}{12} \, x^{2} + \frac {1}{2} \, x \arctan \relax (x) - \frac {1}{4} \, \arctan \relax (x)^{2} - \frac {1}{3} \, \log \left (x^{2} + 1\right ) \]

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)

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maple [A]  time = 0.10, size = 42, normalized size = 0.79

method	result	size
default	\(\frac {x^{2}}{12}+\frac {x \arctan \relax (x )}{2}-\frac {x^{3} \arctan \relax (x )}{6}-\frac {\arctan \relax (x )^{2}}{4}+\frac {x^{4} \arctan \relax (x )^{2}}{4}-\frac {\ln \left (x^{2}+1\right )}{3}\)	\(42\)
risch	\(-\frac {\left (\frac {x^{4}}{4}-\frac {1}{4}\right ) \ln \left (i x +1\right )^{2}}{4}-\frac {\left (-\frac {x^{4} \ln \left (-i x +1\right )}{2}-\frac {i x^{3}}{3}+i x +\frac {\ln \left (-i x +1\right )}{2}\right ) \ln \left (i x +1\right )}{4}-\frac {x^{4} \ln \left (-i x +1\right )^{2}}{16}+\frac {\ln \left (-i x +1\right )^{2}}{16}-\frac {i x^{3} \ln \left (-i x +1\right )}{12}+\frac {i \ln \left (-i x +1\right ) x}{4}+\frac {x^{2}}{12}-\frac {\ln \left (x^{2}+1\right )}{3}\)	\(123\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 1.00, size = 44, normalized size = 0.83 \[ \frac {1}{4} \, x^{4} \arctan \relax (x)^{2} + \frac {1}{12} \, x^{2} - \frac {1}{6} \, {\left (x^{3} - 3 \, x + 3 \, \arctan \relax (x)\right )} \arctan \relax (x) + \frac {1}{4} \, \arctan \relax (x)^{2} - \frac {1}{3} \, \log \left (x^{2} + 1\right ) \]

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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mupad [B]  time = 0.34, size = 41, normalized size = 0.77 \[ \frac {x^4\,{\mathrm {atan}\relax (x)}^2}{4}-\frac {x^3\,\mathrm {atan}\relax (x)}{6}-\frac {{\mathrm {atan}\relax (x)}^2}{4}-\frac {\ln \left (x^2+1\right )}{3}+\frac {x\,\mathrm {atan}\relax (x)}{2}+\frac {x^2}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.62, size = 44, normalized size = 0.83 \[ \frac {x^{4} \operatorname {atan}^{2}{\relax (x )}}{4} - \frac {x^{3} \operatorname {atan}{\relax (x )}}{6} + \frac {x^{2}}{12} + \frac {x \operatorname {atan}{\relax (x )}}{2} - \frac {\log {\left (x^{2} + 1 \right )}}{3} - \frac {\operatorname {atan}^{2}{\relax (x )}}{4} \]

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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