\(\int x^3 \arctan (x)^2 \, dx\) [647]

Optimal result

Integrand size = 8, antiderivative size = 53 \[ \int x^3 \arctan (x)^2 \, dx=\frac {x^2}{12}+\frac {1}{2} x \arctan (x)-\frac {1}{6} x^3 \arctan (x)-\frac {\arctan (x)^2}{4}+\frac {1}{4} x^4 \arctan (x)^2-\frac {1}{3} \log \left (1+x^2\right ) \]

[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)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.875, Rules used = {4946, 5036, 272, 45, 4930, 266, 5004} \[ \int x^3 \arctan (x)^2 \, dx=\frac {1}{4} x^4 \arctan (x)^2-\frac {1}{6} x^3 \arctan (x)+\frac {1}{2} x \arctan (x)-\frac {\arctan (x)^2}{4}+\frac {x^2}{12}-\frac {1}{3} \log \left (x^2+1\right ) \]

[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 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 266

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 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 4930

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

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]

Rule 5004

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 5036

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.70 \[ \int x^3 \arctan (x)^2 \, dx=\frac {1}{12} \left (x^2-2 x \left (-3+x^2\right ) \arctan (x)+3 \left (-1+x^4\right ) \arctan (x)^2-4 \log \left (1+x^2\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.79

method	result	size
default	\(\frac {x^{2}}{12}+\frac {x \arctan \left (x \right )}{2}-\frac {x^{3} \arctan \left (x \right )}{6}-\frac {\arctan \left (x \right )^{2}}{4}+\frac {x^{4} \arctan \left (x \right )^{2}}{4}-\frac {\ln \left (x^{2}+1\right )}{3}\)	\(42\)
parts	\(\frac {x^{2}}{12}+\frac {x \arctan \left (x \right )}{2}-\frac {x^{3} \arctan \left (x \right )}{6}-\frac {\arctan \left (x \right )^{2}}{4}+\frac {x^{4} \arctan \left (x \right )^{2}}{4}-\frac {\ln \left (x^{2}+1\right )}{3}\)	\(42\)
parallelrisch	\(\frac {x^{4} \arctan \left (x \right )^{2}}{4}-\frac {x^{3} \arctan \left (x \right )}{6}+\frac {x^{2}}{12}+\frac {x \arctan \left (x \right )}{2}-\frac {\arctan \left (x \right )^{2}}{4}-\frac {\ln \left (x^{2}+1\right )}{3}-\frac {1}{12}\)	\(43\)
risch	\(-\frac {\left (\frac {x^{4}}{4}-\frac {1}{4}\right ) \ln \left (i x +1\right )^{2}}{4}-\frac {\left (-\frac {x^{4} \ln \left (-i x +1\right )}{2}-\frac {i x^{3}}{3}+i x +\frac {\ln \left (-i x +1\right )}{2}\right ) \ln \left (i x +1\right )}{4}-\frac {x^{4} \ln \left (-i x +1\right )^{2}}{16}+\frac {\ln \left (-i x +1\right )^{2}}{16}-\frac {i x^{3} \ln \left (-i x +1\right )}{12}+\frac {i x \ln \left (-i x +1\right )}{4}+\frac {x^{2}}{12}-\frac {\ln \left (x^{2}+1\right )}{3}\)	\(123\)

[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)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.68 \[ \int x^3 \arctan (x)^2 \, dx=\frac {1}{4} \, {\left (x^{4} - 1\right )} \arctan \left (x\right )^{2} + \frac {1}{12} \, x^{2} - \frac {1}{6} \, {\left (x^{3} - 3 \, x\right )} \arctan \left (x\right ) - \frac {1}{3} \, \log \left (x^{2} + 1\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int x^3 \arctan (x)^2 \, dx=\frac {x^{4} \operatorname {atan}^{2}{\left (x \right )}}{4} - \frac {x^{3} \operatorname {atan}{\left (x \right )}}{6} + \frac {x^{2}}{12} + \frac {x \operatorname {atan}{\left (x \right )}}{2} - \frac {\log {\left (x^{2} + 1 \right )}}{3} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{4} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int x^3 \arctan (x)^2 \, dx=\frac {1}{4} \, x^{4} \arctan \left (x\right )^{2} + \frac {1}{12} \, x^{2} - \frac {1}{6} \, {\left (x^{3} - 3 \, x + 3 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac {1}{4} \, \arctan \left (x\right )^{2} - \frac {1}{3} \, \log \left (x^{2} + 1\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.77 \[ \int x^3 \arctan (x)^2 \, dx=\frac {1}{4} \, x^{4} \arctan \left (x\right )^{2} - \frac {1}{6} \, x^{3} \arctan \left (x\right ) + \frac {1}{12} \, x^{2} + \frac {1}{2} \, x \arctan \left (x\right ) - \frac {1}{4} \, \arctan \left (x\right )^{2} - \frac {1}{3} \, \log \left (x^{2} + 1\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.77 \[ \int x^3 \arctan (x)^2 \, dx=\frac {x^4\,{\mathrm {atan}\left (x\right )}^2}{4}-\frac {x^3\,\mathrm {atan}\left (x\right )}{6}-\frac {{\mathrm {atan}\left (x\right )}^2}{4}-\frac {\ln \left (x^2+1\right )}{3}+\frac {x\,\mathrm {atan}\left (x\right )}{2}+\frac {x^2}{12} \]

[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