3.1277 \(\int x^2 \tan ^{-1}(x) \log (1+x^2) \, dx\)

Optimal. Leaf size=82 \[ -\frac {2}{9} x^3 \tan ^{-1}(x)+\frac {5 x^2}{18}+\frac {1}{12} \log ^2\left (x^2+1\right )-\frac {1}{6} x^2 \log \left (x^2+1\right )-\frac {11}{18} \log \left (x^2+1\right )+\frac {1}{3} x^3 \log \left (x^2+1\right ) \tan ^{-1}(x)+\frac {2}{3} x \tan ^{-1}(x)-\frac {1}{3} \tan ^{-1}(x)^2 \]

[Out]

5/18*x^2+2/3*x*arctan(x)-2/9*x^3*arctan(x)-1/3*arctan(x)^2-11/18*ln(x^2+1)-1/6*x^2*ln(x^2+1)+1/3*x^3*arctan(x)
*ln(x^2+1)+1/12*ln(x^2+1)^2

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Rubi [A]  time = 0.33, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 12, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {4852, 266, 43, 5021, 6725, 4916, 4846, 260, 4884, 2475, 2390, 2301} \[ \frac {5 x^2}{18}+\frac {1}{12} \log ^2\left (x^2+1\right )-\frac {1}{6} x^2 \log \left (x^2+1\right )-\frac {11}{18} \log \left (x^2+1\right )-\frac {2}{9} x^3 \tan ^{-1}(x)+\frac {1}{3} x^3 \log \left (x^2+1\right ) \tan ^{-1}(x)+\frac {2}{3} x \tan ^{-1}(x)-\frac {1}{3} \tan ^{-1}(x)^2 \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTan[x]*Log[1 + x^2],x]

[Out]

(5*x^2)/18 + (2*x*ArcTan[x])/3 - (2*x^3*ArcTan[x])/9 - ArcTan[x]^2/3 - (11*Log[1 + x^2])/18 - (x^2*Log[1 + x^2
])/6 + (x^3*ArcTan[x]*Log[1 + x^2])/3 + Log[1 + x^2]^2/12

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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 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 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]

Rule 5021

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> With
[{u = IntHide[x^m*(a + b*ArcTan[c*x]), x]}, Dist[d + e*Log[f + g*x^2], u, x] - Dist[2*e*g, Int[ExpandIntegrand
[(x*u)/(f + g*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IntegerQ[m] && NeQ[m, -1]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int x^2 \tan ^{-1}(x) \log \left (1+x^2\right ) \, dx &=-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{6} \log ^2\left (1+x^2\right )-2 \int \left (\frac {x^3 \left (-1+2 x \tan ^{-1}(x)\right )}{6 \left (1+x^2\right )}+\frac {x \log \left (1+x^2\right )}{6 \left (1+x^2\right )}\right ) \, dx\\ &=-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{6} \log ^2\left (1+x^2\right )-\frac {1}{3} \int \frac {x^3 \left (-1+2 x \tan ^{-1}(x)\right )}{1+x^2} \, dx-\frac {1}{3} \int \frac {x \log \left (1+x^2\right )}{1+x^2} \, dx\\ &=-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{6} \log ^2\left (1+x^2\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log (1+x)}{1+x} \, dx,x,x^2\right )-\frac {1}{3} \int \left (-\frac {x^3}{1+x^2}+\frac {2 x^4 \tan ^{-1}(x)}{1+x^2}\right ) \, dx\\ &=-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{6} \log ^2\left (1+x^2\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+x^2\right )+\frac {1}{3} \int \frac {x^3}{1+x^2} \, dx-\frac {2}{3} \int \frac {x^4 \tan ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {x}{1+x} \, dx,x,x^2\right )-\frac {2}{3} \int x^2 \tan ^{-1}(x) \, dx+\frac {2}{3} \int \frac {x^2 \tan ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac {2}{9} x^3 \tan ^{-1}(x)-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {1}{6} \operatorname {Subst}\left (\int \left (1+\frac {1}{-1-x}\right ) \, dx,x,x^2\right )+\frac {2}{9} \int \frac {x^3}{1+x^2} \, dx+\frac {2}{3} \int \tan ^{-1}(x) \, dx-\frac {2}{3} \int \frac {\tan ^{-1}(x)}{1+x^2} \, dx\\ &=\frac {x^2}{6}+\frac {2}{3} x \tan ^{-1}(x)-\frac {2}{9} x^3 \tan ^{-1}(x)-\frac {1}{3} \tan ^{-1}(x)^2-\frac {1}{6} \log \left (1+x^2\right )-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {1}{9} \operatorname {Subst}\left (\int \frac {x}{1+x} \, dx,x,x^2\right )-\frac {2}{3} \int \frac {x}{1+x^2} \, dx\\ &=\frac {x^2}{6}+\frac {2}{3} x \tan ^{-1}(x)-\frac {2}{9} x^3 \tan ^{-1}(x)-\frac {1}{3} \tan ^{-1}(x)^2-\frac {1}{2} \log \left (1+x^2\right )-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {1}{9} \operatorname {Subst}\left (\int \left (1+\frac {1}{-1-x}\right ) \, dx,x,x^2\right )\\ &=\frac {5 x^2}{18}+\frac {2}{3} x \tan ^{-1}(x)-\frac {2}{9} x^3 \tan ^{-1}(x)-\frac {1}{3} \tan ^{-1}(x)^2-\frac {11}{18} \log \left (1+x^2\right )-\frac {1}{6} x^2 \log \left (1+x^2\right )+\frac {1}{3} x^3 \tan ^{-1}(x) \log \left (1+x^2\right )+\frac {1}{12} \log ^2\left (1+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 64, normalized size = 0.78 \[ \frac {1}{36} \left (10 x^2+3 \log ^2\left (x^2+1\right )-2 \left (3 x^2+11\right ) \log \left (x^2+1\right )+4 x \left (-2 x^2+3 x^2 \log \left (x^2+1\right )+6\right ) \tan ^{-1}(x)-12 \tan ^{-1}(x)^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTan[x]*Log[1 + x^2],x]

[Out]

(10*x^2 - 12*ArcTan[x]^2 - 2*(11 + 3*x^2)*Log[1 + x^2] + 3*Log[1 + x^2]^2 + 4*x*ArcTan[x]*(6 - 2*x^2 + 3*x^2*L
og[1 + x^2]))/36

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fricas [A]  time = 0.40, size = 55, normalized size = 0.67 \[ \frac {5}{18} \, x^{2} - \frac {2}{9} \, {\left (x^{3} - 3 \, x\right )} \arctan \relax (x) - \frac {1}{3} \, \arctan \relax (x)^{2} + \frac {1}{18} \, {\left (6 \, x^{3} \arctan \relax (x) - 3 \, x^{2} - 11\right )} \log \left (x^{2} + 1\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(x)*log(x^2+1),x, algorithm="fricas")

[Out]

5/18*x^2 - 2/9*(x^3 - 3*x)*arctan(x) - 1/3*arctan(x)^2 + 1/18*(6*x^3*arctan(x) - 3*x^2 - 11)*log(x^2 + 1) + 1/
12*log(x^2 + 1)^2

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giac [B]  time = 0.13, size = 135, normalized size = 1.65 \[ \frac {1}{6} \, \pi x^{3} \log \left (x^{2} + 1\right ) \mathrm {sgn}\relax (x) - \frac {1}{3} \, x^{3} \arctan \left (\frac {1}{x}\right ) \log \left (x^{2} + 1\right ) - \frac {1}{9} \, \pi x^{3} \mathrm {sgn}\relax (x) + \frac {2}{9} \, x^{3} \arctan \left (\frac {1}{x}\right ) - \frac {1}{6} \, x^{2} \log \left (x^{2} + 1\right ) + \frac {1}{6} \, \pi ^{2} \mathrm {sgn}\relax (x) + \frac {1}{3} \, \pi x \mathrm {sgn}\relax (x) + \frac {1}{3} \, \pi \arctan \left (\frac {1}{x}\right ) \mathrm {sgn}\relax (x) - \frac {1}{6} \, \pi ^{2} + \frac {5}{18} \, x^{2} - \frac {1}{3} \, \pi \arctan \relax (x) - \frac {1}{3} \, \pi \arctan \left (\frac {1}{x}\right ) - \frac {2}{3} \, x \arctan \left (\frac {1}{x}\right ) - \frac {1}{3} \, \arctan \left (\frac {1}{x}\right )^{2} + \frac {1}{12} \, \log \left (x^{2} + 1\right )^{2} - \frac {11}{18} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(x)*log(x^2+1),x, algorithm="giac")

[Out]

1/6*pi*x^3*log(x^2 + 1)*sgn(x) - 1/3*x^3*arctan(1/x)*log(x^2 + 1) - 1/9*pi*x^3*sgn(x) + 2/9*x^3*arctan(1/x) -
1/6*x^2*log(x^2 + 1) + 1/6*pi^2*sgn(x) + 1/3*pi*x*sgn(x) + 1/3*pi*arctan(1/x)*sgn(x) - 1/6*pi^2 + 5/18*x^2 - 1
/3*pi*arctan(x) - 1/3*pi*arctan(1/x) - 2/3*x*arctan(1/x) - 1/3*arctan(1/x)^2 + 1/12*log(x^2 + 1)^2 - 11/18*log
(x^2 + 1)

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maple [C]  time = 2.69, size = 3039, normalized size = 37.06 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctan(x)*ln(x^2+1),x)

[Out]

-1/6*I*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1
)^2)*arctan(x)*Pi*x^3+2/3*x*arctan(x)-2/9*x^3*arctan(x)-8/9*I*arctan(x)+5/18-1/3*x^2*ln(2)+1/3*ln((1+I*x)^2/(x
^2+1)+1)*x^2-2/3*ln((1+I*x)^2/(x^2+1)+1)*arctan(x)*x^3+2/3*ln(2)*arctan(x)*x^3-1/12*I*Pi*csgn(I*((1+I*x)^2/(x^
2+1)+1)^2)^3+2/3*I*ln(2)*arctan(x)+1/12*I*Pi*csgn(I*(1+I*x)^2/(x^2+1))^3+1/12*I*Pi*csgn(I*(1+I*x)^2/(x^2+1)/((
1+I*x)^2/(x^2+1)+1)^2)^3+1/6*csgn(I*(1+I*x)^2/(x^2+1))^3*arctan(x)*Pi+1/6*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/
(x^2+1)+1)^2)^3*arctan(x)*Pi-1/6*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^3*arctan(x)*Pi-1/3*(-2*I*arctan(x)-2*x^3*arct
an(x)+2*ln((1+I*x)^2/(x^2+1)+1)+1+x^2)*ln((1+I*x)/(x^2+1)^(1/2))-1/3*csgn(I*(1+I*x)^2/(x^2+1))^2*csgn(I*(1+I*x
)/(x^2+1)^(1/2))*arctan(x)*Pi-1/6*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^
2*arctan(x)*Pi+1/6*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)/(x^2+1)^(1/2))^2*arctan(x)*Pi-1/6*csgn(I*(1+I*x)^2
/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*arctan(x)*Pi-1/6*csgn(I*((1+I*x)^2/(x^2+1)
+1))^2*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)*arctan(x)*Pi+1/3*csgn(I*((1+I*x)^2/(x^2+1)+1))*csgn(I*((1+I*x)^2/(x^2+1
)+1)^2)^2*arctan(x)*Pi-1/6*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^3*arctan(x)*Pi*x^3+1/6*I*csgn(I
*((1+I*x)^2/(x^2+1)+1)^2)^3*arctan(x)*Pi*x^3-1/12*I*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*csgn(I*(1+I*x)^2/(x^2+1)/(
(1+I*x)^2/(x^2+1)+1)^2)^2*Pi*x^2-1/6*I*csgn(I*(1+I*x)^2/(x^2+1))^2*csgn(I*(1+I*x)/(x^2+1)^(1/2))*Pi*x^2-2/3*ln
(2)*ln((1+I*x)^2/(x^2+1)+1)+1/6*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)*cs
gn(I/((1+I*x)^2/(x^2+1)+1)^2)*arctan(x)*Pi+1/6*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)*csgn(I*(1+I
*x)^2/(x^2+1))*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*Pi*ln((1+I*x)^2/(x^2+1)+1)+1/6*I*csgn(I/((1+I*x)^2/(x^2+1)+1)^2
)*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2*arctan(x)*Pi*x^3+1/3*I*csgn(I*(1+I*x)^2/(x^2+1))^2*csgn(
I*(1+I*x)/(x^2+1)^(1/2))*arctan(x)*Pi*x^3+1/6*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/
(x^2+1)+1)^2)^2*arctan(x)*Pi*x^3-1/6*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)/(x^2+1)^(1/2))^2*arctan(x)*Pi*
x^3+1/6*I*csgn(I*((1+I*x)^2/(x^2+1)+1))^2*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)*arctan(x)*Pi*x^3-1/3*I*csgn(I*((1+I*
x)^2/(x^2+1)+1))*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^2*arctan(x)*Pi*x^3+1/12*I*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*csg
n(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)*Pi*x^2+5/18*x^2-1/3*ln(2)+11/9*ln((1+
I*x)^2/(x^2+1)+1)+1/3*ln((1+I*x)^2/(x^2+1)+1)^2-1/6*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2*csgn
(I/((1+I*x)^2/(x^2+1)+1)^2)*Pi*ln((1+I*x)^2/(x^2+1)+1)-1/6*I*csgn(I*(1+I*x)^2/(x^2+1))^3*arctan(x)*Pi*x^3+1/12
*I*Pi*csgn(I*(1+I*x)/(x^2+1)^(1/2))^2*csgn(I*(1+I*x)^2/(x^2+1))-1/12*I*Pi*csgn(I*((1+I*x)^2/(x^2+1)+1))^2*csgn
(I*((1+I*x)^2/(x^2+1)+1)^2)+1/6*I*Pi*csgn(I*((1+I*x)^2/(x^2+1)+1))*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^2+1/12*I*cs
gn(I*(1+I*x)^2/(x^2+1))^3*Pi*x^2+1/12*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^3*Pi*x^2-1/12*I*csgn
(I*((1+I*x)^2/(x^2+1)+1)^2)^3*Pi*x^2-1/6*I*Pi*csgn(I*(1+I*x)/(x^2+1)^(1/2))*csgn(I*(1+I*x)^2/(x^2+1))^2-1/12*I
*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*Pi-1/12*I*Pi*csgn(I*(1+I*
x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2+1/6*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x
^2+1)+1)^2)^3*Pi*ln((1+I*x)^2/(x^2+1)+1)+1/6*I*ln((1+I*x)^2/(x^2+1)+1)*Pi*csgn(I*(1+I*x)^2/(x^2+1))^3-1/6*I*ln
((1+I*x)^2/(x^2+1)+1)*Pi*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^3-1/12*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)^2/(
x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2*Pi*x^2+1/12*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)/(x^2+1)^(1/2))^2*Pi*x
^2-1/12*I*csgn(I*((1+I*x)^2/(x^2+1)+1))^2*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)*Pi*x^2+1/6*I*csgn(I*((1+I*x)^2/(x^2+
1)+1))*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^2*Pi*x^2+1/12*I*Pi*csgn(I/((1+I*x)^2/(x^2+1)+1)^2)*csgn(I*(1+I*x)^2/(x^
2+1))*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)+1/6*I*csgn(I*(1+I*x)^2/(x^2+1))*csgn(I*(1+I*x)/(x^2+1)
^(1/2))^2*Pi*ln((1+I*x)^2/(x^2+1)+1)-1/3*I*csgn(I*(1+I*x)^2/(x^2+1))^2*csgn(I*(1+I*x)/(x^2+1)^(1/2))*Pi*ln((1+
I*x)^2/(x^2+1)+1)-1/6*I*csgn(I*(1+I*x)^2/(x^2+1)/((1+I*x)^2/(x^2+1)+1)^2)^2*csgn(I*(1+I*x)^2/(x^2+1))*Pi*ln((1
+I*x)^2/(x^2+1)+1)-1/6*I*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)*csgn(I*((1+I*x)^2/(x^2+1)+1))^2*Pi*ln((1+I*x)^2/(x^2+
1)+1)+1/3*I*csgn(I*((1+I*x)^2/(x^2+1)+1)^2)^2*csgn(I*((1+I*x)^2/(x^2+1)+1))*Pi*ln((1+I*x)^2/(x^2+1)+1)

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maxima [A]  time = 0.42, size = 65, normalized size = 0.79 \[ \frac {5}{18} \, x^{2} + \frac {1}{9} \, {\left (3 \, x^{3} \log \left (x^{2} + 1\right ) - 2 \, x^{3} + 6 \, x - 6 \, \arctan \relax (x)\right )} \arctan \relax (x) + \frac {1}{3} \, \arctan \relax (x)^{2} - \frac {1}{18} \, {\left (3 \, x^{2} + 11\right )} \log \left (x^{2} + 1\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctan(x)*log(x^2+1),x, algorithm="maxima")

[Out]

5/18*x^2 + 1/9*(3*x^3*log(x^2 + 1) - 2*x^3 + 6*x - 6*arctan(x))*arctan(x) + 1/3*arctan(x)^2 - 1/18*(3*x^2 + 11
)*log(x^2 + 1) + 1/12*log(x^2 + 1)^2

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mupad [B]  time = 0.46, size = 65, normalized size = 0.79 \[ \frac {{\ln \left (x^2+1\right )}^2}{12}-\frac {11\,\ln \left (x^2+1\right )}{18}-\frac {{\mathrm {atan}\relax (x)}^2}{3}-x^2\,\left (\frac {\ln \left (x^2+1\right )}{6}-\frac {5}{18}\right )-x^3\,\left (\frac {2\,\mathrm {atan}\relax (x)}{9}-\frac {\ln \left (x^2+1\right )\,\mathrm {atan}\relax (x)}{3}\right )+\frac {2\,x\,\mathrm {atan}\relax (x)}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*log(x^2 + 1)*atan(x),x)

[Out]

log(x^2 + 1)^2/12 - (11*log(x^2 + 1))/18 - atan(x)^2/3 - x^2*(log(x^2 + 1)/6 - 5/18) - x^3*((2*atan(x))/9 - (l
og(x^2 + 1)*atan(x))/3) + (2*x*atan(x))/3

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sympy [A]  time = 1.80, size = 78, normalized size = 0.95 \[ \frac {x^{3} \log {\left (x^{2} + 1 \right )} \operatorname {atan}{\relax (x )}}{3} - \frac {2 x^{3} \operatorname {atan}{\relax (x )}}{9} - \frac {x^{2} \log {\left (x^{2} + 1 \right )}}{6} + \frac {5 x^{2}}{18} + \frac {2 x \operatorname {atan}{\relax (x )}}{3} + \frac {\log {\left (x^{2} + 1 \right )}^{2}}{12} - \frac {11 \log {\left (x^{2} + 1 \right )}}{18} - \frac {\operatorname {atan}^{2}{\relax (x )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atan(x)*ln(x**2+1),x)

[Out]

x**3*log(x**2 + 1)*atan(x)/3 - 2*x**3*atan(x)/9 - x**2*log(x**2 + 1)/6 + 5*x**2/18 + 2*x*atan(x)/3 + log(x**2
+ 1)**2/12 - 11*log(x**2 + 1)/18 - atan(x)**2/3

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