3.13.77 \(\int x^2 \text {ArcTan}(x) \log (1+x^2) \, dx\) [1277]

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

[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 = 0.23, 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 = {4946, 272, 45, 5141, 6857, 5036, 4930, 266, 5004, 2525, 2437, 2338} \begin {gather*} -\frac {2}{9} x^3 \text {ArcTan}(x)+\frac {1}{3} x^3 \text {ArcTan}(x) \log \left (x^2+1\right )+\frac {2}{3} x \text {ArcTan}(x)-\frac {\text {ArcTan}(x)^2}{3}+\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 ) \end {gather*}

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 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 2338

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 2437

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 2525

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 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]

Rule 5141

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 6857

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} \text {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} \text {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} \text {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} \text {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} \text {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} \text {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 \begin {gather*} \frac {1}{36} \left (10 x^2-12 \text {ArcTan}(x)^2-2 \left (11+3 x^2\right ) \log \left (1+x^2\right )+3 \log ^2\left (1+x^2\right )+4 x \text {ArcTan}(x) \left (6-2 x^2+3 x^2 \log \left (1+x^2\right )\right )\right ) \end {gather*}

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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 4.25, size = 1078, normalized size = 13.15

method	result	size
default	\(\text {Expression too large to display}\)	\(1078\)
risch	\(\text {Expression too large to display}\)	\(5252\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 65, normalized size = 0.79 \begin {gather*} \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 \left (x\right )\right )} \arctan \left (x\right ) + \frac {1}{3} \, \arctan \left (x\right )^{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} \end {gather*}

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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Fricas [A]
time = 1.74, size = 55, normalized size = 0.67 \begin {gather*} \frac {5}{18} \, x^{2} - \frac {2}{9} \, {\left (x^{3} - 3 \, x\right )} \arctan \left (x\right ) - \frac {1}{3} \, \arctan \left (x\right )^{2} + \frac {1}{18} \, {\left (6 \, x^{3} \arctan \left (x\right ) - 3 \, x^{2} - 11\right )} \log \left (x^{2} + 1\right ) + \frac {1}{12} \, \log \left (x^{2} + 1\right )^{2} \end {gather*}

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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Sympy [A]
time = 0.40, size = 78, normalized size = 0.95 \begin {gather*} \frac {x^{3} \log {\left (x^{2} + 1 \right )} \operatorname {atan}{\left (x \right )}}{3} - \frac {2 x^{3} \operatorname {atan}{\left (x \right )}}{9} - \frac {x^{2} \log {\left (x^{2} + 1 \right )}}{6} + \frac {5 x^{2}}{18} + \frac {2 x \operatorname {atan}{\left (x \right )}}{3} + \frac {\log {\left (x^{2} + 1 \right )}^{2}}{12} - \frac {11 \log {\left (x^{2} + 1 \right )}}{18} - \frac {\operatorname {atan}^{2}{\left (x \right )}}{3} \end {gather*}

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 135 vs. \(2 (66) = 132\).
time = 0.40, size = 135, normalized size = 1.65 \begin {gather*} \frac {1}{6} \, \pi x^{3} \log \left (x^{2} + 1\right ) \mathrm {sgn}\left (x\right ) - \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}\left (x\right ) + \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}\left (x\right ) + \frac {1}{3} \, \pi x \mathrm {sgn}\left (x\right ) + \frac {1}{3} \, \pi \arctan \left (\frac {1}{x}\right ) \mathrm {sgn}\left (x\right ) - \frac {1}{6} \, \pi ^{2} + \frac {5}{18} \, x^{2} - \frac {1}{3} \, \pi \arctan \left (x\right ) - \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 ) \end {gather*}

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

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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