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

Optimal. Leaf size=77 \[ \frac {1}{4} \log ^2\left (x^2+1\right )-\frac {3}{2} \log \left (x^2+1\right )-\frac {1}{2} x^2 \tan ^{-1}(x)^2+\frac {1}{2} \left (x^2+1\right ) \log \left (x^2+1\right ) \tan ^{-1}(x)^2-x \log \left (x^2+1\right ) \tan ^{-1}(x)-\frac {3}{2} \tan ^{-1}(x)^2+3 x \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {4852, 4916, 4846, 260, 4884, 5023, 5009, 2475, 2390, 2301} \[ \frac {1}{4} \log ^2\left (x^2+1\right )-\frac {3}{2} \log \left (x^2+1\right )-\frac {1}{2} x^2 \tan ^{-1}(x)^2+\frac {1}{2} \left (x^2+1\right ) \log \left (x^2+1\right ) \tan ^{-1}(x)^2-x \log \left (x^2+1\right ) \tan ^{-1}(x)-\frac {3}{2} \tan ^{-1}(x)^2+3 x \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x*ArcTan[x]^2*Log[1 + x^2],x]

[Out]

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

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 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 5009

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.)), x_Symbol] :> Simp[x*(d + e*L
og[f + g*x^2])*(a + b*ArcTan[c*x]), x] + (-Dist[b*c, Int[(x*(d + e*Log[f + g*x^2]))/(1 + c^2*x^2), x], x] - Di
st[2*e*g, Int[(x^2*(a + b*ArcTan[c*x]))/(f + g*x^2), x], x]) /; FreeQ[{a, b, c, d, e, f, g}, x]

Rule 5023

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^2*((d_.) + Log[(f_) + (g_.)*(x_)^2]*(e_.))*(x_), x_Symbol] :> Simp[((f
+ g*x^2)*(d + e*Log[f + g*x^2])*(a + b*ArcTan[c*x])^2)/(2*g), x] + (-Dist[b/c, Int[(d + e*Log[f + g*x^2])*(a +
 b*ArcTan[c*x]), x], x] + Dist[b*c*e, Int[(x^2*(a + b*ArcTan[c*x]))/(1 + c^2*x^2), x], x] - Simp[(e*x^2*(a + b
*ArcTan[c*x])^2)/2, x]) /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[g, c^2*f]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 58, normalized size = 0.75 \[ \frac {1}{4} \left (\left (\log \left (x^2+1\right )-6\right ) \log \left (x^2+1\right )+2 \left (-x^2+\left (x^2+1\right ) \log \left (x^2+1\right )-3\right ) \tan ^{-1}(x)^2-4 x \left (\log \left (x^2+1\right )-3\right ) \tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 52, normalized size = 0.68 \[ -\frac {1}{2} \, {\left (x^{2} + 3\right )} \arctan \relax (x)^{2} + 3 \, x \arctan \relax (x) + \frac {1}{2} \, {\left ({\left (x^{2} + 1\right )} \arctan \relax (x)^{2} - 2 \, x \arctan \relax (x) - 3\right )} \log \left (x^{2} + 1\right ) + \frac {1}{4} \, \log \left (x^{2} + 1\right )^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \arctan \relax (x)^{2} \log \left (x^{2} + 1\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arctan(x)^2*log(x^2 + 1), x)

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maple [C]  time = 1.16, size = 3134, normalized size = 40.70 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x^2*arctan(x)^2+3*x*arctan(x)-1/2*arctan(x)^2-1/4*I*arctan(x)^2*Pi*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x
^2+1)+1)^2)*csgn(I/((I*x+1)^2/(x^2+1)+1)^2)*csgn(I*(I*x+1)^2/(x^2+1))*x^2+1/2*I*arctan(x)*Pi*csgn(I*(I*x+1)^2/
(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)*csgn(I/((I*x+1)^2/(x^2+1)+1)^2)*csgn(I*(I*x+1)^2/(x^2+1))*x-arctan(x)^2*ln((I
*x+1)^2/(x^2+1)+1)*x^2+1/2*arctan(x)*Pi*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)^3+I*Pi*ln((I*x+1)^2/
(x^2+1)+1)*csgn(I*((I*x+1)^2/(x^2+1)+1))*csgn(I*((I*x+1)^2/(x^2+1)+1)^2)^2+1/2*I*arctan(x)^2*Pi*csgn(I*(I*x+1)
/(x^2+1)^(1/2))*csgn(I*(I*x+1)^2/(x^2+1))^2+1/4*I*arctan(x)^2*Pi*csgn(I*((I*x+1)^2/(x^2+1)+1)^2)^3*x^2-1/2*I*a
rctan(x)*Pi*csgn(I*((I*x+1)^2/(x^2+1)+1)^2)^3*x-1/4*I*arctan(x)^2*Pi*csgn(I*(I*x+1)/(x^2+1)^(1/2))^2*csgn(I*(I
*x+1)^2/(x^2+1))+1/4*I*arctan(x)^2*Pi*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)^2*csgn(I/((I*x+1)^2/(x
^2+1)+1)^2)+1/4*I*arctan(x)^2*Pi*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)^2*csgn(I*(I*x+1)^2/(x^2+1))
+1/4*I*arctan(x)^2*Pi*csgn(I*((I*x+1)^2/(x^2+1)+1))^2*csgn(I*((I*x+1)^2/(x^2+1)+1)^2)-1/4*I*arctan(x)^2*Pi*csg
n(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)^3*x^2-1/4*I*arctan(x)^2*Pi*csgn(I*(I*x+1)^2/(x^2+1))^3*x^2+3*ln
((I*x+1)^2/(x^2+1)+1)-1/2*I*Pi*ln((I*x+1)^2/(x^2+1)+1)*csgn(I*((I*x+1)^2/(x^2+1)+1))^2*csgn(I*((I*x+1)^2/(x^2+
1)+1)^2)-1/2*I*Pi*ln((I*x+1)^2/(x^2+1)+1)*csgn(I*(I*x+1)^2/(x^2+1))*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1
)+1)^2)^2-1/2*I*Pi*ln((I*x+1)^2/(x^2+1)+1)*csgn(I/((I*x+1)^2/(x^2+1)+1)^2)*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2
/(x^2+1)+1)^2)^2+1/2*I*Pi*ln((I*x+1)^2/(x^2+1)+1)*csgn(I*(I*x+1)/(x^2+1)^(1/2))^2*csgn(I*(I*x+1)^2/(x^2+1))-1/
2*I*arctan(x)^2*Pi*csgn(I*((I*x+1)^2/(x^2+1)+1))*csgn(I*((I*x+1)^2/(x^2+1)+1)^2)^2+1/2*I*arctan(x)*Pi*csgn(I*(
I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)^3*x+1/2*I*arctan(x)*Pi*csgn(I*(I*x+1)^2/(x^2+1))^3*x-I*Pi*csgn(I*(I*
x+1)/(x^2+1)^(1/2))*csgn(I*(I*x+1)^2/(x^2+1))^2*ln((I*x+1)^2/(x^2+1)+1)+1/2*arctan(x)*Pi*csgn(I*(I*x+1)^2/(x^2
+1)/((I*x+1)^2/(x^2+1)+1)^2)*csgn(I/((I*x+1)^2/(x^2+1)+1)^2)*csgn(I*(I*x+1)^2/(x^2+1))+ln((I*x+1)^2/(x^2+1)+1)
^2-3*I*arctan(x)+arctan(x)^2*ln(2)-arctan(x)^2*ln((I*x+1)^2/(x^2+1)+1)+(x^2*arctan(x)^2+2*I*arctan(x)+arctan(x
)^2-2*x*arctan(x)-2*ln((I*x+1)^2/(x^2+1)+1))*ln((I*x+1)/(x^2+1)^(1/2))-1/2*arctan(x)*Pi*csgn(I*((I*x+1)^2/(x^2
+1)+1)^2)^3-2*arctan(x)*ln(2)*x+2*arctan(x)*ln((I*x+1)^2/(x^2+1)+1)*x+arctan(x)^2*ln(2)*x^2+2*I*arctan(x)*ln(2
)-2*ln(2)*ln((I*x+1)^2/(x^2+1)+1)+1/2*arctan(x)*Pi*csgn(I*(I*x+1)^2/(x^2+1))^3-1/4*I*arctan(x)^2*Pi*csgn(I*(I*
x+1)^2/(x^2+1))^3+1/2*I*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)^3*ln((I*x+1)^2/(x^2+1)+1)*Pi+1/2*I*c
sgn(I*(I*x+1)^2/(x^2+1))^3*ln((I*x+1)^2/(x^2+1)+1)*Pi-1/2*I*csgn(I*((I*x+1)^2/(x^2+1)+1)^2)^3*ln((I*x+1)^2/(x^
2+1)+1)*Pi-1/2*arctan(x)*Pi*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)^2*csgn(I/((I*x+1)^2/(x^2+1)+1)^2
)-1/2*arctan(x)*Pi*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)^2*csgn(I*(I*x+1)^2/(x^2+1))-1/2*arctan(x)
*Pi*csgn(I*((I*x+1)^2/(x^2+1)+1))^2*csgn(I*((I*x+1)^2/(x^2+1)+1)^2)+arctan(x)*Pi*csgn(I*((I*x+1)^2/(x^2+1)+1))
*csgn(I*((I*x+1)^2/(x^2+1)+1)^2)^2+1/2*arctan(x)*Pi*csgn(I*(I*x+1)/(x^2+1)^(1/2))^2*csgn(I*(I*x+1)^2/(x^2+1))-
arctan(x)*Pi*csgn(I*(I*x+1)/(x^2+1)^(1/2))*csgn(I*(I*x+1)^2/(x^2+1))^2+1/4*I*arctan(x)^2*Pi*csgn(I*((I*x+1)^2/
(x^2+1)+1)^2)^3-1/4*I*arctan(x)^2*Pi*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)^3+1/4*I*arctan(x)^2*Pi*
csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)^2*csgn(I/((I*x+1)^2/(x^2+1)+1)^2)*x^2-1/4*I*arctan(x)^2*Pi*c
sgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)*csgn(I/((I*x+1)^2/(x^2+1)+1)^2)*csgn(I*(I*x+1)^2/(x^2+1))+1/4
*I*arctan(x)^2*Pi*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)^2*csgn(I*(I*x+1)^2/(x^2+1))*x^2+1/4*I*arct
an(x)^2*Pi*csgn(I*((I*x+1)^2/(x^2+1)+1))^2*csgn(I*((I*x+1)^2/(x^2+1)+1)^2)*x^2-1/2*I*arctan(x)^2*Pi*csgn(I*((I
*x+1)^2/(x^2+1)+1))*csgn(I*((I*x+1)^2/(x^2+1)+1)^2)^2*x^2-1/2*I*arctan(x)*Pi*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)
^2/(x^2+1)+1)^2)^2*csgn(I/((I*x+1)^2/(x^2+1)+1)^2)*x-1/2*I*arctan(x)*Pi*csgn(I*(I*x+1)^2/(x^2+1)/((I*x+1)^2/(x
^2+1)+1)^2)^2*csgn(I*(I*x+1)^2/(x^2+1))*x-1/2*I*arctan(x)*Pi*csgn(I*((I*x+1)^2/(x^2+1)+1))^2*csgn(I*((I*x+1)^2
/(x^2+1)+1)^2)*x-I*arctan(x)*Pi*csgn(I*(I*x+1)/(x^2+1)^(1/2))*csgn(I*(I*x+1)^2/(x^2+1))^2*x-1/4*I*arctan(x)^2*
Pi*csgn(I*(I*x+1)/(x^2+1)^(1/2))^2*csgn(I*(I*x+1)^2/(x^2+1))*x^2+1/2*I*arctan(x)^2*Pi*csgn(I*(I*x+1)/(x^2+1)^(
1/2))*csgn(I*(I*x+1)^2/(x^2+1))^2*x^2+1/2*I*arctan(x)*Pi*csgn(I*(I*x+1)/(x^2+1)^(1/2))^2*csgn(I*(I*x+1)^2/(x^2
+1))*x+1/2*I*Pi*ln((I*x+1)^2/(x^2+1)+1)*csgn(I/((I*x+1)^2/(x^2+1)+1)^2)*csgn(I*(I*x+1)^2/(x^2+1))*csgn(I*(I*x+
1)^2/(x^2+1)/((I*x+1)^2/(x^2+1)+1)^2)+I*arctan(x)*Pi*csgn(I*((I*x+1)^2/(x^2+1)+1))*csgn(I*((I*x+1)^2/(x^2+1)+1
)^2)^2*x

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maxima [A]  time = 0.98, size = 67, normalized size = 0.87 \[ -\frac {1}{2} \, {\left (x^{2} - {\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) + 1\right )} \arctan \relax (x)^{2} - {\left (x \log \left (x^{2} + 1\right ) - 3 \, x + 2 \, \arctan \relax (x)\right )} \arctan \relax (x) + \arctan \relax (x)^{2} + \frac {1}{4} \, \log \left (x^{2} + 1\right )^{2} - \frac {3}{2} \, \log \left (x^{2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.26, size = 78, normalized size = 1.01 \[ \frac {{\ln \left (x^2+1\right )}^2}{4}-\frac {3\,\ln \left (x^2+1\right )}{2}-\frac {3\,{\mathrm {atan}\relax (x)}^2}{2}+\frac {\ln \left (x^2+1\right )\,{\mathrm {atan}\relax (x)}^2}{2}+x\,\left (3\,\mathrm {atan}\relax (x)-\ln \left (x^2+1\right )\,\mathrm {atan}\relax (x)\right )-x^2\,\left (\frac {{\mathrm {atan}\relax (x)}^2}{2}-\frac {\ln \left (x^2+1\right )\,{\mathrm {atan}\relax (x)}^2}{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*log(x^2 + 1)*atan(x)^2,x)

[Out]

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

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sympy [A]  time = 2.49, size = 87, normalized size = 1.13 \[ \frac {x^{2} \log {\left (x^{2} + 1 \right )} \operatorname {atan}^{2}{\relax (x )}}{2} - \frac {x^{2} \operatorname {atan}^{2}{\relax (x )}}{2} - x \log {\left (x^{2} + 1 \right )} \operatorname {atan}{\relax (x )} + 3 x \operatorname {atan}{\relax (x )} + \frac {\log {\left (x^{2} + 1 \right )}^{2}}{4} + \frac {\log {\left (x^{2} + 1 \right )} \operatorname {atan}^{2}{\relax (x )}}{2} - \frac {3 \log {\left (x^{2} + 1 \right )}}{2} - \frac {3 \operatorname {atan}^{2}{\relax (x )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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