3.98 \(\int \frac{x^2 \log (1+x+x^2)}{2+3 x+x^2} \, dx\)

Optimal. Leaf size=311 \[ -\text{PolyLog}\left (2,\frac{2 (x+1)}{1-i \sqrt{3}}\right )-\text{PolyLog}\left (2,\frac{2 (x+1)}{1+i \sqrt{3}}\right )+4 \text{PolyLog}\left (2,\frac{2 (x+2)}{3-i \sqrt{3}}\right )+4 \text{PolyLog}\left (2,\frac{2 (x+2)}{3+i \sqrt{3}}\right )+x \log \left (x^2+x+1\right )+\log (2 x+2) \log \left (x^2+x+1\right )-4 \log (2 x+4) \log \left (x^2+x+1\right )+\frac{1}{2} \log \left (x^2+x+1\right )-2 x-\log (2 x+2) \log \left (-\frac{2 x-i \sqrt{3}+1}{1+i \sqrt{3}}\right )+4 \log (2 x+4) \log \left (-\frac{2 x-i \sqrt{3}+1}{3+i \sqrt{3}}\right )-\log (2 x+2) \log \left (-\frac{2 x+i \sqrt{3}+1}{1-i \sqrt{3}}\right )+4 \log (2 x+4) \log \left (-\frac{2 x+i \sqrt{3}+1}{3-i \sqrt{3}}\right )+\sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.477789, antiderivative size = 311, normalized size of antiderivative = 1., number of steps used = 28, number of rules used = 12, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {2528, 2523, 773, 634, 618, 204, 628, 2524, 2418, 2394, 2393, 2391} \[ -\text{PolyLog}\left (2,\frac{2 (x+1)}{1-i \sqrt{3}}\right )-\text{PolyLog}\left (2,\frac{2 (x+1)}{1+i \sqrt{3}}\right )+4 \text{PolyLog}\left (2,\frac{2 (x+2)}{3-i \sqrt{3}}\right )+4 \text{PolyLog}\left (2,\frac{2 (x+2)}{3+i \sqrt{3}}\right )+x \log \left (x^2+x+1\right )+\log (2 x+2) \log \left (x^2+x+1\right )-4 \log (2 x+4) \log \left (x^2+x+1\right )+\frac{1}{2} \log \left (x^2+x+1\right )-2 x-\log (2 x+2) \log \left (-\frac{2 x-i \sqrt{3}+1}{1+i \sqrt{3}}\right )+4 \log (2 x+4) \log \left (-\frac{2 x-i \sqrt{3}+1}{3+i \sqrt{3}}\right )-\log (2 x+2) \log \left (-\frac{2 x+i \sqrt{3}+1}{1-i \sqrt{3}}\right )+4 \log (2 x+4) \log \left (-\frac{2 x+i \sqrt{3}+1}{3-i \sqrt{3}}\right )+\sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Log[1 + x + x^2])/(2 + 3*x + x^2),x]

[Out]

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

Rule 2528

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*(RGx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*
RFx^p])^n, RGx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, p}, x] && RationalFunctionQ[RFx, x] && RationalF
unctionQ[RGx, x] && IGtQ[n, 0]

Rule 2523

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*Log[c*RFx^p])^n, x] - Dist[b*n*p
, Int[SimplifyIntegrand[(x*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, p}, x] &
& RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 773

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/
c, x] + Dist[1/c, Int[(c*d*f - a*e*g + (c*e*f + c*d*g - b*e*g)*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c,
d, e, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 2524

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[d + e*x]*(a + b
*Log[c*RFx^p])^n)/e, x] - Dist[(b*n*p)/e, Int[(Log[d + e*x]*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x
] /; FreeQ[{a, b, c, d, e, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rule 2394

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

Rule 2393

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{x^2 \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx &=\int \left (\log \left (1+x+x^2\right )-\frac{(2+3 x) \log \left (1+x+x^2\right )}{2+3 x+x^2}\right ) \, dx\\ &=\int \log \left (1+x+x^2\right ) \, dx-\int \frac{(2+3 x) \log \left (1+x+x^2\right )}{2+3 x+x^2} \, dx\\ &=x \log \left (1+x+x^2\right )-\int \frac{x (1+2 x)}{1+x+x^2} \, dx-\int \left (-\frac{2 \log \left (1+x+x^2\right )}{2+2 x}+\frac{8 \log \left (1+x+x^2\right )}{4+2 x}\right ) \, dx\\ &=-2 x+x \log \left (1+x+x^2\right )+2 \int \frac{\log \left (1+x+x^2\right )}{2+2 x} \, dx-8 \int \frac{\log \left (1+x+x^2\right )}{4+2 x} \, dx-\int \frac{-2-x}{1+x+x^2} \, dx\\ &=-2 x+x \log \left (1+x+x^2\right )+\log (2+2 x) \log \left (1+x+x^2\right )-4 \log (4+2 x) \log \left (1+x+x^2\right )+\frac{1}{2} \int \frac{1+2 x}{1+x+x^2} \, dx+\frac{3}{2} \int \frac{1}{1+x+x^2} \, dx+4 \int \frac{(1+2 x) \log (4+2 x)}{1+x+x^2} \, dx-\int \frac{(1+2 x) \log (2+2 x)}{1+x+x^2} \, dx\\ &=-2 x+\frac{1}{2} \log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+\log (2+2 x) \log \left (1+x+x^2\right )-4 \log (4+2 x) \log \left (1+x+x^2\right )-3 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x\right )+4 \int \left (\frac{2 \log (4+2 x)}{1-i \sqrt{3}+2 x}+\frac{2 \log (4+2 x)}{1+i \sqrt{3}+2 x}\right ) \, dx-\int \left (\frac{2 \log (2+2 x)}{1-i \sqrt{3}+2 x}+\frac{2 \log (2+2 x)}{1+i \sqrt{3}+2 x}\right ) \, dx\\ &=-2 x+\sqrt{3} \tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )+\frac{1}{2} \log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+\log (2+2 x) \log \left (1+x+x^2\right )-4 \log (4+2 x) \log \left (1+x+x^2\right )-2 \int \frac{\log (2+2 x)}{1-i \sqrt{3}+2 x} \, dx-2 \int \frac{\log (2+2 x)}{1+i \sqrt{3}+2 x} \, dx+8 \int \frac{\log (4+2 x)}{1-i \sqrt{3}+2 x} \, dx+8 \int \frac{\log (4+2 x)}{1+i \sqrt{3}+2 x} \, dx\\ &=-2 x+\sqrt{3} \tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )-\log (2+2 x) \log \left (-\frac{1-i \sqrt{3}+2 x}{1+i \sqrt{3}}\right )+4 \log (4+2 x) \log \left (-\frac{1-i \sqrt{3}+2 x}{3+i \sqrt{3}}\right )-\log (2+2 x) \log \left (-\frac{1+i \sqrt{3}+2 x}{1-i \sqrt{3}}\right )+4 \log (4+2 x) \log \left (-\frac{1+i \sqrt{3}+2 x}{3-i \sqrt{3}}\right )+\frac{1}{2} \log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+\log (2+2 x) \log \left (1+x+x^2\right )-4 \log (4+2 x) \log \left (1+x+x^2\right )+2 \int \frac{\log \left (\frac{2 \left (1-i \sqrt{3}+2 x\right )}{-4+2 \left (1-i \sqrt{3}\right )}\right )}{2+2 x} \, dx+2 \int \frac{\log \left (\frac{2 \left (1+i \sqrt{3}+2 x\right )}{-4+2 \left (1+i \sqrt{3}\right )}\right )}{2+2 x} \, dx-8 \int \frac{\log \left (\frac{2 \left (1-i \sqrt{3}+2 x\right )}{-8+2 \left (1-i \sqrt{3}\right )}\right )}{4+2 x} \, dx-8 \int \frac{\log \left (\frac{2 \left (1+i \sqrt{3}+2 x\right )}{-8+2 \left (1+i \sqrt{3}\right )}\right )}{4+2 x} \, dx\\ &=-2 x+\sqrt{3} \tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )-\log (2+2 x) \log \left (-\frac{1-i \sqrt{3}+2 x}{1+i \sqrt{3}}\right )+4 \log (4+2 x) \log \left (-\frac{1-i \sqrt{3}+2 x}{3+i \sqrt{3}}\right )-\log (2+2 x) \log \left (-\frac{1+i \sqrt{3}+2 x}{1-i \sqrt{3}}\right )+4 \log (4+2 x) \log \left (-\frac{1+i \sqrt{3}+2 x}{3-i \sqrt{3}}\right )+\frac{1}{2} \log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+\log (2+2 x) \log \left (1+x+x^2\right )-4 \log (4+2 x) \log \left (1+x+x^2\right )-4 \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 x}{-8+2 \left (1-i \sqrt{3}\right )}\right )}{x} \, dx,x,4+2 x\right )-4 \operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 x}{-8+2 \left (1+i \sqrt{3}\right )}\right )}{x} \, dx,x,4+2 x\right )+\operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 x}{-4+2 \left (1-i \sqrt{3}\right )}\right )}{x} \, dx,x,2+2 x\right )+\operatorname{Subst}\left (\int \frac{\log \left (1+\frac{2 x}{-4+2 \left (1+i \sqrt{3}\right )}\right )}{x} \, dx,x,2+2 x\right )\\ &=-2 x+\sqrt{3} \tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )-\log (2+2 x) \log \left (-\frac{1-i \sqrt{3}+2 x}{1+i \sqrt{3}}\right )+4 \log (4+2 x) \log \left (-\frac{1-i \sqrt{3}+2 x}{3+i \sqrt{3}}\right )-\log (2+2 x) \log \left (-\frac{1+i \sqrt{3}+2 x}{1-i \sqrt{3}}\right )+4 \log (4+2 x) \log \left (-\frac{1+i \sqrt{3}+2 x}{3-i \sqrt{3}}\right )+\frac{1}{2} \log \left (1+x+x^2\right )+x \log \left (1+x+x^2\right )+\log (2+2 x) \log \left (1+x+x^2\right )-4 \log (4+2 x) \log \left (1+x+x^2\right )-\text{Li}_2\left (\frac{2 (1+x)}{1+i \sqrt{3}}\right )-\text{Li}_2\left (\frac{2 i (1+x)}{i+\sqrt{3}}\right )+4 \text{Li}_2\left (\frac{2 (2+x)}{3-i \sqrt{3}}\right )+4 \text{Li}_2\left (\frac{2 (2+x)}{3+i \sqrt{3}}\right )\\ \end{align*}

Mathematica [A]  time = 0.159707, size = 290, normalized size = 0.93 \[ -\text{PolyLog}\left (2,\frac{2 (x+1)}{1+i \sqrt{3}}\right )-\text{PolyLog}\left (2,\frac{2 i (x+1)}{\sqrt{3}+i}\right )+4 \left (\text{PolyLog}\left (2,\frac{2 (x+2)}{3+i \sqrt{3}}\right )+\text{PolyLog}\left (2,\frac{2 i (x+2)}{\sqrt{3}+3 i}\right )+\left (\log \left (\frac{-2 i x+\sqrt{3}-i}{\sqrt{3}+3 i}\right )+\log \left (\frac{2 i x+\sqrt{3}+i}{\sqrt{3}-3 i}\right )\right ) \log (2 (x+2))\right )+x \log \left (x^2+x+1\right )+\log (2 (x+1)) \log \left (x^2+x+1\right )-4 \log (2 (x+2)) \log \left (x^2+x+1\right )+\frac{1}{2} \log \left (x^2+x+1\right )-2 x-\log \left (\frac{-2 i x+\sqrt{3}-i}{\sqrt{3}+i}\right ) \log (2 (x+1))-\log \left (\frac{2 i x+\sqrt{3}+i}{\sqrt{3}-i}\right ) \log (2 (x+1))+\sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*Log[1 + x + x^2])/(2 + 3*x + x^2),x]

[Out]

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

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Maple [A]  time = 0.021, size = 279, normalized size = 0.9 \begin{align*} x\ln \left ({x}^{2}+x+1 \right ) -2\,x+{\frac{\ln \left ({x}^{2}+x+1 \right ) }{2}}+\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) \sqrt{3}+\ln \left ( 1+x \right ) \ln \left ({x}^{2}+x+1 \right ) -\ln \left ( 1+x \right ) \ln \left ({\frac{-1-2\,x+i\sqrt{3}}{1+i\sqrt{3}}} \right ) -\ln \left ( 1+x \right ) \ln \left ({\frac{1+2\,x+i\sqrt{3}}{i\sqrt{3}-1}} \right ) -{\it dilog} \left ({\frac{-1-2\,x+i\sqrt{3}}{1+i\sqrt{3}}} \right ) -{\it dilog} \left ({\frac{1+2\,x+i\sqrt{3}}{i\sqrt{3}-1}} \right ) -4\,\ln \left ( 2+x \right ) \ln \left ({x}^{2}+x+1 \right ) +4\,\ln \left ( 2+x \right ) \ln \left ({\frac{-1-2\,x+i\sqrt{3}}{3+i\sqrt{3}}} \right ) +4\,\ln \left ( 2+x \right ) \ln \left ({\frac{1+2\,x+i\sqrt{3}}{i\sqrt{3}-3}} \right ) +4\,{\it dilog} \left ({\frac{-1-2\,x+i\sqrt{3}}{3+i\sqrt{3}}} \right ) +4\,{\it dilog} \left ({\frac{1+2\,x+i\sqrt{3}}{i\sqrt{3}-3}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*ln(x^2+x+1)/(x^2+3*x+2),x)

[Out]

x*ln(x^2+x+1)-2*x+1/2*ln(x^2+x+1)+arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)+ln(1+x)*ln(x^2+x+1)-ln(1+x)*ln((-1-2*x+I
*3^(1/2))/(1+I*3^(1/2)))-ln(1+x)*ln((1+2*x+I*3^(1/2))/(I*3^(1/2)-1))-dilog((-1-2*x+I*3^(1/2))/(1+I*3^(1/2)))-d
ilog((1+2*x+I*3^(1/2))/(I*3^(1/2)-1))-4*ln(2+x)*ln(x^2+x+1)+4*ln(2+x)*ln((-1-2*x+I*3^(1/2))/(3+I*3^(1/2)))+4*l
n(2+x)*ln((1+2*x+I*3^(1/2))/(I*3^(1/2)-3))+4*dilog((-1-2*x+I*3^(1/2))/(3+I*3^(1/2)))+4*dilog((1+2*x+I*3^(1/2))
/(I*3^(1/2)-3))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \log \left (x^{2} + x + 1\right )}{x^{2} + 3 \, x + 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*log(x^2+x+1)/(x^2+3*x+2),x, algorithm="maxima")

[Out]

integrate(x^2*log(x^2 + x + 1)/(x^2 + 3*x + 2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{2} \log \left (x^{2} + x + 1\right )}{x^{2} + 3 \, x + 2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*log(x^2+x+1)/(x^2+3*x+2),x, algorithm="fricas")

[Out]

integral(x^2*log(x^2 + x + 1)/(x^2 + 3*x + 2), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*ln(x**2+x+1)/(x**2+3*x+2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{2} \log \left (x^{2} + x + 1\right )}{x^{2} + 3 \, x + 2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*log(x^2+x+1)/(x^2+3*x+2),x, algorithm="giac")

[Out]

integrate(x^2*log(x^2 + x + 1)/(x^2 + 3*x + 2), x)