Integrand size = 116, antiderivative size = 22 \[ \int \frac {1}{9} \left (8 x-6 \log (3)+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)+e^x \left ((-12-48 x) \log (3)+36 \log ^2(3)\right )+e^{2 x} \left ((-6-48 x) \log (3)+108 \log ^2(3)\right )+\left (-30 x+18 \log (3)+e^{2 x} (18+36 x) \log (3)+e^x (36+36 x) \log (3)\right ) \log (x)+18 x \log ^2(x)\right ) \, dx=\left (-\frac {4 x}{3}+\left (1+e^x\right )^2 \log (3)+x \log (x)\right )^2 \] Output:
(ln(3)*(1+exp(x))^2-4/3*x+x*ln(x))^2
Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {1}{9} \left (8 x-6 \log (3)+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)+e^x \left ((-12-48 x) \log (3)+36 \log ^2(3)\right )+e^{2 x} \left ((-6-48 x) \log (3)+108 \log ^2(3)\right )+\left (-30 x+18 \log (3)+e^{2 x} (18+36 x) \log (3)+e^x (36+36 x) \log (3)\right ) \log (x)+18 x \log ^2(x)\right ) \, dx=\frac {1}{9} \left (-4 x+6 e^x \log (3)+\log (27)+e^{2 x} \log (27)+3 x \log (x)\right )^2 \] Input:
Integrate[(8*x - 6*Log[3] + 108*E^(3*x)*Log[3]^2 + 36*E^(4*x)*Log[3]^2 + E ^x*((-12 - 48*x)*Log[3] + 36*Log[3]^2) + E^(2*x)*((-6 - 48*x)*Log[3] + 108 *Log[3]^2) + (-30*x + 18*Log[3] + E^(2*x)*(18 + 36*x)*Log[3] + E^x*(36 + 3 6*x)*Log[3])*Log[x] + 18*x*Log[x]^2)/9,x]
Output:
(-4*x + 6*E^x*Log[3] + Log[27] + E^(2*x)*Log[27] + 3*x*Log[x])^2/9
Leaf count is larger than twice the leaf count of optimal. \(177\) vs. \(2(22)=44\).
Time = 0.43 (sec) , antiderivative size = 177, normalized size of antiderivative = 8.05, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {27, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{9} \left (8 x+18 x \log ^2(x)+e^x \left ((-48 x-12) \log (3)+36 \log ^2(3)\right )+e^{2 x} \left ((-48 x-6) \log (3)+108 \log ^2(3)\right )+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)+\left (-30 x+e^{2 x} (36 x+18) \log (3)+e^x (36 x+36) \log (3)+18 \log (3)\right ) \log (x)-6 \log (3)\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int \left (18 x \log ^2(x)-6 \left (5 x-6 e^x (x+1) \log (3)-3 e^{2 x} (2 x+1) \log (3)-3 \log (3)\right ) \log (x)+8 x-6 e^{2 x} \left ((8 x+1) \log (3)-18 \log ^2(3)\right )-12 e^x \left ((4 x+1) \log (3)-3 \log ^2(3)\right )+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)-6 \log (3)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{9} \left (16 x^2+9 x^2 \log ^2(x)-24 x^2 \log (x)+54 e^{2 x} \log ^2(3)+36 e^{3 x} \log ^2(3)+9 e^{4 x} \log ^2(3)+18 x \log (3) \log (x)-12 e^x x \log (81)-24 x \log (3)-36 e^x \log (3) \log (x)-9 e^{2 x} \log (3) \log (x)+36 e^x (x+1) \log (3) \log (x)+9 e^{2 x} (2 x+1) \log (3) \log (x)+12 e^x \log (81)-36 e^x \log (3)+3 e^{2 x} \log (3)-3 e^{2 x} (8 x+1) \log (3)-12 e^x (1-3 \log (3)) \log (3)\right )\) |
Input:
Int[(8*x - 6*Log[3] + 108*E^(3*x)*Log[3]^2 + 36*E^(4*x)*Log[3]^2 + E^x*((- 12 - 48*x)*Log[3] + 36*Log[3]^2) + E^(2*x)*((-6 - 48*x)*Log[3] + 108*Log[3 ]^2) + (-30*x + 18*Log[3] + E^(2*x)*(18 + 36*x)*Log[3] + E^x*(36 + 36*x)*L og[3])*Log[x] + 18*x*Log[x]^2)/9,x]
Output:
(16*x^2 - 36*E^x*Log[3] + 3*E^(2*x)*Log[3] - 24*x*Log[3] - 3*E^(2*x)*(1 + 8*x)*Log[3] - 12*E^x*(1 - 3*Log[3])*Log[3] + 54*E^(2*x)*Log[3]^2 + 36*E^(3 *x)*Log[3]^2 + 9*E^(4*x)*Log[3]^2 + 12*E^x*Log[81] - 12*E^x*x*Log[81] - 24 *x^2*Log[x] - 36*E^x*Log[3]*Log[x] - 9*E^(2*x)*Log[3]*Log[x] + 18*x*Log[3] *Log[x] + 36*E^x*(1 + x)*Log[3]*Log[x] + 9*E^(2*x)*(1 + 2*x)*Log[3]*Log[x] + 9*x^2*Log[x]^2)/9
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Leaf count of result is larger than twice the leaf count of optimal. \(106\) vs. \(2(19)=38\).
Time = 0.03 (sec) , antiderivative size = 107, normalized size of antiderivative = 4.86
\[x^{2} \ln \left (x \right )^{2}-\frac {8 x^{2} \ln \left (x \right )}{3}+\frac {16 x^{2}}{9}+2 \ln \left (x \right ) \ln \left (3\right ) {\mathrm e}^{2 x} x +4 x \ln \left (3\right ) {\mathrm e}^{x} \ln \left (x \right )+2 \ln \left (x \right ) \ln \left (3\right ) x -\frac {8 x \ln \left (3\right )}{3}+\ln \left (3\right )^{2} {\mathrm e}^{4 x}+4 \ln \left (3\right )^{2} {\mathrm e}^{3 x}+6 \ln \left (3\right )^{2} {\mathrm e}^{2 x}-\frac {8 \ln \left (3\right ) {\mathrm e}^{2 x} x}{3}+4 \ln \left (3\right )^{2} {\mathrm e}^{x}-\frac {16 x \ln \left (3\right ) {\mathrm e}^{x}}{3}\]
Input:
int(2*x*ln(x)^2+1/9*((36*x+18)*ln(3)*exp(x)^2+(36*x+36)*ln(3)*exp(x)+18*ln (3)-30*x)*ln(x)+4*ln(3)^2*exp(x)^4+12*ln(3)^2*exp(x)^3+1/9*(108*ln(3)^2+(- 48*x-6)*ln(3))*exp(x)^2+1/9*(36*ln(3)^2+(-48*x-12)*ln(3))*exp(x)-2/3*ln(3) +8/9*x,x)
Output:
x^2*ln(x)^2-8/3*x^2*ln(x)+16/9*x^2+2*ln(x)*ln(3)*exp(x)^2*x+4*x*ln(3)*exp( x)*ln(x)+2*ln(x)*ln(3)*x-8/3*x*ln(3)+ln(3)^2*exp(x)^4+4*ln(3)^2*exp(x)^3+6 *ln(3)^2*exp(x)^2-8/3*ln(3)*exp(x)^2*x+4*ln(3)^2*exp(x)-16/3*x*ln(3)*exp(x )
Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (23) = 46\).
Time = 0.07 (sec) , antiderivative size = 103, normalized size of antiderivative = 4.68 \[ \int \frac {1}{9} \left (8 x-6 \log (3)+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)+e^x \left ((-12-48 x) \log (3)+36 \log ^2(3)\right )+e^{2 x} \left ((-6-48 x) \log (3)+108 \log ^2(3)\right )+\left (-30 x+18 \log (3)+e^{2 x} (18+36 x) \log (3)+e^x (36+36 x) \log (3)\right ) \log (x)+18 x \log ^2(x)\right ) \, dx=x^{2} \log \left (x\right )^{2} + e^{\left (4 \, x\right )} \log \left (3\right )^{2} + 4 \, e^{\left (3 \, x\right )} \log \left (3\right )^{2} + \frac {16}{9} \, x^{2} - \frac {2}{3} \, {\left (4 \, x \log \left (3\right ) - 9 \, \log \left (3\right )^{2}\right )} e^{\left (2 \, x\right )} - \frac {4}{3} \, {\left (4 \, x \log \left (3\right ) - 3 \, \log \left (3\right )^{2}\right )} e^{x} - \frac {8}{3} \, x \log \left (3\right ) + \frac {2}{3} \, {\left (3 \, x e^{\left (2 \, x\right )} \log \left (3\right ) + 6 \, x e^{x} \log \left (3\right ) - 4 \, x^{2} + 3 \, x \log \left (3\right )\right )} \log \left (x\right ) \] Input:
integrate(2*x*log(x)^2+1/9*((36*x+18)*log(3)*exp(x)^2+(36*x+36)*log(3)*exp (x)+18*log(3)-30*x)*log(x)+4*log(3)^2*exp(x)^4+12*log(3)^2*exp(x)^3+1/9*(1 08*log(3)^2+(-48*x-6)*log(3))*exp(x)^2+1/9*(36*log(3)^2+(-48*x-12)*log(3)) *exp(x)-2/3*log(3)+8/9*x,x, algorithm="fricas")
Output:
x^2*log(x)^2 + e^(4*x)*log(3)^2 + 4*e^(3*x)*log(3)^2 + 16/9*x^2 - 2/3*(4*x *log(3) - 9*log(3)^2)*e^(2*x) - 4/3*(4*x*log(3) - 3*log(3)^2)*e^x - 8/3*x* log(3) + 2/3*(3*x*e^(2*x)*log(3) + 6*x*e^x*log(3) - 4*x^2 + 3*x*log(3))*lo g(x)
Leaf count of result is larger than twice the leaf count of optimal. 121 vs. \(2 (20) = 40\).
Time = 0.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 5.50 \[ \int \frac {1}{9} \left (8 x-6 \log (3)+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)+e^x \left ((-12-48 x) \log (3)+36 \log ^2(3)\right )+e^{2 x} \left ((-6-48 x) \log (3)+108 \log ^2(3)\right )+\left (-30 x+18 \log (3)+e^{2 x} (18+36 x) \log (3)+e^x (36+36 x) \log (3)\right ) \log (x)+18 x \log ^2(x)\right ) \, dx=x^{2} \log {\left (x \right )}^{2} + \frac {16 x^{2}}{9} - \frac {8 x \log {\left (3 \right )}}{3} + \left (- \frac {8 x^{2}}{3} + 2 x \log {\left (3 \right )}\right ) \log {\left (x \right )} + \frac {\left (18 x \log {\left (3 \right )} \log {\left (x \right )} - 24 x \log {\left (3 \right )} + 54 \log {\left (3 \right )}^{2}\right ) e^{2 x}}{9} + \frac {\left (36 x \log {\left (3 \right )} \log {\left (x \right )} - 48 x \log {\left (3 \right )} + 36 \log {\left (3 \right )}^{2}\right ) e^{x}}{9} + e^{4 x} \log {\left (3 \right )}^{2} + 4 e^{3 x} \log {\left (3 \right )}^{2} \] Input:
integrate(2*x*ln(x)**2+1/9*((36*x+18)*ln(3)*exp(x)**2+(36*x+36)*ln(3)*exp( x)+18*ln(3)-30*x)*ln(x)+4*ln(3)**2*exp(x)**4+12*ln(3)**2*exp(x)**3+1/9*(10 8*ln(3)**2+(-48*x-6)*ln(3))*exp(x)**2+1/9*(36*ln(3)**2+(-48*x-12)*ln(3))*e xp(x)-2/3*ln(3)+8/9*x,x)
Output:
x**2*log(x)**2 + 16*x**2/9 - 8*x*log(3)/3 + (-8*x**2/3 + 2*x*log(3))*log(x ) + (18*x*log(3)*log(x) - 24*x*log(3) + 54*log(3)**2)*exp(2*x)/9 + (36*x*l og(3)*log(x) - 48*x*log(3) + 36*log(3)**2)*exp(x)/9 + exp(4*x)*log(3)**2 + 4*exp(3*x)*log(3)**2
Leaf count of result is larger than twice the leaf count of optimal. 134 vs. \(2 (23) = 46\).
Time = 0.11 (sec) , antiderivative size = 134, normalized size of antiderivative = 6.09 \[ \int \frac {1}{9} \left (8 x-6 \log (3)+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)+e^x \left ((-12-48 x) \log (3)+36 \log ^2(3)\right )+e^{2 x} \left ((-6-48 x) \log (3)+108 \log ^2(3)\right )+\left (-30 x+18 \log (3)+e^{2 x} (18+36 x) \log (3)+e^x (36+36 x) \log (3)\right ) \log (x)+18 x \log ^2(x)\right ) \, dx=\frac {1}{2} \, {\left (2 \, \log \left (x\right )^{2} - 2 \, \log \left (x\right ) + 1\right )} x^{2} + e^{\left (4 \, x\right )} \log \left (3\right )^{2} + 4 \, e^{\left (3 \, x\right )} \log \left (3\right )^{2} + \frac {23}{18} \, x^{2} - \frac {1}{3} \, {\left (8 \, x \log \left (3\right ) - 18 \, \log \left (3\right )^{2} - 3 \, \log \left (3\right )\right )} e^{\left (2 \, x\right )} - \frac {4}{3} \, {\left (4 \, x \log \left (3\right ) - 3 \, \log \left (3\right )^{2} - 3 \, \log \left (3\right )\right )} e^{x} - \frac {8}{3} \, x \log \left (3\right ) - e^{\left (2 \, x\right )} \log \left (3\right ) - 4 \, e^{x} \log \left (3\right ) + \frac {1}{3} \, {\left (6 \, x e^{\left (2 \, x\right )} \log \left (3\right ) + 12 \, x e^{x} \log \left (3\right ) - 5 \, x^{2} + 6 \, x \log \left (3\right )\right )} \log \left (x\right ) \] Input:
integrate(2*x*log(x)^2+1/9*((36*x+18)*log(3)*exp(x)^2+(36*x+36)*log(3)*exp (x)+18*log(3)-30*x)*log(x)+4*log(3)^2*exp(x)^4+12*log(3)^2*exp(x)^3+1/9*(1 08*log(3)^2+(-48*x-6)*log(3))*exp(x)^2+1/9*(36*log(3)^2+(-48*x-12)*log(3)) *exp(x)-2/3*log(3)+8/9*x,x, algorithm="maxima")
Output:
1/2*(2*log(x)^2 - 2*log(x) + 1)*x^2 + e^(4*x)*log(3)^2 + 4*e^(3*x)*log(3)^ 2 + 23/18*x^2 - 1/3*(8*x*log(3) - 18*log(3)^2 - 3*log(3))*e^(2*x) - 4/3*(4 *x*log(3) - 3*log(3)^2 - 3*log(3))*e^x - 8/3*x*log(3) - e^(2*x)*log(3) - 4 *e^x*log(3) + 1/3*(6*x*e^(2*x)*log(3) + 12*x*e^x*log(3) - 5*x^2 + 6*x*log( 3))*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 132 vs. \(2 (23) = 46\).
Time = 0.11 (sec) , antiderivative size = 132, normalized size of antiderivative = 6.00 \[ \int \frac {1}{9} \left (8 x-6 \log (3)+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)+e^x \left ((-12-48 x) \log (3)+36 \log ^2(3)\right )+e^{2 x} \left ((-6-48 x) \log (3)+108 \log ^2(3)\right )+\left (-30 x+18 \log (3)+e^{2 x} (18+36 x) \log (3)+e^x (36+36 x) \log (3)\right ) \log (x)+18 x \log ^2(x)\right ) \, dx=x^{2} \log \left (x\right )^{2} + e^{\left (4 \, x\right )} \log \left (3\right )^{2} + 4 \, e^{\left (3 \, x\right )} \log \left (3\right )^{2} - x^{2} \log \left (x\right ) + \frac {16}{9} \, x^{2} - \frac {1}{3} \, {\left (8 \, x \log \left (3\right ) - 18 \, \log \left (3\right )^{2} - 3 \, \log \left (3\right )\right )} e^{\left (2 \, x\right )} - \frac {4}{3} \, {\left (4 \, x \log \left (3\right ) - 3 \, \log \left (3\right )^{2} - 3 \, \log \left (3\right )\right )} e^{x} - \frac {8}{3} \, x \log \left (3\right ) - e^{\left (2 \, x\right )} \log \left (3\right ) - 4 \, e^{x} \log \left (3\right ) + \frac {1}{3} \, {\left (6 \, x e^{\left (2 \, x\right )} \log \left (3\right ) + 12 \, x e^{x} \log \left (3\right ) - 5 \, x^{2} + 6 \, x \log \left (3\right )\right )} \log \left (x\right ) \] Input:
integrate(2*x*log(x)^2+1/9*((36*x+18)*log(3)*exp(x)^2+(36*x+36)*log(3)*exp (x)+18*log(3)-30*x)*log(x)+4*log(3)^2*exp(x)^4+12*log(3)^2*exp(x)^3+1/9*(1 08*log(3)^2+(-48*x-6)*log(3))*exp(x)^2+1/9*(36*log(3)^2+(-48*x-12)*log(3)) *exp(x)-2/3*log(3)+8/9*x,x, algorithm="giac")
Output:
x^2*log(x)^2 + e^(4*x)*log(3)^2 + 4*e^(3*x)*log(3)^2 - x^2*log(x) + 16/9*x ^2 - 1/3*(8*x*log(3) - 18*log(3)^2 - 3*log(3))*e^(2*x) - 4/3*(4*x*log(3) - 3*log(3)^2 - 3*log(3))*e^x - 8/3*x*log(3) - e^(2*x)*log(3) - 4*e^x*log(3) + 1/3*(6*x*e^(2*x)*log(3) + 12*x*e^x*log(3) - 5*x^2 + 6*x*log(3))*log(x)
Time = 2.27 (sec) , antiderivative size = 106, normalized size of antiderivative = 4.82 \[ \int \frac {1}{9} \left (8 x-6 \log (3)+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)+e^x \left ((-12-48 x) \log (3)+36 \log ^2(3)\right )+e^{2 x} \left ((-6-48 x) \log (3)+108 \log ^2(3)\right )+\left (-30 x+18 \log (3)+e^{2 x} (18+36 x) \log (3)+e^x (36+36 x) \log (3)\right ) \log (x)+18 x \log ^2(x)\right ) \, dx=4\,{\mathrm {e}}^x\,{\ln \left (3\right )}^2-\frac {8\,x^2\,\ln \left (x\right )}{3}-\frac {8\,x\,\ln \left (3\right )}{3}+6\,{\mathrm {e}}^{2\,x}\,{\ln \left (3\right )}^2+4\,{\mathrm {e}}^{3\,x}\,{\ln \left (3\right )}^2+{\mathrm {e}}^{4\,x}\,{\ln \left (3\right )}^2+x^2\,{\ln \left (x\right )}^2+\frac {16\,x^2}{9}-\frac {16\,x\,{\mathrm {e}}^x\,\ln \left (3\right )}{3}+2\,x\,\ln \left (3\right )\,\ln \left (x\right )-\frac {8\,x\,{\mathrm {e}}^{2\,x}\,\ln \left (3\right )}{3}+4\,x\,{\mathrm {e}}^x\,\ln \left (3\right )\,\ln \left (x\right )+2\,x\,{\mathrm {e}}^{2\,x}\,\ln \left (3\right )\,\ln \left (x\right ) \] Input:
int((8*x)/9 - (2*log(3))/3 + 2*x*log(x)^2 - (exp(2*x)*(log(3)*(48*x + 6) - 108*log(3)^2))/9 + (log(x)*(18*log(3) - 30*x + exp(2*x)*log(3)*(36*x + 18 ) + exp(x)*log(3)*(36*x + 36)))/9 + 12*exp(3*x)*log(3)^2 + 4*exp(4*x)*log( 3)^2 - (exp(x)*(log(3)*(48*x + 12) - 36*log(3)^2))/9,x)
Output:
4*exp(x)*log(3)^2 - (8*x^2*log(x))/3 - (8*x*log(3))/3 + 6*exp(2*x)*log(3)^ 2 + 4*exp(3*x)*log(3)^2 + exp(4*x)*log(3)^2 + x^2*log(x)^2 + (16*x^2)/9 - (16*x*exp(x)*log(3))/3 + 2*x*log(3)*log(x) - (8*x*exp(2*x)*log(3))/3 + 4*x *exp(x)*log(3)*log(x) + 2*x*exp(2*x)*log(3)*log(x)
Time = 0.19 (sec) , antiderivative size = 114, normalized size of antiderivative = 5.18 \[ \int \frac {1}{9} \left (8 x-6 \log (3)+108 e^{3 x} \log ^2(3)+36 e^{4 x} \log ^2(3)+e^x \left ((-12-48 x) \log (3)+36 \log ^2(3)\right )+e^{2 x} \left ((-6-48 x) \log (3)+108 \log ^2(3)\right )+\left (-30 x+18 \log (3)+e^{2 x} (18+36 x) \log (3)+e^x (36+36 x) \log (3)\right ) \log (x)+18 x \log ^2(x)\right ) \, dx=e^{4 x} \mathrm {log}\left (3\right )^{2}+4 e^{3 x} \mathrm {log}\left (3\right )^{2}+2 e^{2 x} \mathrm {log}\left (x \right ) \mathrm {log}\left (3\right ) x +6 e^{2 x} \mathrm {log}\left (3\right )^{2}-\frac {8 e^{2 x} \mathrm {log}\left (3\right ) x}{3}+4 e^{x} \mathrm {log}\left (x \right ) \mathrm {log}\left (3\right ) x +4 e^{x} \mathrm {log}\left (3\right )^{2}-\frac {16 e^{x} \mathrm {log}\left (3\right ) x}{3}+\mathrm {log}\left (x \right )^{2} x^{2}+2 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (3\right ) x -\frac {8 \,\mathrm {log}\left (x \right ) x^{2}}{3}-\frac {8 \,\mathrm {log}\left (3\right ) x}{3}+\frac {16 x^{2}}{9} \] Input:
int(2*x*log(x)^2+1/9*((36*x+18)*log(3)*exp(x)^2+(36*x+36)*log(3)*exp(x)+18 *log(3)-30*x)*log(x)+4*log(3)^2*exp(x)^4+12*log(3)^2*exp(x)^3+1/9*(108*log (3)^2+(-48*x-6)*log(3))*exp(x)^2+1/9*(36*log(3)^2+(-48*x-12)*log(3))*exp(x )-2/3*log(3)+8/9*x,x)
Output:
(9*e**(4*x)*log(3)**2 + 36*e**(3*x)*log(3)**2 + 18*e**(2*x)*log(x)*log(3)* x + 54*e**(2*x)*log(3)**2 - 24*e**(2*x)*log(3)*x + 36*e**x*log(x)*log(3)*x + 36*e**x*log(3)**2 - 48*e**x*log(3)*x + 9*log(x)**2*x**2 + 18*log(x)*log (3)*x - 24*log(x)*x**2 - 24*log(3)*x + 16*x**2)/9