Integrand size = 106, antiderivative size = 25 \[ \int \frac {-5 x^2+e \left (5 x^2+10 x^3\right ) \log (3)+\left (-4 x+10 e^4 x-10 x^2+e \left (10 x^2+10 x^3\right ) \log (3)\right ) \log \left (\frac {1}{5} \left (2-5 e^4+5 x+e \left (-5 x-5 x^2\right ) \log (3)\right )\right )}{-2+5 e^4-5 x+e \left (5 x+5 x^2\right ) \log (3)} \, dx=x^2 \log \left (\frac {2}{5}+x-e \left (e^3+\left (x+x^2\right ) \log (3)\right )\right ) \] Output:
x^2*ln(x+2/5-exp(1)*((x^2+x)*ln(3)+exp(3)))
Leaf count is larger than twice the leaf count of optimal. \(385\) vs. \(2(25)=50\).
Time = 0.58 (sec) , antiderivative size = 385, normalized size of antiderivative = 15.40 \[ \int \frac {-5 x^2+e \left (5 x^2+10 x^3\right ) \log (3)+\left (-4 x+10 e^4 x-10 x^2+e \left (10 x^2+10 x^3\right ) \log (3)\right ) \log \left (\frac {1}{5} \left (2-5 e^4+5 x+e \left (-5 x-5 x^2\right ) \log (3)\right )\right )}{-2+5 e^4-5 x+e \left (5 x+5 x^2\right ) \log (3)} \, dx=\frac {1}{10} \left (-\frac {5 x^2 \log (9)}{\log (3)}-\frac {2 \arctan \left (\sqrt {\frac {5}{-5+20 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (9)}} (1-e (\log (3)+x \log (9)))\right ) (-1+e \log (3)) \sqrt {5 \left (-5-5 e^2 \log ^2(3)+e \log (9)+5 e^5 \log (81)\right )}}{e^2 \log ^2(3)}+\frac {2 \arctan \left (\sqrt {\frac {5}{-5+20 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (9)}} (1-e (\log (3)+x \log (9)))\right ) (-1+e \log (3)) \sqrt {5 \left (-5-5 e^2 \log ^2(3)+e \log (9)+5 e^5 \log (81)\right )} \left (-4 e \log ^2(3)-5 e^2 \log ^3(3)+e \log (9) \log (27)+5 e^5 \log (3) \log (81)-\log (243)\right )}{e^2 \log ^3(3) \left (-5+20 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (9)\right )}+\frac {\left (-5 e^2 \log ^3(3)+5 e^5 \log (3) \log (9)+e \log (9) \log (27)-\log (243)\right ) \log \left (2-5 e^4+5 x-5 e x (1+x) \log (3)\right )}{e^2 \log ^3(3)}-\frac {\left (-5+10 e^5 \log (3)-5 e^2 \log ^2(3)+e \log (729)\right ) \log \left (2-5 e^4+5 x-5 e x (1+x) \log (3)\right )}{e^2 \log ^2(3)}+10 x^2 \left (1+\log \left (\frac {2}{5}-e^4+x-e x (1+x) \log (3)\right )\right )\right ) \] Input:
Integrate[(-5*x^2 + E*(5*x^2 + 10*x^3)*Log[3] + (-4*x + 10*E^4*x - 10*x^2 + E*(10*x^2 + 10*x^3)*Log[3])*Log[(2 - 5*E^4 + 5*x + E*(-5*x - 5*x^2)*Log[ 3])/5])/(-2 + 5*E^4 - 5*x + E*(5*x + 5*x^2)*Log[3]),x]
Output:
((-5*x^2*Log[9])/Log[3] - (2*ArcTan[Sqrt[5/(-5 + 20*E^5*Log[3] - 5*E^2*Log [3]^2 + E*Log[9])]*(1 - E*(Log[3] + x*Log[9]))]*(-1 + E*Log[3])*Sqrt[5*(-5 - 5*E^2*Log[3]^2 + E*Log[9] + 5*E^5*Log[81])])/(E^2*Log[3]^2) + (2*ArcTan [Sqrt[5/(-5 + 20*E^5*Log[3] - 5*E^2*Log[3]^2 + E*Log[9])]*(1 - E*(Log[3] + x*Log[9]))]*(-1 + E*Log[3])*Sqrt[5*(-5 - 5*E^2*Log[3]^2 + E*Log[9] + 5*E^ 5*Log[81])]*(-4*E*Log[3]^2 - 5*E^2*Log[3]^3 + E*Log[9]*Log[27] + 5*E^5*Log [3]*Log[81] - Log[243]))/(E^2*Log[3]^3*(-5 + 20*E^5*Log[3] - 5*E^2*Log[3]^ 2 + E*Log[9])) + ((-5*E^2*Log[3]^3 + 5*E^5*Log[3]*Log[9] + E*Log[9]*Log[27 ] - Log[243])*Log[2 - 5*E^4 + 5*x - 5*E*x*(1 + x)*Log[3]])/(E^2*Log[3]^3) - ((-5 + 10*E^5*Log[3] - 5*E^2*Log[3]^2 + E*Log[729])*Log[2 - 5*E^4 + 5*x - 5*E*x*(1 + x)*Log[3]])/(E^2*Log[3]^2) + 10*x^2*(1 + Log[2/5 - E^4 + x - E*x*(1 + x)*Log[3]]))/10
Time = 1.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.36, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {7292, 7239, 2010, 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.
| \(\displaystyle \int \frac {-5 x^2+\left (-10 x^2+e \left (10 x^3+10 x^2\right ) \log (3)+10 e^4 x-4 x\right ) \log \left (\frac {1}{5} \left (e \left (-5 x^2-5 x\right ) \log (3)+5 x-5 e^4+2\right )\right )+e \left (10 x^3+5 x^2\right ) \log (3)}{e \left (5 x^2+5 x\right ) \log (3)-5 x+5 e^4-2} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {5 x^2-\left (-10 x^2+e \left (10 x^3+10 x^2\right ) \log (3)+10 e^4 x-4 x\right ) \log \left (\frac {1}{5} \left (e \left (-5 x^2-5 x\right ) \log (3)+5 x-5 e^4+2\right )\right )-e \left (10 x^3+5 x^2\right ) \log (3)}{-5 e x^2 \log (3)+5 x (1-e \log (3))-5 e^4+2}dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int x \left (\frac {5 x (e (x \log (9)+\log (3))-1)}{-5 x+5 e (x+1) x \log (3)+5 e^4-2}+2 \log \left (x-e (x+1) x \log (3)-e^4+\frac {2}{5}\right )\right )dx\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle \int \left (\frac {5 x^2 (-e x \log (9)+1-e \log (3))}{-5 e x^2 \log (3)+5 x (1-e \log (3))-5 e^4+2}+2 x \log \left (-e x^2 \log (3)+x (1-e \log (3))+\frac {1}{5} \left (2-5 e^4\right )\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle x^2 \log \left (-e x^2 \log (3)+x (1-e \log (3))+\frac {1}{5} \left (2-5 e^4\right )\right )\) |
Input:
Int[(-5*x^2 + E*(5*x^2 + 10*x^3)*Log[3] + (-4*x + 10*E^4*x - 10*x^2 + E*(1 0*x^2 + 10*x^3)*Log[3])*Log[(2 - 5*E^4 + 5*x + E*(-5*x - 5*x^2)*Log[3])/5] )/(-2 + 5*E^4 - 5*x + E*(5*x + 5*x^2)*Log[3]),x]
Output:
x^2*Log[(2 - 5*E^4)/5 - E*x^2*Log[3] + x*(1 - E*Log[3])]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 10.54 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
| method | result | size |
| risch | \(x^{2} \ln \left (\frac {\left (-5 x^{2}-5 x \right ) {\mathrm e} \ln \left (3\right )}{5}-{\mathrm e}^{4}+x +\frac {2}{5}\right )\) | \(28\) |
| norman | \(x^{2} \ln \left (\frac {\left (-5 x^{2}-5 x \right ) {\mathrm e} \ln \left (3\right )}{5}-{\mathrm e} \,{\mathrm e}^{3}+x +\frac {2}{5}\right )\) | \(30\) |
| parallelrisch | \(x^{2} \ln \left (\frac {\left (-5 x^{2}-5 x \right ) {\mathrm e} \ln \left (3\right )}{5}-{\mathrm e} \,{\mathrm e}^{3}+x +\frac {2}{5}\right )\) | \(30\) |
| default | \(-x^{2} \ln \left (5\right )+x^{2} \ln \left (-5 \ln \left (3\right ) {\mathrm e} x^{2}-5 x \,{\mathrm e} \ln \left (3\right )-5 \,{\mathrm e} \,{\mathrm e}^{3}+5 x +2\right )\) | \(41\) |
| parts | \(-x^{2} \ln \left (5\right )+x^{2} \ln \left (-5 \ln \left (3\right ) {\mathrm e} x^{2}-5 x \,{\mathrm e} \ln \left (3\right )-5 \,{\mathrm e} \,{\mathrm e}^{3}+5 x +2\right )\) | \(41\) |
Input:
int((((10*x^3+10*x^2)*exp(1)*ln(3)+10*x*exp(1)*exp(3)-10*x^2-4*x)*ln(1/5*( -5*x^2-5*x)*exp(1)*ln(3)-exp(1)*exp(3)+x+2/5)+(10*x^3+5*x^2)*exp(1)*ln(3)- 5*x^2)/((5*x^2+5*x)*exp(1)*ln(3)+5*exp(1)*exp(3)-5*x-2),x,method=_RETURNVE RBOSE)
Output:
x^2*ln(1/5*(-5*x^2-5*x)*exp(1)*ln(3)-exp(4)+x+2/5)
Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-5 x^2+e \left (5 x^2+10 x^3\right ) \log (3)+\left (-4 x+10 e^4 x-10 x^2+e \left (10 x^2+10 x^3\right ) \log (3)\right ) \log \left (\frac {1}{5} \left (2-5 e^4+5 x+e \left (-5 x-5 x^2\right ) \log (3)\right )\right )}{-2+5 e^4-5 x+e \left (5 x+5 x^2\right ) \log (3)} \, dx=x^{2} \log \left (-{\left (x^{2} + x\right )} e \log \left (3\right ) + x - e^{4} + \frac {2}{5}\right ) \] Input:
integrate((((10*x^3+10*x^2)*exp(1)*log(3)+10*x*exp(1)*exp(3)-10*x^2-4*x)*l og(1/5*(-5*x^2-5*x)*exp(1)*log(3)-exp(1)*exp(3)+x+2/5)+(10*x^3+5*x^2)*exp( 1)*log(3)-5*x^2)/((5*x^2+5*x)*exp(1)*log(3)+5*exp(1)*exp(3)-5*x-2),x, algo rithm="fricas")
Output:
x^2*log(-(x^2 + x)*e*log(3) + x - e^4 + 2/5)
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-5 x^2+e \left (5 x^2+10 x^3\right ) \log (3)+\left (-4 x+10 e^4 x-10 x^2+e \left (10 x^2+10 x^3\right ) \log (3)\right ) \log \left (\frac {1}{5} \left (2-5 e^4+5 x+e \left (-5 x-5 x^2\right ) \log (3)\right )\right )}{-2+5 e^4-5 x+e \left (5 x+5 x^2\right ) \log (3)} \, dx=x^{2} \log {\left (x + e \left (- x^{2} - x\right ) \log {\left (3 \right )} - e^{4} + \frac {2}{5} \right )} \] Input:
integrate((((10*x**3+10*x**2)*exp(1)*ln(3)+10*x*exp(1)*exp(3)-10*x**2-4*x) *ln(1/5*(-5*x**2-5*x)*exp(1)*ln(3)-exp(1)*exp(3)+x+2/5)+(10*x**3+5*x**2)*e xp(1)*ln(3)-5*x**2)/((5*x**2+5*x)*exp(1)*ln(3)+5*exp(1)*exp(3)-5*x-2),x)
Output:
x**2*log(x + E*(-x**2 - x)*log(3) - exp(4) + 2/5)
Leaf count of result is larger than twice the leaf count of optimal. 705 vs. \(2 (23) = 46\).
Time = 1.52 (sec) , antiderivative size = 705, normalized size of antiderivative = 28.20 \[ \int \frac {-5 x^2+e \left (5 x^2+10 x^3\right ) \log (3)+\left (-4 x+10 e^4 x-10 x^2+e \left (10 x^2+10 x^3\right ) \log (3)\right ) \log \left (\frac {1}{5} \left (2-5 e^4+5 x+e \left (-5 x-5 x^2\right ) \log (3)\right )\right )}{-2+5 e^4-5 x+e \left (5 x+5 x^2\right ) \log (3)} \, dx=\text {Too large to display} \] Input:
integrate((((10*x^3+10*x^2)*exp(1)*log(3)+10*x*exp(1)*exp(3)-10*x^2-4*x)*l og(1/5*(-5*x^2-5*x)*exp(1)*log(3)-exp(1)*exp(3)+x+2/5)+(10*x^3+5*x^2)*exp( 1)*log(3)-5*x^2)/((5*x^2+5*x)*exp(1)*log(3)+5*exp(1)*exp(3)-5*x-2),x, algo rithm="maxima")
Output:
1/2*(2*x*e^(-1)/log(3) - (e*log(3) - 1)*e^(-2)*log(5*x^2*e*log(3) + 5*(e*l og(3) - 1)*x + 5*e^4 - 2)/log(3)^2 + 2*(5*e^2*log(3)^2 - 2*(5*e^5 + 3*e)*l og(3) + 5)*arctan(5*(2*x*e*log(3) + e*log(3) - 1)/sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*log(3) - 25))*e^(-2)/(sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*log(3) - 25)*log(3)^2))*e*log(3) + 1/5*(5*(x^2*e*log(3) - 2*(e*log(3) - 1)*x)*e^(-2)/log(3)^2 + (5*e^2*log(3)^2 - (5*e^5 + 8*e)*log(3) + 5)*e^(- 3)*log(5*x^2*e*log(3) + 5*(e*log(3) - 1)*x + 5*e^4 - 2)/log(3)^3 - 10*(5*e ^3*log(3)^3 - 3*(5*e^6 + 3*e^2)*log(3)^2 + 3*(5*e^5 + 3*e)*log(3) - 5)*arc tan(5*(2*x*e*log(3) + e*log(3) - 1)/sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e )*log(3) - 25))*e^(-3)/(sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*log(3) - 2 5)*log(3)^3))*e*log(3) - x*e^(-1)/log(3) + 1/2*(e*log(3) - 1)*e^(-2)*log(5 *x^2*e*log(3) + 5*(e*log(3) - 1)*x + 5*e^4 - 2)/log(3)^2 + (5*e^3*log(3)^3 - 20*e^6*log(3)^2 - 7*e^2*log(3)^2 + 20*e^5*log(3) + 7*e*log(3) - 5)*arct an(5*(2*x*e*log(3) + e*log(3) - 1)/sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e) *log(3) - 25))*e^(-2)/(sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*log(3) - 25 )*log(3)^2) - (5*e^2*log(3)^2 - 2*(5*e^5 + 3*e)*log(3) + 5)*arctan(5*(2*x* e*log(3) + e*log(3) - 1)/sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*log(3) - 25))*e^(-2)/(sqrt(-25*e^2*log(3)^2 + 10*(10*e^5 + e)*log(3) - 25)*log(3)^2 ) - 1/10*(10*x^2*(log(5) + 1)*e^2*log(3)^2 - 10*(e^2*log(3)^2 - e*log(3))* x - (10*x^2*e^2*log(3)^2 - 5*e^2*log(3)^2 + 10*e^5*log(3) + 6*e*log(3) ...
Time = 0.15 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {-5 x^2+e \left (5 x^2+10 x^3\right ) \log (3)+\left (-4 x+10 e^4 x-10 x^2+e \left (10 x^2+10 x^3\right ) \log (3)\right ) \log \left (\frac {1}{5} \left (2-5 e^4+5 x+e \left (-5 x-5 x^2\right ) \log (3)\right )\right )}{-2+5 e^4-5 x+e \left (5 x+5 x^2\right ) \log (3)} \, dx=-x^{2} \log \left (5\right ) + x^{2} \log \left (5 \, x^{2} e \log \left (3\right ) + 5 \, x e \log \left (3\right ) - 5 \, x + 5 \, e^{4} - 2\right ) \] Input:
integrate((((10*x^3+10*x^2)*exp(1)*log(3)+10*x*exp(1)*exp(3)-10*x^2-4*x)*l og(1/5*(-5*x^2-5*x)*exp(1)*log(3)-exp(1)*exp(3)+x+2/5)+(10*x^3+5*x^2)*exp( 1)*log(3)-5*x^2)/((5*x^2+5*x)*exp(1)*log(3)+5*exp(1)*exp(3)-5*x-2),x, algo rithm="giac")
Output:
-x^2*log(5) + x^2*log(5*x^2*e*log(3) + 5*x*e*log(3) - 5*x + 5*e^4 - 2)
Time = 4.60 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {-5 x^2+e \left (5 x^2+10 x^3\right ) \log (3)+\left (-4 x+10 e^4 x-10 x^2+e \left (10 x^2+10 x^3\right ) \log (3)\right ) \log \left (\frac {1}{5} \left (2-5 e^4+5 x+e \left (-5 x-5 x^2\right ) \log (3)\right )\right )}{-2+5 e^4-5 x+e \left (5 x+5 x^2\right ) \log (3)} \, dx=x^2\,\ln \left (x-{\mathrm {e}}^4-\frac {\mathrm {e}\,\ln \left (3\right )\,\left (5\,x^2+5\,x\right )}{5}+\frac {2}{5}\right ) \] Input:
int((log(x - exp(4) - (exp(1)*log(3)*(5*x + 5*x^2))/5 + 2/5)*(4*x - 10*x*e xp(4) + 10*x^2 - exp(1)*log(3)*(10*x^2 + 10*x^3)) + 5*x^2 - exp(1)*log(3)* (5*x^2 + 10*x^3))/(5*x - 5*exp(4) - exp(1)*log(3)*(5*x + 5*x^2) + 2),x)
Output:
x^2*log(x - exp(4) - (exp(1)*log(3)*(5*x + 5*x^2))/5 + 2/5)
Time = 0.17 (sec) , antiderivative size = 283, normalized size of antiderivative = 11.32 \[ \int \frac {-5 x^2+e \left (5 x^2+10 x^3\right ) \log (3)+\left (-4 x+10 e^4 x-10 x^2+e \left (10 x^2+10 x^3\right ) \log (3)\right ) \log \left (\frac {1}{5} \left (2-5 e^4+5 x+e \left (-5 x-5 x^2\right ) \log (3)\right )\right )}{-2+5 e^4-5 x+e \left (5 x+5 x^2\right ) \log (3)} \, dx=\frac {5 \,\mathrm {log}\left (5 \,\mathrm {log}\left (3\right ) e \,x^{2}+5 \,\mathrm {log}\left (3\right ) e x +5 e^{4}-5 x -2\right ) \mathrm {log}\left (3\right )^{2} e^{2}-10 \,\mathrm {log}\left (5 \,\mathrm {log}\left (3\right ) e \,x^{2}+5 \,\mathrm {log}\left (3\right ) e x +5 e^{4}-5 x -2\right ) \mathrm {log}\left (3\right ) e^{5}-6 \,\mathrm {log}\left (5 \,\mathrm {log}\left (3\right ) e \,x^{2}+5 \,\mathrm {log}\left (3\right ) e x +5 e^{4}-5 x -2\right ) \mathrm {log}\left (3\right ) e +5 \,\mathrm {log}\left (5 \,\mathrm {log}\left (3\right ) e \,x^{2}+5 \,\mathrm {log}\left (3\right ) e x +5 e^{4}-5 x -2\right )+10 \,\mathrm {log}\left (-\mathrm {log}\left (3\right ) e \,x^{2}-\mathrm {log}\left (3\right ) e x -e^{4}+x +\frac {2}{5}\right ) \mathrm {log}\left (3\right )^{2} e^{2} x^{2}-5 \,\mathrm {log}\left (-\mathrm {log}\left (3\right ) e \,x^{2}-\mathrm {log}\left (3\right ) e x -e^{4}+x +\frac {2}{5}\right ) \mathrm {log}\left (3\right )^{2} e^{2}+10 \,\mathrm {log}\left (-\mathrm {log}\left (3\right ) e \,x^{2}-\mathrm {log}\left (3\right ) e x -e^{4}+x +\frac {2}{5}\right ) \mathrm {log}\left (3\right ) e^{5}+6 \,\mathrm {log}\left (-\mathrm {log}\left (3\right ) e \,x^{2}-\mathrm {log}\left (3\right ) e x -e^{4}+x +\frac {2}{5}\right ) \mathrm {log}\left (3\right ) e -5 \,\mathrm {log}\left (-\mathrm {log}\left (3\right ) e \,x^{2}-\mathrm {log}\left (3\right ) e x -e^{4}+x +\frac {2}{5}\right )}{10 \mathrm {log}\left (3\right )^{2} e^{2}} \] Input:
int((((10*x^3+10*x^2)*exp(1)*log(3)+10*x*exp(1)*exp(3)-10*x^2-4*x)*log(1/5 *(-5*x^2-5*x)*exp(1)*log(3)-exp(1)*exp(3)+x+2/5)+(10*x^3+5*x^2)*exp(1)*log (3)-5*x^2)/((5*x^2+5*x)*exp(1)*log(3)+5*exp(1)*exp(3)-5*x-2),x)
Output:
(5*log(5*log(3)*e*x**2 + 5*log(3)*e*x + 5*e**4 - 5*x - 2)*log(3)**2*e**2 - 10*log(5*log(3)*e*x**2 + 5*log(3)*e*x + 5*e**4 - 5*x - 2)*log(3)*e**5 - 6 *log(5*log(3)*e*x**2 + 5*log(3)*e*x + 5*e**4 - 5*x - 2)*log(3)*e + 5*log(5 *log(3)*e*x**2 + 5*log(3)*e*x + 5*e**4 - 5*x - 2) + 10*log(( - 5*log(3)*e* x**2 - 5*log(3)*e*x - 5*e**4 + 5*x + 2)/5)*log(3)**2*e**2*x**2 - 5*log(( - 5*log(3)*e*x**2 - 5*log(3)*e*x - 5*e**4 + 5*x + 2)/5)*log(3)**2*e**2 + 10 *log(( - 5*log(3)*e*x**2 - 5*log(3)*e*x - 5*e**4 + 5*x + 2)/5)*log(3)*e**5 + 6*log(( - 5*log(3)*e*x**2 - 5*log(3)*e*x - 5*e**4 + 5*x + 2)/5)*log(3)* e - 5*log(( - 5*log(3)*e*x**2 - 5*log(3)*e*x - 5*e**4 + 5*x + 2)/5))/(10*l og(3)**2*e**2)