Integrand size = 203, antiderivative size = 31 \[ \int \frac {e (1-x)+e^6 \left (-x+2 x^2\right )-e x \log (x)+\left (e^6 \left (2+x-x^2\right )+e (-1+x) \log (x)\right ) \log \left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )+\left (e^{10} \left (2+x-x^2\right )+e^5 \left (-6+x+5 x^2-2 x^3\right )+\left (3+e^5 (-1+x)-5 x+2 x^2\right ) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )}{\left (e^5 \left (2+x-x^2\right )+(-1+x) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )} \, dx=x \left (-3+e^5+x+\frac {e}{\log \left (-2+(1-x) \left (-x+\frac {\log (x)}{e^5}\right )\right )}\right ) \] Output:
x*(exp(1)/ln((ln(x)/exp(5)-x)*(1-x)-2)+x-3+exp(5))
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e (1-x)+e^6 \left (-x+2 x^2\right )-e x \log (x)+\left (e^6 \left (2+x-x^2\right )+e (-1+x) \log (x)\right ) \log \left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )+\left (e^{10} \left (2+x-x^2\right )+e^5 \left (-6+x+5 x^2-2 x^3\right )+\left (3+e^5 (-1+x)-5 x+2 x^2\right ) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )}{\left (e^5 \left (2+x-x^2\right )+(-1+x) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )} \, dx=x \left (-3+e^5+x+\frac {e}{\log \left (-2-x+x^2-\frac {(-1+x) \log (x)}{e^5}\right )}\right ) \] Input:
Integrate[(E*(1 - x) + E^6*(-x + 2*x^2) - E*x*Log[x] + (E^6*(2 + x - x^2) + E*(-1 + x)*Log[x])*Log[(E^5*(-2 - x + x^2) + (1 - x)*Log[x])/E^5] + (E^1 0*(2 + x - x^2) + E^5*(-6 + x + 5*x^2 - 2*x^3) + (3 + E^5*(-1 + x) - 5*x + 2*x^2)*Log[x])*Log[(E^5*(-2 - x + x^2) + (1 - x)*Log[x])/E^5]^2)/((E^5*(2 + x - x^2) + (-1 + x)*Log[x])*Log[(E^5*(-2 - x + x^2) + (1 - x)*Log[x])/E ^5]^2),x]
Output:
x*(-3 + E^5 + x + E/Log[-2 - x + x^2 - ((-1 + x)*Log[x])/E^5])
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.
| \(\displaystyle \int \frac {e^6 \left (2 x^2-x\right )+\left (e^6 \left (-x^2+x+2\right )+e (x-1) \log (x)\right ) \log \left (\frac {e^5 \left (x^2-x-2\right )+(1-x) \log (x)}{e^5}\right )+\left (e^{10} \left (-x^2+x+2\right )+\left (2 x^2-5 x+e^5 (x-1)+3\right ) \log (x)+e^5 \left (-2 x^3+5 x^2+x-6\right )\right ) \log ^2\left (\frac {e^5 \left (x^2-x-2\right )+(1-x) \log (x)}{e^5}\right )+e (1-x)-e x \log (x)}{\left (e^5 \left (-x^2+x+2\right )+(x-1) \log (x)\right ) \log ^2\left (\frac {e^5 \left (x^2-x-2\right )+(1-x) \log (x)}{e^5}\right )} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {e \left (2 e^5 x^2-\left (1+e^5\right ) x-x \log (x)+1\right )}{\left (-e^5 x^2+e^5 x+x \log (x)-\log (x)+2 e^5\right ) \log ^2\left (x^2-x-\frac {(x-1) \log (x)}{e^5}-2\right )}+\frac {e}{\log \left (x^2-x-\frac {(x-1) \log (x)}{e^5}-2\right )}+2 x-3 \left (1-\frac {e^5}{3}\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -e \int \frac {1}{\left (e^5 x^2-\log (x) x-e^5 x+\log (x)-2 e^5\right ) \log ^2\left (x^2-x-\frac {(x-1) \log (x)}{e^5}-2\right )}dx+e \left (1+e^5\right ) \int \frac {x}{\left (e^5 x^2-\log (x) x-e^5 x+\log (x)-2 e^5\right ) \log ^2\left (x^2-x-\frac {(x-1) \log (x)}{e^5}-2\right )}dx-2 e^6 \int \frac {x^2}{\left (e^5 x^2-\log (x) x-e^5 x+\log (x)-2 e^5\right ) \log ^2\left (x^2-x-\frac {(x-1) \log (x)}{e^5}-2\right )}dx+e \int \frac {x \log (x)}{\left (e^5 x^2-\log (x) x-e^5 x+\log (x)-2 e^5\right ) \log ^2\left (x^2-x-\frac {(x-1) \log (x)}{e^5}-2\right )}dx+e \int \frac {1}{\log \left (x^2-x-\frac {(x-1) \log (x)}{e^5}-2\right )}dx+x^2-\left (3-e^5\right ) x\) |
Input:
Int[(E*(1 - x) + E^6*(-x + 2*x^2) - E*x*Log[x] + (E^6*(2 + x - x^2) + E*(- 1 + x)*Log[x])*Log[(E^5*(-2 - x + x^2) + (1 - x)*Log[x])/E^5] + (E^10*(2 + x - x^2) + E^5*(-6 + x + 5*x^2 - 2*x^3) + (3 + E^5*(-1 + x) - 5*x + 2*x^2 )*Log[x])*Log[(E^5*(-2 - x + x^2) + (1 - x)*Log[x])/E^5]^2)/((E^5*(2 + x - x^2) + (-1 + x)*Log[x])*Log[(E^5*(-2 - x + x^2) + (1 - x)*Log[x])/E^5]^2) ,x]
Output:
$Aborted
Time = 2.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35
| method | result | size |
| risch | \(x \,{\mathrm e}^{5}+x^{2}-3 x +\frac {{\mathrm e} x}{\ln \left (\left (\left (1-x \right ) \ln \left (x \right )+\left (x^{2}-x -2\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-5}\right )}\) | \(42\) |
| default | \({\mathrm e}^{5+\ln \left (x \right )}+x^{2}-3 x +\frac {{\mathrm e}^{\ln \left (x \right )+1}}{-5+\ln \left ({\mathrm e}^{2 \ln \left (x \right )+5}-x \ln \left (x \right )-{\mathrm e}^{5+\ln \left (x \right )}+\ln \left (x \right )-2 \,{\mathrm e}^{5}\right )}\) | \(50\) |
| parallelrisch | \(\frac {{\mathrm e}^{5} x \ln \left (\left (\left (1-x \right ) \ln \left (x \right )+\left (x^{2}-x -2\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-5}\right )+\ln \left (\left (\left (1-x \right ) \ln \left (x \right )+\left (x^{2}-x -2\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-5}\right ) x^{2}+2 \,{\mathrm e}^{5} \ln \left (\left (\left (1-x \right ) \ln \left (x \right )+\left (x^{2}-x -2\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-5}\right )+x \,{\mathrm e}-3 x \ln \left (\left (\left (1-x \right ) \ln \left (x \right )+\left (x^{2}-x -2\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-5}\right )-7 \ln \left (\left (\left (1-x \right ) \ln \left (x \right )+\left (x^{2}-x -2\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-5}\right )}{\ln \left (\left (\left (1-x \right ) \ln \left (x \right )+\left (x^{2}-x -2\right ) {\mathrm e}^{5}\right ) {\mathrm e}^{-5}\right )}\) | \(182\) |
Input:
int(((((-1+x)*exp(5)+2*x^2-5*x+3)*ln(x)+(-x^2+x+2)*exp(5)^2+(-2*x^3+5*x^2+ x-6)*exp(5))*ln(((1-x)*ln(x)+(x^2-x-2)*exp(5))/exp(5))^2+((-1+x)*exp(1)*ln (x)+(-x^2+x+2)*exp(1)*exp(5))*ln(((1-x)*ln(x)+(x^2-x-2)*exp(5))/exp(5))-x* exp(1)*ln(x)+(2*x^2-x)*exp(1)*exp(5)+(1-x)*exp(1))/((-1+x)*ln(x)+(-x^2+x+2 )*exp(5))/ln(((1-x)*ln(x)+(x^2-x-2)*exp(5))/exp(5))^2,x,method=_RETURNVERB OSE)
Output:
x*exp(5)+x^2-3*x+exp(1)*x/ln(((1-x)*ln(x)+(x^2-x-2)*exp(5))*exp(-5))
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (29) = 58\).
Time = 0.10 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.13 \[ \int \frac {e (1-x)+e^6 \left (-x+2 x^2\right )-e x \log (x)+\left (e^6 \left (2+x-x^2\right )+e (-1+x) \log (x)\right ) \log \left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )+\left (e^{10} \left (2+x-x^2\right )+e^5 \left (-6+x+5 x^2-2 x^3\right )+\left (3+e^5 (-1+x)-5 x+2 x^2\right ) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )}{\left (e^5 \left (2+x-x^2\right )+(-1+x) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )} \, dx=\frac {x e + {\left (x^{2} + x e^{5} - 3 \, x\right )} \log \left ({\left ({\left (x^{2} - x - 2\right )} e^{5} - {\left (x - 1\right )} \log \left (x\right )\right )} e^{\left (-5\right )}\right )}{\log \left ({\left ({\left (x^{2} - x - 2\right )} e^{5} - {\left (x - 1\right )} \log \left (x\right )\right )} e^{\left (-5\right )}\right )} \] Input:
integrate(((((-1+x)*exp(5)+2*x^2-5*x+3)*log(x)+(-x^2+x+2)*exp(5)^2+(-2*x^3 +5*x^2+x-6)*exp(5))*log(((1-x)*log(x)+(x^2-x-2)*exp(5))/exp(5))^2+((-1+x)* exp(1)*log(x)+(-x^2+x+2)*exp(1)*exp(5))*log(((1-x)*log(x)+(x^2-x-2)*exp(5) )/exp(5))-x*exp(1)*log(x)+(2*x^2-x)*exp(1)*exp(5)+(1-x)*exp(1))/((-1+x)*lo g(x)+(-x^2+x+2)*exp(5))/log(((1-x)*log(x)+(x^2-x-2)*exp(5))/exp(5))^2,x, a lgorithm="fricas")
Output:
(x*e + (x^2 + x*e^5 - 3*x)*log(((x^2 - x - 2)*e^5 - (x - 1)*log(x))*e^(-5) ))/log(((x^2 - x - 2)*e^5 - (x - 1)*log(x))*e^(-5))
Time = 0.29 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {e (1-x)+e^6 \left (-x+2 x^2\right )-e x \log (x)+\left (e^6 \left (2+x-x^2\right )+e (-1+x) \log (x)\right ) \log \left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )+\left (e^{10} \left (2+x-x^2\right )+e^5 \left (-6+x+5 x^2-2 x^3\right )+\left (3+e^5 (-1+x)-5 x+2 x^2\right ) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )}{\left (e^5 \left (2+x-x^2\right )+(-1+x) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )} \, dx=x^{2} + x \left (-3 + e^{5}\right ) + \frac {e x}{\log {\left (\frac {\left (1 - x\right ) \log {\left (x \right )} + \left (x^{2} - x - 2\right ) e^{5}}{e^{5}} \right )}} \] Input:
integrate(((((-1+x)*exp(5)+2*x**2-5*x+3)*ln(x)+(-x**2+x+2)*exp(5)**2+(-2*x **3+5*x**2+x-6)*exp(5))*ln(((1-x)*ln(x)+(x**2-x-2)*exp(5))/exp(5))**2+((-1 +x)*exp(1)*ln(x)+(-x**2+x+2)*exp(1)*exp(5))*ln(((1-x)*ln(x)+(x**2-x-2)*exp (5))/exp(5))-x*exp(1)*ln(x)+(2*x**2-x)*exp(1)*exp(5)+(1-x)*exp(1))/((-1+x) *ln(x)+(-x**2+x+2)*exp(5))/ln(((1-x)*ln(x)+(x**2-x-2)*exp(5))/exp(5))**2,x )
Output:
x**2 + x*(-3 + exp(5)) + E*x/log(((1 - x)*log(x) + (x**2 - x - 2)*exp(5))* exp(-5))
Leaf count of result is larger than twice the leaf count of optimal. 84 vs. \(2 (29) = 58\).
Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 2.71 \[ \int \frac {e (1-x)+e^6 \left (-x+2 x^2\right )-e x \log (x)+\left (e^6 \left (2+x-x^2\right )+e (-1+x) \log (x)\right ) \log \left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )+\left (e^{10} \left (2+x-x^2\right )+e^5 \left (-6+x+5 x^2-2 x^3\right )+\left (3+e^5 (-1+x)-5 x+2 x^2\right ) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )}{\left (e^5 \left (2+x-x^2\right )+(-1+x) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )} \, dx=-\frac {5 \, x^{2} + x {\left (5 \, e^{5} - e - 15\right )} - {\left (x^{2} + x {\left (e^{5} - 3\right )}\right )} \log \left (x^{2} e^{5} - x e^{5} - {\left (x - 1\right )} \log \left (x\right ) - 2 \, e^{5}\right )}{\log \left (x^{2} e^{5} - x e^{5} - {\left (x - 1\right )} \log \left (x\right ) - 2 \, e^{5}\right ) - 5} \] Input:
integrate(((((-1+x)*exp(5)+2*x^2-5*x+3)*log(x)+(-x^2+x+2)*exp(5)^2+(-2*x^3 +5*x^2+x-6)*exp(5))*log(((1-x)*log(x)+(x^2-x-2)*exp(5))/exp(5))^2+((-1+x)* exp(1)*log(x)+(-x^2+x+2)*exp(1)*exp(5))*log(((1-x)*log(x)+(x^2-x-2)*exp(5) )/exp(5))-x*exp(1)*log(x)+(2*x^2-x)*exp(1)*exp(5)+(1-x)*exp(1))/((-1+x)*lo g(x)+(-x^2+x+2)*exp(5))/log(((1-x)*log(x)+(x^2-x-2)*exp(5))/exp(5))^2,x, a lgorithm="maxima")
Output:
-(5*x^2 + x*(5*e^5 - e - 15) - (x^2 + x*(e^5 - 3))*log(x^2*e^5 - x*e^5 - ( x - 1)*log(x) - 2*e^5))/(log(x^2*e^5 - x*e^5 - (x - 1)*log(x) - 2*e^5) - 5 )
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (29) = 58\).
Time = 0.60 (sec) , antiderivative size = 130, normalized size of antiderivative = 4.19 \[ \int \frac {e (1-x)+e^6 \left (-x+2 x^2\right )-e x \log (x)+\left (e^6 \left (2+x-x^2\right )+e (-1+x) \log (x)\right ) \log \left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )+\left (e^{10} \left (2+x-x^2\right )+e^5 \left (-6+x+5 x^2-2 x^3\right )+\left (3+e^5 (-1+x)-5 x+2 x^2\right ) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )}{\left (e^5 \left (2+x-x^2\right )+(-1+x) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )} \, dx=\frac {x^{2} \log \left (x^{2} e^{5} - x e^{5} - x \log \left (x\right ) - 2 \, e^{5} + \log \left (x\right )\right ) + x e^{5} \log \left (x^{2} e^{5} - x e^{5} - x \log \left (x\right ) - 2 \, e^{5} + \log \left (x\right )\right ) - 5 \, x^{2} - 5 \, x e^{5} + x e - 3 \, x \log \left (x^{2} e^{5} - x e^{5} - x \log \left (x\right ) - 2 \, e^{5} + \log \left (x\right )\right ) + 15 \, x}{\log \left (x^{2} e^{5} - x e^{5} - x \log \left (x\right ) - 2 \, e^{5} + \log \left (x\right )\right ) - 5} \] Input:
integrate(((((-1+x)*exp(5)+2*x^2-5*x+3)*log(x)+(-x^2+x+2)*exp(5)^2+(-2*x^3 +5*x^2+x-6)*exp(5))*log(((1-x)*log(x)+(x^2-x-2)*exp(5))/exp(5))^2+((-1+x)* exp(1)*log(x)+(-x^2+x+2)*exp(1)*exp(5))*log(((1-x)*log(x)+(x^2-x-2)*exp(5) )/exp(5))-x*exp(1)*log(x)+(2*x^2-x)*exp(1)*exp(5)+(1-x)*exp(1))/((-1+x)*lo g(x)+(-x^2+x+2)*exp(5))/log(((1-x)*log(x)+(x^2-x-2)*exp(5))/exp(5))^2,x, a lgorithm="giac")
Output:
(x^2*log(x^2*e^5 - x*e^5 - x*log(x) - 2*e^5 + log(x)) + x*e^5*log(x^2*e^5 - x*e^5 - x*log(x) - 2*e^5 + log(x)) - 5*x^2 - 5*x*e^5 + x*e - 3*x*log(x^2 *e^5 - x*e^5 - x*log(x) - 2*e^5 + log(x)) + 15*x)/(log(x^2*e^5 - x*e^5 - x *log(x) - 2*e^5 + log(x)) - 5)
Time = 4.18 (sec) , antiderivative size = 165, normalized size of antiderivative = 5.32 \[ \int \frac {e (1-x)+e^6 \left (-x+2 x^2\right )-e x \log (x)+\left (e^6 \left (2+x-x^2\right )+e (-1+x) \log (x)\right ) \log \left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )+\left (e^{10} \left (2+x-x^2\right )+e^5 \left (-6+x+5 x^2-2 x^3\right )+\left (3+e^5 (-1+x)-5 x+2 x^2\right ) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )}{\left (e^5 \left (2+x-x^2\right )+(-1+x) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )} \, dx=\frac {2\,{\mathrm {e}}^6+x\,{\mathrm {e}}^6-\ln \left (x\right )\,\left (\mathrm {e}-x\,\mathrm {e}\right )-x^2\,{\mathrm {e}}^6}{\ln \left (x\right )+\frac {x+x\,{\mathrm {e}}^5-2\,x^2\,{\mathrm {e}}^5-1}{x}}+\frac {x\,\mathrm {e}-\frac {x\,\mathrm {e}\,\ln \left (-{\mathrm {e}}^{-5}\,\left (\ln \left (x\right )\,\left (x-1\right )+{\mathrm {e}}^5\,\left (-x^2+x+2\right )\right )\right )\,\left (2\,{\mathrm {e}}^5-\ln \left (x\right )+x\,{\mathrm {e}}^5-x^2\,{\mathrm {e}}^5+x\,\ln \left (x\right )\right )}{x+x\,{\mathrm {e}}^5-2\,x^2\,{\mathrm {e}}^5+x\,\ln \left (x\right )-1}}{\ln \left (-{\mathrm {e}}^{-5}\,\left (\ln \left (x\right )\,\left (x-1\right )+{\mathrm {e}}^5\,\left (-x^2+x+2\right )\right )\right )}+x\,\left ({\mathrm {e}}^5-3\right )+x^2 \] Input:
int(-(exp(1)*(x - 1) - log(-exp(-5)*(log(x)*(x - 1) + exp(5)*(x - x^2 + 2) ))*(exp(6)*(x - x^2 + 2) + exp(1)*log(x)*(x - 1)) - log(-exp(-5)*(log(x)*( x - 1) + exp(5)*(x - x^2 + 2)))^2*(exp(5)*(x + 5*x^2 - 2*x^3 - 6) + log(x) *(exp(5)*(x - 1) - 5*x + 2*x^2 + 3) + exp(10)*(x - x^2 + 2)) + exp(6)*(x - 2*x^2) + x*exp(1)*log(x))/(log(-exp(-5)*(log(x)*(x - 1) + exp(5)*(x - x^2 + 2)))^2*(log(x)*(x - 1) + exp(5)*(x - x^2 + 2))),x)
Output:
(2*exp(6) + x*exp(6) - log(x)*(exp(1) - x*exp(1)) - x^2*exp(6))/(log(x) + (x + x*exp(5) - 2*x^2*exp(5) - 1)/x) + (x*exp(1) - (x*exp(1)*log(-exp(-5)* (log(x)*(x - 1) + exp(5)*(x - x^2 + 2)))*(2*exp(5) - log(x) + x*exp(5) - x ^2*exp(5) + x*log(x)))/(x + x*exp(5) - 2*x^2*exp(5) + x*log(x) - 1))/log(- exp(-5)*(log(x)*(x - 1) + exp(5)*(x - x^2 + 2))) + x*(exp(5) - 3) + x^2
Time = 0.46 (sec) , antiderivative size = 138, normalized size of antiderivative = 4.45 \[ \int \frac {e (1-x)+e^6 \left (-x+2 x^2\right )-e x \log (x)+\left (e^6 \left (2+x-x^2\right )+e (-1+x) \log (x)\right ) \log \left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )+\left (e^{10} \left (2+x-x^2\right )+e^5 \left (-6+x+5 x^2-2 x^3\right )+\left (3+e^5 (-1+x)-5 x+2 x^2\right ) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )}{\left (e^5 \left (2+x-x^2\right )+(-1+x) \log (x)\right ) \log ^2\left (\frac {e^5 \left (-2-x+x^2\right )+(1-x) \log (x)}{e^5}\right )} \, dx=\frac {x \left (\mathrm {log}\left (\frac {-\mathrm {log}\left (x \right ) x +\mathrm {log}\left (x \right )+e^{5} x^{2}-e^{5} x -2 e^{5}}{e^{5}}\right ) e^{5}+\mathrm {log}\left (\frac {-\mathrm {log}\left (x \right ) x +\mathrm {log}\left (x \right )+e^{5} x^{2}-e^{5} x -2 e^{5}}{e^{5}}\right ) x -3 \,\mathrm {log}\left (\frac {-\mathrm {log}\left (x \right ) x +\mathrm {log}\left (x \right )+e^{5} x^{2}-e^{5} x -2 e^{5}}{e^{5}}\right )+e \right )}{\mathrm {log}\left (\frac {-\mathrm {log}\left (x \right ) x +\mathrm {log}\left (x \right )+e^{5} x^{2}-e^{5} x -2 e^{5}}{e^{5}}\right )} \] Input:
int(((((-1+x)*exp(5)+2*x^2-5*x+3)*log(x)+(-x^2+x+2)*exp(5)^2+(-2*x^3+5*x^2 +x-6)*exp(5))*log(((1-x)*log(x)+(x^2-x-2)*exp(5))/exp(5))^2+((-1+x)*exp(1) *log(x)+(-x^2+x+2)*exp(1)*exp(5))*log(((1-x)*log(x)+(x^2-x-2)*exp(5))/exp( 5))-x*exp(1)*log(x)+(2*x^2-x)*exp(1)*exp(5)+(1-x)*exp(1))/((-1+x)*log(x)+( -x^2+x+2)*exp(5))/log(((1-x)*log(x)+(x^2-x-2)*exp(5))/exp(5))^2,x)
Output:
(x*(log(( - log(x)*x + log(x) + e**5*x**2 - e**5*x - 2*e**5)/e**5)*e**5 + log(( - log(x)*x + log(x) + e**5*x**2 - e**5*x - 2*e**5)/e**5)*x - 3*log(( - log(x)*x + log(x) + e**5*x**2 - e**5*x - 2*e**5)/e**5) + e))/log(( - lo g(x)*x + log(x) + e**5*x**2 - e**5*x - 2*e**5)/e**5)