Integrand size = 209, antiderivative size = 35 \[ \int \frac {\left (80-480 x+e^x \left (-40+216 x+120 x^2\right )+\left (-16+80 x+e^x \left (8-32 x-40 x^2\right )\right ) \log \left (x^2-5 x^3\right )\right ) \log \left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )+\left (48-240 x+e^x (-24+120 x)+\left (-16+e^x (8-40 x)+80 x\right ) \log \left (x^2-5 x^3\right )\right ) \log ^2\left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )}{-6 x^3+30 x^4+e^x \left (3 x^3-15 x^4\right )+\left (2 x^3-10 x^4+e^x \left (-x^3+5 x^4\right )\right ) \log \left (x^2-5 x^3\right )} \, dx=\frac {4 \log ^2\left (\frac {3-\log \left (x \left (x-5 x^2\right )\right )}{\left (2-e^x\right ) x}\right )}{x^2} \] Output:
4*ln((3-ln((-5*x^2+x)*x))/x/(-exp(x)+2))^2/x^2
\[ \int \frac {\left (80-480 x+e^x \left (-40+216 x+120 x^2\right )+\left (-16+80 x+e^x \left (8-32 x-40 x^2\right )\right ) \log \left (x^2-5 x^3\right )\right ) \log \left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )+\left (48-240 x+e^x (-24+120 x)+\left (-16+e^x (8-40 x)+80 x\right ) \log \left (x^2-5 x^3\right )\right ) \log ^2\left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )}{-6 x^3+30 x^4+e^x \left (3 x^3-15 x^4\right )+\left (2 x^3-10 x^4+e^x \left (-x^3+5 x^4\right )\right ) \log \left (x^2-5 x^3\right )} \, dx=\int \frac {\left (80-480 x+e^x \left (-40+216 x+120 x^2\right )+\left (-16+80 x+e^x \left (8-32 x-40 x^2\right )\right ) \log \left (x^2-5 x^3\right )\right ) \log \left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )+\left (48-240 x+e^x (-24+120 x)+\left (-16+e^x (8-40 x)+80 x\right ) \log \left (x^2-5 x^3\right )\right ) \log ^2\left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )}{-6 x^3+30 x^4+e^x \left (3 x^3-15 x^4\right )+\left (2 x^3-10 x^4+e^x \left (-x^3+5 x^4\right )\right ) \log \left (x^2-5 x^3\right )} \, dx \] Input:
Integrate[((80 - 480*x + E^x*(-40 + 216*x + 120*x^2) + (-16 + 80*x + E^x*( 8 - 32*x - 40*x^2))*Log[x^2 - 5*x^3])*Log[(-3 + Log[x^2 - 5*x^3])/(-2*x + E^x*x)] + (48 - 240*x + E^x*(-24 + 120*x) + (-16 + E^x*(8 - 40*x) + 80*x)* Log[x^2 - 5*x^3])*Log[(-3 + Log[x^2 - 5*x^3])/(-2*x + E^x*x)]^2)/(-6*x^3 + 30*x^4 + E^x*(3*x^3 - 15*x^4) + (2*x^3 - 10*x^4 + E^x*(-x^3 + 5*x^4))*Log [x^2 - 5*x^3]),x]
Output:
Integrate[((80 - 480*x + E^x*(-40 + 216*x + 120*x^2) + (-16 + 80*x + E^x*( 8 - 32*x - 40*x^2))*Log[x^2 - 5*x^3])*Log[(-3 + Log[x^2 - 5*x^3])/(-2*x + E^x*x)] + (48 - 240*x + E^x*(-24 + 120*x) + (-16 + E^x*(8 - 40*x) + 80*x)* Log[x^2 - 5*x^3])*Log[(-3 + Log[x^2 - 5*x^3])/(-2*x + E^x*x)]^2)/(-6*x^3 + 30*x^4 + E^x*(3*x^3 - 15*x^4) + (2*x^3 - 10*x^4 + E^x*(-x^3 + 5*x^4))*Log [x^2 - 5*x^3]), x]
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.
| \(\displaystyle \int \frac {\left (\left (e^x (8-40 x)+80 x-16\right ) \log \left (x^2-5 x^3\right )-240 x+e^x (120 x-24)+48\right ) \log ^2\left (\frac {\log \left (x^2-5 x^3\right )-3}{e^x x-2 x}\right )+\left (e^x \left (120 x^2+216 x-40\right )+\left (e^x \left (-40 x^2-32 x+8\right )+80 x-16\right ) \log \left (x^2-5 x^3\right )-480 x+80\right ) \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{e^x x-2 x}\right )}{30 x^4-6 x^3+e^x \left (3 x^3-15 x^4\right )+\left (-10 x^4+2 x^3+e^x \left (5 x^4-x^3\right )\right ) \log \left (x^2-5 x^3\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {-\left (\left (\left (e^x (8-40 x)+80 x-16\right ) \log \left (x^2-5 x^3\right )-240 x+e^x (120 x-24)+48\right ) \log ^2\left (\frac {\log \left (x^2-5 x^3\right )-3}{e^x x-2 x}\right )\right )-\left (e^x \left (120 x^2+216 x-40\right )+\left (e^x \left (-40 x^2-32 x+8\right )+80 x-16\right ) \log \left (x^2-5 x^3\right )-480 x+80\right ) \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{e^x x-2 x}\right )}{\left (2-e^x\right ) (1-5 x) x^3 \left (3-\log \left (x^2-5 x^3\right )\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {8 \left (-15 x^2+5 x^2 \log \left (x^2-5 x^3\right )+4 x \log \left (x^2-5 x^3\right )+5 x \log \left (x^2-5 x^3\right ) \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )-15 x \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )-\log \left (x^2-5 x^3\right )-\log \left (x^2-5 x^3\right ) \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )+3 \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )-27 x+5\right ) \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )}{x^3 (5 x-1) \left (\log \left (x^2-5 x^3\right )-3\right )}-\frac {16 \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )}{\left (e^x-2\right ) x^2}\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (-\frac {8 \left (-15 x^2+5 x^2 \log \left (x^2-5 x^3\right )+4 x \log \left (x^2-5 x^3\right )+5 x \log \left (x^2-5 x^3\right ) \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )-15 x \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )-\log \left (x^2-5 x^3\right )-\log \left (x^2-5 x^3\right ) \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )+3 \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )-27 x+5\right ) \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )}{x^3 (5 x-1) \left (\log \left (x^2-5 x^3\right )-3\right )}-\frac {16 \log \left (\frac {\log \left (x^2-5 x^3\right )-3}{\left (e^x-2\right ) x}\right )}{\left (e^x-2\right ) x^2}\right )dx\) |
Input:
Int[((80 - 480*x + E^x*(-40 + 216*x + 120*x^2) + (-16 + 80*x + E^x*(8 - 32 *x - 40*x^2))*Log[x^2 - 5*x^3])*Log[(-3 + Log[x^2 - 5*x^3])/(-2*x + E^x*x) ] + (48 - 240*x + E^x*(-24 + 120*x) + (-16 + E^x*(8 - 40*x) + 80*x)*Log[x^ 2 - 5*x^3])*Log[(-3 + Log[x^2 - 5*x^3])/(-2*x + E^x*x)]^2)/(-6*x^3 + 30*x^ 4 + E^x*(3*x^3 - 15*x^4) + (2*x^3 - 10*x^4 + E^x*(-x^3 + 5*x^4))*Log[x^2 - 5*x^3]),x]
Output:
$Aborted
Time = 28.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89
| method | result | size |
| parallelrisch | \(\frac {4 {\ln \left (\frac {\ln \left (-5 x^{3}+x^{2}\right )-3}{x \left ({\mathrm e}^{x}-2\right )}\right )}^{2}}{x^{2}}\) | \(31\) |
| risch | \(\text {Expression too large to display}\) | \(47593\) |
Input:
int(((((-40*x+8)*exp(x)+80*x-16)*ln(-5*x^3+x^2)+(120*x-24)*exp(x)-240*x+48 )*ln((ln(-5*x^3+x^2)-3)/(exp(x)*x-2*x))^2+(((-40*x^2-32*x+8)*exp(x)+80*x-1 6)*ln(-5*x^3+x^2)+(120*x^2+216*x-40)*exp(x)-480*x+80)*ln((ln(-5*x^3+x^2)-3 )/(exp(x)*x-2*x)))/(((5*x^4-x^3)*exp(x)-10*x^4+2*x^3)*ln(-5*x^3+x^2)+(-15* x^4+3*x^3)*exp(x)+30*x^4-6*x^3),x,method=_RETURNVERBOSE)
Output:
4*ln((ln(-5*x^3+x^2)-3)/x/(exp(x)-2))^2/x^2
Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {\left (80-480 x+e^x \left (-40+216 x+120 x^2\right )+\left (-16+80 x+e^x \left (8-32 x-40 x^2\right )\right ) \log \left (x^2-5 x^3\right )\right ) \log \left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )+\left (48-240 x+e^x (-24+120 x)+\left (-16+e^x (8-40 x)+80 x\right ) \log \left (x^2-5 x^3\right )\right ) \log ^2\left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )}{-6 x^3+30 x^4+e^x \left (3 x^3-15 x^4\right )+\left (2 x^3-10 x^4+e^x \left (-x^3+5 x^4\right )\right ) \log \left (x^2-5 x^3\right )} \, dx=\frac {4 \, \log \left (\frac {\log \left (-5 \, x^{3} + x^{2}\right ) - 3}{x e^{x} - 2 \, x}\right )^{2}}{x^{2}} \] Input:
integrate(((((-40*x+8)*exp(x)+80*x-16)*log(-5*x^3+x^2)+(120*x-24)*exp(x)-2 40*x+48)*log((log(-5*x^3+x^2)-3)/(exp(x)*x-2*x))^2+(((-40*x^2-32*x+8)*exp( x)+80*x-16)*log(-5*x^3+x^2)+(120*x^2+216*x-40)*exp(x)-480*x+80)*log((log(- 5*x^3+x^2)-3)/(exp(x)*x-2*x)))/(((5*x^4-x^3)*exp(x)-10*x^4+2*x^3)*log(-5*x ^3+x^2)+(-15*x^4+3*x^3)*exp(x)+30*x^4-6*x^3),x, algorithm="fricas")
Output:
4*log((log(-5*x^3 + x^2) - 3)/(x*e^x - 2*x))^2/x^2
Time = 21.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {\left (80-480 x+e^x \left (-40+216 x+120 x^2\right )+\left (-16+80 x+e^x \left (8-32 x-40 x^2\right )\right ) \log \left (x^2-5 x^3\right )\right ) \log \left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )+\left (48-240 x+e^x (-24+120 x)+\left (-16+e^x (8-40 x)+80 x\right ) \log \left (x^2-5 x^3\right )\right ) \log ^2\left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )}{-6 x^3+30 x^4+e^x \left (3 x^3-15 x^4\right )+\left (2 x^3-10 x^4+e^x \left (-x^3+5 x^4\right )\right ) \log \left (x^2-5 x^3\right )} \, dx=\frac {4 \log {\left (\frac {\log {\left (- 5 x^{3} + x^{2} \right )} - 3}{x e^{x} - 2 x} \right )}^{2}}{x^{2}} \] Input:
integrate(((((-40*x+8)*exp(x)+80*x-16)*ln(-5*x**3+x**2)+(120*x-24)*exp(x)- 240*x+48)*ln((ln(-5*x**3+x**2)-3)/(exp(x)*x-2*x))**2+(((-40*x**2-32*x+8)*e xp(x)+80*x-16)*ln(-5*x**3+x**2)+(120*x**2+216*x-40)*exp(x)-480*x+80)*ln((l n(-5*x**3+x**2)-3)/(exp(x)*x-2*x)))/(((5*x**4-x**3)*exp(x)-10*x**4+2*x**3) *ln(-5*x**3+x**2)+(-15*x**4+3*x**3)*exp(x)+30*x**4-6*x**3),x)
Output:
4*log((log(-5*x**3 + x**2) - 3)/(x*exp(x) - 2*x))**2/x**2
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.83 \[ \int \frac {\left (80-480 x+e^x \left (-40+216 x+120 x^2\right )+\left (-16+80 x+e^x \left (8-32 x-40 x^2\right )\right ) \log \left (x^2-5 x^3\right )\right ) \log \left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )+\left (48-240 x+e^x (-24+120 x)+\left (-16+e^x (8-40 x)+80 x\right ) \log \left (x^2-5 x^3\right )\right ) \log ^2\left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )}{-6 x^3+30 x^4+e^x \left (3 x^3-15 x^4\right )+\left (2 x^3-10 x^4+e^x \left (-x^3+5 x^4\right )\right ) \log \left (x^2-5 x^3\right )} \, dx=\frac {4 \, {\left (\log \left (x\right )^{2} + 2 \, \log \left (x\right ) \log \left (e^{x} - 2\right ) + \log \left (e^{x} - 2\right )^{2} - 2 \, {\left (\log \left (x\right ) + \log \left (e^{x} - 2\right )\right )} \log \left (2 \, \log \left (x\right ) + \log \left (-5 \, x + 1\right ) - 3\right ) + \log \left (2 \, \log \left (x\right ) + \log \left (-5 \, x + 1\right ) - 3\right )^{2}\right )}}{x^{2}} \] Input:
integrate(((((-40*x+8)*exp(x)+80*x-16)*log(-5*x^3+x^2)+(120*x-24)*exp(x)-2 40*x+48)*log((log(-5*x^3+x^2)-3)/(exp(x)*x-2*x))^2+(((-40*x^2-32*x+8)*exp( x)+80*x-16)*log(-5*x^3+x^2)+(120*x^2+216*x-40)*exp(x)-480*x+80)*log((log(- 5*x^3+x^2)-3)/(exp(x)*x-2*x)))/(((5*x^4-x^3)*exp(x)-10*x^4+2*x^3)*log(-5*x ^3+x^2)+(-15*x^4+3*x^3)*exp(x)+30*x^4-6*x^3),x, algorithm="maxima")
Output:
4*(log(x)^2 + 2*log(x)*log(e^x - 2) + log(e^x - 2)^2 - 2*(log(x) + log(e^x - 2))*log(2*log(x) + log(-5*x + 1) - 3) + log(2*log(x) + log(-5*x + 1) - 3)^2)/x^2
\[ \int \frac {\left (80-480 x+e^x \left (-40+216 x+120 x^2\right )+\left (-16+80 x+e^x \left (8-32 x-40 x^2\right )\right ) \log \left (x^2-5 x^3\right )\right ) \log \left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )+\left (48-240 x+e^x (-24+120 x)+\left (-16+e^x (8-40 x)+80 x\right ) \log \left (x^2-5 x^3\right )\right ) \log ^2\left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )}{-6 x^3+30 x^4+e^x \left (3 x^3-15 x^4\right )+\left (2 x^3-10 x^4+e^x \left (-x^3+5 x^4\right )\right ) \log \left (x^2-5 x^3\right )} \, dx=\int { \frac {8 \, {\left ({\left (3 \, {\left (5 \, x - 1\right )} e^{x} - {\left ({\left (5 \, x - 1\right )} e^{x} - 10 \, x + 2\right )} \log \left (-5 \, x^{3} + x^{2}\right ) - 30 \, x + 6\right )} \log \left (\frac {\log \left (-5 \, x^{3} + x^{2}\right ) - 3}{x e^{x} - 2 \, x}\right )^{2} + {\left ({\left (15 \, x^{2} + 27 \, x - 5\right )} e^{x} - {\left ({\left (5 \, x^{2} + 4 \, x - 1\right )} e^{x} - 10 \, x + 2\right )} \log \left (-5 \, x^{3} + x^{2}\right ) - 60 \, x + 10\right )} \log \left (\frac {\log \left (-5 \, x^{3} + x^{2}\right ) - 3}{x e^{x} - 2 \, x}\right )\right )}}{30 \, x^{4} - 6 \, x^{3} - 3 \, {\left (5 \, x^{4} - x^{3}\right )} e^{x} - {\left (10 \, x^{4} - 2 \, x^{3} - {\left (5 \, x^{4} - x^{3}\right )} e^{x}\right )} \log \left (-5 \, x^{3} + x^{2}\right )} \,d x } \] Input:
integrate(((((-40*x+8)*exp(x)+80*x-16)*log(-5*x^3+x^2)+(120*x-24)*exp(x)-2 40*x+48)*log((log(-5*x^3+x^2)-3)/(exp(x)*x-2*x))^2+(((-40*x^2-32*x+8)*exp( x)+80*x-16)*log(-5*x^3+x^2)+(120*x^2+216*x-40)*exp(x)-480*x+80)*log((log(- 5*x^3+x^2)-3)/(exp(x)*x-2*x)))/(((5*x^4-x^3)*exp(x)-10*x^4+2*x^3)*log(-5*x ^3+x^2)+(-15*x^4+3*x^3)*exp(x)+30*x^4-6*x^3),x, algorithm="giac")
Output:
integrate(8*((3*(5*x - 1)*e^x - ((5*x - 1)*e^x - 10*x + 2)*log(-5*x^3 + x^ 2) - 30*x + 6)*log((log(-5*x^3 + x^2) - 3)/(x*e^x - 2*x))^2 + ((15*x^2 + 2 7*x - 5)*e^x - ((5*x^2 + 4*x - 1)*e^x - 10*x + 2)*log(-5*x^3 + x^2) - 60*x + 10)*log((log(-5*x^3 + x^2) - 3)/(x*e^x - 2*x)))/(30*x^4 - 6*x^3 - 3*(5* x^4 - x^3)*e^x - (10*x^4 - 2*x^3 - (5*x^4 - x^3)*e^x)*log(-5*x^3 + x^2)), x)
Time = 1.78 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {\left (80-480 x+e^x \left (-40+216 x+120 x^2\right )+\left (-16+80 x+e^x \left (8-32 x-40 x^2\right )\right ) \log \left (x^2-5 x^3\right )\right ) \log \left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )+\left (48-240 x+e^x (-24+120 x)+\left (-16+e^x (8-40 x)+80 x\right ) \log \left (x^2-5 x^3\right )\right ) \log ^2\left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )}{-6 x^3+30 x^4+e^x \left (3 x^3-15 x^4\right )+\left (2 x^3-10 x^4+e^x \left (-x^3+5 x^4\right )\right ) \log \left (x^2-5 x^3\right )} \, dx=\frac {4\,{\ln \left (-\frac {\ln \left (x^2-5\,x^3\right )-3}{2\,x-x\,{\mathrm {e}}^x}\right )}^2}{x^2} \] Input:
int(-(log(-(log(x^2 - 5*x^3) - 3)/(2*x - x*exp(x)))^2*(240*x - exp(x)*(120 *x - 24) + log(x^2 - 5*x^3)*(exp(x)*(40*x - 8) - 80*x + 16) - 48) + log(-( log(x^2 - 5*x^3) - 3)/(2*x - x*exp(x)))*(480*x + log(x^2 - 5*x^3)*(exp(x)* (32*x + 40*x^2 - 8) - 80*x + 16) - exp(x)*(216*x + 120*x^2 - 40) - 80))/(e xp(x)*(3*x^3 - 15*x^4) - log(x^2 - 5*x^3)*(exp(x)*(x^3 - 5*x^4) - 2*x^3 + 10*x^4) - 6*x^3 + 30*x^4),x)
Output:
(4*log(-(log(x^2 - 5*x^3) - 3)/(2*x - x*exp(x)))^2)/x^2
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int \frac {\left (80-480 x+e^x \left (-40+216 x+120 x^2\right )+\left (-16+80 x+e^x \left (8-32 x-40 x^2\right )\right ) \log \left (x^2-5 x^3\right )\right ) \log \left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )+\left (48-240 x+e^x (-24+120 x)+\left (-16+e^x (8-40 x)+80 x\right ) \log \left (x^2-5 x^3\right )\right ) \log ^2\left (\frac {-3+\log \left (x^2-5 x^3\right )}{-2 x+e^x x}\right )}{-6 x^3+30 x^4+e^x \left (3 x^3-15 x^4\right )+\left (2 x^3-10 x^4+e^x \left (-x^3+5 x^4\right )\right ) \log \left (x^2-5 x^3\right )} \, dx=\frac {4 {\mathrm {log}\left (\frac {\mathrm {log}\left (-5 x^{3}+x^{2}\right )-3}{e^{x} x -2 x}\right )}^{2}}{x^{2}} \] Input:
int(((((-40*x+8)*exp(x)+80*x-16)*log(-5*x^3+x^2)+(120*x-24)*exp(x)-240*x+4 8)*log((log(-5*x^3+x^2)-3)/(exp(x)*x-2*x))^2+(((-40*x^2-32*x+8)*exp(x)+80* x-16)*log(-5*x^3+x^2)+(120*x^2+216*x-40)*exp(x)-480*x+80)*log((log(-5*x^3+ x^2)-3)/(exp(x)*x-2*x)))/(((5*x^4-x^3)*exp(x)-10*x^4+2*x^3)*log(-5*x^3+x^2 )+(-15*x^4+3*x^3)*exp(x)+30*x^4-6*x^3),x)
Output:
(4*log((log( - 5*x**3 + x**2) - 3)/(e**x*x - 2*x))**2)/x**2