Integrand size = 68, antiderivative size = 34 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=\frac {-x+625 \log ^2\left (\frac {e^{2 e^x}}{x-x^2}\right )}{5+4 e} \]
Leaf count is larger than twice the leaf count of optimal. \(143\) vs. \(2(34)=68\).
Time = 0.32 (sec) , antiderivative size = 143, normalized size of antiderivative = 4.21 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=\frac {-2500 e^{2 x}-x+625 \log ^2\left (\frac {1}{(-1+x) x}\right )+2500 e^x \log (x)+1250 \log \left (\frac {1}{(-1+x) x}\right ) \log (x)+1250 \log (1-x) \left (2 e^x+\log \left (\frac {1}{(-1+x) x}\right )-\log \left (\frac {e^{2 e^x}}{x-x^2}\right )\right )+2500 e^x \log \left (\frac {e^{2 e^x}}{x-x^2}\right )-1250 \log (x) \log \left (\frac {e^{2 e^x}}{x-x^2}\right )}{5+4 e} \]
Integrate[(x - x^2 + (1250 - 2500*x + E^x*(-2500*x + 2500*x^2))*Log[-(E^(2 *E^x)/(-x + x^2))])/(-5*x + 5*x^2 + E*(-4*x + 4*x^2)),x]
(-2500*E^(2*x) - x + 625*Log[1/((-1 + x)*x)]^2 + 2500*E^x*Log[x] + 1250*Lo g[1/((-1 + x)*x)]*Log[x] + 1250*Log[1 - x]*(2*E^x + Log[1/((-1 + x)*x)] - Log[E^(2*E^x)/(x - x^2)]) + 2500*E^x*Log[E^(2*E^x)/(x - x^2)] - 1250*Log[x ]*Log[E^(2*E^x)/(x - x^2)])/(5 + 4*E)
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.
\(\displaystyle \int \frac {-x^2+\left (e^x \left (2500 x^2-2500 x\right )-2500 x+1250\right ) \log \left (-\frac {e^{2 e^x}}{x^2-x}\right )+x}{5 x^2+e \left (4 x^2-4 x\right )-5 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-x^2+\left (e^x \left (2500 x^2-2500 x\right )-2500 x+1250\right ) \log \left (-\frac {e^{2 e^x}}{x^2-x}\right )+x}{x ((5+4 e) x-4 e-5)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {x}{(5+4 e) (x-1)}+\frac {1}{(5+4 e) (x-1)}+\frac {2500 \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{(5+4 e) (1-x)}+\frac {2500 e^x \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{5+4 e}+\frac {1250 \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{(-5-4 e) (1-x) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1250 \int \frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{1-x}dx}{5+4 e}-\frac {1250 \int \frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{x}dx}{5+4 e}+\frac {2500 e \operatorname {ExpIntegralEi}(x-1)}{5+4 e}+\frac {2500 \operatorname {ExpIntegralEi}(x)}{5+4 e}-\frac {x}{5+4 e}-\frac {2500 e^{2 x}}{5+4 e}+\frac {2500 e^x \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{5+4 e}\) |
Int[(x - x^2 + (1250 - 2500*x + E^x*(-2500*x + 2500*x^2))*Log[-(E^(2*E^x)/ (-x + x^2))])/(-5*x + 5*x^2 + E*(-4*x + 4*x^2)),x]
3.10.75.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 3.70 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03
method | result | size |
parallelrisch | \(\frac {-2+625 \ln \left (-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x \left (-1+x \right )}\right )^{2}-x}{5+4 \,{\mathrm e}}\) | \(35\) |
default | \(\frac {\left (2500 \ln \left (-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x^{2}-x}\right )+2500 \ln \left (x^{2}-x \right )-5000 \,{\mathrm e}^{x}\right ) {\mathrm e}^{x}-2500 \ln \left (x^{2}-x \right ) {\mathrm e}^{x}}{5+4 \,{\mathrm e}}+\frac {2500 \,{\mathrm e}^{2 x}}{5+4 \,{\mathrm e}}-\frac {x +\left (1250 \ln \left (-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x^{2}-x}\right )+1250 \ln \left (x^{2}-x \right )-2500 \,{\mathrm e}^{x}\right ) \ln \left (x \right )+\left (1250 \ln \left (-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x^{2}-x}\right )+1250 \ln \left (x^{2}-x \right )-2500 \,{\mathrm e}^{x}\right ) \ln \left (-1+x \right )}{5+4 \,{\mathrm e}}+\frac {1250 \ln \left (x \right ) \ln \left (x^{2}-x \right )-625 \ln \left (x \right )^{2}+1250 \operatorname {dilog}\left (x \right )}{5+4 \,{\mathrm e}}+\frac {1250 \ln \left (-1+x \right ) \ln \left (x^{2}-x \right )-1250 \operatorname {dilog}\left (x \right )-1250 \ln \left (x \right ) \ln \left (-1+x \right )-625 \ln \left (-1+x \right )^{2}}{5+4 \,{\mathrm e}}\) | \(237\) |
int((((2500*x^2-2500*x)*exp(x)-2500*x+1250)*ln(-exp(exp(x))^2/(x^2-x))-x^2 +x)/((4*x^2-4*x)*exp(1)+5*x^2-5*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=\frac {625 \, \log \left (-\frac {e^{\left (2 \, e^{x}\right )}}{x^{2} - x}\right )^{2} - x}{4 \, e + 5} \]
integrate((((2500*x^2-2500*x)*exp(x)-2500*x+1250)*log(-exp(exp(x))^2/(x^2- x))-x^2+x)/((4*x^2-4*x)*exp(1)+5*x^2-5*x),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=- \frac {x}{5 + 4 e} + \frac {625 \log {\left (- \frac {e^{2 e^{x}}}{x^{2} - x} \right )}^{2}}{5 + 4 e} \]
integrate((((2500*x**2-2500*x)*exp(x)-2500*x+1250)*ln(-exp(exp(x))**2/(x** 2-x))-x**2+x)/((4*x**2-4*x)*exp(1)+5*x**2-5*x),x)
Time = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.00 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=-\frac {625 \, {\left (4 \, e^{x} \log \left (x\right ) - \log \left (x\right )^{2} + 2 \, {\left (2 \, e^{x} - \log \left (x\right )\right )} \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} - 4 \, e^{\left (2 \, x\right )}\right )}}{4 \, e + 5} - \frac {x}{4 \, e + 5} \]
integrate((((2500*x^2-2500*x)*exp(x)-2500*x+1250)*log(-exp(exp(x))^2/(x^2- x))-x^2+x)/((4*x^2-4*x)*exp(1)+5*x^2-5*x),x, algorithm=\
-625*(4*e^x*log(x) - log(x)^2 + 2*(2*e^x - log(x))*log(-x + 1) - log(-x + 1)^2 - 4*e^(2*x))/(4*e + 5) - x/(4*e + 5)
\[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=\int { -\frac {x^{2} - 1250 \, {\left (2 \, {\left (x^{2} - x\right )} e^{x} - 2 \, x + 1\right )} \log \left (-\frac {e^{\left (2 \, e^{x}\right )}}{x^{2} - x}\right ) - x}{5 \, x^{2} + 4 \, {\left (x^{2} - x\right )} e - 5 \, x} \,d x } \]
integrate((((2500*x^2-2500*x)*exp(x)-2500*x+1250)*log(-exp(exp(x))^2/(x^2- x))-x^2+x)/((4*x^2-4*x)*exp(1)+5*x^2-5*x),x, algorithm=\
integrate(-(x^2 - 1250*(2*(x^2 - x)*e^x - 2*x + 1)*log(-e^(2*e^x)/(x^2 - x )) - x)/(5*x^2 + 4*(x^2 - x)*e - 5*x), x)
Time = 12.69 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=\frac {625\,{\ln \left (\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^x}}{x-x^2}\right )}^2}{4\,\mathrm {e}+5}-\frac {x}{4\,\mathrm {e}+5} \]