\(\int \frac {x-x^2+(1250-2500 x+e^x (-2500 x+2500 x^2)) \log (-\frac {e^{2 e^x}}{-x+x^2})}{-5 x+5 x^2+e (-4 x+4 x^2)} \, dx\) [6975]

Optimal result

Integrand size = 68, antiderivative size = 34 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=\frac {-x+625 \log ^2\left (\frac {e^{2 e^x}}{x-x^2}\right )}{5+4 e} \]

[Out]

(625*ln(exp(exp(x))^2/(-x^2+x))^2-x)/(5+4*exp(1))

Rubi [F]

\[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=\int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx \]

[In]

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]

[Out]

(-2500*E^(2*x))/(5 + 4*E) - x/(5 + 4*E) + (2500*E*ExpIntegralEi[-1 + x])/(5 + 4*E) + (2500*ExpIntegralEi[x])/(
5 + 4*E) + (2500*E^x*Log[E^(2*E^x)/((1 - x)*x)])/(5 + 4*E) + (1250*Defer[Int][Log[E^(2*E^x)/((1 - x)*x)]/(1 -
x), x])/(5 + 4*E) - (1250*Defer[Int][Log[E^(2*E^x)/((1 - x)*x)]/x, x])/(5 + 4*E)

Rubi steps \begin{align*} \text {integral}& = \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 )}{x (5+4 e-(5+4 e) x)} \, dx \\ & = \int \left (\frac {1}{(5+4 e) (-1+x)}-\frac {x}{(5+4 e) (-1+x)}+\frac {2500 e^x \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{5+4 e}+\frac {2500 \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{(5+4 e) (1-x)}+\frac {1250 \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{(-5-4 e) (1-x) x}\right ) \, dx \\ & = \frac {\log (1-x)}{5+4 e}-\frac {\int \frac {x}{-1+x} \, dx}{5+4 e}-\frac {1250 \int \frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{(1-x) x} \, dx}{5+4 e}+\frac {2500 \int e^x \log \left (\frac {e^{2 e^x}}{(1-x) x}\right ) \, dx}{5+4 e}+\frac {2500 \int \frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{1-x} \, dx}{5+4 e} \\ & = \frac {\log (1-x)}{5+4 e}+\frac {2500 e^x \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{5+4 e}-\frac {\int \left (1+\frac {1}{-1+x}\right ) \, dx}{5+4 e}-\frac {1250 \int \left (\frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{1-x}+\frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{x}\right ) \, dx}{5+4 e}-\frac {2500 \int \frac {e^x \left (-1+2 \left (1+e^x\right ) x-2 e^x x^2\right )}{(1-x) x} \, dx}{5+4 e}+\frac {2500 \int \frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{1-x} \, dx}{5+4 e} \\ & = -\frac {x}{5+4 e}+\frac {2500 e^x \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{5+4 e}-\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 \int \left (2 e^{2 x}+\frac {e^x (1-2 x)}{(-1+x) x}\right ) \, dx}{5+4 e}+\frac {2500 \int \frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{1-x} \, dx}{5+4 e} \\ & = -\frac {x}{5+4 e}+\frac {2500 e^x \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{5+4 e}-\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 \int \frac {e^x (1-2 x)}{(-1+x) x} \, dx}{5+4 e}+\frac {2500 \int \frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{1-x} \, dx}{5+4 e}-\frac {5000 \int e^{2 x} \, dx}{5+4 e} \\ & = -\frac {2500 e^{2 x}}{5+4 e}-\frac {x}{5+4 e}+\frac {2500 e^x \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{5+4 e}-\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 \int \left (\frac {e^x}{1-x}-\frac {e^x}{x}\right ) \, dx}{5+4 e}+\frac {2500 \int \frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{1-x} \, dx}{5+4 e} \\ & = -\frac {2500 e^{2 x}}{5+4 e}-\frac {x}{5+4 e}+\frac {2500 e^x \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{5+4 e}-\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 \int \frac {e^x}{1-x} \, dx}{5+4 e}+\frac {2500 \int \frac {e^x}{x} \, dx}{5+4 e}+\frac {2500 \int \frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{1-x} \, dx}{5+4 e} \\ & = -\frac {2500 e^{2 x}}{5+4 e}-\frac {x}{5+4 e}+\frac {2500 e \text {Ei}(-1+x)}{5+4 e}+\frac {2500 \text {Ei}(x)}{5+4 e}+\frac {2500 e^x \log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{5+4 e}-\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 \int \frac {\log \left (\frac {e^{2 e^x}}{(1-x) x}\right )}{1-x} \, dx}{5+4 e} \\ \end{align*}

Mathematica [B] (verified)

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 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=\frac {-2500 e^{2 x}-x+625 \log ^2\left (\frac {1}{(-1+x) x}\right )+2500 e^x \log (x)+1250 \log \left (\frac {1}{(-1+x) x}\right ) \log (x)+1250 \log (1-x) \left (2 e^x+\log \left (\frac {1}{(-1+x) x}\right )-\log \left (\frac {e^{2 e^x}}{x-x^2}\right )\right )+2500 e^x \log \left (\frac {e^{2 e^x}}{x-x^2}\right )-1250 \log (x) \log \left (\frac {e^{2 e^x}}{x-x^2}\right )}{5+4 e} \]

[In]

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]

[Out]

(-2500*E^(2*x) - x + 625*Log[1/((-1 + x)*x)]^2 + 2500*E^x*Log[x] + 1250*Log[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)

Maple [A] (verified)

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	\(\frac {\left (2500 \ln \left (-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x^{2}-x}\right )+2500 \ln \left (x^{2}-x \right )-5000 \,{\mathrm e}^{x}\right ) {\mathrm e}^{x}-2500 \ln \left (x^{2}-x \right ) {\mathrm e}^{x}}{5+4 \,{\mathrm e}}+\frac {2500 \,{\mathrm e}^{2 x}}{5+4 \,{\mathrm e}}-\frac {x +\left (1250 \ln \left (-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x^{2}-x}\right )+1250 \ln \left (x^{2}-x \right )-2500 \,{\mathrm e}^{x}\right ) \ln \left (x \right )+\left (1250 \ln \left (-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x^{2}-x}\right )+1250 \ln \left (x^{2}-x \right )-2500 \,{\mathrm e}^{x}\right ) \ln \left (-1+x \right )}{5+4 \,{\mathrm e}}+\frac {1250 \ln \left (x \right ) \ln \left (x^{2}-x \right )-625 \ln \left (x \right )^{2}+1250 \operatorname {dilog}\left (x \right )}{5+4 \,{\mathrm e}}+\frac {1250 \ln \left (-1+x \right ) \ln \left (x^{2}-x \right )-1250 \operatorname {dilog}\left (x \right )-1250 \ln \left (x \right ) \ln \left (-1+x \right )-625 \ln \left (-1+x \right )^{2}}{5+4 \,{\mathrm e}}\)	\(237\)

[In]

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)

[Out]

(-2+625*ln(-exp(exp(x))^2/x/(-1+x))^2-x)/(5+4*exp(1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=\frac {625 \, \log \left (-\frac {e^{\left (2 \, e^{x}\right )}}{x^{2} - x}\right )^{2} - x}{4 \, e + 5} \]

[In]

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="fricas")

[Out]

(625*log(-e^(2*e^x)/(x^2 - x))^2 - x)/(4*e + 5)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=- \frac {x}{5 + 4 e} + \frac {625 \log {\left (- \frac {e^{2 e^{x}}}{x^{2} - x} \right )}^{2}}{5 + 4 e} \]

[In]

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)

[Out]

-x/(5 + 4*E) + 625*log(-exp(2*exp(x))/(x**2 - x))**2/(5 + 4*E)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.00 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=-\frac {625 \, {\left (4 \, e^{x} \log \left (x\right ) - \log \left (x\right )^{2} + 2 \, {\left (2 \, e^{x} - \log \left (x\right )\right )} \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} - 4 \, e^{\left (2 \, x\right )}\right )}}{4 \, e + 5} - \frac {x}{4 \, e + 5} \]

[In]

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="maxima")

[Out]

-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)

Giac [F]

\[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=\int { -\frac {x^{2} - 1250 \, {\left (2 \, {\left (x^{2} - x\right )} e^{x} - 2 \, x + 1\right )} \log \left (-\frac {e^{\left (2 \, e^{x}\right )}}{x^{2} - x}\right ) - x}{5 \, x^{2} + 4 \, {\left (x^{2} - x\right )} e - 5 \, x} \,d x } \]

[In]

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="giac")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 12.69 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.18 \[ \int \frac {x-x^2+\left (1250-2500 x+e^x \left (-2500 x+2500 x^2\right )\right ) \log \left (-\frac {e^{2 e^x}}{-x+x^2}\right )}{-5 x+5 x^2+e \left (-4 x+4 x^2\right )} \, dx=\frac {625\,{\ln \left (\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^x}}{x-x^2}\right )}^2}{4\,\mathrm {e}+5}-\frac {x}{4\,\mathrm {e}+5} \]

[In]

int((log(exp(2*exp(x))/(x - x^2))*(2500*x + exp(x)*(2500*x - 2500*x^2) - 1250) - x + x^2)/(5*x + exp(1)*(4*x -
4*x^2) - 5*x^2),x)

[Out]

(625*log(exp(2*exp(x))/(x - x^2))^2)/(4*exp(1) + 5) - x/(4*exp(1) + 5)