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

Optimal. Leaf size=34 \[ \frac {-x+625 \log ^2\left (\frac {e^{2 e^x}}{x-x^2}\right )}{5+4 e} \]

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Rubi [F]  time = 1.40, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [B]  time = 0.23, size = 143, normalized size = 4.21 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.93, size = 34, normalized size = 1.00 \begin {gather*} \frac {625 \, \log \left (-\frac {e^{\left (2 \, e^{x}\right )}}{x^{2} - x}\right )^{2} - x}{4 \, e + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [B]  time = 0.75, size = 334, normalized size = 9.82

method	result	size
default	\(-\frac {2500 \ln \left (x^{2}-x \right ) {\mathrm e}^{x}}{5+4 \,{\mathrm e}}+\frac {5000 \,{\mathrm e}^{x} \left (\ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )-{\mathrm e}^{x}\right )}{5+4 \,{\mathrm e}}+\frac {2500 \left (\ln \left (-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x^{2}-x}\right )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )+\ln \left (x^{2}-x \right )\right ) {\mathrm e}^{x}}{5+4 \,{\mathrm e}}+\frac {2500 \,{\mathrm e}^{2 x}}{5+4 \,{\mathrm e}}-\frac {2500 \ln \relax (x ) \left (\ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )-{\mathrm e}^{x}\right )}{5+4 \,{\mathrm e}}-\frac {1250 \ln \relax (x ) \left (\ln \left (-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x^{2}-x}\right )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )+\ln \left (x^{2}-x \right )\right )}{5+4 \,{\mathrm e}}-\frac {2500 \ln \left (x -1\right ) \left (\ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )-{\mathrm e}^{x}\right )}{5+4 \,{\mathrm e}}-\frac {1250 \ln \left (x -1\right ) \left (\ln \left (-\frac {{\mathrm e}^{2 \,{\mathrm e}^{x}}}{x^{2}-x}\right )-2 \ln \left ({\mathrm e}^{{\mathrm e}^{x}}\right )+\ln \left (x^{2}-x \right )\right )}{5+4 \,{\mathrm e}}-\frac {x}{5+4 \,{\mathrm e}}+\frac {1250 \ln \relax (x ) \ln \left (x^{2}-x \right )}{5+4 \,{\mathrm e}}-\frac {625 \ln \relax (x )^{2}}{5+4 \,{\mathrm e}}+\frac {1250 \ln \left (x -1\right ) \ln \left (x^{2}-x \right )}{5+4 \,{\mathrm e}}-\frac {625 \ln \left (x -1\right )^{2}}{5+4 \,{\mathrm e}}-\frac {1250 \ln \left (x -1\right ) \ln \relax (x )}{5+4 \,{\mathrm e}}\)	\(334\)

Verification of antiderivative is not currently implemented for this CAS.

[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]

-2500/(5+4*exp(1))*ln(x^2-x)*exp(x)+5000/(5+4*exp(1))*exp(x)*(ln(exp(exp(x)))-exp(x))+2500*(ln(-exp(exp(x))^2/
(x^2-x))-2*ln(exp(exp(x)))+ln(x^2-x))/(5+4*exp(1))*exp(x)+2500/(5+4*exp(1))*exp(x)^2-2500/(5+4*exp(1))*ln(x)*(
ln(exp(exp(x)))-exp(x))-1250/(5+4*exp(1))*ln(x)*(ln(-exp(exp(x))^2/(x^2-x))-2*ln(exp(exp(x)))+ln(x^2-x))-2500/
(5+4*exp(1))*ln(x-1)*(ln(exp(exp(x)))-exp(x))-1250/(5+4*exp(1))*ln(x-1)*(ln(-exp(exp(x))^2/(x^2-x))-2*ln(exp(e
xp(x)))+ln(x^2-x))-1/(5+4*exp(1))*x+1250/(5+4*exp(1))*ln(x)*ln(x^2-x)-625/(5+4*exp(1))*ln(x)^2+1250/(5+4*exp(1
))*ln(x-1)*ln(x^2-x)-625/(5+4*exp(1))*ln(x-1)^2-1250/(5+4*exp(1))*ln(x-1)*ln(x)

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maxima [A]  time = 0.43, size = 68, normalized size = 2.00 \begin {gather*} -\frac {625 \, {\left (4 \, e^{x} \log \relax (x) - \log \relax (x)^{2} + 2 \, {\left (2 \, e^{x} - \log \relax (x)\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.46, size = 40, normalized size = 1.18 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

Timed out

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