\(\int \frac {e^{\frac {-1+\log (\frac {4}{x})}{-5+2 e^x+x}} (-5+e^x (2-2 x)+(x+2 e^x x) \log (\frac {4}{x}))}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x (-20 x+4 x^2)} \, dx\) [3608]

Optimal result

Integrand size = 83, antiderivative size = 25 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=5-e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx \]

[In]

Int[(E^((-1 + Log[4/x])/(-5 + 2*E^x + x))*(-5 + E^x*(2 - 2*x) + (x + 2*E^x*x)*Log[4/x]))/(25*x + 4*E^(2*x)*x -
10*x^2 + x^3 + E^x*(-20*x + 4*x^2)),x]

[Out]

-6*Defer[Int][E^((-1 + Log[4/x])/(-5 + 2*E^x + x))/(-5 + 2*E^x + x)^2, x] + Defer[Int][(E^((-1 + Log[4/x])/(-5
+ 2*E^x + x))*x)/(-5 + 2*E^x + x)^2, x] - Defer[Int][E^((-1 + Log[4/x])/(-5 + 2*E^x + x))/(-5 + 2*E^x + x), x
] + Defer[Int][E^((-1 + Log[4/x])/(-5 + 2*E^x + x))/(x*(-5 + 2*E^x + x)), x] + 6*Defer[Int][(E^((-1 + Log[4/x]
)/(-5 + 2*E^x + x))*Log[4/x])/(-5 + 2*E^x + x)^2, x] - Defer[Int][(E^((-1 + Log[4/x])/(-5 + 2*E^x + x))*x*Log[
4/x])/(-5 + 2*E^x + x)^2, x] + Defer[Int][(E^((-1 + Log[4/x])/(-5 + 2*E^x + x))*Log[4/x])/(-5 + 2*E^x + x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{\left (5-2 e^x-x\right )^2 x} \, dx \\ & = \int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} (-6+x) \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (1-x+x \log \left (\frac {4}{x}\right )\right )}{x \left (-5+2 e^x+x\right )}\right ) \, dx \\ & = -\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} (-6+x) \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (1-x+x \log \left (\frac {4}{x}\right )\right )}{x \left (-5+2 e^x+x\right )} \, dx \\ & = -\int \left (-\frac {6 e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2}\right ) \, dx+\int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x}\right ) \, dx \\ & = 6 \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \left (-1+\log \left (\frac {4}{x}\right )\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x} \, dx \\ & = 6 \int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2}\right ) \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x} \, dx-\int \left (-\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x}{\left (-5+2 e^x+x\right )^2}+\frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2}\right ) \, dx \\ & = -\left (6 \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{\left (-5+2 e^x+x\right )^2} \, dx\right )+6 \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x}{\left (-5+2 e^x+x\right )^2} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{-5+2 e^x+x} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}}}{x \left (-5+2 e^x+x\right )} \, dx-\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} x \log \left (\frac {4}{x}\right )}{\left (-5+2 e^x+x\right )^2} \, dx+\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \log \left (\frac {4}{x}\right )}{-5+2 e^x+x} \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=\int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx \]

[In]

Integrate[(E^((-1 + Log[4/x])/(-5 + 2*E^x + x))*(-5 + E^x*(2 - 2*x) + (x + 2*E^x*x)*Log[4/x]))/(25*x + 4*E^(2*
x)*x - 10*x^2 + x^3 + E^x*(-20*x + 4*x^2)),x]

[Out]

Integrate[(E^((-1 + Log[4/x])/(-5 + 2*E^x + x))*(-5 + E^x*(2 - 2*x) + (x + 2*E^x*x)*Log[4/x]))/(25*x + 4*E^(2*
x)*x - 10*x^2 + x^3 + E^x*(-20*x + 4*x^2)), x]

Maple [A] (verified)

Time = 3.56 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(-{\mathrm e}^{\frac {\ln \left (\frac {4}{x}\right )-1}{2 \,{\mathrm e}^{x}+x -5}}\)	\(22\)
risch	\(-{\mathrm e}^{\frac {2 \ln \left (2\right )-\ln \left (x \right )-1}{2 \,{\mathrm e}^{x}+x -5}}\)	\(24\)

[In]

int(((2*exp(x)*x+x)*ln(4/x)+(2-2*x)*exp(x)-5)*exp((ln(4/x)-1)/(2*exp(x)+x-5))/(4*x*exp(x)^2+(4*x^2-20*x)*exp(x
)+x^3-10*x^2+25*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=-e^{\left (\frac {\log \left (\frac {4}{x}\right ) - 1}{x + 2 \, e^{x} - 5}\right )} \]

[In]

integrate(((2*exp(x)*x+x)*log(4/x)+(2-2*x)*exp(x)-5)*exp((log(4/x)-1)/(2*exp(x)+x-5))/(4*x*exp(x)^2+(4*x^2-20*
x)*exp(x)+x^3-10*x^2+25*x),x, algorithm="fricas")

[Out]

-e^((log(4/x) - 1)/(x + 2*e^x - 5))

Sympy [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=- e^{\frac {\log {\left (\frac {4}{x} \right )} - 1}{x + 2 e^{x} - 5}} \]

[In]

integrate(((2*exp(x)*x+x)*ln(4/x)+(2-2*x)*exp(x)-5)*exp((ln(4/x)-1)/(2*exp(x)+x-5))/(4*x*exp(x)**2+(4*x**2-20*
x)*exp(x)+x**3-10*x**2+25*x),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=-e^{\left (\frac {2 \, \log \left (2\right )}{x + 2 \, e^{x} - 5} - \frac {\log \left (x\right )}{x + 2 \, e^{x} - 5} - \frac {1}{x + 2 \, e^{x} - 5}\right )} \]

[In]

integrate(((2*exp(x)*x+x)*log(4/x)+(2-2*x)*exp(x)-5)*exp((log(4/x)-1)/(2*exp(x)+x-5))/(4*x*exp(x)^2+(4*x^2-20*
x)*exp(x)+x^3-10*x^2+25*x),x, algorithm="maxima")

[Out]

-e^(2*log(2)/(x + 2*e^x - 5) - log(x)/(x + 2*e^x - 5) - 1/(x + 2*e^x - 5))

Giac [F]

\[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=\int { -\frac {{\left (2 \, {\left (x - 1\right )} e^{x} - {\left (2 \, x e^{x} + x\right )} \log \left (\frac {4}{x}\right ) + 5\right )} e^{\left (\frac {\log \left (\frac {4}{x}\right ) - 1}{x + 2 \, e^{x} - 5}\right )}}{x^{3} - 10 \, x^{2} + 4 \, x e^{\left (2 \, x\right )} + 4 \, {\left (x^{2} - 5 \, x\right )} e^{x} + 25 \, x} \,d x } \]

[In]

integrate(((2*exp(x)*x+x)*log(4/x)+(2-2*x)*exp(x)-5)*exp((log(4/x)-1)/(2*exp(x)+x-5))/(4*x*exp(x)^2+(4*x^2-20*
x)*exp(x)+x^3-10*x^2+25*x),x, algorithm="giac")

[Out]

integrate(-(2*(x - 1)*e^x - (2*x*e^x + x)*log(4/x) + 5)*e^((log(4/x) - 1)/(x + 2*e^x - 5))/(x^3 - 10*x^2 + 4*x
*e^(2*x) + 4*(x^2 - 5*x)*e^x + 25*x), x)

Mupad [B] (verification not implemented)

Time = 9.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \left (-5+e^x (2-2 x)+\left (x+2 e^x x\right ) \log \left (\frac {4}{x}\right )\right )}{25 x+4 e^{2 x} x-10 x^2+x^3+e^x \left (-20 x+4 x^2\right )} \, dx=-{\mathrm {e}}^{-\frac {1}{x+2\,{\mathrm {e}}^x-5}}\,{\left (\frac {4}{x}\right )}^{\frac {1}{x+2\,{\mathrm {e}}^x-5}} \]

[In]

int(-(exp((log(4/x) - 1)/(x + 2*exp(x) - 5))*(exp(x)*(2*x - 2) - log(4/x)*(x + 2*x*exp(x)) + 5))/(25*x + 4*x*e
xp(2*x) - exp(x)*(20*x - 4*x^2) - 10*x^2 + x^3),x)

[Out]

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