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

Optimal. Leaf size=25 \[ 5-e^{\frac {-1+\log \left (\frac {4}{x}\right )}{-5+2 e^x+x}} \]

________________________________________________________________________________________

Rubi [F]  time = 9.14, 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 {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 \end {gather*}

Verification is not applicable to the result.

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

________________________________________________________________________________________

Mathematica [F]  time = 3.45, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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]

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 21, normalized size = 0.84 \begin {gather*} -e^{\left (\frac {\log \left (\frac {4}{x}\right ) - 1}{x + 2 \, e^{x} - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+x)*log(4/x)+(-2*x+2)*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))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+x)*log(4/x)+(-2*x+2)*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)

________________________________________________________________________________________

maple [A]  time = 0.20, size = 24, normalized size = 0.96

method	result	size
risch	\(-{\mathrm e}^{\frac {2 \ln \relax (2)-\ln \relax (x )-1}{2 \,{\mathrm e}^{x}+x -5}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

maxima [A]  time = 1.07, size = 41, normalized size = 1.64 \begin {gather*} -e^{\left (\frac {2 \, \log \relax (2)}{x + 2 \, e^{x} - 5} - \frac {\log \relax (x)}{x + 2 \, e^{x} - 5} - \frac {1}{x + 2 \, e^{x} - 5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+x)*log(4/x)+(-2*x+2)*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))

________________________________________________________________________________________

mupad [B]  time = 2.26, size = 29, normalized size = 1.16 \begin {gather*} -{\mathrm {e}}^{-\frac {1}{x+2\,{\mathrm {e}}^x-5}}\,{\left (\frac {4}{x}\right )}^{\frac {1}{x+2\,{\mathrm {e}}^x-5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

________________________________________________________________________________________

sympy [A]  time = 0.93, size = 17, normalized size = 0.68 \begin {gather*} - e^{\frac {\log {\left (\frac {4}{x} \right )} - 1}{x + 2 e^{x} - 5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x)*x+x)*ln(4/x)+(-2*x+2)*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))

________________________________________________________________________________________