3.67.34 \(\int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5))))}{(-4 x^2+2 x^3+x^2 \log (5)) \log ^2(4-2 x-\log (5))} \, dx\)

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

________________________________________________________________________________________

Rubi [F]  time = 8.41, 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 {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-8 + x - 2*Log[x] + 10*x*Log[4 - 2*x - Log[5]])/(2*x*Log[4 - 2*x - Log[5]]))*(8*x - x^2 + (-12 + 6*x
+ 3*Log[5])*Log[4 - 2*x - Log[5]] + Log[x]*(2*x + (-4 + 2*x + Log[5])*Log[4 - 2*x - Log[5]])))/((-4*x^2 + 2*x^
3 + x^2*Log[5])*Log[4 - 2*x - Log[5]]^2),x]

[Out]

(-8*Defer[Int][E^((-8 + x - 2*Log[x] + 10*x*Log[4 - 2*x - Log[5]])/(2*x*Log[4 - 2*x - Log[5]]))/(x*Log[4 - 2*x
- Log[5]]^2), x])/(4 - Log[5]) - Defer[Int][E^((-8 + x - 2*Log[x] + 10*x*Log[4 - 2*x - Log[5]])/(2*x*Log[4 -
2*x - Log[5]]))/((-4 + 2*x + Log[5])*Log[4 - 2*x - Log[5]]^2), x] + (16*Defer[Int][E^((-8 + x - 2*Log[x] + 10*
x*Log[4 - 2*x - Log[5]])/(2*x*Log[4 - 2*x - Log[5]]))/((-4 + 2*x + Log[5])*Log[4 - 2*x - Log[5]]^2), x])/(4 -
Log[5]) - (2*Defer[Int][(E^((-8 + x - 2*Log[x] + 10*x*Log[4 - 2*x - Log[5]])/(2*x*Log[4 - 2*x - Log[5]]))*Log[
x])/(x*Log[4 - 2*x - Log[5]]^2), x])/(4 - Log[5]) + (4*Defer[Int][(E^((-8 + x - 2*Log[x] + 10*x*Log[4 - 2*x -
Log[5]])/(2*x*Log[4 - 2*x - Log[5]]))*Log[x])/((-4 + 2*x + Log[5])*Log[4 - 2*x - Log[5]]^2), x])/(4 - Log[5])
+ 3*Defer[Int][E^((-8 + x - 2*Log[x] + 10*x*Log[4 - 2*x - Log[5]])/(2*x*Log[4 - 2*x - Log[5]]))/(x^2*Log[4 - 2
*x - Log[5]]), x] + Defer[Int][(E^((-8 + x - 2*Log[x] + 10*x*Log[4 - 2*x - Log[5]])/(2*x*Log[4 - 2*x - Log[5]]
))*Log[x])/(x^2*Log[4 - 2*x - Log[5]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (2 x^3+x^2 (-4+\log (5))\right ) \log ^2(4-2 x-\log (5))} \, dx\\ &=\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{x^2 (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx\\ &=\int \left (\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (3+\log (x))}{x^2 \log (4-2 x-\log (5))}\right ) \, dx\\ &=\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (3+\log (x))}{x^2 \log (4-2 x-\log (5))} \, dx\\ &=\int \left (\frac {2 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (-8+x-2 \log (x))}{(-4+\log (5)) (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x (-4+\log (5)) \log ^2(4-2 x-\log (5))}\right ) \, dx+\int \left (\frac {3 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))}\right ) \, dx\\ &=3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (-8+x-2 \log (x))}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (8-x+2 \log (x))}{x \log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx\\ &=3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \left (-\frac {8 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) x}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}-\frac {2 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}\right ) \, dx}{4-\log (5)}+\frac {\int \left (-\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))}+\frac {8 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))}+\frac {2 \exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))}\right ) \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx\\ &=3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) x}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {4 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {8 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {16 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx\\ &=3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {2 \int \left (\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{2 \log ^2(4-2 x-\log (5))}+\frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) (4-\log (5))}{2 (-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))}\right ) \, dx}{4-\log (5)}-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {4 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {8 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {16 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx\\ &=3 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x^2 \log (4-2 x-\log (5))} \, dx-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {2 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {4 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {8 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{x \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}+\frac {16 \int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx}{4-\log (5)}-\frac {\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{\log ^2(4-2 x-\log (5))} \, dx}{-4+\log (5)}-\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right )}{(-4+2 x+\log (5)) \log ^2(4-2 x-\log (5))} \, dx+\int \frac {\exp \left (\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}\right ) \log (x)}{x^2 \log (4-2 x-\log (5))} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 30, normalized size = 0.88 \begin {gather*} e^{5+\frac {-8+x-2 \log (x)}{2 x \log (4-2 x-\log (5))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-8 + x - 2*Log[x] + 10*x*Log[4 - 2*x - Log[5]])/(2*x*Log[4 - 2*x - Log[5]]))*(8*x - x^2 + (-12
+ 6*x + 3*Log[5])*Log[4 - 2*x - Log[5]] + Log[x]*(2*x + (-4 + 2*x + Log[5])*Log[4 - 2*x - Log[5]])))/((-4*x^2
+ 2*x^3 + x^2*Log[5])*Log[4 - 2*x - Log[5]]^2),x]

[Out]

E^(5 + (-8 + x - 2*Log[x])/(2*x*Log[4 - 2*x - Log[5]]))

________________________________________________________________________________________

fricas [A]  time = 0.50, size = 38, normalized size = 1.12 \begin {gather*} e^{\left (\frac {10 \, x \log \left (-2 \, x - \log \relax (5) + 4\right ) + x - 2 \, \log \relax (x) - 8}{2 \, x \log \left (-2 \, x - \log \relax (5) + 4\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((log(5)+2*x-4)*log(-log(5)+4-2*x)+2*x)*log(x)+(3*log(5)+6*x-12)*log(-log(5)+4-2*x)-x^2+8*x)*exp(1/
2*(-2*log(x)+10*x*log(-log(5)+4-2*x)-8+x)/x/log(-log(5)+4-2*x))/(x^2*log(5)+2*x^3-4*x^2)/log(-log(5)+4-2*x)^2,
x, algorithm="fricas")

[Out]

e^(1/2*(10*x*log(-2*x - log(5) + 4) + x - 2*log(x) - 8)/(x*log(-2*x - log(5) + 4)))

________________________________________________________________________________________

giac [A]  time = 0.81, size = 53, normalized size = 1.56 \begin {gather*} e^{\left (-\frac {\log \relax (x)}{x \log \left (-2 \, x - \log \relax (5) + 4\right )} + \frac {1}{2 \, \log \left (-2 \, x - \log \relax (5) + 4\right )} - \frac {4}{x \log \left (-2 \, x - \log \relax (5) + 4\right )} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((log(5)+2*x-4)*log(-log(5)+4-2*x)+2*x)*log(x)+(3*log(5)+6*x-12)*log(-log(5)+4-2*x)-x^2+8*x)*exp(1/
2*(-2*log(x)+10*x*log(-log(5)+4-2*x)-8+x)/x/log(-log(5)+4-2*x))/(x^2*log(5)+2*x^3-4*x^2)/log(-log(5)+4-2*x)^2,
x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.10, size = 41, normalized size = 1.21

method	result	size
risch	\({\mathrm e}^{-\frac {-10 x \ln \left (-\ln \relax (5)+4-2 x \right )+2 \ln \relax (x )-x +8}{2 x \ln \left (-\ln \relax (5)+4-2 x \right )}}\)	\(41\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((ln(5)+2*x-4)*ln(-ln(5)+4-2*x)+2*x)*ln(x)+(3*ln(5)+6*x-12)*ln(-ln(5)+4-2*x)-x^2+8*x)*exp(1/2*(-2*ln(x)+1
0*x*ln(-ln(5)+4-2*x)-8+x)/x/ln(-ln(5)+4-2*x))/(x^2*ln(5)+2*x^3-4*x^2)/ln(-ln(5)+4-2*x)^2,x,method=_RETURNVERBO
SE)

[Out]

exp(-1/2*(-10*x*ln(-ln(5)+4-2*x)+2*ln(x)-x+8)/x/ln(-ln(5)+4-2*x))

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((log(5)+2*x-4)*log(-log(5)+4-2*x)+2*x)*log(x)+(3*log(5)+6*x-12)*log(-log(5)+4-2*x)-x^2+8*x)*exp(1/
2*(-2*log(x)+10*x*log(-log(5)+4-2*x)-8+x)/x/log(-log(5)+4-2*x))/(x^2*log(5)+2*x^3-4*x^2)/log(-log(5)+4-2*x)^2,
x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

________________________________________________________________________________________

mupad [B]  time = 4.85, size = 56, normalized size = 1.65 \begin {gather*} \frac {{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {4}{x\,\ln \left (4-\ln \relax (5)-2\,x\right )}}\,{\mathrm {e}}^{\frac {1}{2\,\ln \left (4-\ln \relax (5)-2\,x\right )}}}{x^{\frac {1}{x\,\ln \left (4-\ln \relax (5)-2\,x\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((x/2 - log(x) + 5*x*log(4 - log(5) - 2*x) - 4)/(x*log(4 - log(5) - 2*x)))*(8*x + log(4 - log(5) - 2*x
)*(6*x + 3*log(5) - 12) - x^2 + log(x)*(2*x + log(4 - log(5) - 2*x)*(2*x + log(5) - 4))))/(log(4 - log(5) - 2*
x)^2*(x^2*log(5) - 4*x^2 + 2*x^3)),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 1.69, size = 34, normalized size = 1.00 \begin {gather*} e^{\frac {5 x \log {\left (- 2 x - \log {\relax (5 )} + 4 \right )} + \frac {x}{2} - \log {\relax (x )} - 4}{x \log {\left (- 2 x - \log {\relax (5 )} + 4 \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((ln(5)+2*x-4)*ln(-ln(5)+4-2*x)+2*x)*ln(x)+(3*ln(5)+6*x-12)*ln(-ln(5)+4-2*x)-x**2+8*x)*exp(1/2*(-2*
ln(x)+10*x*ln(-ln(5)+4-2*x)-8+x)/x/ln(-ln(5)+4-2*x))/(x**2*ln(5)+2*x**3-4*x**2)/ln(-ln(5)+4-2*x)**2,x)

[Out]

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

________________________________________________________________________________________