Integrand size = 195, antiderivative size = 29 \[ \int \frac {e^{2 x} (-2+8 x)+4 \log (5)+e^x (-4+8 x+(2-4 x) \log (5))}{\left (e^{2 x} x-2 x \log (5)+e^x (2 x-x \log (5))\right ) \log \left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right ) \log \left (\log ^2\left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right )\right )} \, dx=\log \left (\log \left (\log ^2\left (\frac {\left (2+e^x\right )^2 \left (e^x-\log (5)\right )^2}{2 x}\right )\right )\right ) \] Output:
ln(ln(ln(1/2*(2+exp(x))^2*(exp(x)-ln(5))^2/x)^2))
\[ \int \frac {e^{2 x} (-2+8 x)+4 \log (5)+e^x (-4+8 x+(2-4 x) \log (5))}{\left (e^{2 x} x-2 x \log (5)+e^x (2 x-x \log (5))\right ) \log \left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right ) \log \left (\log ^2\left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right )\right )} \, dx=\int \frac {e^{2 x} (-2+8 x)+4 \log (5)+e^x (-4+8 x+(2-4 x) \log (5))}{\left (e^{2 x} x-2 x \log (5)+e^x (2 x-x \log (5))\right ) \log \left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right ) \log \left (\log ^2\left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right )\right )} \, dx \] Input:
Integrate[(E^(2*x)*(-2 + 8*x) + 4*Log[5] + E^x*(-4 + 8*x + (2 - 4*x)*Log[5 ]))/((E^(2*x)*x - 2*x*Log[5] + E^x*(2*x - x*Log[5]))*Log[(E^(4*x) + E^(3*x )*(4 - 2*Log[5]) + 4*Log[5]^2 + E^(2*x)*(4 - 8*Log[5] + Log[5]^2) + E^x*(- 8*Log[5] + 4*Log[5]^2))/(2*x)]*Log[Log[(E^(4*x) + E^(3*x)*(4 - 2*Log[5]) + 4*Log[5]^2 + E^(2*x)*(4 - 8*Log[5] + Log[5]^2) + E^x*(-8*Log[5] + 4*Log[5 ]^2))/(2*x)]^2]),x]
Output:
Integrate[(E^(2*x)*(-2 + 8*x) + 4*Log[5] + E^x*(-4 + 8*x + (2 - 4*x)*Log[5 ]))/((E^(2*x)*x - 2*x*Log[5] + E^x*(2*x - x*Log[5]))*Log[(E^(4*x) + E^(3*x )*(4 - 2*Log[5]) + 4*Log[5]^2 + E^(2*x)*(4 - 8*Log[5] + Log[5]^2) + E^x*(- 8*Log[5] + 4*Log[5]^2))/(2*x)]*Log[Log[(E^(4*x) + E^(3*x)*(4 - 2*Log[5]) + 4*Log[5]^2 + E^(2*x)*(4 - 8*Log[5] + Log[5]^2) + E^x*(-8*Log[5] + 4*Log[5 ]^2))/(2*x)]^2]), x]
Leaf count is larger than twice the leaf count of optimal. \(66\) vs. \(2(29)=58\).
Time = 0.89 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.28, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.005, Rules used = {7235}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {e^{2 x} (8 x-2)+e^x (8 x+(2-4 x) \log (5)-4)+4 \log (5)}{\left (e^{2 x} x-2 x \log (5)+e^x (2 x-x \log (5))\right ) \log \left (\frac {e^{4 x}+e^{2 x} \left (4+\log ^2(5)-8 \log (5)\right )+e^x \left (4 \log ^2(5)-8 \log (5)\right )+e^{3 x} (4-2 \log (5))+4 \log ^2(5)}{2 x}\right ) \log \left (\log ^2\left (\frac {e^{4 x}+e^{2 x} \left (4+\log ^2(5)-8 \log (5)\right )+e^x \left (4 \log ^2(5)-8 \log (5)\right )+e^{3 x} (4-2 \log (5))+4 \log ^2(5)}{2 x}\right )\right )} \, dx\) |
| \(\Big \downarrow \) 7235 |
| \(\displaystyle \log \left (\log \left (\log ^2\left (\frac {e^{4 x}+e^{2 x} \left (4+\log ^2(5)-8 \log (5)\right )+2 e^{3 x} (2-\log (5))-4 e^x (2-\log (5)) \log (5)+4 \log ^2(5)}{2 x}\right )\right )\right )\) |
Input:
Int[(E^(2*x)*(-2 + 8*x) + 4*Log[5] + E^x*(-4 + 8*x + (2 - 4*x)*Log[5]))/(( E^(2*x)*x - 2*x*Log[5] + E^x*(2*x - x*Log[5]))*Log[(E^(4*x) + E^(3*x)*(4 - 2*Log[5]) + 4*Log[5]^2 + E^(2*x)*(4 - 8*Log[5] + Log[5]^2) + E^x*(-8*Log[ 5] + 4*Log[5]^2))/(2*x)]*Log[Log[(E^(4*x) + E^(3*x)*(4 - 2*Log[5]) + 4*Log [5]^2 + E^(2*x)*(4 - 8*Log[5] + Log[5]^2) + E^x*(-8*Log[5] + 4*Log[5]^2))/ (2*x)]^2]),x]
Output:
Log[Log[Log[(E^(4*x) + 2*E^(3*x)*(2 - Log[5]) - 4*E^x*(2 - Log[5])*Log[5] + 4*Log[5]^2 + E^(2*x)*(4 - 8*Log[5] + Log[5]^2))/(2*x)]^2]]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Leaf count of result is larger than twice the leaf count of optimal. \(61\) vs. \(2(25)=50\).
Time = 49.49 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.14
| method | result | size |
| parallelrisch | \(\ln \left (\ln \left (\ln \left (\frac {{\mathrm e}^{4 x}+\left (-2 \ln \left (5\right )+4\right ) {\mathrm e}^{3 x}+\left (\ln \left (5\right )^{2}-8 \ln \left (5\right )+4\right ) {\mathrm e}^{2 x}+\left (4 \ln \left (5\right )^{2}-8 \ln \left (5\right )\right ) {\mathrm e}^{x}+4 \ln \left (5\right )^{2}}{2 x}\right )^{2}\right )\right )\) | \(62\) |
| risch | \(\text {Expression too large to display}\) | \(3312\) |
Input:
int(((8*x-2)*exp(x)^2+((-4*x+2)*ln(5)+8*x-4)*exp(x)+4*ln(5))/(x*exp(x)^2+( -x*ln(5)+2*x)*exp(x)-2*x*ln(5))/ln(1/2*(exp(x)^4+(-2*ln(5)+4)*exp(x)^3+(ln (5)^2-8*ln(5)+4)*exp(x)^2+(4*ln(5)^2-8*ln(5))*exp(x)+4*ln(5)^2)/x)/ln(ln(1 /2*(exp(x)^4+(-2*ln(5)+4)*exp(x)^3+(ln(5)^2-8*ln(5)+4)*exp(x)^2+(4*ln(5)^2 -8*ln(5))*exp(x)+4*ln(5)^2)/x)^2),x,method=_RETURNVERBOSE)
Output:
ln(ln(ln(1/2*(exp(x)^4+(-2*ln(5)+4)*exp(x)^3+(ln(5)^2-8*ln(5)+4)*exp(x)^2+ (4*ln(5)^2-8*ln(5))*exp(x)+4*ln(5)^2)/x)^2))
Leaf count of result is larger than twice the leaf count of optimal. 62 vs. \(2 (25) = 50\).
Time = 0.07 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.14 \[ \int \frac {e^{2 x} (-2+8 x)+4 \log (5)+e^x (-4+8 x+(2-4 x) \log (5))}{\left (e^{2 x} x-2 x \log (5)+e^x (2 x-x \log (5))\right ) \log \left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right ) \log \left (\log ^2\left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right )\right )} \, dx=\log \left (\log \left (\log \left (-\frac {2 \, {\left (\log \left (5\right ) - 2\right )} e^{\left (3 \, x\right )} - {\left (\log \left (5\right )^{2} - 8 \, \log \left (5\right ) + 4\right )} e^{\left (2 \, x\right )} - 4 \, {\left (\log \left (5\right )^{2} - 2 \, \log \left (5\right )\right )} e^{x} - 4 \, \log \left (5\right )^{2} - e^{\left (4 \, x\right )}}{2 \, x}\right )^{2}\right )\right ) \] Input:
integrate(((8*x-2)*exp(x)^2+((-4*x+2)*log(5)+8*x-4)*exp(x)+4*log(5))/(x*ex p(x)^2+(-x*log(5)+2*x)*exp(x)-2*x*log(5))/log(1/2*(exp(x)^4+(-2*log(5)+4)* exp(x)^3+(log(5)^2-8*log(5)+4)*exp(x)^2+(4*log(5)^2-8*log(5))*exp(x)+4*log (5)^2)/x)/log(log(1/2*(exp(x)^4+(-2*log(5)+4)*exp(x)^3+(log(5)^2-8*log(5)+ 4)*exp(x)^2+(4*log(5)^2-8*log(5))*exp(x)+4*log(5)^2)/x)^2),x, algorithm="f ricas")
Output:
log(log(log(-1/2*(2*(log(5) - 2)*e^(3*x) - (log(5)^2 - 8*log(5) + 4)*e^(2* x) - 4*(log(5)^2 - 2*log(5))*e^x - 4*log(5)^2 - e^(4*x))/x)^2))
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (24) = 48\).
Time = 6.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.41 \[ \int \frac {e^{2 x} (-2+8 x)+4 \log (5)+e^x (-4+8 x+(2-4 x) \log (5))}{\left (e^{2 x} x-2 x \log (5)+e^x (2 x-x \log (5))\right ) \log \left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right ) \log \left (\log ^2\left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right )\right )} \, dx=\log {\left (\log {\left (\log {\left (\frac {\frac {e^{4 x}}{2} + \frac {\left (4 - 2 \log {\left (5 \right )}\right ) e^{3 x}}{2} + \frac {\left (- 8 \log {\left (5 \right )} + \log {\left (5 \right )}^{2} + 4\right ) e^{2 x}}{2} + \frac {\left (- 8 \log {\left (5 \right )} + 4 \log {\left (5 \right )}^{2}\right ) e^{x}}{2} + 2 \log {\left (5 \right )}^{2}}{x} \right )}^{2} \right )} \right )} \] Input:
integrate(((8*x-2)*exp(x)**2+((-4*x+2)*ln(5)+8*x-4)*exp(x)+4*ln(5))/(x*exp (x)**2+(-x*ln(5)+2*x)*exp(x)-2*x*ln(5))/ln(1/2*(exp(x)**4+(-2*ln(5)+4)*exp (x)**3+(ln(5)**2-8*ln(5)+4)*exp(x)**2+(4*ln(5)**2-8*ln(5))*exp(x)+4*ln(5)* *2)/x)/ln(ln(1/2*(exp(x)**4+(-2*ln(5)+4)*exp(x)**3+(ln(5)**2-8*ln(5)+4)*ex p(x)**2+(4*ln(5)**2-8*ln(5))*exp(x)+4*ln(5)**2)/x)**2),x)
Output:
log(log(log((exp(4*x)/2 + (4 - 2*log(5))*exp(3*x)/2 + (-8*log(5) + log(5)* *2 + 4)*exp(2*x)/2 + (-8*log(5) + 4*log(5)**2)*exp(x)/2 + 2*log(5)**2)/x)* *2))
Time = 3.50 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{2 x} (-2+8 x)+4 \log (5)+e^x (-4+8 x+(2-4 x) \log (5))}{\left (e^{2 x} x-2 x \log (5)+e^x (2 x-x \log (5))\right ) \log \left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right ) \log \left (\log ^2\left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right )\right )} \, dx=\log \left (\log \left (-\log \left (2\right ) - \log \left (x\right ) + 2 \, \log \left (e^{x} - \log \left (5\right )\right ) + 2 \, \log \left (e^{x} + 2\right )\right )\right ) \] Input:
integrate(((8*x-2)*exp(x)^2+((-4*x+2)*log(5)+8*x-4)*exp(x)+4*log(5))/(x*ex p(x)^2+(-x*log(5)+2*x)*exp(x)-2*x*log(5))/log(1/2*(exp(x)^4+(-2*log(5)+4)* exp(x)^3+(log(5)^2-8*log(5)+4)*exp(x)^2+(4*log(5)^2-8*log(5))*exp(x)+4*log (5)^2)/x)/log(log(1/2*(exp(x)^4+(-2*log(5)+4)*exp(x)^3+(log(5)^2-8*log(5)+ 4)*exp(x)^2+(4*log(5)^2-8*log(5))*exp(x)+4*log(5)^2)/x)^2),x, algorithm="m axima")
Output:
log(log(-log(2) - log(x) + 2*log(e^x - log(5)) + 2*log(e^x + 2)))
Leaf count of result is larger than twice the leaf count of optimal. 143 vs. \(2 (25) = 50\).
Time = 2.47 (sec) , antiderivative size = 143, normalized size of antiderivative = 4.93 \[ \int \frac {e^{2 x} (-2+8 x)+4 \log (5)+e^x (-4+8 x+(2-4 x) \log (5))}{\left (e^{2 x} x-2 x \log (5)+e^x (2 x-x \log (5))\right ) \log \left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right ) \log \left (\log ^2\left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right )\right )} \, dx=\log \left (\log \left (\log \left (e^{\left (2 \, x\right )} \log \left (5\right )^{2} + 4 \, e^{x} \log \left (5\right )^{2} - 2 \, e^{\left (3 \, x\right )} \log \left (5\right ) - 8 \, e^{\left (2 \, x\right )} \log \left (5\right ) - 8 \, e^{x} \log \left (5\right ) + 4 \, \log \left (5\right )^{2} + e^{\left (4 \, x\right )} + 4 \, e^{\left (3 \, x\right )} + 4 \, e^{\left (2 \, x\right )}\right )^{2} - 2 \, \log \left (e^{\left (2 \, x\right )} \log \left (5\right )^{2} + 4 \, e^{x} \log \left (5\right )^{2} - 2 \, e^{\left (3 \, x\right )} \log \left (5\right ) - 8 \, e^{\left (2 \, x\right )} \log \left (5\right ) - 8 \, e^{x} \log \left (5\right ) + 4 \, \log \left (5\right )^{2} + e^{\left (4 \, x\right )} + 4 \, e^{\left (3 \, x\right )} + 4 \, e^{\left (2 \, x\right )}\right ) \log \left (2 \, x\right ) + \log \left (2 \, x\right )^{2}\right )\right ) \] Input:
integrate(((8*x-2)*exp(x)^2+((-4*x+2)*log(5)+8*x-4)*exp(x)+4*log(5))/(x*ex p(x)^2+(-x*log(5)+2*x)*exp(x)-2*x*log(5))/log(1/2*(exp(x)^4+(-2*log(5)+4)* exp(x)^3+(log(5)^2-8*log(5)+4)*exp(x)^2+(4*log(5)^2-8*log(5))*exp(x)+4*log (5)^2)/x)/log(log(1/2*(exp(x)^4+(-2*log(5)+4)*exp(x)^3+(log(5)^2-8*log(5)+ 4)*exp(x)^2+(4*log(5)^2-8*log(5))*exp(x)+4*log(5)^2)/x)^2),x, algorithm="g iac")
Output:
log(log(log(e^(2*x)*log(5)^2 + 4*e^x*log(5)^2 - 2*e^(3*x)*log(5) - 8*e^(2* x)*log(5) - 8*e^x*log(5) + 4*log(5)^2 + e^(4*x) + 4*e^(3*x) + 4*e^(2*x))^2 - 2*log(e^(2*x)*log(5)^2 + 4*e^x*log(5)^2 - 2*e^(3*x)*log(5) - 8*e^(2*x)* log(5) - 8*e^x*log(5) + 4*log(5)^2 + e^(4*x) + 4*e^(3*x) + 4*e^(2*x))*log( 2*x) + log(2*x)^2))
Time = 3.87 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.24 \[ \int \frac {e^{2 x} (-2+8 x)+4 \log (5)+e^x (-4+8 x+(2-4 x) \log (5))}{\left (e^{2 x} x-2 x \log (5)+e^x (2 x-x \log (5))\right ) \log \left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right ) \log \left (\log ^2\left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right )\right )} \, dx=\ln \left (\ln \left ({\ln \left (\frac {\frac {{\mathrm {e}}^{4\,x}}{2}-\frac {{\mathrm {e}}^x\,\left (8\,\ln \left (5\right )-4\,{\ln \left (5\right )}^2\right )}{2}+\frac {{\mathrm {e}}^{2\,x}\,\left ({\ln \left (5\right )}^2-8\,\ln \left (5\right )+4\right )}{2}-\frac {{\mathrm {e}}^{3\,x}\,\left (2\,\ln \left (5\right )-4\right )}{2}+2\,{\ln \left (5\right )}^2}{x}\right )}^2\right )\right ) \] Input:
int((4*log(5) - exp(x)*(log(5)*(4*x - 2) - 8*x + 4) + exp(2*x)*(8*x - 2))/ (log(log((exp(4*x)/2 - (exp(x)*(8*log(5) - 4*log(5)^2))/2 + (exp(2*x)*(log (5)^2 - 8*log(5) + 4))/2 - (exp(3*x)*(2*log(5) - 4))/2 + 2*log(5)^2)/x)^2) *log((exp(4*x)/2 - (exp(x)*(8*log(5) - 4*log(5)^2))/2 + (exp(2*x)*(log(5)^ 2 - 8*log(5) + 4))/2 - (exp(3*x)*(2*log(5) - 4))/2 + 2*log(5)^2)/x)*(x*exp (2*x) - 2*x*log(5) + exp(x)*(2*x - x*log(5)))),x)
Output:
log(log(log((exp(4*x)/2 - (exp(x)*(8*log(5) - 4*log(5)^2))/2 + (exp(2*x)*( log(5)^2 - 8*log(5) + 4))/2 - (exp(3*x)*(2*log(5) - 4))/2 + 2*log(5)^2)/x) ^2))
Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.76 \[ \int \frac {e^{2 x} (-2+8 x)+4 \log (5)+e^x (-4+8 x+(2-4 x) \log (5))}{\left (e^{2 x} x-2 x \log (5)+e^x (2 x-x \log (5))\right ) \log \left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right ) \log \left (\log ^2\left (\frac {e^{4 x}+e^{3 x} (4-2 \log (5))+4 \log ^2(5)+e^{2 x} \left (4-8 \log (5)+\log ^2(5)\right )+e^x \left (-8 \log (5)+4 \log ^2(5)\right )}{2 x}\right )\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\frac {e^{4 x}-2 e^{3 x} \mathrm {log}\left (5\right )+4 e^{3 x}+e^{2 x} \mathrm {log}\left (5\right )^{2}-8 e^{2 x} \mathrm {log}\left (5\right )+4 e^{2 x}+4 e^{x} \mathrm {log}\left (5\right )^{2}-8 e^{x} \mathrm {log}\left (5\right )+4 \mathrm {log}\left (5\right )^{2}}{2 x}\right )^{2}\right )\right ) \] Input:
int(((8*x-2)*exp(x)^2+((-4*x+2)*log(5)+8*x-4)*exp(x)+4*log(5))/(x*exp(x)^2 +(-x*log(5)+2*x)*exp(x)-2*x*log(5))/log(1/2*(exp(x)^4+(-2*log(5)+4)*exp(x) ^3+(log(5)^2-8*log(5)+4)*exp(x)^2+(4*log(5)^2-8*log(5))*exp(x)+4*log(5)^2) /x)/log(log(1/2*(exp(x)^4+(-2*log(5)+4)*exp(x)^3+(log(5)^2-8*log(5)+4)*exp (x)^2+(4*log(5)^2-8*log(5))*exp(x)+4*log(5)^2)/x)^2),x)
Output:
log(log(log((e**(4*x) - 2*e**(3*x)*log(5) + 4*e**(3*x) + e**(2*x)*log(5)** 2 - 8*e**(2*x)*log(5) + 4*e**(2*x) + 4*e**x*log(5)**2 - 8*e**x*log(5) + 4* log(5)**2)/(2*x))**2))