Integrand size = 125, antiderivative size = 20 \[ \int \frac {e^{\frac {25+(10+10 x) \log (2 x)+\left (1+2 x+x^2\right ) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}} \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx=1+e^{\left (1+\frac {1}{x+\frac {5}{\log (2 x)}}\right )^2} \] Output:
exp((1/(x+5/ln(2*x))+1)^2)+1
\[ \int \frac {e^{\frac {25+(10+10 x) \log (2 x)+\left (1+2 x+x^2\right ) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}} \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx=\int \frac {e^{\frac {25+(10+10 x) \log (2 x)+\left (1+2 x+x^2\right ) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}} \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx \] Input:
Integrate[(E^((25 + (10 + 10*x)*Log[2*x] + (1 + 2*x + x^2)*Log[2*x]^2)/(25 + 10*x*Log[2*x] + x^2*Log[2*x]^2))*(50 + (10 + 10*x)*Log[2*x] - 10*x*Log[ 2*x]^2 + (-2*x - 2*x^2)*Log[2*x]^3))/(125*x + 75*x^2*Log[2*x] + 15*x^3*Log [2*x]^2 + x^4*Log[2*x]^3),x]
Output:
Integrate[(E^((25 + (10 + 10*x)*Log[2*x] + (1 + 2*x + x^2)*Log[2*x]^2)/(25 + 10*x*Log[2*x] + x^2*Log[2*x]^2))*(50 + (10 + 10*x)*Log[2*x] - 10*x*Log[ 2*x]^2 + (-2*x - 2*x^2)*Log[2*x]^3))/(125*x + 75*x^2*Log[2*x] + 15*x^3*Log [2*x]^2 + x^4*Log[2*x]^3), x]
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.
| \(\displaystyle \int \frac {\left (\left (-2 x^2-2 x\right ) \log ^3(2 x)-10 x \log ^2(2 x)+(10 x+10) \log (2 x)+50\right ) \exp \left (\frac {\left (x^2+2 x+1\right ) \log ^2(2 x)+(10 x+10) \log (2 x)+25}{x^2 \log ^2(2 x)+10 x \log (2 x)+25}\right )}{x^4 \log ^3(2 x)+15 x^3 \log ^2(2 x)+75 x^2 \log (2 x)+125 x} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (\left (-2 x^2-2 x\right ) \log ^3(2 x)-10 x \log ^2(2 x)+(10 x+10) \log (2 x)+50\right ) \exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x (x \log (2 x)+5)^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {50 (x-5) \exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^3 (x \log (2 x)+5)^3}-\frac {2 (x+1) \exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^3}+\frac {10 (2 x+3) \exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^3 (x \log (2 x)+5)}+\frac {10 \left (x^2-4 x-15\right ) \exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^3 (x \log (2 x)+5)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 \int \frac {\exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^3}dx+250 \int \frac {\exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^3 (x \log (2 x)+5)^3}dx-150 \int \frac {\exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^3 (x \log (2 x)+5)^2}dx+30 \int \frac {\exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^3 (x \log (2 x)+5)}dx-2 \int \frac {\exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^2}dx-50 \int \frac {\exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^2 (x \log (2 x)+5)^3}dx-40 \int \frac {\exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^2 (x \log (2 x)+5)^2}dx+20 \int \frac {\exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x^2 (x \log (2 x)+5)}dx+10 \int \frac {\exp \left (\frac {(x \log (2 x)+\log (2 x)+5)^2}{(x \log (2 x)+5)^2}\right )}{x (x \log (2 x)+5)^2}dx\) |
Input:
Int[(E^((25 + (10 + 10*x)*Log[2*x] + (1 + 2*x + x^2)*Log[2*x]^2)/(25 + 10* x*Log[2*x] + x^2*Log[2*x]^2))*(50 + (10 + 10*x)*Log[2*x] - 10*x*Log[2*x]^2 + (-2*x - 2*x^2)*Log[2*x]^3))/(125*x + 75*x^2*Log[2*x] + 15*x^3*Log[2*x]^ 2 + x^4*Log[2*x]^3),x]
Output:
$Aborted
Time = 3.45 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35
| method | result | size |
| risch | \({\mathrm e}^{\frac {\left (x \ln \left (2 x \right )+\ln \left (2 x \right )+5\right )^{2}}{\left (x \ln \left (2 x \right )+5\right )^{2}}}\) | \(27\) |
| parallelrisch | \({\mathrm e}^{\frac {\left (x^{2}+2 x +1\right ) \ln \left (2 x \right )^{2}+\left (10 x +10\right ) \ln \left (2 x \right )+25}{x^{2} \ln \left (2 x \right )^{2}+10 x \ln \left (2 x \right )+25}}\) | \(51\) |
Input:
int(((-2*x^2-2*x)*ln(2*x)^3-10*x*ln(2*x)^2+(10*x+10)*ln(2*x)+50)*exp(((x^2 +2*x+1)*ln(2*x)^2+(10*x+10)*ln(2*x)+25)/(x^2*ln(2*x)^2+10*x*ln(2*x)+25))/( x^4*ln(2*x)^3+15*x^3*ln(2*x)^2+75*x^2*ln(2*x)+125*x),x,method=_RETURNVERBO SE)
Output:
exp((x*ln(2*x)+ln(2*x)+5)^2/(x*ln(2*x)+5)^2)
Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (19) = 38\).
Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.45 \[ \int \frac {e^{\frac {25+(10+10 x) \log (2 x)+\left (1+2 x+x^2\right ) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}} \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx=e^{\left (\frac {{\left (x^{2} + 2 \, x + 1\right )} \log \left (2 \, x\right )^{2} + 10 \, {\left (x + 1\right )} \log \left (2 \, x\right ) + 25}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25}\right )} \] Input:
integrate(((-2*x^2-2*x)*log(2*x)^3-10*x*log(2*x)^2+(10*x+10)*log(2*x)+50)* exp(((x^2+2*x+1)*log(2*x)^2+(10*x+10)*log(2*x)+25)/(x^2*log(2*x)^2+10*x*lo g(2*x)+25))/(x^4*log(2*x)^3+15*x^3*log(2*x)^2+75*x^2*log(2*x)+125*x),x, al gorithm="fricas")
Output:
e^(((x^2 + 2*x + 1)*log(2*x)^2 + 10*(x + 1)*log(2*x) + 25)/(x^2*log(2*x)^2 + 10*x*log(2*x) + 25))
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (15) = 30\).
Time = 1.14 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.40 \[ \int \frac {e^{\frac {25+(10+10 x) \log (2 x)+\left (1+2 x+x^2\right ) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}} \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx=e^{\frac {\left (10 x + 10\right ) \log {\left (2 x \right )} + \left (x^{2} + 2 x + 1\right ) \log {\left (2 x \right )}^{2} + 25}{x^{2} \log {\left (2 x \right )}^{2} + 10 x \log {\left (2 x \right )} + 25}} \] Input:
integrate(((-2*x**2-2*x)*ln(2*x)**3-10*x*ln(2*x)**2+(10*x+10)*ln(2*x)+50)* exp(((x**2+2*x+1)*ln(2*x)**2+(10*x+10)*ln(2*x)+25)/(x**2*ln(2*x)**2+10*x*l n(2*x)+25))/(x**4*ln(2*x)**3+15*x**3*ln(2*x)**2+75*x**2*ln(2*x)+125*x),x)
Output:
exp(((10*x + 10)*log(2*x) + (x**2 + 2*x + 1)*log(2*x)**2 + 25)/(x**2*log(2 *x)**2 + 10*x*log(2*x) + 25))
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (19) = 38\).
Time = 0.52 (sec) , antiderivative size = 168, normalized size of antiderivative = 8.40 \[ \int \frac {e^{\frac {25+(10+10 x) \log (2 x)+\left (1+2 x+x^2\right ) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}} \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx=e^{\left (\frac {\log \left (2\right )^{2}}{x^{2} \log \left (2\right )^{2} + x^{2} \log \left (x\right )^{2} + 10 \, x \log \left (2\right ) + 2 \, {\left (x^{2} \log \left (2\right ) + 5 \, x\right )} \log \left (x\right ) + 25} + \frac {2 \, \log \left (2\right ) \log \left (x\right )}{x^{2} \log \left (2\right )^{2} + x^{2} \log \left (x\right )^{2} + 10 \, x \log \left (2\right ) + 2 \, {\left (x^{2} \log \left (2\right ) + 5 \, x\right )} \log \left (x\right ) + 25} + \frac {\log \left (x\right )^{2}}{x^{2} \log \left (2\right )^{2} + x^{2} \log \left (x\right )^{2} + 10 \, x \log \left (2\right ) + 2 \, {\left (x^{2} \log \left (2\right ) + 5 \, x\right )} \log \left (x\right ) + 25} + \frac {2 \, \log \left (2\right )}{x \log \left (2\right ) + x \log \left (x\right ) + 5} + \frac {2 \, \log \left (x\right )}{x \log \left (2\right ) + x \log \left (x\right ) + 5} + 1\right )} \] Input:
integrate(((-2*x^2-2*x)*log(2*x)^3-10*x*log(2*x)^2+(10*x+10)*log(2*x)+50)* exp(((x^2+2*x+1)*log(2*x)^2+(10*x+10)*log(2*x)+25)/(x^2*log(2*x)^2+10*x*lo g(2*x)+25))/(x^4*log(2*x)^3+15*x^3*log(2*x)^2+75*x^2*log(2*x)+125*x),x, al gorithm="maxima")
Output:
e^(log(2)^2/(x^2*log(2)^2 + x^2*log(x)^2 + 10*x*log(2) + 2*(x^2*log(2) + 5 *x)*log(x) + 25) + 2*log(2)*log(x)/(x^2*log(2)^2 + x^2*log(x)^2 + 10*x*log (2) + 2*(x^2*log(2) + 5*x)*log(x) + 25) + log(x)^2/(x^2*log(2)^2 + x^2*log (x)^2 + 10*x*log(2) + 2*(x^2*log(2) + 5*x)*log(x) + 25) + 2*log(2)/(x*log( 2) + x*log(x) + 5) + 2*log(x)/(x*log(2) + x*log(x) + 5) + 1)
Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (19) = 38\).
Time = 1.35 (sec) , antiderivative size = 169, normalized size of antiderivative = 8.45 \[ \int \frac {e^{\frac {25+(10+10 x) \log (2 x)+\left (1+2 x+x^2\right ) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}} \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx=e^{\left (\frac {x^{2} \log \left (2 \, x\right )^{2}}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25} + \frac {2 \, x \log \left (2 \, x\right )^{2}}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25} + \frac {10 \, x \log \left (2 \, x\right )}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25} + \frac {\log \left (2 \, x\right )^{2}}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25} + \frac {10 \, \log \left (2 \, x\right )}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25} + \frac {25}{x^{2} \log \left (2 \, x\right )^{2} + 10 \, x \log \left (2 \, x\right ) + 25}\right )} \] Input:
integrate(((-2*x^2-2*x)*log(2*x)^3-10*x*log(2*x)^2+(10*x+10)*log(2*x)+50)* exp(((x^2+2*x+1)*log(2*x)^2+(10*x+10)*log(2*x)+25)/(x^2*log(2*x)^2+10*x*lo g(2*x)+25))/(x^4*log(2*x)^3+15*x^3*log(2*x)^2+75*x^2*log(2*x)+125*x),x, al gorithm="giac")
Output:
e^(x^2*log(2*x)^2/(x^2*log(2*x)^2 + 10*x*log(2*x) + 25) + 2*x*log(2*x)^2/( x^2*log(2*x)^2 + 10*x*log(2*x) + 25) + 10*x*log(2*x)/(x^2*log(2*x)^2 + 10* x*log(2*x) + 25) + log(2*x)^2/(x^2*log(2*x)^2 + 10*x*log(2*x) + 25) + 10*l og(2*x)/(x^2*log(2*x)^2 + 10*x*log(2*x) + 25) + 25/(x^2*log(2*x)^2 + 10*x* log(2*x) + 25))
Time = 2.88 (sec) , antiderivative size = 429, normalized size of antiderivative = 21.45 \[ \int \frac {e^{\frac {25+(10+10 x) \log (2 x)+\left (1+2 x+x^2\right ) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}} \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx={1024}^{\frac {x+1}{x^2\,{\ln \left (x\right )}^2+2\,\ln \left (2\right )\,x^2\,\ln \left (x\right )+{\ln \left (2\right )}^2\,x^2+10\,x\,\ln \left (x\right )+10\,\ln \left (2\right )\,x+25}}\,x^{\frac {2\,\left (5\,x+\ln \left (2\right )+2\,x\,\ln \left (2\right )+x^2\,\ln \left (2\right )+5\right )}{x^2\,{\ln \left (x\right )}^2+2\,\ln \left (2\right )\,x^2\,\ln \left (x\right )+{\ln \left (2\right )}^2\,x^2+10\,x\,\ln \left (x\right )+10\,\ln \left (2\right )\,x+25}}\,{\mathrm {e}}^{\frac {x^2\,{\ln \left (2\right )}^2}{x^2\,{\ln \left (x\right )}^2+2\,\ln \left (2\right )\,x^2\,\ln \left (x\right )+{\ln \left (2\right )}^2\,x^2+10\,x\,\ln \left (x\right )+10\,\ln \left (2\right )\,x+25}}\,{\mathrm {e}}^{\frac {2\,x\,{\ln \left (x\right )}^2}{x^2\,{\ln \left (x\right )}^2+2\,\ln \left (2\right )\,x^2\,\ln \left (x\right )+{\ln \left (2\right )}^2\,x^2+10\,x\,\ln \left (x\right )+10\,\ln \left (2\right )\,x+25}}\,{\mathrm {e}}^{\frac {{\ln \left (2\right )}^2}{x^2\,{\ln \left (x\right )}^2+2\,\ln \left (2\right )\,x^2\,\ln \left (x\right )+{\ln \left (2\right )}^2\,x^2+10\,x\,\ln \left (x\right )+10\,\ln \left (2\right )\,x+25}}\,{\mathrm {e}}^{\frac {25}{x^2\,{\ln \left (x\right )}^2+2\,\ln \left (2\right )\,x^2\,\ln \left (x\right )+{\ln \left (2\right )}^2\,x^2+10\,x\,\ln \left (x\right )+10\,\ln \left (2\right )\,x+25}}\,{\mathrm {e}}^{\frac {x^2\,{\ln \left (x\right )}^2}{x^2\,{\ln \left (x\right )}^2+2\,\ln \left (2\right )\,x^2\,\ln \left (x\right )+{\ln \left (2\right )}^2\,x^2+10\,x\,\ln \left (x\right )+10\,\ln \left (2\right )\,x+25}}\,{\mathrm {e}}^{\frac {{\ln \left (x\right )}^2}{x^2\,{\ln \left (x\right )}^2+2\,\ln \left (2\right )\,x^2\,\ln \left (x\right )+{\ln \left (2\right )}^2\,x^2+10\,x\,\ln \left (x\right )+10\,\ln \left (2\right )\,x+25}}\,{\mathrm {e}}^{\frac {2\,x\,{\ln \left (2\right )}^2}{x^2\,{\ln \left (x\right )}^2+2\,\ln \left (2\right )\,x^2\,\ln \left (x\right )+{\ln \left (2\right )}^2\,x^2+10\,x\,\ln \left (x\right )+10\,\ln \left (2\right )\,x+25}} \] Input:
int(-(exp((log(2*x)^2*(2*x + x^2 + 1) + log(2*x)*(10*x + 10) + 25)/(10*x*l og(2*x) + x^2*log(2*x)^2 + 25))*(log(2*x)^3*(2*x + 2*x^2) + 10*x*log(2*x)^ 2 - log(2*x)*(10*x + 10) - 50))/(125*x + 75*x^2*log(2*x) + 15*x^3*log(2*x) ^2 + x^4*log(2*x)^3),x)
Output:
1024^((x + 1)/(x^2*log(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2 *x^2*log(2)*log(x) + 25))*x^((2*(5*x + log(2) + 2*x*log(2) + x^2*log(2) + 5))/(x^2*log(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2 )*log(x) + 25))*exp((x^2*log(2)^2)/(x^2*log(2)^2 + 10*x*log(2) + x^2*log(x )^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))*exp((2*x*log(x)^2)/(x^2*log (2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 2 5))*exp(log(2)^2/(x^2*log(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))*exp(25/(x^2*log(2)^2 + 10*x*log(2) + x^2*log( x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))*exp((x^2*log(x)^2)/(x^2*lo g(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))*exp(log(x)^2/(x^2*log(2)^2 + 10*x*log(2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))*exp((2*x*log(2)^2)/(x^2*log(2)^2 + 10*x*log( 2) + x^2*log(x)^2 + 10*x*log(x) + 2*x^2*log(2)*log(x) + 25))
Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.40 \[ \int \frac {e^{\frac {25+(10+10 x) \log (2 x)+\left (1+2 x+x^2\right ) \log ^2(2 x)}{25+10 x \log (2 x)+x^2 \log ^2(2 x)}} \left (50+(10+10 x) \log (2 x)-10 x \log ^2(2 x)+\left (-2 x-2 x^2\right ) \log ^3(2 x)\right )}{125 x+75 x^2 \log (2 x)+15 x^3 \log ^2(2 x)+x^4 \log ^3(2 x)} \, dx=e^{\frac {2 \mathrm {log}\left (2 x \right )^{2} x +\mathrm {log}\left (2 x \right )^{2}+10 \,\mathrm {log}\left (2 x \right )}{\mathrm {log}\left (2 x \right )^{2} x^{2}+10 \,\mathrm {log}\left (2 x \right ) x +25}} e \] Input:
int(((-2*x^2-2*x)*log(2*x)^3-10*x*log(2*x)^2+(10*x+10)*log(2*x)+50)*exp((( x^2+2*x+1)*log(2*x)^2+(10*x+10)*log(2*x)+25)/(x^2*log(2*x)^2+10*x*log(2*x) +25))/(x^4*log(2*x)^3+15*x^3*log(2*x)^2+75*x^2*log(2*x)+125*x),x)
Output:
e**((2*log(2*x)**2*x + log(2*x)**2 + 10*log(2*x))/(log(2*x)**2*x**2 + 10*l og(2*x)*x + 25))*e