Integrand size = 215, antiderivative size = 32 \[ \int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx=\left (15-\frac {e^{e^{5-\frac {5}{2 x}}}}{-\log (4)+\log \left (\frac {x}{5}\right )}\right )^2 \] Output:
(15-exp(exp(5-5/2/x))/(ln(1/5*x)-2*ln(2)))^2
\[ \int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx=\int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx \] Input:
Integrate[(E^(2*E^((-5 + 10*x)/(2*x)))*(-2*x - 5*E^((-5 + 10*x)/(2*x))*Log [4] + 5*E^((-5 + 10*x)/(2*x))*Log[x/5]) + E^E^((-5 + 10*x)/(2*x))*(-30*x*L og[4] - 75*E^((-5 + 10*x)/(2*x))*Log[4]^2 + (30*x + 150*E^((-5 + 10*x)/(2* x))*Log[4])*Log[x/5] - 75*E^((-5 + 10*x)/(2*x))*Log[x/5]^2))/(-(x^2*Log[4] ^3) + 3*x^2*Log[4]^2*Log[x/5] - 3*x^2*Log[4]*Log[x/5]^2 + x^2*Log[x/5]^3), x]
Output:
Integrate[(E^(2*E^((-5 + 10*x)/(2*x)))*(-2*x - 5*E^((-5 + 10*x)/(2*x))*Log [4] + 5*E^((-5 + 10*x)/(2*x))*Log[x/5]) + E^E^((-5 + 10*x)/(2*x))*(-30*x*L og[4] - 75*E^((-5 + 10*x)/(2*x))*Log[4]^2 + (30*x + 150*E^((-5 + 10*x)/(2* x))*Log[4])*Log[x/5] - 75*E^((-5 + 10*x)/(2*x))*Log[x/5]^2))/(-(x^2*Log[4] ^3) + 3*x^2*Log[4]^2*Log[x/5] - 3*x^2*Log[4]*Log[x/5]^2 + x^2*Log[x/5]^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.
| \(\displaystyle \int \frac {e^{e^{\frac {10 x-5}{2 x}}} \left (-75 e^{\frac {10 x-5}{2 x}} \log ^2\left (\frac {x}{5}\right )-75 e^{\frac {10 x-5}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {10 x-5}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-30 x \log (4)\right )+e^{2 e^{\frac {10 x-5}{2 x}}} \left (-2 x+5 e^{\frac {10 x-5}{2 x}} \log \left (\frac {x}{5}\right )-5 e^{\frac {10 x-5}{2 x}} \log (4)\right )}{x^2 \log ^3\left (\frac {x}{5}\right )-x^2 \log ^3(4)-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^{e^{5-\frac {5}{2 x}}-\frac {5}{2 x}} \left (e^{e^{5-\frac {5}{2 x}}}-15 \log \left (\frac {x}{5}\right )+15 \log (4)\right ) \left (-2 e^{\left .\frac {5}{2}\right /x} x+5 e^5 \log \left (\frac {x}{5}\right )-5 e^5 \log (4)\right )}{x^2 \log ^3\left (\frac {x}{20}\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {5 e^{e^{5-\frac {5}{2 x}}-\frac {5}{2 x}+5} (\log (20)-\log (x)) \left (e^{e^{5-\frac {5}{2 x}}}-15 \log \left (\frac {x}{5}\right )+15 \log (4)\right )}{x^2 \log ^3\left (\frac {x}{20}\right )}-\frac {2 e^{e^{5-\frac {5}{2 x}}} \left (e^{e^{5-\frac {5}{2 x}}}-15 \log \left (\frac {x}{5}\right )+15 \log (4)\right )}{x \log ^3\left (\frac {x}{20}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 5 \int \frac {e^{2 e^{5-\frac {5}{2 x}}+5-\frac {5}{2 x}}}{x^2 \log ^2\left (\frac {x}{20}\right )}dx-75 \int \frac {e^{e^{5-\frac {5}{2 x}}+5-\frac {5}{2 x}}}{x^2 \log \left (\frac {x}{20}\right )}dx-2 \int \frac {e^{2 e^{5-\frac {5}{2 x}}}}{x \log ^3\left (\frac {x}{20}\right )}dx+30 \int \frac {e^{e^{5-\frac {5}{2 x}}}}{x \log ^2\left (\frac {x}{20}\right )}dx\) |
Input:
Int[(E^(2*E^((-5 + 10*x)/(2*x)))*(-2*x - 5*E^((-5 + 10*x)/(2*x))*Log[4] + 5*E^((-5 + 10*x)/(2*x))*Log[x/5]) + E^E^((-5 + 10*x)/(2*x))*(-30*x*Log[4] - 75*E^((-5 + 10*x)/(2*x))*Log[4]^2 + (30*x + 150*E^((-5 + 10*x)/(2*x))*Lo g[4])*Log[x/5] - 75*E^((-5 + 10*x)/(2*x))*Log[x/5]^2))/(-(x^2*Log[4]^3) + 3*x^2*Log[4]^2*Log[x/5] - 3*x^2*Log[4]*Log[x/5]^2 + x^2*Log[x/5]^3),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(76\) vs. \(2(26)=52\).
Time = 6.98 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.41
| method | result | size |
| parallelrisch | \(-\frac {-60 \ln \left (2\right ) {\mathrm e}^{{\mathrm e}^{\frac {5 x -\frac {5}{2}}{x}}}+30 \ln \left (\frac {x}{5}\right ) {\mathrm e}^{{\mathrm e}^{\frac {5 x -\frac {5}{2}}{x}}}-{\mathrm e}^{2 \,{\mathrm e}^{\frac {5 x -\frac {5}{2}}{x}}}}{4 \ln \left (2\right )^{2}-4 \ln \left (\frac {x}{5}\right ) \ln \left (2\right )+\ln \left (\frac {x}{5}\right )^{2}}\) | \(77\) |
Input:
int(((5*exp(1/2*(10*x-5)/x)*ln(1/5*x)-10*ln(2)*exp(1/2*(10*x-5)/x)-2*x)*ex p(exp(1/2*(10*x-5)/x))^2+(-75*exp(1/2*(10*x-5)/x)*ln(1/5*x)^2+(300*ln(2)*e xp(1/2*(10*x-5)/x)+30*x)*ln(1/5*x)-300*ln(2)^2*exp(1/2*(10*x-5)/x)-60*x*ln (2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*ln(1/5*x)^3-6*x^2*ln(2)*ln(1/5*x)^2+12 *x^2*ln(2)^2*ln(1/5*x)-8*x^2*ln(2)^3),x,method=_RETURNVERBOSE)
Output:
-(-60*ln(2)*exp(exp(5/2*(-1+2*x)/x))+30*ln(1/5*x)*exp(exp(5/2*(-1+2*x)/x)) -exp(exp(5/2*(-1+2*x)/x))^2)/(4*ln(2)^2-4*ln(1/5*x)*ln(2)+ln(1/5*x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (27) = 54\).
Time = 0.15 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.00 \[ \int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx=\frac {30 \, {\left (2 \, \log \left (2\right ) - \log \left (\frac {1}{5} \, x\right )\right )} e^{\left (e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )} + e^{\left (2 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) \log \left (\frac {1}{5} \, x\right ) + \log \left (\frac {1}{5} \, x\right )^{2}} \] Input:
integrate(((5*exp(1/2*(10*x-5)/x)*log(1/5*x)-10*log(2)*exp(1/2*(10*x-5)/x) -2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-75*exp(1/2*(10*x-5)/x)*log(1/5*x)^2+(30 0*log(2)*exp(1/2*(10*x-5)/x)+30*x)*log(1/5*x)-300*log(2)^2*exp(1/2*(10*x-5 )/x)-60*x*log(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*log(1/5*x)^3-6*x^2*log(2) *log(1/5*x)^2+12*x^2*log(2)^2*log(1/5*x)-8*x^2*log(2)^3),x, algorithm="fri cas")
Output:
(30*(2*log(2) - log(1/5*x))*e^(e^(5/2*(2*x - 1)/x)) + e^(2*e^(5/2*(2*x - 1 )/x)))/(4*log(2)^2 - 4*log(2)*log(1/5*x) + log(1/5*x)^2)
Leaf count of result is larger than twice the leaf count of optimal. 97 vs. \(2 (22) = 44\).
Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.03 \[ \int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx=\frac {\left (\log {\left (\frac {x}{5} \right )} - 2 \log {\left (2 \right )}\right ) e^{2 e^{\frac {5 x - \frac {5}{2}}{x}}} + \left (- 30 \log {\left (\frac {x}{5} \right )}^{2} + 120 \log {\left (2 \right )} \log {\left (\frac {x}{5} \right )} - 120 \log {\left (2 \right )}^{2}\right ) e^{e^{\frac {5 x - \frac {5}{2}}{x}}}}{\log {\left (\frac {x}{5} \right )}^{3} - 6 \log {\left (2 \right )} \log {\left (\frac {x}{5} \right )}^{2} + 12 \log {\left (2 \right )}^{2} \log {\left (\frac {x}{5} \right )} - 8 \log {\left (2 \right )}^{3}} \] Input:
integrate(((5*exp(1/2*(10*x-5)/x)*ln(1/5*x)-10*ln(2)*exp(1/2*(10*x-5)/x)-2 *x)*exp(exp(1/2*(10*x-5)/x))**2+(-75*exp(1/2*(10*x-5)/x)*ln(1/5*x)**2+(300 *ln(2)*exp(1/2*(10*x-5)/x)+30*x)*ln(1/5*x)-300*ln(2)**2*exp(1/2*(10*x-5)/x )-60*x*ln(2))*exp(exp(1/2*(10*x-5)/x)))/(x**2*ln(1/5*x)**3-6*x**2*ln(2)*ln (1/5*x)**2+12*x**2*ln(2)**2*ln(1/5*x)-8*x**2*ln(2)**3),x)
Output:
((log(x/5) - 2*log(2))*exp(2*exp((5*x - 5/2)/x)) + (-30*log(x/5)**2 + 120* log(2)*log(x/5) - 120*log(2)**2)*exp(exp((5*x - 5/2)/x)))/(log(x/5)**3 - 6 *log(2)*log(x/5)**2 + 12*log(2)**2*log(x/5) - 8*log(2)**3)
\[ \int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx=\int { \frac {{\left (10 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (2\right ) - 5 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (\frac {1}{5} \, x\right ) + 2 \, x\right )} e^{\left (2 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )} + 15 \, {\left (20 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (2\right )^{2} + 5 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (\frac {1}{5} \, x\right )^{2} + 4 \, x \log \left (2\right ) - 2 \, {\left (10 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (2\right ) + x\right )} \log \left (\frac {1}{5} \, x\right )\right )} e^{\left (e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )}}{8 \, x^{2} \log \left (2\right )^{3} - 12 \, x^{2} \log \left (2\right )^{2} \log \left (\frac {1}{5} \, x\right ) + 6 \, x^{2} \log \left (2\right ) \log \left (\frac {1}{5} \, x\right )^{2} - x^{2} \log \left (\frac {1}{5} \, x\right )^{3}} \,d x } \] Input:
integrate(((5*exp(1/2*(10*x-5)/x)*log(1/5*x)-10*log(2)*exp(1/2*(10*x-5)/x) -2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-75*exp(1/2*(10*x-5)/x)*log(1/5*x)^2+(30 0*log(2)*exp(1/2*(10*x-5)/x)+30*x)*log(1/5*x)-300*log(2)^2*exp(1/2*(10*x-5 )/x)-60*x*log(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*log(1/5*x)^3-6*x^2*log(2) *log(1/5*x)^2+12*x^2*log(2)^2*log(1/5*x)-8*x^2*log(2)^3),x, algorithm="max ima")
Output:
integrate(((10*e^(5/2*(2*x - 1)/x)*log(2) - 5*e^(5/2*(2*x - 1)/x)*log(1/5* x) + 2*x)*e^(2*e^(5/2*(2*x - 1)/x)) + 15*(20*e^(5/2*(2*x - 1)/x)*log(2)^2 + 5*e^(5/2*(2*x - 1)/x)*log(1/5*x)^2 + 4*x*log(2) - 2*(10*e^(5/2*(2*x - 1) /x)*log(2) + x)*log(1/5*x))*e^(e^(5/2*(2*x - 1)/x)))/(8*x^2*log(2)^3 - 12* x^2*log(2)^2*log(1/5*x) + 6*x^2*log(2)*log(1/5*x)^2 - x^2*log(1/5*x)^3), x )
\[ \int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx=\int { \frac {{\left (10 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (2\right ) - 5 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (\frac {1}{5} \, x\right ) + 2 \, x\right )} e^{\left (2 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )} + 15 \, {\left (20 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (2\right )^{2} + 5 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (\frac {1}{5} \, x\right )^{2} + 4 \, x \log \left (2\right ) - 2 \, {\left (10 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )} \log \left (2\right ) + x\right )} \log \left (\frac {1}{5} \, x\right )\right )} e^{\left (e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )}}{8 \, x^{2} \log \left (2\right )^{3} - 12 \, x^{2} \log \left (2\right )^{2} \log \left (\frac {1}{5} \, x\right ) + 6 \, x^{2} \log \left (2\right ) \log \left (\frac {1}{5} \, x\right )^{2} - x^{2} \log \left (\frac {1}{5} \, x\right )^{3}} \,d x } \] Input:
integrate(((5*exp(1/2*(10*x-5)/x)*log(1/5*x)-10*log(2)*exp(1/2*(10*x-5)/x) -2*x)*exp(exp(1/2*(10*x-5)/x))^2+(-75*exp(1/2*(10*x-5)/x)*log(1/5*x)^2+(30 0*log(2)*exp(1/2*(10*x-5)/x)+30*x)*log(1/5*x)-300*log(2)^2*exp(1/2*(10*x-5 )/x)-60*x*log(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*log(1/5*x)^3-6*x^2*log(2) *log(1/5*x)^2+12*x^2*log(2)^2*log(1/5*x)-8*x^2*log(2)^3),x, algorithm="gia c")
Output:
integrate(((10*e^(5/2*(2*x - 1)/x)*log(2) - 5*e^(5/2*(2*x - 1)/x)*log(1/5* x) + 2*x)*e^(2*e^(5/2*(2*x - 1)/x)) + 15*(20*e^(5/2*(2*x - 1)/x)*log(2)^2 + 5*e^(5/2*(2*x - 1)/x)*log(1/5*x)^2 + 4*x*log(2) - 2*(10*e^(5/2*(2*x - 1) /x)*log(2) + x)*log(1/5*x))*e^(e^(5/2*(2*x - 1)/x)))/(8*x^2*log(2)^3 - 12* x^2*log(2)^2*log(1/5*x) + 6*x^2*log(2)*log(1/5*x)^2 - x^2*log(1/5*x)^3), x )
Timed out. \[ \int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx=-\int -\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}}\,\left (2\,x-5\,\ln \left (\frac {x}{5}\right )\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}+10\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,\ln \left (2\right )\right )+{\mathrm {e}}^{{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}}\,\left (75\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,{\ln \left (\frac {x}{5}\right )}^2+\left (-30\,x-300\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,\ln \left (2\right )\right )\,\ln \left (\frac {x}{5}\right )+60\,x\,\ln \left (2\right )+300\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,{\ln \left (2\right )}^2\right )}{-x^2\,{\ln \left (\frac {x}{5}\right )}^3+6\,\ln \left (2\right )\,x^2\,{\ln \left (\frac {x}{5}\right )}^2-12\,{\ln \left (2\right )}^2\,x^2\,\ln \left (\frac {x}{5}\right )+8\,{\ln \left (2\right )}^3\,x^2} \,d x \] Input:
int((exp(2*exp((5*x - 5/2)/x))*(2*x - 5*log(x/5)*exp((5*x - 5/2)/x) + 10*e xp((5*x - 5/2)/x)*log(2)) + exp(exp((5*x - 5/2)/x))*(60*x*log(2) - log(x/5 )*(30*x + 300*exp((5*x - 5/2)/x)*log(2)) + 75*log(x/5)^2*exp((5*x - 5/2)/x ) + 300*exp((5*x - 5/2)/x)*log(2)^2))/(8*x^2*log(2)^3 - x^2*log(x/5)^3 - 1 2*x^2*log(x/5)*log(2)^2 + 6*x^2*log(x/5)^2*log(2)),x)
Output:
-int(-(exp(2*exp((5*x - 5/2)/x))*(2*x - 5*log(x/5)*exp((5*x - 5/2)/x) + 10 *exp((5*x - 5/2)/x)*log(2)) + exp(exp((5*x - 5/2)/x))*(60*x*log(2) - log(x /5)*(30*x + 300*exp((5*x - 5/2)/x)*log(2)) + 75*log(x/5)^2*exp((5*x - 5/2) /x) + 300*exp((5*x - 5/2)/x)*log(2)^2))/(8*x^2*log(2)^3 - x^2*log(x/5)^3 - 12*x^2*log(x/5)*log(2)^2 + 6*x^2*log(x/5)^2*log(2)), x)
Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.22 \[ \int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx=\frac {e^{\frac {e^{5}}{e^{\frac {5}{2 x}}}} \left (e^{\frac {e^{5}}{e^{\frac {5}{2 x}}}}+30 \,\mathrm {log}\left (\frac {5}{x}\right )+60 \,\mathrm {log}\left (2\right )\right )}{\mathrm {log}\left (\frac {5}{x}\right )^{2}+4 \,\mathrm {log}\left (\frac {5}{x}\right ) \mathrm {log}\left (2\right )+4 \mathrm {log}\left (2\right )^{2}} \] Input:
int(((5*exp(1/2*(10*x-5)/x)*log(1/5*x)-10*log(2)*exp(1/2*(10*x-5)/x)-2*x)* exp(exp(1/2*(10*x-5)/x))^2+(-75*exp(1/2*(10*x-5)/x)*log(1/5*x)^2+(300*log( 2)*exp(1/2*(10*x-5)/x)+30*x)*log(1/5*x)-300*log(2)^2*exp(1/2*(10*x-5)/x)-6 0*x*log(2))*exp(exp(1/2*(10*x-5)/x)))/(x^2*log(1/5*x)^3-6*x^2*log(2)*log(1 /5*x)^2+12*x^2*log(2)^2*log(1/5*x)-8*x^2*log(2)^3),x)
Output:
(e**(e**5/e**(5/(2*x)))*(e**(e**5/e**(5/(2*x))) + 30*log(5/x) + 60*log(2)) )/(log(5/x)**2 + 4*log(5/x)*log(2) + 4*log(2)**2)