Integrand size = 234, antiderivative size = 30 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\frac {5}{\log \left (\frac {\log \left (\log \left (\left (5-e^x\right )^2+(3-x) x \log (x)\right )\right )}{x}\right )} \] Output:
5/ln(ln(ln((5-exp(x))^2+(3-x)*x*ln(x)))/x)
Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\frac {5}{\log \left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(-3+x) x \log (x)\right )\right )}{x}\right )} \] Input:
Integrate[(15*x - 50*E^x*x + 10*E^(2*x)*x - 5*x^2 + (15*x - 10*x^2)*Log[x] + (-125 + 50*E^x - 5*E^(2*x) + (-15*x + 5*x^2)*Log[x])*Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]*Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)* Log[x]]])/((-25*x + 10*E^x*x - E^(2*x)*x + (-3*x^2 + x^3)*Log[x])*Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]*Log[Log[25 - 10*E^x + E^(2*x) + (3 *x - x^2)*Log[x]]]*Log[Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]] ]/x]^2),x]
Output:
5/Log[Log[Log[(-5 + E^x)^2 - (-3 + x)*x*Log[x]]]/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.
| \(\displaystyle \int \frac {-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (\left (5 x^2-15 x\right ) \log (x)+50 e^x-5 e^{2 x}-125\right ) \log \left (\left (3 x-x^2\right ) \log (x)-10 e^x+e^{2 x}+25\right ) \log \left (\log \left (\left (3 x-x^2\right ) \log (x)-10 e^x+e^{2 x}+25\right )\right )-50 e^x x+10 e^{2 x} x+15 x}{\left (\left (x^3-3 x^2\right ) \log (x)+10 e^x x-e^{2 x} x-25 x\right ) \log \left (\left (3 x-x^2\right ) \log (x)-10 e^x+e^{2 x}+25\right ) \log \left (\log \left (\left (3 x-x^2\right ) \log (x)-10 e^x+e^{2 x}+25\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (3 x-x^2\right ) \log (x)-10 e^x+e^{2 x}+25\right )\right )}{x}\right )} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (-\frac {5 \left (2 x^2 \log (x)-x+10 e^x-8 x \log (x)+3 \log (x)-47\right )}{\left (x^2 (-\log (x))-10 e^x+e^{2 x}+3 x \log (x)+25\right ) \log \left (\left (e^x-5\right )^2-(x-3) x \log (x)\right ) \log \left (\log \left (\left (e^x-5\right )^2-(x-3) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (e^x-5\right )^2-(x-3) x \log (x)\right )\right )}{x}\right )}-\frac {5 \left (2 x-\log \left (\left (e^x-5\right )^2-(x-3) x \log (x)\right ) \log \left (\log \left (\left (e^x-5\right )^2-(x-3) x \log (x)\right )\right )\right )}{x \log \left (\left (e^x-5\right )^2-(x-3) x \log (x)\right ) \log \left (\log \left (\left (e^x-5\right )^2-(x-3) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (e^x-5\right )^2-(x-3) x \log (x)\right )\right )}{x}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 235 \int \frac {1}{\left (-\log (x) x^2+3 \log (x) x-10 e^x+e^{2 x}+25\right ) \log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right )}{x}\right )}dx-50 \int \frac {e^x}{\left (-\log (x) x^2+3 \log (x) x-10 e^x+e^{2 x}+25\right ) \log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right )}{x}\right )}dx-5 \int \frac {x}{\left (\log (x) x^2-3 \log (x) x+10 e^x-e^{2 x}-25\right ) \log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right )}{x}\right )}dx+15 \int \frac {\log (x)}{\left (\log (x) x^2-3 \log (x) x+10 e^x-e^{2 x}-25\right ) \log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right )}{x}\right )}dx-40 \int \frac {x \log (x)}{\left (\log (x) x^2-3 \log (x) x+10 e^x-e^{2 x}-25\right ) \log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right )}{x}\right )}dx+10 \int \frac {x^2 \log (x)}{\left (\log (x) x^2-3 \log (x) x+10 e^x-e^{2 x}-25\right ) \log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right )}{x}\right )}dx+5 \int \frac {1}{x \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right )}{x}\right )}dx-10 \int \frac {1}{\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right ) \log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (\left (-5+e^x\right )^2-(x-3) x \log (x)\right )\right )}{x}\right )}dx\) |
Input:
Int[(15*x - 50*E^x*x + 10*E^(2*x)*x - 5*x^2 + (15*x - 10*x^2)*Log[x] + (-1 25 + 50*E^x - 5*E^(2*x) + (-15*x + 5*x^2)*Log[x])*Log[25 - 10*E^x + E^(2*x ) + (3*x - x^2)*Log[x]]*Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x] ]])/((-25*x + 10*E^x*x - E^(2*x)*x + (-3*x^2 + x^3)*Log[x])*Log[25 - 10*E^ x + E^(2*x) + (3*x - x^2)*Log[x]]*Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x ^2)*Log[x]]]*Log[Log[Log[25 - 10*E^x + E^(2*x) + (3*x - x^2)*Log[x]]]/x]^2 ),x]
Output:
$Aborted
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.10 (sec) , antiderivative size = 250, normalized size of antiderivative = 8.33
\[\frac {10 i}{\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )\right ) \operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right )-\pi \,\operatorname {csgn}\left (\frac {i}{x}\right ) {\operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )\right ) {\operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right )}^{2}+\pi {\operatorname {csgn}\left (\frac {i \ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )}{x}\right )}^{3}-2 i \ln \left (x \right )+2 i \ln \left (\ln \left (\ln \left (\left (-x^{2}+3 x \right ) \ln \left (x \right )+{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+25\right )\right )\right )}\]
Input:
int((((5*x^2-15*x)*ln(x)-5*exp(x)^2+50*exp(x)-125)*ln((-x^2+3*x)*ln(x)+exp (x)^2-10*exp(x)+25)*ln(ln((-x^2+3*x)*ln(x)+exp(x)^2-10*exp(x)+25))+(-10*x^ 2+15*x)*ln(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3*x^2)*ln(x)-x*e xp(x)^2+10*exp(x)*x-25*x)/ln((-x^2+3*x)*ln(x)+exp(x)^2-10*exp(x)+25)/ln(ln ((-x^2+3*x)*ln(x)+exp(x)^2-10*exp(x)+25))/ln(ln(ln((-x^2+3*x)*ln(x)+exp(x) ^2-10*exp(x)+25))/x)^2,x)
Output:
10*I/(Pi*csgn(I/x)*csgn(I*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))* csgn(I/x*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))-Pi*csgn(I/x)*csgn (I/x*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))^2-Pi*csgn(I*ln(ln((-x ^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25)))*csgn(I/x*ln(ln((-x^2+3*x)*ln(x)+exp (2*x)-10*exp(x)+25)))^2+Pi*csgn(I/x*ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp (x)+25)))^3-2*I*ln(x)+2*I*ln(ln(ln((-x^2+3*x)*ln(x)+exp(2*x)-10*exp(x)+25) )))
Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\frac {5}{\log \left (\frac {\log \left (\log \left (-{\left (x^{2} - 3 \, x\right )} \log \left (x\right ) + e^{\left (2 \, x\right )} - 10 \, e^{x} + 25\right )\right )}{x}\right )} \] Input:
integrate((((5*x^2-15*x)*log(x)-5*exp(x)^2+50*exp(x)-125)*log((-x^2+3*x)*l og(x)+exp(x)^2-10*exp(x)+25)*log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+ 25))+(-10*x^2+15*x)*log(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3*x ^2)*log(x)-x*exp(x)^2+10*exp(x)*x-25*x)/log((-x^2+3*x)*log(x)+exp(x)^2-10* exp(x)+25)/log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/log(log(log(( -x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/x)^2,x, algorithm="fricas")
Output:
5/log(log(log(-(x^2 - 3*x)*log(x) + e^(2*x) - 10*e^x + 25))/x)
Timed out. \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\text {Timed out} \] Input:
integrate((((5*x**2-15*x)*ln(x)-5*exp(x)**2+50*exp(x)-125)*ln((-x**2+3*x)* ln(x)+exp(x)**2-10*exp(x)+25)*ln(ln((-x**2+3*x)*ln(x)+exp(x)**2-10*exp(x)+ 25))+(-10*x**2+15*x)*ln(x)+10*x*exp(x)**2-50*exp(x)*x-5*x**2+15*x)/((x**3- 3*x**2)*ln(x)-x*exp(x)**2+10*exp(x)*x-25*x)/ln((-x**2+3*x)*ln(x)+exp(x)**2 -10*exp(x)+25)/ln(ln((-x**2+3*x)*ln(x)+exp(x)**2-10*exp(x)+25))/ln(ln(ln(( -x**2+3*x)*ln(x)+exp(x)**2-10*exp(x)+25))/x)**2,x)
Output:
Timed out
Time = 0.55 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=-\frac {5}{\log \left (x\right ) - \log \left (\log \left (\log \left (-{\left (x^{2} - 3 \, x\right )} \log \left (x\right ) + e^{\left (2 \, x\right )} - 10 \, e^{x} + 25\right )\right )\right )} \] Input:
integrate((((5*x^2-15*x)*log(x)-5*exp(x)^2+50*exp(x)-125)*log((-x^2+3*x)*l og(x)+exp(x)^2-10*exp(x)+25)*log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+ 25))+(-10*x^2+15*x)*log(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3*x ^2)*log(x)-x*exp(x)^2+10*exp(x)*x-25*x)/log((-x^2+3*x)*log(x)+exp(x)^2-10* exp(x)+25)/log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/log(log(log(( -x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/x)^2,x, algorithm="maxima")
Output:
-5/(log(x) - log(log(log(-(x^2 - 3*x)*log(x) + e^(2*x) - 10*e^x + 25))))
Time = 3.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=-\frac {5}{\log \left (x\right ) - \log \left (\log \left (\log \left (-x^{2} \log \left (x\right ) + 3 \, x \log \left (x\right ) + e^{\left (2 \, x\right )} - 10 \, e^{x} + 25\right )\right )\right )} \] Input:
integrate((((5*x^2-15*x)*log(x)-5*exp(x)^2+50*exp(x)-125)*log((-x^2+3*x)*l og(x)+exp(x)^2-10*exp(x)+25)*log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+ 25))+(-10*x^2+15*x)*log(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3*x ^2)*log(x)-x*exp(x)^2+10*exp(x)*x-25*x)/log((-x^2+3*x)*log(x)+exp(x)^2-10* exp(x)+25)/log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/log(log(log(( -x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/x)^2,x, algorithm="giac")
Output:
-5/(log(x) - log(log(log(-x^2*log(x) + 3*x*log(x) + e^(2*x) - 10*e^x + 25) )))
Time = 3.96 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\frac {5}{\ln \left (\frac {\ln \left (\ln \left ({\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^x+\ln \left (x\right )\,\left (3\,x-x^2\right )+25\right )\right )}{x}\right )} \] Input:
int(-(15*x + 10*x*exp(2*x) + log(x)*(15*x - 10*x^2) - 50*x*exp(x) - 5*x^2 - log(log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25))*log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25)*(5*exp(2*x) - 50*exp(x) + log(x)*(15* x - 5*x^2) + 125))/(log(log(log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25))/x)^2*log(log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25))*log(e xp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25)*(25*x + log(x)*(3*x^2 - x^3 ) + x*exp(2*x) - 10*x*exp(x))),x)
Output:
5/log(log(log(exp(2*x) - 10*exp(x) + log(x)*(3*x - x^2) + 25))/x)
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {15 x-50 e^x x+10 e^{2 x} x-5 x^2+\left (15 x-10 x^2\right ) \log (x)+\left (-125+50 e^x-5 e^{2 x}+\left (-15 x+5 x^2\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{\left (-25 x+10 e^x x-e^{2 x} x+\left (-3 x^2+x^3\right ) \log (x)\right ) \log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right ) \log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right ) \log ^2\left (\frac {\log \left (\log \left (25-10 e^x+e^{2 x}+\left (3 x-x^2\right ) \log (x)\right )\right )}{x}\right )} \, dx=\frac {5}{\mathrm {log}\left (\frac {\mathrm {log}\left (\mathrm {log}\left (e^{2 x}-10 e^{x}-\mathrm {log}\left (x \right ) x^{2}+3 \,\mathrm {log}\left (x \right ) x +25\right )\right )}{x}\right )} \] Input:
int((((5*x^2-15*x)*log(x)-5*exp(x)^2+50*exp(x)-125)*log((-x^2+3*x)*log(x)+ exp(x)^2-10*exp(x)+25)*log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))+( -10*x^2+15*x)*log(x)+10*x*exp(x)^2-50*exp(x)*x-5*x^2+15*x)/((x^3-3*x^2)*lo g(x)-x*exp(x)^2+10*exp(x)*x-25*x)/log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x) +25)/log(log((-x^2+3*x)*log(x)+exp(x)^2-10*exp(x)+25))/log(log(log((-x^2+3 *x)*log(x)+exp(x)^2-10*exp(x)+25))/x)^2,x)
Output:
5/log(log(log(e**(2*x) - 10*e**x - log(x)*x**2 + 3*log(x)*x + 25))/x)