Integrand size = 233, antiderivative size = 31 \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=1+\left (3-\frac {\log \left (\log \left (\left (4+2 \left (3+e^x-x\right )\right ) x+\log (x)\right )\right )}{\log (x)}\right )^2 \] Output:
1+(3-ln(ln((10+2*exp(x)-2*x)*x+ln(x)))/ln(x))^2
Time = 0.12 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=2 \left (-\frac {3 \log \left (\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )\right )}{\log (x)}+\frac {\log ^2\left (\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )\right )}{2 \log ^2(x)}\right ) \] Input:
Integrate[((-6 - 60*x + 24*x^2 + E^x*(-12*x - 12*x^2))*Log[x]^2 + ((2 + 20 *x - 8*x^2 + E^x*(4*x + 4*x^2))*Log[x] + ((60*x + 12*E^x*x - 12*x^2)*Log[x ] + 6*Log[x]^2)*Log[10*x + 2*E^x*x - 2*x^2 + Log[x]])*Log[Log[10*x + 2*E^x *x - 2*x^2 + Log[x]]] + (-20*x - 4*E^x*x + 4*x^2 - 2*Log[x])*Log[10*x + 2* E^x*x - 2*x^2 + Log[x]]*Log[Log[10*x + 2*E^x*x - 2*x^2 + Log[x]]]^2)/(((10 *x^2 + 2*E^x*x^2 - 2*x^3)*Log[x]^3 + x*Log[x]^4)*Log[10*x + 2*E^x*x - 2*x^ 2 + Log[x]]),x]
Output:
2*((-3*Log[Log[-2*x*(-5 - E^x + x) + Log[x]]])/Log[x] + Log[Log[-2*x*(-5 - E^x + x) + Log[x]]]^2/(2*Log[x]^2))
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.
| \(\displaystyle \int \frac {\left (24 x^2+e^x \left (-12 x^2-12 x\right )-60 x-6\right ) \log ^2(x)+\left (4 x^2-4 e^x x-20 x-2 \log (x)\right ) \log \left (-2 x^2+2 e^x x+10 x+\log (x)\right ) \log ^2\left (\log \left (-2 x^2+2 e^x x+10 x+\log (x)\right )\right )+\left (\left (\left (-12 x^2+12 e^x x+60 x\right ) \log (x)+6 \log ^2(x)\right ) \log \left (-2 x^2+2 e^x x+10 x+\log (x)\right )+\left (-8 x^2+e^x \left (4 x^2+4 x\right )+20 x+2\right ) \log (x)\right ) \log \left (\log \left (-2 x^2+2 e^x x+10 x+\log (x)\right )\right )}{\left (\left (-2 x^3+2 e^x x^2+10 x^2\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (-2 x^2+2 e^x x+10 x+\log (x)\right )} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {\left (24 x^2+e^x \left (-12 x^2-12 x\right )-60 x-6\right ) \log ^2(x)+\left (4 x^2-4 e^x x-20 x-2 \log (x)\right ) \log \left (-2 x^2+2 e^x x+10 x+\log (x)\right ) \log ^2\left (\log \left (-2 x^2+2 e^x x+10 x+\log (x)\right )\right )+\left (\left (\left (-12 x^2+12 e^x x+60 x\right ) \log (x)+6 \log ^2(x)\right ) \log \left (-2 x^2+2 e^x x+10 x+\log (x)\right )+\left (-8 x^2+e^x \left (4 x^2+4 x\right )+20 x+2\right ) \log (x)\right ) \log \left (\log \left (-2 x^2+2 e^x x+10 x+\log (x)\right )\right )}{x \log ^3(x) \left (-2 x^2+2 e^x x+10 x+\log (x)\right ) \log \left (-2 x^2+2 e^x x+10 x+\log (x)\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 \left (2 x^3-12 x^2-x \log (x)-\log (x)+1\right ) \left (3 \log (x)-\log \left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )\right )}{x \left (2 x^2-2 e^x x-10 x-\log (x)\right ) \log ^2(x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}-\frac {2 \left (3 \log (x)-\log \left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )\right ) \left (x \log (x)+\log (x)-\log \left (\log (x)-2 x \left (x-e^x-5\right )\right ) \log \left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )\right )}{x \log ^3(x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -2 \int \frac {\log \left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )}{x \left (2 x^2-2 e^x x-10 x-\log (x)\right ) \log ^2(x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx+24 \int \frac {x \log \left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )}{\left (2 x^2-2 e^x x-10 x-\log (x)\right ) \log ^2(x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx-4 \int \frac {x^2 \log \left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )}{\left (2 x^2-2 e^x x-10 x-\log (x)\right ) \log ^2(x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx-6 \int \frac {1}{\left (2 x^2-2 e^x x-10 x-\log (x)\right ) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx-6 \int \frac {1}{x \left (2 x^2-2 e^x x-10 x-\log (x)\right ) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx+6 \int \frac {1}{x \left (2 x^2-2 e^x x-10 x-\log (x)\right ) \log (x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx-72 \int \frac {x}{\left (2 x^2-2 e^x x-10 x-\log (x)\right ) \log (x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx+12 \int \frac {x^2}{\left (2 x^2-2 e^x x-10 x-\log (x)\right ) \log (x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx+2 \int \frac {\log \left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )}{\left (2 x^2-2 e^x x-10 x-\log (x)\right ) \log (x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx+2 \int \frac {\log \left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )}{x \left (2 x^2-2 e^x x-10 x-\log (x)\right ) \log (x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx+6 \int \frac {\log \left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )}{x \log ^2(x)}dx+2 \int \frac {\log \left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )}{\log ^2(x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx+2 \int \frac {\log \left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )}{x \log ^2(x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx-2 \int \frac {\log ^2\left (\log \left (\log (x)-2 x \left (x-e^x-5\right )\right )\right )}{x \log ^3(x)}dx-6 \int \frac {1}{\log (x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx-6 \int \frac {1}{x \log (x) \log \left (\log (x)-2 x \left (x-e^x-5\right )\right )}dx\) |
Input:
Int[((-6 - 60*x + 24*x^2 + E^x*(-12*x - 12*x^2))*Log[x]^2 + ((2 + 20*x - 8 *x^2 + E^x*(4*x + 4*x^2))*Log[x] + ((60*x + 12*E^x*x - 12*x^2)*Log[x] + 6* Log[x]^2)*Log[10*x + 2*E^x*x - 2*x^2 + Log[x]])*Log[Log[10*x + 2*E^x*x - 2 *x^2 + Log[x]]] + (-20*x - 4*E^x*x + 4*x^2 - 2*Log[x])*Log[10*x + 2*E^x*x - 2*x^2 + Log[x]]*Log[Log[10*x + 2*E^x*x - 2*x^2 + Log[x]]]^2)/(((10*x^2 + 2*E^x*x^2 - 2*x^3)*Log[x]^3 + x*Log[x]^4)*Log[10*x + 2*E^x*x - 2*x^2 + Lo g[x]]),x]
Output:
$Aborted
Time = 198.98 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65
| method | result | size |
| risch | \(\frac {{\ln \left (\ln \left (\ln \left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )\right )}^{2}}{\ln \left (x \right )^{2}}-\frac {6 \ln \left (\ln \left (\ln \left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )\right )}{\ln \left (x \right )}\) | \(51\) |
| parallelrisch | \(\frac {-24 \ln \left (x \right ) \ln \left (\ln \left (\ln \left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )\right )+4 {\ln \left (\ln \left (\ln \left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )\right )}^{2}}{4 \ln \left (x \right )^{2}}\) | \(52\) |
Input:
int(((-2*ln(x)-4*exp(x)*x+4*x^2-20*x)*ln(ln(x)+2*exp(x)*x-2*x^2+10*x)*ln(l n(ln(x)+2*exp(x)*x-2*x^2+10*x))^2+((6*ln(x)^2+(12*exp(x)*x-12*x^2+60*x)*ln (x))*ln(ln(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)-8*x^2+20*x+2)*ln( x))*ln(ln(ln(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x)*exp(x)+24*x^2-60*x -6)*ln(x)^2)/(x*ln(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)*ln(x)^3)/ln(ln(x)+2*ex p(x)*x-2*x^2+10*x),x,method=_RETURNVERBOSE)
Output:
1/ln(x)^2*ln(ln(ln(x)+2*exp(x)*x-2*x^2+10*x))^2-6/ln(x)*ln(ln(ln(x)+2*exp( x)*x-2*x^2+10*x))
Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=-\frac {6 \, \log \left (x\right ) \log \left (\log \left (-2 \, x^{2} + 2 \, x e^{x} + 10 \, x + \log \left (x\right )\right )\right ) - \log \left (\log \left (-2 \, x^{2} + 2 \, x e^{x} + 10 \, x + \log \left (x\right )\right )\right )^{2}}{\log \left (x\right )^{2}} \] Input:
integrate(((-2*log(x)-4*exp(x)*x+4*x^2-20*x)*log(log(x)+2*exp(x)*x-2*x^2+1 0*x)*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))^2+((6*log(x)^2+(12*exp(x)*x-12 *x^2+60*x)*log(x))*log(log(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)-8 *x^2+20*x+2)*log(x))*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x )*exp(x)+24*x^2-60*x-6)*log(x)^2)/(x*log(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)* log(x)^3)/log(log(x)+2*exp(x)*x-2*x^2+10*x),x, algorithm="fricas")
Output:
-(6*log(x)*log(log(-2*x^2 + 2*x*e^x + 10*x + log(x))) - log(log(-2*x^2 + 2 *x*e^x + 10*x + log(x)))^2)/log(x)^2
Timed out. \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=\text {Timed out} \] Input:
integrate(((-2*ln(x)-4*exp(x)*x+4*x**2-20*x)*ln(ln(x)+2*exp(x)*x-2*x**2+10 *x)*ln(ln(ln(x)+2*exp(x)*x-2*x**2+10*x))**2+((6*ln(x)**2+(12*exp(x)*x-12*x **2+60*x)*ln(x))*ln(ln(x)+2*exp(x)*x-2*x**2+10*x)+((4*x**2+4*x)*exp(x)-8*x **2+20*x+2)*ln(x))*ln(ln(ln(x)+2*exp(x)*x-2*x**2+10*x))+((-12*x**2-12*x)*e xp(x)+24*x**2-60*x-6)*ln(x)**2)/(x*ln(x)**4+(2*exp(x)*x**2-2*x**3+10*x**2) *ln(x)**3)/ln(ln(x)+2*exp(x)*x-2*x**2+10*x),x)
Output:
Timed out
Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=-\frac {6 \, \log \left (x\right ) \log \left (\log \left (-2 \, x^{2} + 2 \, x e^{x} + 10 \, x + \log \left (x\right )\right )\right ) - \log \left (\log \left (-2 \, x^{2} + 2 \, x e^{x} + 10 \, x + \log \left (x\right )\right )\right )^{2}}{\log \left (x\right )^{2}} \] Input:
integrate(((-2*log(x)-4*exp(x)*x+4*x^2-20*x)*log(log(x)+2*exp(x)*x-2*x^2+1 0*x)*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))^2+((6*log(x)^2+(12*exp(x)*x-12 *x^2+60*x)*log(x))*log(log(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)-8 *x^2+20*x+2)*log(x))*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x )*exp(x)+24*x^2-60*x-6)*log(x)^2)/(x*log(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)* log(x)^3)/log(log(x)+2*exp(x)*x-2*x^2+10*x),x, algorithm="maxima")
Output:
-(6*log(x)*log(log(-2*x^2 + 2*x*e^x + 10*x + log(x))) - log(log(-2*x^2 + 2 *x*e^x + 10*x + log(x)))^2)/log(x)^2
Exception generated. \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(((-2*log(x)-4*exp(x)*x+4*x^2-20*x)*log(log(x)+2*exp(x)*x-2*x^2+1 0*x)*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))^2+((6*log(x)^2+(12*exp(x)*x-12 *x^2+60*x)*log(x))*log(log(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)-8 *x^2+20*x+2)*log(x))*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x )*exp(x)+24*x^2-60*x-6)*log(x)^2)/(x*log(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)* log(x)^3)/log(log(x)+2*exp(x)*x-2*x^2+10*x),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Sign error %%%{ln(w),0%%%}Sign erro r %%%{ln(w),0%%%}Sign error %%%{ln(w),0%%%}Done
Time = 3.64 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=\frac {\ln \left (\ln \left (10\,x+\ln \left (x\right )+2\,x\,{\mathrm {e}}^x-2\,x^2\right )\right )\,\left (\ln \left (\ln \left (10\,x+\ln \left (x\right )+2\,x\,{\mathrm {e}}^x-2\,x^2\right )\right )-6\,\ln \left (x\right )\right )}{{\ln \left (x\right )}^2} \] Input:
int(-(log(x)^2*(60*x + exp(x)*(12*x + 12*x^2) - 24*x^2 + 6) - log(log(10*x + log(x) + 2*x*exp(x) - 2*x^2))*(log(10*x + log(x) + 2*x*exp(x) - 2*x^2)* (6*log(x)^2 + log(x)*(60*x + 12*x*exp(x) - 12*x^2)) + log(x)*(20*x + exp(x )*(4*x + 4*x^2) - 8*x^2 + 2)) + log(10*x + log(x) + 2*x*exp(x) - 2*x^2)*lo g(log(10*x + log(x) + 2*x*exp(x) - 2*x^2))^2*(20*x + 2*log(x) + 4*x*exp(x) - 4*x^2))/(log(10*x + log(x) + 2*x*exp(x) - 2*x^2)*(x*log(x)^4 + log(x)^3 *(2*x^2*exp(x) + 10*x^2 - 2*x^3))),x)
Output:
(log(log(10*x + log(x) + 2*x*exp(x) - 2*x^2))*(log(log(10*x + log(x) + 2*x *exp(x) - 2*x^2)) - 6*log(x)))/log(x)^2
\[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=\int \frac {\left (-2 \,\mathrm {log}\left (x \right )-4 \,{\mathrm e}^{x} x +4 x^{2}-20 x \right ) \mathrm {log}\left (\mathrm {log}\left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right ) {\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )\right )}^{2}+\left (\left (6 \mathrm {log}\left (x \right )^{2}+\left (12 \,{\mathrm e}^{x} x -12 x^{2}+60 x \right ) \mathrm {log}\left (x \right )\right ) \mathrm {log}\left (\mathrm {log}\left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )+\left (\left (4 x^{2}+4 x \right ) {\mathrm e}^{x}-8 x^{2}+20 x +2\right ) \mathrm {log}\left (x \right )\right ) \mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )\right )+\left (\left (-12 x^{2}-12 x \right ) {\mathrm e}^{x}+24 x^{2}-60 x -6\right ) \mathrm {log}\left (x \right )^{2}}{\left (x \mathrm {log}\left (x \right )^{4}+\left (2 \,{\mathrm e}^{x} x^{2}-2 x^{3}+10 x^{2}\right ) \mathrm {log}\left (x \right )^{3}\right ) \mathrm {log}\left (\mathrm {log}\left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )}d x \] Input:
int(((-2*log(x)-4*exp(x)*x+4*x^2-20*x)*log(log(x)+2*exp(x)*x-2*x^2+10*x)*l og(log(log(x)+2*exp(x)*x-2*x^2+10*x))^2+((6*log(x)^2+(12*exp(x)*x-12*x^2+6 0*x)*log(x))*log(log(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)-8*x^2+2 0*x+2)*log(x))*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x)*exp( x)+24*x^2-60*x-6)*log(x)^2)/(x*log(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)*log(x) ^3)/log(log(x)+2*exp(x)*x-2*x^2+10*x),x)
Output:
int(((-2*log(x)-4*exp(x)*x+4*x^2-20*x)*log(log(x)+2*exp(x)*x-2*x^2+10*x)*l og(log(log(x)+2*exp(x)*x-2*x^2+10*x))^2+((6*log(x)^2+(12*exp(x)*x-12*x^2+6 0*x)*log(x))*log(log(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)-8*x^2+2 0*x+2)*log(x))*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x)*exp( x)+24*x^2-60*x-6)*log(x)^2)/(x*log(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)*log(x) ^3)/log(log(x)+2*exp(x)*x-2*x^2+10*x),x)