Integrand size = 94, antiderivative size = 19 \[ \int \frac {e^x (40-20 x)+e^x (10-5 x) \log (-2+x)+\frac {e^{4+\log ^2(4+\log (-2+x))} \left (e^x (-12+4 x)+e^x (-2+x) \log (-2+x)+2 e^x \log (4+\log (-2+x))\right )}{(4+\log (-2+x))^4}}{-8+4 x+(-2+x) \log (-2+x)} \, dx=e^x \left (-5+e^{(-2+\log (4+\log (-2+x)))^2}\right ) \] Output:
exp(x)*(exp((-2+ln(ln(-2+x)+4))^2)-5)
Time = 5.11 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.53 \[ \int \frac {e^x (40-20 x)+e^x (10-5 x) \log (-2+x)+\frac {e^{4+\log ^2(4+\log (-2+x))} \left (e^x (-12+4 x)+e^x (-2+x) \log (-2+x)+2 e^x \log (4+\log (-2+x))\right )}{(4+\log (-2+x))^4}}{-8+4 x+(-2+x) \log (-2+x)} \, dx=-5 e^x+\frac {e^{4+x+\log ^2(4+\log (-2+x))}}{(4+\log (-2+x))^4} \] Input:
Integrate[(E^x*(40 - 20*x) + E^x*(10 - 5*x)*Log[-2 + x] + (E^(4 + Log[4 + Log[-2 + x]]^2)*(E^x*(-12 + 4*x) + E^x*(-2 + x)*Log[-2 + x] + 2*E^x*Log[4 + Log[-2 + x]]))/(4 + Log[-2 + x])^4)/(-8 + 4*x + (-2 + x)*Log[-2 + x]),x]
Output:
-5*E^x + E^(4 + x + Log[4 + Log[-2 + x]]^2)/(4 + Log[-2 + x])^4
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.
\(\displaystyle \int \frac {e^x (40-20 x)+\frac {e^{\log ^2(\log (x-2)+4)+4} \left (e^x (4 x-12)+e^x (x-2) \log (x-2)+2 e^x \log (\log (x-2)+4)\right )}{(\log (x-2)+4)^4}+e^x (10-5 x) \log (x-2)}{4 x+(x-2) \log (x-2)-8} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-e^x (40-20 x)-\frac {e^{\log ^2(\log (x-2)+4)+4} \left (e^x (4 x-12)+e^x (x-2) \log (x-2)+2 e^x \log (\log (x-2)+4)\right )}{(\log (x-2)+4)^4}-e^x (10-5 x) \log (x-2)}{(2-x) (\log (x-2)+4)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x+\log ^2(\log (x-2)+4)+4} (4 x+x \log (x-2)-2 \log (x-2)+2 \log (\log (x-2)+4)-12)}{(x-2) (\log (x-2)+4)^5}-5 e^x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 \int \frac {e^{\log ^2(\log (x-2)+4)+x+4}}{(x-2) (\log (x-2)+4)^5}dx+\int \frac {e^{\log ^2(\log (x-2)+4)+x+4}}{(\log (x-2)+4)^4}dx+2 \int \frac {e^{\log ^2(\log (x-2)+4)+x+4} \log (\log (x-2)+4)}{(x-2) (\log (x-2)+4)^5}dx-5 e^x\) |
Input:
Int[(E^x*(40 - 20*x) + E^x*(10 - 5*x)*Log[-2 + x] + (E^(4 + Log[4 + Log[-2 + x]]^2)*(E^x*(-12 + 4*x) + E^x*(-2 + x)*Log[-2 + x] + 2*E^x*Log[4 + Log[ -2 + x]]))/(4 + Log[-2 + x])^4)/(-8 + 4*x + (-2 + x)*Log[-2 + x]),x]
Output:
$Aborted
Time = 3.18 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.47
method | result | size |
risch | \(\frac {{\mathrm e}^{x +\ln \left (\ln \left (-2+x \right )+4\right )^{2}+4}}{\left (\ln \left (-2+x \right )+4\right )^{4}}-5 \,{\mathrm e}^{x}\) | \(28\) |
parallelrisch | \({\mathrm e}^{x} {\mathrm e}^{\ln \left (\ln \left (-2+x \right )+4\right )^{2}-4 \ln \left (\ln \left (-2+x \right )+4\right )+4}-5 \,{\mathrm e}^{x}\) | \(30\) |
Input:
int(((2*exp(x)*ln(ln(-2+x)+4)+(-2+x)*exp(x)*ln(-2+x)+(4*x-12)*exp(x))*exp( ln(ln(-2+x)+4)^2-4*ln(ln(-2+x)+4)+4)+(-5*x+10)*exp(x)*ln(-2+x)+(-20*x+40)* exp(x))/((-2+x)*ln(-2+x)+4*x-8),x,method=_RETURNVERBOSE)
Output:
1/(ln(-2+x)+4)^4*exp(x+ln(ln(-2+x)+4)^2+4)-5*exp(x)
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {e^x (40-20 x)+e^x (10-5 x) \log (-2+x)+\frac {e^{4+\log ^2(4+\log (-2+x))} \left (e^x (-12+4 x)+e^x (-2+x) \log (-2+x)+2 e^x \log (4+\log (-2+x))\right )}{(4+\log (-2+x))^4}}{-8+4 x+(-2+x) \log (-2+x)} \, dx=e^{\left (\log \left (\log \left (x - 2\right ) + 4\right )^{2} + x - 4 \, \log \left (\log \left (x - 2\right ) + 4\right ) + 4\right )} - 5 \, e^{x} \] Input:
integrate(((2*exp(x)*log(log(-2+x)+4)+(-2+x)*exp(x)*log(-2+x)+(4*x-12)*exp (x))*exp(log(log(-2+x)+4)^2-4*log(log(-2+x)+4)+4)+(-5*x+10)*exp(x)*log(-2+ x)+(-20*x+40)*exp(x))/((-2+x)*log(-2+x)+4*x-8),x, algorithm="fricas")
Output:
e^(log(log(x - 2) + 4)^2 + x - 4*log(log(x - 2) + 4) + 4) - 5*e^x
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (17) = 34\).
Time = 0.48 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.79 \[ \int \frac {e^x (40-20 x)+e^x (10-5 x) \log (-2+x)+\frac {e^{4+\log ^2(4+\log (-2+x))} \left (e^x (-12+4 x)+e^x (-2+x) \log (-2+x)+2 e^x \log (4+\log (-2+x))\right )}{(4+\log (-2+x))^4}}{-8+4 x+(-2+x) \log (-2+x)} \, dx=- 5 e^{x} + \frac {e^{x} e^{\log {\left (\log {\left (x - 2 \right )} + 4 \right )}^{2} + 4}}{\log {\left (x - 2 \right )}^{4} + 16 \log {\left (x - 2 \right )}^{3} + 96 \log {\left (x - 2 \right )}^{2} + 256 \log {\left (x - 2 \right )} + 256} \] Input:
integrate(((2*exp(x)*ln(ln(-2+x)+4)+(-2+x)*exp(x)*ln(-2+x)+(4*x-12)*exp(x) )*exp(ln(ln(-2+x)+4)**2-4*ln(ln(-2+x)+4)+4)+(-5*x+10)*exp(x)*ln(-2+x)+(-20 *x+40)*exp(x))/((-2+x)*ln(-2+x)+4*x-8),x)
Output:
-5*exp(x) + exp(x)*exp(log(log(x - 2) + 4)**2 + 4)/(log(x - 2)**4 + 16*log (x - 2)**3 + 96*log(x - 2)**2 + 256*log(x - 2) + 256)
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (17) = 34\).
Time = 0.09 (sec) , antiderivative size = 92, normalized size of antiderivative = 4.84 \[ \int \frac {e^x (40-20 x)+e^x (10-5 x) \log (-2+x)+\frac {e^{4+\log ^2(4+\log (-2+x))} \left (e^x (-12+4 x)+e^x (-2+x) \log (-2+x)+2 e^x \log (4+\log (-2+x))\right )}{(4+\log (-2+x))^4}}{-8+4 x+(-2+x) \log (-2+x)} \, dx=-\frac {5 \, e^{x} \log \left (x - 2\right )^{4} + 80 \, e^{x} \log \left (x - 2\right )^{3} + 480 \, e^{x} \log \left (x - 2\right )^{2} + 1280 \, e^{x} \log \left (x - 2\right ) - e^{\left (\log \left (\log \left (x - 2\right ) + 4\right )^{2} + x + 4\right )} + 1280 \, e^{x}}{\log \left (x - 2\right )^{4} + 16 \, \log \left (x - 2\right )^{3} + 96 \, \log \left (x - 2\right )^{2} + 256 \, \log \left (x - 2\right ) + 256} \] Input:
integrate(((2*exp(x)*log(log(-2+x)+4)+(-2+x)*exp(x)*log(-2+x)+(4*x-12)*exp (x))*exp(log(log(-2+x)+4)^2-4*log(log(-2+x)+4)+4)+(-5*x+10)*exp(x)*log(-2+ x)+(-20*x+40)*exp(x))/((-2+x)*log(-2+x)+4*x-8),x, algorithm="maxima")
Output:
-(5*e^x*log(x - 2)^4 + 80*e^x*log(x - 2)^3 + 480*e^x*log(x - 2)^2 + 1280*e ^x*log(x - 2) - e^(log(log(x - 2) + 4)^2 + x + 4) + 1280*e^x)/(log(x - 2)^ 4 + 16*log(x - 2)^3 + 96*log(x - 2)^2 + 256*log(x - 2) + 256)
Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (17) = 34\).
Time = 0.16 (sec) , antiderivative size = 92, normalized size of antiderivative = 4.84 \[ \int \frac {e^x (40-20 x)+e^x (10-5 x) \log (-2+x)+\frac {e^{4+\log ^2(4+\log (-2+x))} \left (e^x (-12+4 x)+e^x (-2+x) \log (-2+x)+2 e^x \log (4+\log (-2+x))\right )}{(4+\log (-2+x))^4}}{-8+4 x+(-2+x) \log (-2+x)} \, dx=-\frac {5 \, e^{x} \log \left (x - 2\right )^{4} + 80 \, e^{x} \log \left (x - 2\right )^{3} + 480 \, e^{x} \log \left (x - 2\right )^{2} + 1280 \, e^{x} \log \left (x - 2\right ) - e^{\left (\log \left (\log \left (x - 2\right ) + 4\right )^{2} + x + 4\right )} + 1280 \, e^{x}}{\log \left (x - 2\right )^{4} + 16 \, \log \left (x - 2\right )^{3} + 96 \, \log \left (x - 2\right )^{2} + 256 \, \log \left (x - 2\right ) + 256} \] Input:
integrate(((2*exp(x)*log(log(-2+x)+4)+(-2+x)*exp(x)*log(-2+x)+(4*x-12)*exp (x))*exp(log(log(-2+x)+4)^2-4*log(log(-2+x)+4)+4)+(-5*x+10)*exp(x)*log(-2+ x)+(-20*x+40)*exp(x))/((-2+x)*log(-2+x)+4*x-8),x, algorithm="giac")
Output:
-(5*e^x*log(x - 2)^4 + 80*e^x*log(x - 2)^3 + 480*e^x*log(x - 2)^2 + 1280*e ^x*log(x - 2) - e^(log(log(x - 2) + 4)^2 + x + 4) + 1280*e^x)/(log(x - 2)^ 4 + 16*log(x - 2)^3 + 96*log(x - 2)^2 + 256*log(x - 2) + 256)
Time = 4.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 3.05 \[ \int \frac {e^x (40-20 x)+e^x (10-5 x) \log (-2+x)+\frac {e^{4+\log ^2(4+\log (-2+x))} \left (e^x (-12+4 x)+e^x (-2+x) \log (-2+x)+2 e^x \log (4+\log (-2+x))\right )}{(4+\log (-2+x))^4}}{-8+4 x+(-2+x) \log (-2+x)} \, dx=-\frac {{\mathrm {e}}^x\,\left (1280\,\ln \left (x-2\right )-{\mathrm {e}}^4\,{\mathrm {e}}^{{\ln \left (\ln \left (x-2\right )+4\right )}^2}+480\,{\ln \left (x-2\right )}^2+80\,{\ln \left (x-2\right )}^3+5\,{\ln \left (x-2\right )}^4+1280\right )}{{\left (\ln \left (x-2\right )+4\right )}^4} \] Input:
int(-(exp(x)*(20*x - 40) - exp(log(log(x - 2) + 4)^2 - 4*log(log(x - 2) + 4) + 4)*(2*log(log(x - 2) + 4)*exp(x) + exp(x)*(4*x - 12) + log(x - 2)*exp (x)*(x - 2)) + log(x - 2)*exp(x)*(5*x - 10))/(4*x + log(x - 2)*(x - 2) - 8 ),x)
Output:
-(exp(x)*(1280*log(x - 2) - exp(4)*exp(log(log(x - 2) + 4)^2) + 480*log(x - 2)^2 + 80*log(x - 2)^3 + 5*log(x - 2)^4 + 1280))/(log(x - 2) + 4)^4
Time = 0.16 (sec) , antiderivative size = 83, normalized size of antiderivative = 4.37 \[ \int \frac {e^x (40-20 x)+e^x (10-5 x) \log (-2+x)+\frac {e^{4+\log ^2(4+\log (-2+x))} \left (e^x (-12+4 x)+e^x (-2+x) \log (-2+x)+2 e^x \log (4+\log (-2+x))\right )}{(4+\log (-2+x))^4}}{-8+4 x+(-2+x) \log (-2+x)} \, dx=\frac {e^{x} \left (e^{\mathrm {log}\left (\mathrm {log}\left (x -2\right )+4\right )^{2}} e^{4}-5 \mathrm {log}\left (x -2\right )^{4}-80 \mathrm {log}\left (x -2\right )^{3}-480 \mathrm {log}\left (x -2\right )^{2}-1280 \,\mathrm {log}\left (x -2\right )-1280\right )}{\mathrm {log}\left (x -2\right )^{4}+16 \mathrm {log}\left (x -2\right )^{3}+96 \mathrm {log}\left (x -2\right )^{2}+256 \,\mathrm {log}\left (x -2\right )+256} \] Input:
int(((2*exp(x)*log(log(-2+x)+4)+(-2+x)*exp(x)*log(-2+x)+(4*x-12)*exp(x))*e xp(log(log(-2+x)+4)^2-4*log(log(-2+x)+4)+4)+(-5*x+10)*exp(x)*log(-2+x)+(-2 0*x+40)*exp(x))/((-2+x)*log(-2+x)+4*x-8),x)
Output:
(e**x*(e**(log(log(x - 2) + 4)**2)*e**4 - 5*log(x - 2)**4 - 80*log(x - 2)* *3 - 480*log(x - 2)**2 - 1280*log(x - 2) - 1280))/(log(x - 2)**4 + 16*log( x - 2)**3 + 96*log(x - 2)**2 + 256*log(x - 2) + 256)