Integrand size = 93, antiderivative size = 22 \[ \int \frac {20+e^2 (5-5 x)+e^{10} (5-5 x)-25 x+5 x^2+\left (-100+e^2 (-25-5 x)+e^{10} (-25-5 x)+30 x+10 x^2+\left (20+5 e^2+5 e^{10}-10 x\right ) \log (x)\right ) \log (5+x-\log (x))}{-5-x+\log (x)} \, dx=5 \left (4+e^2+e^{10}-x\right ) x \log (5+x-\log (x)) \] Output:
5*(exp(2)-x+exp(5)^2+4)*x*ln(x-ln(x)+5)
Time = 0.63 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {20+e^2 (5-5 x)+e^{10} (5-5 x)-25 x+5 x^2+\left (-100+e^2 (-25-5 x)+e^{10} (-25-5 x)+30 x+10 x^2+\left (20+5 e^2+5 e^{10}-10 x\right ) \log (x)\right ) \log (5+x-\log (x))}{-5-x+\log (x)} \, dx=5 \left (4+e^2+e^{10}-x\right ) x \log (5+x-\log (x)) \] Input:
Integrate[(20 + E^2*(5 - 5*x) + E^10*(5 - 5*x) - 25*x + 5*x^2 + (-100 + E^ 2*(-25 - 5*x) + E^10*(-25 - 5*x) + 30*x + 10*x^2 + (20 + 5*E^2 + 5*E^10 - 10*x)*Log[x])*Log[5 + x - Log[x]])/(-5 - x + Log[x]),x]
Output:
5*(4 + E^2 + E^10 - x)*x*Log[5 + x - Log[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.
\(\displaystyle \int \frac {5 x^2+\left (10 x^2+30 x+e^{10} (-5 x-25)+e^2 (-5 x-25)+\left (-10 x+5 e^{10}+5 e^2+20\right ) \log (x)-100\right ) \log (x-\log (x)+5)-25 x+e^{10} (5-5 x)+e^2 (5-5 x)+20}{-x+\log (x)-5} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {5 x^2+\left (10 x^2+30 x+e^{10} (-5 x-25)+e^2 (-5 x-25)+\left (-10 x+5 e^{10}+5 e^2+20\right ) \log (x)-100\right ) \log (x-\log (x)+5)-25 x+\left (e^2+e^{10}\right ) (5-5 x)+20}{-x+\log (x)-5}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {5 \left (-x+e^{10}+e^2+4\right ) (x-1)}{x-\log (x)+5}+5 \left (-2 x+e^{10}+e^2+4\right ) \log (x-\log (x)+5)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \int \frac {x^2}{x-\log (x)+5}dx-5 \left (4+e^2+e^{10}\right ) \int \frac {1}{x-\log (x)+5}dx+5 \left (5+e^2+e^{10}\right ) \int \frac {x}{x-\log (x)+5}dx+5 \left (4+e^2+e^{10}\right ) \int \log (x-\log (x)+5)dx-10 \int x \log (x-\log (x)+5)dx\) |
Input:
Int[(20 + E^2*(5 - 5*x) + E^10*(5 - 5*x) - 25*x + 5*x^2 + (-100 + E^2*(-25 - 5*x) + E^10*(-25 - 5*x) + 30*x + 10*x^2 + (20 + 5*E^2 + 5*E^10 - 10*x)* Log[x])*Log[5 + x - Log[x]])/(-5 - x + Log[x]),x]
Output:
$Aborted
Time = 1.56 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32
method | result | size |
risch | \(\left (5 x \,{\mathrm e}^{10}+5 \,{\mathrm e}^{2} x -5 x^{2}+20 x \right ) \ln \left (x -\ln \left (x \right )+5\right )\) | \(29\) |
norman | \(\left (5 \,{\mathrm e}^{10}+5 \,{\mathrm e}^{2}+20\right ) x \ln \left (x -\ln \left (x \right )+5\right )-5 \ln \left (x -\ln \left (x \right )+5\right ) x^{2}\) | \(37\) |
parallelrisch | \(5 \ln \left (x -\ln \left (x \right )+5\right ) {\mathrm e}^{10} x +5 \,{\mathrm e}^{2} \ln \left (x -\ln \left (x \right )+5\right ) x -5 \ln \left (x -\ln \left (x \right )+5\right ) x^{2}+20 \ln \left (x -\ln \left (x \right )+5\right ) x\) | \(54\) |
Input:
int((((5*exp(5)^2+5*exp(2)+20-10*x)*ln(x)+(-5*x-25)*exp(5)^2+(-5*x-25)*exp (2)+10*x^2+30*x-100)*ln(x-ln(x)+5)+(-5*x+5)*exp(5)^2+(-5*x+5)*exp(2)+5*x^2 -25*x+20)/(ln(x)-5-x),x,method=_RETURNVERBOSE)
Output:
(5*x*exp(10)+5*exp(2)*x-5*x^2+20*x)*ln(x-ln(x)+5)
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {20+e^2 (5-5 x)+e^{10} (5-5 x)-25 x+5 x^2+\left (-100+e^2 (-25-5 x)+e^{10} (-25-5 x)+30 x+10 x^2+\left (20+5 e^2+5 e^{10}-10 x\right ) \log (x)\right ) \log (5+x-\log (x))}{-5-x+\log (x)} \, dx=-5 \, {\left (x^{2} - x e^{10} - x e^{2} - 4 \, x\right )} \log \left (x - \log \left (x\right ) + 5\right ) \] Input:
integrate((((5*exp(5)^2+5*exp(2)+20-10*x)*log(x)+(-5*x-25)*exp(5)^2+(-5*x- 25)*exp(2)+10*x^2+30*x-100)*log(x-log(x)+5)+(-5*x+5)*exp(5)^2+(-5*x+5)*exp (2)+5*x^2-25*x+20)/(log(x)-5-x),x, algorithm="fricas")
Output:
-5*(x^2 - x*e^10 - x*e^2 - 4*x)*log(x - log(x) + 5)
Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {20+e^2 (5-5 x)+e^{10} (5-5 x)-25 x+5 x^2+\left (-100+e^2 (-25-5 x)+e^{10} (-25-5 x)+30 x+10 x^2+\left (20+5 e^2+5 e^{10}-10 x\right ) \log (x)\right ) \log (5+x-\log (x))}{-5-x+\log (x)} \, dx=\left (- 5 x^{2} + 20 x + 5 x e^{2} + 5 x e^{10}\right ) \log {\left (x - \log {\left (x \right )} + 5 \right )} \] Input:
integrate((((5*exp(5)**2+5*exp(2)+20-10*x)*ln(x)+(-5*x-25)*exp(5)**2+(-5*x -25)*exp(2)+10*x**2+30*x-100)*ln(x-ln(x)+5)+(-5*x+5)*exp(5)**2+(-5*x+5)*ex p(2)+5*x**2-25*x+20)/(ln(x)-5-x),x)
Output:
(-5*x**2 + 20*x + 5*x*exp(2) + 5*x*exp(10))*log(x - log(x) + 5)
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {20+e^2 (5-5 x)+e^{10} (5-5 x)-25 x+5 x^2+\left (-100+e^2 (-25-5 x)+e^{10} (-25-5 x)+30 x+10 x^2+\left (20+5 e^2+5 e^{10}-10 x\right ) \log (x)\right ) \log (5+x-\log (x))}{-5-x+\log (x)} \, dx=-5 \, {\left (x^{2} - x {\left (e^{10} + e^{2} + 4\right )}\right )} \log \left (x - \log \left (x\right ) + 5\right ) \] Input:
integrate((((5*exp(5)^2+5*exp(2)+20-10*x)*log(x)+(-5*x-25)*exp(5)^2+(-5*x- 25)*exp(2)+10*x^2+30*x-100)*log(x-log(x)+5)+(-5*x+5)*exp(5)^2+(-5*x+5)*exp (2)+5*x^2-25*x+20)/(log(x)-5-x),x, algorithm="maxima")
Output:
-5*(x^2 - x*(e^10 + e^2 + 4))*log(x - log(x) + 5)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (22) = 44\).
Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.32 \[ \int \frac {20+e^2 (5-5 x)+e^{10} (5-5 x)-25 x+5 x^2+\left (-100+e^2 (-25-5 x)+e^{10} (-25-5 x)+30 x+10 x^2+\left (20+5 e^2+5 e^{10}-10 x\right ) \log (x)\right ) \log (5+x-\log (x))}{-5-x+\log (x)} \, dx=-5 \, x^{2} \log \left (x - \log \left (x\right ) + 5\right ) + 5 \, x e^{10} \log \left (x - \log \left (x\right ) + 5\right ) + 5 \, x e^{2} \log \left (x - \log \left (x\right ) + 5\right ) + 20 \, x \log \left (x - \log \left (x\right ) + 5\right ) \] Input:
integrate((((5*exp(5)^2+5*exp(2)+20-10*x)*log(x)+(-5*x-25)*exp(5)^2+(-5*x- 25)*exp(2)+10*x^2+30*x-100)*log(x-log(x)+5)+(-5*x+5)*exp(5)^2+(-5*x+5)*exp (2)+5*x^2-25*x+20)/(log(x)-5-x),x, algorithm="giac")
Output:
-5*x^2*log(x - log(x) + 5) + 5*x*e^10*log(x - log(x) + 5) + 5*x*e^2*log(x - log(x) + 5) + 20*x*log(x - log(x) + 5)
Time = 2.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {20+e^2 (5-5 x)+e^{10} (5-5 x)-25 x+5 x^2+\left (-100+e^2 (-25-5 x)+e^{10} (-25-5 x)+30 x+10 x^2+\left (20+5 e^2+5 e^{10}-10 x\right ) \log (x)\right ) \log (5+x-\log (x))}{-5-x+\log (x)} \, dx=5\,x\,\ln \left (x-\ln \left (x\right )+5\right )\,\left ({\mathrm {e}}^2-x+{\mathrm {e}}^{10}+4\right ) \] Input:
int((25*x - log(x - log(x) + 5)*(30*x + log(x)*(5*exp(2) - 10*x + 5*exp(10 ) + 20) + 10*x^2 - exp(2)*(5*x + 25) - exp(10)*(5*x + 25) - 100) - 5*x^2 + exp(2)*(5*x - 5) + exp(10)*(5*x - 5) - 20)/(x - log(x) + 5),x)
Output:
5*x*log(x - log(x) + 5)*(exp(2) - x + exp(10) + 4)
Time = 0.19 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {20+e^2 (5-5 x)+e^{10} (5-5 x)-25 x+5 x^2+\left (-100+e^2 (-25-5 x)+e^{10} (-25-5 x)+30 x+10 x^2+\left (20+5 e^2+5 e^{10}-10 x\right ) \log (x)\right ) \log (5+x-\log (x))}{-5-x+\log (x)} \, dx=5 \,\mathrm {log}\left (-\mathrm {log}\left (x \right )+x +5\right ) x \left (e^{10}+e^{2}-x +4\right ) \] Input:
int((((5*exp(5)^2+5*exp(2)+20-10*x)*log(x)+(-5*x-25)*exp(5)^2+(-5*x-25)*ex p(2)+10*x^2+30*x-100)*log(x-log(x)+5)+(-5*x+5)*exp(5)^2+(-5*x+5)*exp(2)+5* x^2-25*x+20)/(log(x)-5-x),x)
Output:
5*log( - log(x) + x + 5)*x*(e**10 + e**2 - x + 4)