Integrand size = 66, antiderivative size = 28 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {e^x \left (2 x-\left (16+e^5\right ) x\right )}{-1+(5-x) \log (x)} \] Output:
exp(x)*(2*x-x*(exp(5)+16))/(ln(x)*(5-x)-1)
Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {e^x \left (14+e^5\right ) x}{1-5 \log (x)+x \log (x)} \] Input:
Integrate[(E^x*(84 + 6*E^5) + E^x*(-70 - 70*x + 14*x^2 + E^5*(-5 - 5*x + x ^2))*Log[x])/(1 + (-10 + 2*x)*Log[x] + (25 - 10*x + x^2)*Log[x]^2),x]
Output:
(E^x*(14 + E^5)*x)/(1 - 5*Log[x] + 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.
\(\displaystyle \int \frac {e^x \left (14 x^2+e^5 \left (x^2-5 x-5\right )-70 x-70\right ) \log (x)+\left (84+6 e^5\right ) e^x}{\left (x^2-10 x+25\right ) \log ^2(x)+(2 x-10) \log (x)+1} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\left (14+e^5\right ) e^x \left (\left (x^2-5 x-5\right ) \log (x)+6\right )}{((x-5) \log (x)+1)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \left (14+e^5\right ) \int \frac {e^x \left (6-\left (-x^2+5 x+5\right ) \log (x)\right )}{(1-(5-x) \log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \left (14+e^5\right ) \int \left (\frac {e^x \left (-x^2+11 x-25\right )}{(x-5) (x \log (x)-5 \log (x)+1)^2}+\frac {e^x \left (x^2-5 x-5\right )}{(x-5) (x \log (x)-5 \log (x)+1)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \left (14+e^5\right ) \left (6 \int \frac {e^x}{(x \log (x)-5 \log (x)+1)^2}dx+5 \int \frac {e^x}{(x-5) (x \log (x)-5 \log (x)+1)^2}dx-\int \frac {e^x x}{(x \log (x)-5 \log (x)+1)^2}dx-5 \int \frac {e^x}{(x-5) (x \log (x)-5 \log (x)+1)}dx+\int \frac {e^x x}{x \log (x)-5 \log (x)+1}dx\right )\) |
Input:
Int[(E^x*(84 + 6*E^5) + E^x*(-70 - 70*x + 14*x^2 + E^5*(-5 - 5*x + x^2))*L og[x])/(1 + (-10 + 2*x)*Log[x] + (25 - 10*x + x^2)*Log[x]^2),x]
Output:
$Aborted
Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75
method | result | size |
norman | \(\frac {\left ({\mathrm e}^{5}+14\right ) x \,{\mathrm e}^{x}}{x \ln \left (x \right )-5 \ln \left (x \right )+1}\) | \(21\) |
risch | \(\frac {\left ({\mathrm e}^{5}+14\right ) x \,{\mathrm e}^{x}}{x \ln \left (x \right )-5 \ln \left (x \right )+1}\) | \(21\) |
parallelrisch | \(\frac {{\mathrm e}^{x} {\mathrm e}^{5} x +14 \,{\mathrm e}^{x} x}{x \ln \left (x \right )-5 \ln \left (x \right )+1}\) | \(26\) |
Input:
int((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*ln(x)+(6*exp(5)+84)*exp(x) )/((x^2-10*x+25)*ln(x)^2+(2*x-10)*ln(x)+1),x,method=_RETURNVERBOSE)
Output:
(exp(5)+14)*x*exp(x)/(x*ln(x)-5*ln(x)+1)
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {{\left (x e^{5} + 14 \, x\right )} e^{x}}{{\left (x - 5\right )} \log \left (x\right ) + 1} \] Input:
integrate((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*log(x)+(6*exp(5)+84) *exp(x))/((x^2-10*x+25)*log(x)^2+(2*x-10)*log(x)+1),x, algorithm="fricas")
Output:
(x*e^5 + 14*x)*e^x/((x - 5)*log(x) + 1)
Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {\left (14 x + x e^{5}\right ) e^{x}}{x \log {\left (x \right )} - 5 \log {\left (x \right )} + 1} \] Input:
integrate((((x**2-5*x-5)*exp(5)+14*x**2-70*x-70)*exp(x)*ln(x)+(6*exp(5)+84 )*exp(x))/((x**2-10*x+25)*ln(x)**2+(2*x-10)*ln(x)+1),x)
Output:
(14*x + x*exp(5))*exp(x)/(x*log(x) - 5*log(x) + 1)
Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {x {\left (e^{5} + 14\right )} e^{x}}{{\left (x - 5\right )} \log \left (x\right ) + 1} \] Input:
integrate((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*log(x)+(6*exp(5)+84) *exp(x))/((x^2-10*x+25)*log(x)^2+(2*x-10)*log(x)+1),x, algorithm="maxima")
Output:
x*(e^5 + 14)*e^x/((x - 5)*log(x) + 1)
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {x e^{\left (x + 5\right )} + 14 \, x e^{x}}{x \log \left (x\right ) - 5 \, \log \left (x\right ) + 1} \] Input:
integrate((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*log(x)+(6*exp(5)+84) *exp(x))/((x^2-10*x+25)*log(x)^2+(2*x-10)*log(x)+1),x, algorithm="giac")
Output:
(x*e^(x + 5) + 14*x*e^x)/(x*log(x) - 5*log(x) + 1)
Timed out. \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\int \frac {{\mathrm {e}}^x\,\left (6\,{\mathrm {e}}^5+84\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (70\,x+{\mathrm {e}}^5\,\left (-x^2+5\,x+5\right )-14\,x^2+70\right )}{\left (x^2-10\,x+25\right )\,{\ln \left (x\right )}^2+\left (2\,x-10\right )\,\ln \left (x\right )+1} \,d x \] Input:
int((exp(x)*(6*exp(5) + 84) - exp(x)*log(x)*(70*x + exp(5)*(5*x - x^2 + 5) - 14*x^2 + 70))/(log(x)*(2*x - 10) + log(x)^2*(x^2 - 10*x + 25) + 1),x)
Output:
int((exp(x)*(6*exp(5) + 84) - exp(x)*log(x)*(70*x + exp(5)*(5*x - x^2 + 5) - 14*x^2 + 70))/(log(x)*(2*x - 10) + log(x)^2*(x^2 - 10*x + 25) + 1), x)
Time = 0.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^x \left (84+6 e^5\right )+e^x \left (-70-70 x+14 x^2+e^5 \left (-5-5 x+x^2\right )\right ) \log (x)}{1+(-10+2 x) \log (x)+\left (25-10 x+x^2\right ) \log ^2(x)} \, dx=\frac {e^{x} x \left (e^{5}+14\right )}{\mathrm {log}\left (x \right ) x -5 \,\mathrm {log}\left (x \right )+1} \] Input:
int((((x^2-5*x-5)*exp(5)+14*x^2-70*x-70)*exp(x)*log(x)+(6*exp(5)+84)*exp(x ))/((x^2-10*x+25)*log(x)^2+(2*x-10)*log(x)+1),x)
Output:
(e**x*x*(e**5 + 14))/(log(x)*x - 5*log(x) + 1)