Integrand size = 99, antiderivative size = 25 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=x \left (1+\frac {5 e^{-4-x-x^2} x}{1+\log (x)}\right ) \] Output:
(1+5*x/(1+ln(x))/exp(x^2+x+4))*x
Time = 0.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\frac {x \left (1+5 e^{-4-x-x^2} x+\log (x)\right )}{1+\log (x)} \] Input:
Integrate[(E^(4 + x + x^2) + 5*x - 5*x^2 - 10*x^3 + (2*E^(4 + x + x^2) + 1 0*x - 5*x^2 - 10*x^3)*Log[x] + E^(4 + x + x^2)*Log[x]^2)/(E^(4 + x + x^2) + 2*E^(4 + x + x^2)*Log[x] + E^(4 + x + x^2)*Log[x]^2),x]
Output:
(x*(1 + 5*E^(-4 - x - x^2)*x + Log[x]))/(1 + 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.
\(\displaystyle \int \frac {-10 x^3-5 x^2+e^{x^2+x+4}+e^{x^2+x+4} \log ^2(x)+\left (-10 x^3-5 x^2+2 e^{x^2+x+4}+10 x\right ) \log (x)+5 x}{e^{x^2+x+4}+e^{x^2+x+4} \log ^2(x)+2 e^{x^2+x+4} \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x^2-x-4} \left (-10 x^3-5 x^2+e^{x^2+x+4}+e^{x^2+x+4} \log ^2(x)+\left (-10 x^3-5 x^2+2 e^{x^2+x+4}+10 x\right ) \log (x)+5 x\right )}{(\log (x)+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {5 e^{-x^2-x-4} x^2}{(\log (x)+1)^2}-\frac {5 e^{-x^2-x-4} x^2 \log (x)}{(\log (x)+1)^2}+\frac {5 e^{-x^2-x-4} x}{(\log (x)+1)^2}+\frac {10 e^{-x^2-x-4} x \log (x)}{(\log (x)+1)^2}-\frac {10 e^{-x^2-x-4} x^3}{(\log (x)+1)^2}-\frac {10 e^{-x^2-x-4} x^3 \log (x)}{(\log (x)+1)^2}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \int \frac {e^{-x^2-x-4} x}{(\log (x)+1)^2}dx+10 \int \frac {e^{-x^2-x-4} x}{\log (x)+1}dx-5 \int \frac {e^{-x^2-x-4} x^2}{\log (x)+1}dx-10 \int \frac {e^{-x^2-x-4} x^3}{\log (x)+1}dx+x\) |
Input:
Int[(E^(4 + x + x^2) + 5*x - 5*x^2 - 10*x^3 + (2*E^(4 + x + x^2) + 10*x - 5*x^2 - 10*x^3)*Log[x] + E^(4 + x + x^2)*Log[x]^2)/(E^(4 + x + x^2) + 2*E^ (4 + x + x^2)*Log[x] + E^(4 + x + x^2)*Log[x]^2),x]
Output:
$Aborted
Time = 9.53 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
method | result | size |
risch | \(x +\frac {5 x^{2} {\mathrm e}^{-x^{2}-x -4}}{\ln \left (x \right )+1}\) | \(25\) |
parallelrisch | \(\frac {\left (\ln \left (x \right ) {\mathrm e}^{x^{2}+x +4} x +5 x^{2}+{\mathrm e}^{x^{2}+x +4} x \right ) {\mathrm e}^{-x^{2}-x -4}}{\ln \left (x \right )+1}\) | \(43\) |
Input:
int((exp(x^2+x+4)*ln(x)^2+(2*exp(x^2+x+4)-10*x^3-5*x^2+10*x)*ln(x)+exp(x^2 +x+4)-10*x^3-5*x^2+5*x)/(exp(x^2+x+4)*ln(x)^2+2*exp(x^2+x+4)*ln(x)+exp(x^2 +x+4)),x,method=_RETURNVERBOSE)
Output:
x+5*x^2*exp(-x^2-x-4)/(ln(x)+1)
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\frac {x e^{\left (x^{2} + x + 4\right )} \log \left (x\right ) + 5 \, x^{2} + x e^{\left (x^{2} + x + 4\right )}}{e^{\left (x^{2} + x + 4\right )} \log \left (x\right ) + e^{\left (x^{2} + x + 4\right )}} \] Input:
integrate((exp(x^2+x+4)*log(x)^2+(2*exp(x^2+x+4)-10*x^3-5*x^2+10*x)*log(x) +exp(x^2+x+4)-10*x^3-5*x^2+5*x)/(exp(x^2+x+4)*log(x)^2+2*exp(x^2+x+4)*log( x)+exp(x^2+x+4)),x, algorithm="fricas")
Output:
(x*e^(x^2 + x + 4)*log(x) + 5*x^2 + x*e^(x^2 + x + 4))/(e^(x^2 + x + 4)*lo g(x) + e^(x^2 + x + 4))
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\frac {5 x^{2} e^{- x^{2} - x - 4}}{\log {\left (x \right )} + 1} + x \] Input:
integrate((exp(x**2+x+4)*ln(x)**2+(2*exp(x**2+x+4)-10*x**3-5*x**2+10*x)*ln (x)+exp(x**2+x+4)-10*x**3-5*x**2+5*x)/(exp(x**2+x+4)*ln(x)**2+2*exp(x**2+x +4)*ln(x)+exp(x**2+x+4)),x)
Output:
5*x**2*exp(-x**2 - x - 4)/(log(x) + 1) + x
Time = 0.10 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\frac {{\left (5 \, x^{2} e^{\left (-x^{2}\right )} + {\left (x e^{4} \log \left (x\right ) + x e^{4}\right )} e^{x}\right )} e^{\left (-x\right )}}{e^{4} \log \left (x\right ) + e^{4}} \] Input:
integrate((exp(x^2+x+4)*log(x)^2+(2*exp(x^2+x+4)-10*x^3-5*x^2+10*x)*log(x) +exp(x^2+x+4)-10*x^3-5*x^2+5*x)/(exp(x^2+x+4)*log(x)^2+2*exp(x^2+x+4)*log( x)+exp(x^2+x+4)),x, algorithm="maxima")
Output:
(5*x^2*e^(-x^2) + (x*e^4*log(x) + x*e^4)*e^x)*e^(-x)/(e^4*log(x) + e^4)
Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\frac {5 \, x^{2} e^{\left (-x^{2} - x\right )} + x e^{4} \log \left (x\right ) + x e^{4}}{e^{4} \log \left (x\right ) + e^{4}} \] Input:
integrate((exp(x^2+x+4)*log(x)^2+(2*exp(x^2+x+4)-10*x^3-5*x^2+10*x)*log(x) +exp(x^2+x+4)-10*x^3-5*x^2+5*x)/(exp(x^2+x+4)*log(x)^2+2*exp(x^2+x+4)*log( x)+exp(x^2+x+4)),x, algorithm="giac")
Output:
(5*x^2*e^(-x^2 - x) + x*e^4*log(x) + x*e^4)/(e^4*log(x) + e^4)
Time = 2.54 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=x+\frac {5\,x^2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^{-x^2}}{\ln \left (x\right )+1} \] Input:
int((5*x + exp(x + x^2 + 4) + log(x)*(10*x + 2*exp(x + x^2 + 4) - 5*x^2 - 10*x^3) + exp(x + x^2 + 4)*log(x)^2 - 5*x^2 - 10*x^3)/(exp(x + x^2 + 4) + 2*exp(x + x^2 + 4)*log(x) + exp(x + x^2 + 4)*log(x)^2),x)
Output:
x + (5*x^2*exp(-x)*exp(-4)*exp(-x^2))/(log(x) + 1)
Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {e^{4+x+x^2}+5 x-5 x^2-10 x^3+\left (2 e^{4+x+x^2}+10 x-5 x^2-10 x^3\right ) \log (x)+e^{4+x+x^2} \log ^2(x)}{e^{4+x+x^2}+2 e^{4+x+x^2} \log (x)+e^{4+x+x^2} \log ^2(x)} \, dx=\frac {x \left (e^{x^{2}+x} \mathrm {log}\left (x \right ) e^{4}+e^{x^{2}+x} e^{4}+5 x \right )}{e^{x^{2}+x} e^{4} \left (\mathrm {log}\left (x \right )+1\right )} \] Input:
int((exp(x^2+x+4)*log(x)^2+(2*exp(x^2+x+4)-10*x^3-5*x^2+10*x)*log(x)+exp(x ^2+x+4)-10*x^3-5*x^2+5*x)/(exp(x^2+x+4)*log(x)^2+2*exp(x^2+x+4)*log(x)+exp (x^2+x+4)),x)
Output:
(x*(e**(x**2 + x)*log(x)*e**4 + e**(x**2 + x)*e**4 + 5*x))/(e**(x**2 + x)* e**4*(log(x) + 1))