Integrand size = 97, antiderivative size = 32 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=x \left (2 x+\frac {2+e^{-x} (1-x)}{4 e^2 (x+\log (x))}\right ) \] Output:
x*(2*x+1/4*(2+(1-x)/exp(x))/(x+ln(x))/exp(2))
Time = 5.06 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=\frac {1}{4} x \left (8 x+\frac {e^{-2-x} \left (1+2 e^x-x\right )}{x+\log (x)}\right ) \] Input:
Integrate[(-1 + x - 2*x^2 + x^3 + E^x*(-2 + 16*E^2*x^3) + (1 - 3*x + x^2 + E^x*(2 + 32*E^2*x^2))*Log[x] + 16*E^(2 + x)*x*Log[x]^2)/(4*E^(2 + x)*x^2 + 8*E^(2 + x)*x*Log[x] + 4*E^(2 + x)*Log[x]^2),x]
Output:
(x*(8*x + (E^(-2 - x)*(1 + 2*E^x - x))/(x + Log[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.
\(\displaystyle \int \frac {x^3+e^x \left (16 e^2 x^3-2\right )-2 x^2+\left (x^2+e^x \left (32 e^2 x^2+2\right )-3 x+1\right ) \log (x)+x+16 e^{x+2} x \log ^2(x)-1}{4 e^{x+2} x^2+4 e^{x+2} \log ^2(x)+8 e^{x+2} x \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x-2} \left (x^3+e^x \left (16 e^2 x^3-2\right )-2 x^2+\left (x^2+e^x \left (32 e^2 x^2+2\right )-3 x+1\right ) \log (x)+x+16 e^{x+2} x \log ^2(x)-1\right )}{4 (x+\log (x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {e^{-x-2} \left (-x^3+2 x^2-16 e^{x+2} \log ^2(x) x-x+2 e^x \left (1-8 e^2 x^3\right )-\left (x^2-3 x+2 e^x \left (16 e^2 x^2+1\right )+1\right ) \log (x)+1\right )}{(x+\log (x))^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {e^{-x-2} \left (-x^3+2 x^2-16 e^{x+2} \log ^2(x) x-x+2 e^x \left (1-8 e^2 x^3\right )-\left (x^2-3 x+2 e^x \left (16 e^2 x^2+1\right )+1\right ) \log (x)+1\right )}{(x+\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{4} \int \left (-\frac {e^{-x-2} x^3}{(x+\log (x))^2}+\frac {2 e^{-x-2} x^2}{(x+\log (x))^2}-\frac {e^{-x-2} \log (x) x^2}{(x+\log (x))^2}-\frac {e^{-x-2} x}{(x+\log (x))^2}+\frac {3 e^{-x-2} \log (x) x}{(x+\log (x))^2}-\frac {2 \left (8 e^2 x^3+16 e^2 \log (x) x^2+8 e^2 \log ^2(x) x+\log (x)-1\right )}{e^2 (x+\log (x))^2}+\frac {e^{-x-2}}{(x+\log (x))^2}-\frac {e^{-x-2} \log (x)}{(x+\log (x))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (\int \frac {e^{-x-2} x^2}{(x+\log (x))^2}dx+\int \frac {e^{-x-2} x^2}{x+\log (x)}dx-\frac {2 \int \frac {1}{(x+\log (x))^2}dx}{e^2}-\int \frac {e^{-x-2}}{(x+\log (x))^2}dx-\frac {2 \int \frac {x}{(x+\log (x))^2}dx}{e^2}+\frac {2 \int \frac {1}{x+\log (x)}dx}{e^2}+\int \frac {e^{-x-2}}{x+\log (x)}dx-3 \int \frac {e^{-x-2} x}{x+\log (x)}dx+8 x^2\right )\) |
Input:
Int[(-1 + x - 2*x^2 + x^3 + E^x*(-2 + 16*E^2*x^3) + (1 - 3*x + x^2 + E^x*( 2 + 32*E^2*x^2))*Log[x] + 16*E^(2 + x)*x*Log[x]^2)/(4*E^(2 + x)*x^2 + 8*E^ (2 + x)*x*Log[x] + 4*E^(2 + x)*Log[x]^2),x]
Output:
$Aborted
Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91
\[2 x^{2}-\frac {x \left (x -2 \,{\mathrm e}^{x}-1\right ) {\mathrm e}^{-2-x}}{4 \left (x +\ln \left (x \right )\right )}\]
Input:
int((16*x*exp(2)*exp(x)*ln(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1)*ln(x) +(16*x^3*exp(2)-2)*exp(x)+x^3-2*x^2+x-1)/(4*exp(2)*exp(x)*ln(x)^2+8*x*exp( 2)*exp(x)*ln(x)+4*x^2*exp(2)*exp(x)),x)
Output:
2*x^2-1/4/(x+ln(x))*x*(x-2*exp(x)-1)*exp(-2-x)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (27) = 54\).
Time = 0.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.75 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=\frac {8 \, x^{2} e^{\left (x + 4\right )} \log \left (x\right ) - {\left (x^{2} - x\right )} e^{2} + 2 \, {\left (4 \, x^{3} e^{2} + x\right )} e^{\left (x + 2\right )}}{4 \, {\left (x e^{\left (x + 4\right )} + e^{\left (x + 4\right )} \log \left (x\right )\right )}} \] Input:
integrate((16*x*exp(2)*exp(x)*log(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1 )*log(x)+(16*x^3*exp(2)-2)*exp(x)+x^3-2*x^2+x-1)/(4*exp(2)*exp(x)*log(x)^2 +8*x*exp(2)*exp(x)*log(x)+4*x^2*exp(2)*exp(x)),x, algorithm="fricas")
Output:
1/4*(8*x^2*e^(x + 4)*log(x) - (x^2 - x)*e^2 + 2*(4*x^3*e^2 + x)*e^(x + 2)) /(x*e^(x + 4) + e^(x + 4)*log(x))
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.38 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=2 x^{2} + \frac {x}{2 x e^{2} + 2 e^{2} \log {\left (x \right )}} + \frac {\left (- x^{2} + x\right ) e^{- x}}{4 x e^{2} + 4 e^{2} \log {\left (x \right )}} \] Input:
integrate((16*x*exp(2)*exp(x)*ln(x)**2+((32*x**2*exp(2)+2)*exp(x)+x**2-3*x +1)*ln(x)+(16*x**3*exp(2)-2)*exp(x)+x**3-2*x**2+x-1)/(4*exp(2)*exp(x)*ln(x )**2+8*x*exp(2)*exp(x)*ln(x)+4*x**2*exp(2)*exp(x)),x)
Output:
2*x**2 + x/(2*x*exp(2) + 2*exp(2)*log(x)) + (-x**2 + x)*exp(-x)/(4*x*exp(2 ) + 4*exp(2)*log(x))
Time = 0.10 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.47 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=\frac {8 \, x^{3} e^{2} + 8 \, x^{2} e^{2} \log \left (x\right ) - {\left (x^{2} - x\right )} e^{\left (-x\right )} + 2 \, x}{4 \, {\left (x e^{2} + e^{2} \log \left (x\right )\right )}} \] Input:
integrate((16*x*exp(2)*exp(x)*log(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1 )*log(x)+(16*x^3*exp(2)-2)*exp(x)+x^3-2*x^2+x-1)/(4*exp(2)*exp(x)*log(x)^2 +8*x*exp(2)*exp(x)*log(x)+4*x^2*exp(2)*exp(x)),x, algorithm="maxima")
Output:
1/4*(8*x^3*e^2 + 8*x^2*e^2*log(x) - (x^2 - x)*e^(-x) + 2*x)/(x*e^2 + e^2*l og(x))
Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=\frac {8 \, x^{3} e^{2} + 8 \, x^{2} e^{2} \log \left (x\right ) - x^{2} e^{\left (-x\right )} + x e^{\left (-x\right )} + 2 \, x}{4 \, {\left (x e^{2} + e^{2} \log \left (x\right )\right )}} \] Input:
integrate((16*x*exp(2)*exp(x)*log(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1 )*log(x)+(16*x^3*exp(2)-2)*exp(x)+x^3-2*x^2+x-1)/(4*exp(2)*exp(x)*log(x)^2 +8*x*exp(2)*exp(x)*log(x)+4*x^2*exp(2)*exp(x)),x, algorithm="giac")
Output:
1/4*(8*x^3*e^2 + 8*x^2*e^2*log(x) - x^2*e^(-x) + x*e^(-x) + 2*x)/(x*e^2 + e^2*log(x))
Time = 4.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 3.81 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=\frac {\frac {x\,{\mathrm {e}}^{-x-2}\,\left (2\,{\mathrm {e}}^x-x+2\,x^2-x^3+1\right )}{4\,\left (x+1\right )}-\frac {x\,{\mathrm {e}}^{-x-2}\,\ln \left (x\right )\,\left (2\,{\mathrm {e}}^x-3\,x+x^2+1\right )}{4\,\left (x+1\right )}}{x+\ln \left (x\right )}-\frac {1}{2\,{\mathrm {e}}^2+2\,x\,{\mathrm {e}}^2}+2\,x^2+\frac {{\mathrm {e}}^{-x}\,\left (\frac {{\mathrm {e}}^{-2}\,x^3}{4}-\frac {3\,{\mathrm {e}}^{-2}\,x^2}{4}+\frac {{\mathrm {e}}^{-2}\,x}{4}\right )}{x+1} \] Input:
int((x + exp(x)*(16*x^3*exp(2) - 2) + log(x)*(exp(x)*(32*x^2*exp(2) + 2) - 3*x + x^2 + 1) - 2*x^2 + x^3 + 16*x*exp(2)*exp(x)*log(x)^2 - 1)/(4*x^2*ex p(2)*exp(x) + 4*exp(2)*exp(x)*log(x)^2 + 8*x*exp(2)*exp(x)*log(x)),x)
Output:
((x*exp(- x - 2)*(2*exp(x) - x + 2*x^2 - x^3 + 1))/(4*(x + 1)) - (x*exp(- x - 2)*log(x)*(2*exp(x) - 3*x + x^2 + 1))/(4*(x + 1)))/(x + log(x)) - 1/(2 *exp(2) + 2*x*exp(2)) + 2*x^2 + (exp(-x)*((x*exp(-2))/4 - (3*x^2*exp(-2))/ 4 + (x^3*exp(-2))/4))/(x + 1)
Time = 0.17 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69 \[ \int \frac {-1+x-2 x^2+x^3+e^x \left (-2+16 e^2 x^3\right )+\left (1-3 x+x^2+e^x \left (2+32 e^2 x^2\right )\right ) \log (x)+16 e^{2+x} x \log ^2(x)}{4 e^{2+x} x^2+8 e^{2+x} x \log (x)+4 e^{2+x} \log ^2(x)} \, dx=\frac {8 e^{x} \mathrm {log}\left (x \right ) e^{2} x^{2}-2 e^{x} \mathrm {log}\left (x \right )+8 e^{x} e^{2} x^{3}-x^{2}+x}{4 e^{x} e^{2} \left (\mathrm {log}\left (x \right )+x \right )} \] Input:
int((16*x*exp(2)*exp(x)*log(x)^2+((32*x^2*exp(2)+2)*exp(x)+x^2-3*x+1)*log( x)+(16*x^3*exp(2)-2)*exp(x)+x^3-2*x^2+x-1)/(4*exp(2)*exp(x)*log(x)^2+8*x*e xp(2)*exp(x)*log(x)+4*x^2*exp(2)*exp(x)),x)
Output:
(8*e**x*log(x)*e**2*x**2 - 2*e**x*log(x) + 8*e**x*e**2*x**3 - x**2 + x)/(4 *e**x*e**2*(log(x) + x))