Integrand size = 88, antiderivative size = 30 \[ \int \frac {1+8 x+e^{e^{x^2}} \left (-4+\left (-4-8 e^{x^2} x^2\right ) \log (x)\right )}{25 x^2+200 x^3+400 x^4+e^{e^{x^2}} \left (-200 x^2-800 x^3\right ) \log (x)+400 e^{2 e^{x^2}} x^2 \log ^2(x)} \, dx=\log (2)+\frac {1}{25 \left (-x+4 x \left (-x+e^{e^{x^2}} \log (x)\right )\right )} \] Output:
ln(2)+1/25/(4*(exp(exp(x^2))*ln(x)-x)*x-x)
Time = 0.31 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {1+8 x+e^{e^{x^2}} \left (-4+\left (-4-8 e^{x^2} x^2\right ) \log (x)\right )}{25 x^2+200 x^3+400 x^4+e^{e^{x^2}} \left (-200 x^2-800 x^3\right ) \log (x)+400 e^{2 e^{x^2}} x^2 \log ^2(x)} \, dx=\frac {1}{25 x \left (-1-4 x+4 e^{e^{x^2}} \log (x)\right )} \] Input:
Integrate[(1 + 8*x + E^E^x^2*(-4 + (-4 - 8*E^x^2*x^2)*Log[x]))/(25*x^2 + 2 00*x^3 + 400*x^4 + E^E^x^2*(-200*x^2 - 800*x^3)*Log[x] + 400*E^(2*E^x^2)*x ^2*Log[x]^2),x]
Output:
1/(25*x*(-1 - 4*x + 4*E^E^x^2*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.
\(\displaystyle \int \frac {e^{e^{x^2}} \left (\left (-8 e^{x^2} x^2-4\right ) \log (x)-4\right )+8 x+1}{400 x^4+200 x^3+25 x^2+400 e^{2 e^{x^2}} x^2 \log ^2(x)+e^{e^{x^2}} \left (-800 x^3-200 x^2\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{e^{x^2}} \left (\left (-8 e^{x^2} x^2-4\right ) \log (x)-4\right )+8 x+1}{25 x^2 \left (-4 e^{e^{x^2}} \log (x)+4 x+1\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \int \frac {8 x-4 e^{e^{x^2}} \left (\left (2 e^{x^2} x^2+1\right ) \log (x)+1\right )+1}{x^2 \left (4 x-4 e^{e^{x^2}} \log (x)+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{25} \int \left (-\frac {8 e^{x^2+e^{x^2}} \log (x)}{\left (-4 x+4 e^{e^{x^2}} \log (x)-1\right )^2}-\frac {4 e^{e^{x^2}} \log (x)}{x^2 \left (-4 x+4 e^{e^{x^2}} \log (x)-1\right )^2}+\frac {8}{x \left (4 x-4 e^{e^{x^2}} \log (x)+1\right )^2}+\frac {1}{x^2 \left (4 x-4 e^{e^{x^2}} \log (x)+1\right )^2}-\frac {4 e^{e^{x^2}}}{x^2 \left (-4 x+4 e^{e^{x^2}} \log (x)-1\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{25} \left (\int \frac {1}{x^2 \left (4 x-4 e^{e^{x^2}} \log (x)+1\right )^2}dx+8 \int \frac {1}{x \left (4 x-4 e^{e^{x^2}} \log (x)+1\right )^2}dx-4 \int \frac {e^{e^{x^2}}}{x^2 \left (-4 x+4 e^{e^{x^2}} \log (x)-1\right )^2}dx-8 \int \frac {e^{x^2+e^{x^2}} \log (x)}{\left (-4 x+4 e^{e^{x^2}} \log (x)-1\right )^2}dx-4 \int \frac {e^{e^{x^2}} \log (x)}{x^2 \left (-4 x+4 e^{e^{x^2}} \log (x)-1\right )^2}dx\right )\) |
Input:
Int[(1 + 8*x + E^E^x^2*(-4 + (-4 - 8*E^x^2*x^2)*Log[x]))/(25*x^2 + 200*x^3 + 400*x^4 + E^E^x^2*(-200*x^2 - 800*x^3)*Log[x] + 400*E^(2*E^x^2)*x^2*Log [x]^2),x]
Output:
$Aborted
Time = 2.77 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73
method | result | size |
risch | \(-\frac {1}{25 x \left (-4 \,{\mathrm e}^{{\mathrm e}^{x^{2}}} \ln \left (x \right )+4 x +1\right )}\) | \(22\) |
parallelrisch | \(-\frac {1}{25 x \left (-4 \,{\mathrm e}^{{\mathrm e}^{x^{2}}} \ln \left (x \right )+4 x +1\right )}\) | \(22\) |
Input:
int((((-8*x^2*exp(x^2)-4)*ln(x)-4)*exp(exp(x^2))+8*x+1)/(400*x^2*ln(x)^2*e xp(exp(x^2))^2+(-800*x^3-200*x^2)*ln(x)*exp(exp(x^2))+400*x^4+200*x^3+25*x ^2),x,method=_RETURNVERBOSE)
Output:
-1/25/x/(-4*exp(exp(x^2))*ln(x)+4*x+1)
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {1+8 x+e^{e^{x^2}} \left (-4+\left (-4-8 e^{x^2} x^2\right ) \log (x)\right )}{25 x^2+200 x^3+400 x^4+e^{e^{x^2}} \left (-200 x^2-800 x^3\right ) \log (x)+400 e^{2 e^{x^2}} x^2 \log ^2(x)} \, dx=\frac {1}{25 \, {\left (4 \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) - 4 \, x^{2} - x\right )}} \] Input:
integrate((((-8*exp(x^2)*x^2-4)*log(x)-4)*exp(exp(x^2))+8*x+1)/(400*x^2*lo g(x)^2*exp(exp(x^2))^2+(-800*x^3-200*x^2)*log(x)*exp(exp(x^2))+400*x^4+200 *x^3+25*x^2),x, algorithm="fricas")
Output:
1/25/(4*x*e^(e^(x^2))*log(x) - 4*x^2 - x)
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.73 \[ \int \frac {1+8 x+e^{e^{x^2}} \left (-4+\left (-4-8 e^{x^2} x^2\right ) \log (x)\right )}{25 x^2+200 x^3+400 x^4+e^{e^{x^2}} \left (-200 x^2-800 x^3\right ) \log (x)+400 e^{2 e^{x^2}} x^2 \log ^2(x)} \, dx=\frac {1}{- 100 x^{2} + 100 x e^{e^{x^{2}}} \log {\left (x \right )} - 25 x} \] Input:
integrate((((-8*exp(x**2)*x**2-4)*ln(x)-4)*exp(exp(x**2))+8*x+1)/(400*x**2 *ln(x)**2*exp(exp(x**2))**2+(-800*x**3-200*x**2)*ln(x)*exp(exp(x**2))+400* x**4+200*x**3+25*x**2),x)
Output:
1/(-100*x**2 + 100*x*exp(exp(x**2))*log(x) - 25*x)
Time = 0.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {1+8 x+e^{e^{x^2}} \left (-4+\left (-4-8 e^{x^2} x^2\right ) \log (x)\right )}{25 x^2+200 x^3+400 x^4+e^{e^{x^2}} \left (-200 x^2-800 x^3\right ) \log (x)+400 e^{2 e^{x^2}} x^2 \log ^2(x)} \, dx=\frac {1}{25 \, {\left (4 \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) - 4 \, x^{2} - x\right )}} \] Input:
integrate((((-8*exp(x^2)*x^2-4)*log(x)-4)*exp(exp(x^2))+8*x+1)/(400*x^2*lo g(x)^2*exp(exp(x^2))^2+(-800*x^3-200*x^2)*log(x)*exp(exp(x^2))+400*x^4+200 *x^3+25*x^2),x, algorithm="maxima")
Output:
1/25/(4*x*e^(e^(x^2))*log(x) - 4*x^2 - x)
Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {1+8 x+e^{e^{x^2}} \left (-4+\left (-4-8 e^{x^2} x^2\right ) \log (x)\right )}{25 x^2+200 x^3+400 x^4+e^{e^{x^2}} \left (-200 x^2-800 x^3\right ) \log (x)+400 e^{2 e^{x^2}} x^2 \log ^2(x)} \, dx=\frac {1}{25 \, {\left (4 \, x e^{\left (e^{\left (x^{2}\right )}\right )} \log \left (x\right ) - 4 \, x^{2} - x\right )}} \] Input:
integrate((((-8*exp(x^2)*x^2-4)*log(x)-4)*exp(exp(x^2))+8*x+1)/(400*x^2*lo g(x)^2*exp(exp(x^2))^2+(-800*x^3-200*x^2)*log(x)*exp(exp(x^2))+400*x^4+200 *x^3+25*x^2),x, algorithm="giac")
Output:
1/25/(4*x*e^(e^(x^2))*log(x) - 4*x^2 - x)
Time = 2.70 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.87 \[ \int \frac {1+8 x+e^{e^{x^2}} \left (-4+\left (-4-8 e^{x^2} x^2\right ) \log (x)\right )}{25 x^2+200 x^3+400 x^4+e^{e^{x^2}} \left (-200 x^2-800 x^3\right ) \log (x)+400 e^{2 e^{x^2}} x^2 \log ^2(x)} \, dx=-\frac {\frac {4\,x}{25}+\ln \left (x\right )\,\left (\frac {2\,x^2\,{\mathrm {e}}^{x^2}}{25}-\frac {4\,x}{25}+\frac {8\,x^3\,{\mathrm {e}}^{x^2}}{25}\right )+\frac {1}{25}}{\left (4\,x-4\,{\mathrm {e}}^{{\mathrm {e}}^{x^2}}\,\ln \left (x\right )+1\right )\,\left (x-4\,x^2\,\ln \left (x\right )+4\,x^2+2\,x^3\,{\mathrm {e}}^{x^2}\,\ln \left (x\right )+8\,x^4\,{\mathrm {e}}^{x^2}\,\ln \left (x\right )\right )} \] Input:
int((8*x - exp(exp(x^2))*(log(x)*(8*x^2*exp(x^2) + 4) + 4) + 1)/(25*x^2 + 200*x^3 + 400*x^4 + 400*x^2*exp(2*exp(x^2))*log(x)^2 - exp(exp(x^2))*log(x )*(200*x^2 + 800*x^3)),x)
Output:
-((4*x)/25 + log(x)*((2*x^2*exp(x^2))/25 - (4*x)/25 + (8*x^3*exp(x^2))/25) + 1/25)/((4*x - 4*exp(exp(x^2))*log(x) + 1)*(x - 4*x^2*log(x) + 4*x^2 + 2 *x^3*exp(x^2)*log(x) + 8*x^4*exp(x^2)*log(x)))
Time = 0.18 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {1+8 x+e^{e^{x^2}} \left (-4+\left (-4-8 e^{x^2} x^2\right ) \log (x)\right )}{25 x^2+200 x^3+400 x^4+e^{e^{x^2}} \left (-200 x^2-800 x^3\right ) \log (x)+400 e^{2 e^{x^2}} x^2 \log ^2(x)} \, dx=\frac {1}{25 x \left (4 e^{e^{x^{2}}} \mathrm {log}\left (x \right )-4 x -1\right )} \] Input:
int((((-8*exp(x^2)*x^2-4)*log(x)-4)*exp(exp(x^2))+8*x+1)/(400*x^2*log(x)^2 *exp(exp(x^2))^2+(-800*x^3-200*x^2)*log(x)*exp(exp(x^2))+400*x^4+200*x^3+2 5*x^2),x)
Output:
1/(25*x*(4*e**(e**(x**2))*log(x) - 4*x - 1))