Integrand size = 98, antiderivative size = 30 \[ \int \frac {24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} \left (-24+x^2-2 x^3\right ) \log (x)+6 \log ^2(x)}{36 e^{-2 x+2 x^2} x^2-36 e^{-x+x^2} x^2 \log (x)+9 x^2 \log ^2(x)} \, dx=\frac {-2+x-\frac {x}{-6+3 e^{x-x^2} \log (x)}}{3 x} \] Output:
1/3*(x-x/(3*ln(x)/exp(x^2-x)-6)-2)/x
Time = 0.40 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} \left (-24+x^2-2 x^3\right ) \log (x)+6 \log ^2(x)}{36 e^{-2 x+2 x^2} x^2-36 e^{-x+x^2} x^2 \log (x)+9 x^2 \log ^2(x)} \, dx=\frac {1}{9} \left (-\frac {6}{x}-\frac {e^{x^2}}{-2 e^{x^2}+e^x \log (x)}\right ) \] Input:
Integrate[(24*E^(-2*x + 2*x^2) + E^(-x + x^2)*x + E^(-x + x^2)*(-24 + x^2 - 2*x^3)*Log[x] + 6*Log[x]^2)/(36*E^(-2*x + 2*x^2)*x^2 - 36*E^(-x + x^2)*x ^2*Log[x] + 9*x^2*Log[x]^2),x]
Output:
(-6/x - E^x^2/(-2*E^x^2 + E^x*Log[x]))/9
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.
\(\displaystyle \int \frac {24 e^{2 x^2-2 x}+e^{x^2-x} x+e^{x^2-x} \left (-2 x^3+x^2-24\right ) \log (x)+6 \log ^2(x)}{36 e^{2 x^2-2 x} x^2+9 x^2 \log ^2(x)-36 e^{x^2-x} x^2 \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x} \left (24 e^{2 x^2-2 x}+e^{x^2-x} x+e^{x^2-x} \left (-2 x^3+x^2-24\right ) \log (x)+6 \log ^2(x)\right )}{9 x^2 \left (2 e^{x^2}-e^x \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int \frac {e^{2 x} \left (6 \log ^2(x)-e^{x^2-x} \left (2 x^3-x^2+24\right ) \log (x)+24 e^{2 x^2-2 x}+e^{x^2-x} x\right )}{x^2 \left (2 e^{x^2}-e^x \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{9} \int \left (\frac {e^x \left (2 \log (x) x^2-\log (x) x-1\right )}{2 x \left (e^x \log (x)-2 e^{x^2}\right )}-\frac {e^{2 x} \log (x) \left (2 \log (x) x^2-\log (x) x-1\right )}{2 x \left (e^x \log (x)-2 e^{x^2}\right )^2}+\frac {6}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{9} \left (\frac {1}{2} \int \frac {e^{2 x} \log ^2(x)}{\left (e^x \log (x)-2 e^{x^2}\right )^2}dx-\int \frac {e^{2 x} x \log ^2(x)}{\left (e^x \log (x)-2 e^{x^2}\right )^2}dx+\frac {1}{2} \int \frac {e^x}{x \left (2 e^{x^2}-e^x \log (x)\right )}dx+\frac {1}{2} \int \frac {e^{2 x} \log (x)}{x \left (e^x \log (x)-2 e^{x^2}\right )^2}dx-\frac {1}{2} \int \frac {e^x \log (x)}{e^x \log (x)-2 e^{x^2}}dx+\int \frac {e^x x \log (x)}{e^x \log (x)-2 e^{x^2}}dx-\frac {6}{x}\right )\) |
Input:
Int[(24*E^(-2*x + 2*x^2) + E^(-x + x^2)*x + E^(-x + x^2)*(-24 + x^2 - 2*x^ 3)*Log[x] + 6*Log[x]^2)/(36*E^(-2*x + 2*x^2)*x^2 - 36*E^(-x + x^2)*x^2*Log [x] + 9*x^2*Log[x]^2),x]
Output:
$Aborted
Time = 0.43 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-\frac {2}{3 x}+\frac {{\mathrm e}^{x \left (-1+x \right )}}{18 \,{\mathrm e}^{x \left (-1+x \right )}-9 \ln \left (x \right )}\) | \(30\) |
parallelrisch | \(\frac {-x \ln \left (x \right )-12 \ln \left (x \right )+24 \,{\mathrm e}^{x^{2}-x}}{18 x \left (\ln \left (x \right )-2 \,{\mathrm e}^{x^{2}-x}\right )}\) | \(41\) |
Input:
int((6*ln(x)^2+(-2*x^3+x^2-24)*exp(x^2-x)*ln(x)+24*exp(x^2-x)^2+x*exp(x^2- x))/(9*x^2*ln(x)^2-36*x^2*exp(x^2-x)*ln(x)+36*x^2*exp(x^2-x)^2),x,method=_ RETURNVERBOSE)
Output:
-2/3/x+1/9*exp(x*(-1+x))/(2*exp(x*(-1+x))-ln(x))
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} \left (-24+x^2-2 x^3\right ) \log (x)+6 \log ^2(x)}{36 e^{-2 x+2 x^2} x^2-36 e^{-x+x^2} x^2 \log (x)+9 x^2 \log ^2(x)} \, dx=\frac {{\left (x - 12\right )} e^{\left (x^{2} - x\right )} + 6 \, \log \left (x\right )}{9 \, {\left (2 \, x e^{\left (x^{2} - x\right )} - x \log \left (x\right )\right )}} \] Input:
integrate((6*log(x)^2+(-2*x^3+x^2-24)*exp(x^2-x)*log(x)+24*exp(x^2-x)^2+x* exp(x^2-x))/(9*x^2*log(x)^2-36*x^2*exp(x^2-x)*log(x)+36*x^2*exp(x^2-x)^2), x, algorithm="fricas")
Output:
1/9*((x - 12)*e^(x^2 - x) + 6*log(x))/(2*x*e^(x^2 - x) - x*log(x))
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.67 \[ \int \frac {24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} \left (-24+x^2-2 x^3\right ) \log (x)+6 \log ^2(x)}{36 e^{-2 x+2 x^2} x^2-36 e^{-x+x^2} x^2 \log (x)+9 x^2 \log ^2(x)} \, dx=\frac {\log {\left (x \right )}}{36 e^{x^{2} - x} - 18 \log {\left (x \right )}} - \frac {2}{3 x} \] Input:
integrate((6*ln(x)**2+(-2*x**3+x**2-24)*exp(x**2-x)*ln(x)+24*exp(x**2-x)** 2+x*exp(x**2-x))/(9*x**2*ln(x)**2-36*x**2*exp(x**2-x)*ln(x)+36*x**2*exp(x* *2-x)**2),x)
Output:
log(x)/(36*exp(x**2 - x) - 18*log(x)) - 2/(3*x)
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.10 \[ \int \frac {24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} \left (-24+x^2-2 x^3\right ) \log (x)+6 \log ^2(x)}{36 e^{-2 x+2 x^2} x^2-36 e^{-x+x^2} x^2 \log (x)+9 x^2 \log ^2(x)} \, dx=-\frac {{\left (x + 12\right )} e^{x} \log \left (x\right ) - 24 \, e^{\left (x^{2}\right )}}{18 \, {\left (x e^{x} \log \left (x\right ) - 2 \, x e^{\left (x^{2}\right )}\right )}} \] Input:
integrate((6*log(x)^2+(-2*x^3+x^2-24)*exp(x^2-x)*log(x)+24*exp(x^2-x)^2+x* exp(x^2-x))/(9*x^2*log(x)^2-36*x^2*exp(x^2-x)*log(x)+36*x^2*exp(x^2-x)^2), x, algorithm="maxima")
Output:
-1/18*((x + 12)*e^x*log(x) - 24*e^(x^2))/(x*e^x*log(x) - 2*x*e^(x^2))
Time = 0.14 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.33 \[ \int \frac {24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} \left (-24+x^2-2 x^3\right ) \log (x)+6 \log ^2(x)}{36 e^{-2 x+2 x^2} x^2-36 e^{-x+x^2} x^2 \log (x)+9 x^2 \log ^2(x)} \, dx=\frac {x \log \left (x\right ) - 24 \, e^{\left (x^{2} - x\right )} + 12 \, \log \left (x\right )}{18 \, {\left (2 \, x e^{\left (x^{2} - x\right )} - x \log \left (x\right )\right )}} \] Input:
integrate((6*log(x)^2+(-2*x^3+x^2-24)*exp(x^2-x)*log(x)+24*exp(x^2-x)^2+x* exp(x^2-x))/(9*x^2*log(x)^2-36*x^2*exp(x^2-x)*log(x)+36*x^2*exp(x^2-x)^2), x, algorithm="giac")
Output:
1/18*(x*log(x) - 24*e^(x^2 - x) + 12*log(x))/(2*x*e^(x^2 - x) - x*log(x))
Time = 2.94 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} \left (-24+x^2-2 x^3\right ) \log (x)+6 \log ^2(x)}{36 e^{-2 x+2 x^2} x^2-36 e^{-x+x^2} x^2 \log (x)+9 x^2 \log ^2(x)} \, dx=\frac {\ln \left (x\right )}{18\,\left (2\,{\mathrm {e}}^{x^2-x}-\ln \left (x\right )\right )}-\frac {2}{3\,x} \] Input:
int((24*exp(2*x^2 - 2*x) + 6*log(x)^2 + x*exp(x^2 - x) - exp(x^2 - x)*log( x)*(2*x^3 - x^2 + 24))/(36*x^2*exp(2*x^2 - 2*x) + 9*x^2*log(x)^2 - 36*x^2* exp(x^2 - x)*log(x)),x)
Output:
log(x)/(18*(2*exp(x^2 - x) - log(x))) - 2/(3*x)
Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47 \[ \int \frac {24 e^{-2 x+2 x^2}+e^{-x+x^2} x+e^{-x+x^2} \left (-24+x^2-2 x^3\right ) \log (x)+6 \log ^2(x)}{36 e^{-2 x+2 x^2} x^2-36 e^{-x+x^2} x^2 \log (x)+9 x^2 \log ^2(x)} \, dx=\frac {e^{x^{2}} x -12 e^{x^{2}}+6 e^{x} \mathrm {log}\left (x \right )}{9 x \left (2 e^{x^{2}}-e^{x} \mathrm {log}\left (x \right )\right )} \] Input:
int((6*log(x)^2+(-2*x^3+x^2-24)*exp(x^2-x)*log(x)+24*exp(x^2-x)^2+x*exp(x^ 2-x))/(9*x^2*log(x)^2-36*x^2*exp(x^2-x)*log(x)+36*x^2*exp(x^2-x)^2),x)
Output:
(e**(x**2)*x - 12*e**(x**2) + 6*e**x*log(x))/(9*x*(2*e**(x**2) - e**x*log( x)))