Integrand size = 94, antiderivative size = 28 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-4+\frac {\log (x) \left (-4 e^{-x^3}-\log (\log (4))\right )}{\log (3-x)} \] Output:
(-ln(2*ln(2))-4/exp(x^3))*ln(x)/ln(3-x)-4
Time = 5.05 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-\frac {\log (x) \left (4 e^{-x^3}+\log (\log (4))\right )}{\log (3-x)} \] Input:
Integrate[((12 - 4*x)*Log[3 - x] + (4*x + (-36*x^3 + 12*x^4)*Log[3 - x])*L og[x] + (E^x^3*(3 - x)*Log[3 - x] + E^x^3*x*Log[x])*Log[Log[4]])/(E^x^3*(- 3*x + x^2)*Log[3 - x]^2),x]
Output:
-((Log[x]*(4/E^x^3 + Log[Log[4]]))/Log[3 - 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.
\(\displaystyle \int \frac {e^{-x^3} \left (\log (\log (4)) \left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right )+\left (\left (12 x^4-36 x^3\right ) \log (3-x)+4 x\right ) \log (x)+(12-4 x) \log (3-x)\right )}{\left (x^2-3 x\right ) \log ^2(3-x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x^3} \left (\log (\log (4)) \left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right )+\left (\left (12 x^4-36 x^3\right ) \log (3-x)+4 x\right ) \log (x)+(12-4 x) \log (3-x)\right )}{(x-3) x \log ^2(3-x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 e^{-x^3} \left (3 x^4 \log (3-x) \log (x)-9 x^3 \log (3-x) \log (x)-x \log (3-x)+x \log (x)+3 \log (3-x)\right )}{(x-3) x \log ^2(3-x)}-\frac {\log (\log (4)) (x \log (3-x)-3 \log (3-x)-x \log (x))}{(x-3) x \log ^2(3-x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log (\log (4)) \text {Subst}\left (\int \frac {\log (3-x)}{x \log ^2(x)}dx,x,3-x\right )-\log (\log (4)) \int \frac {1}{x \log (3-x)}dx-\frac {4 e^{-x^3} \left (3 x^3 \log (3-x) \log (x)-x^4 \log (3-x) \log (x)\right )}{(3-x) x^3 \log ^2(3-x)}\) |
Input:
Int[((12 - 4*x)*Log[3 - x] + (4*x + (-36*x^3 + 12*x^4)*Log[3 - x])*Log[x] + (E^x^3*(3 - x)*Log[3 - x] + E^x^3*x*Log[x])*Log[Log[4]])/(E^x^3*(-3*x + x^2)*Log[3 - x]^2),x]
Output:
$Aborted
Time = 13.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(-\frac {\left (\ln \left (2 \ln \left (2\right )\right ) \ln \left (x \right ) {\mathrm e}^{x^{3}}+4 \ln \left (x \right )\right ) {\mathrm e}^{-x^{3}}}{\ln \left (-x +3\right )}\) | \(34\) |
risch | \(-\frac {\ln \left (x \right ) \left (\ln \left (2\right ) {\mathrm e}^{x^{3}}+\ln \left (\ln \left (2\right )\right ) {\mathrm e}^{x^{3}}+4\right ) {\mathrm e}^{-x^{3}}}{\ln \left (-x +3\right )}\) | \(36\) |
Input:
int(((x*exp(x^3)*ln(x)+(-x+3)*exp(x^3)*ln(-x+3))*ln(2*ln(2))+((12*x^4-36*x ^3)*ln(-x+3)+4*x)*ln(x)+(-4*x+12)*ln(-x+3))/(x^2-3*x)/exp(x^3)/ln(-x+3)^2, x,method=_RETURNVERBOSE)
Output:
-(ln(2*ln(2))*ln(x)*exp(x^3)+4*ln(x))/ln(-x+3)/exp(x^3)
Time = 0.10 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-\frac {{\left (e^{\left (x^{3}\right )} \log \left (x\right ) \log \left (2 \, \log \left (2\right )\right ) + 4 \, \log \left (x\right )\right )} e^{\left (-x^{3}\right )}}{\log \left (-x + 3\right )} \] Input:
integrate(((x*exp(x^3)*log(x)+(3-x)*exp(x^3)*log(3-x))*log(2*log(2))+((12* x^4-36*x^3)*log(3-x)+4*x)*log(x)+(-4*x+12)*log(3-x))/(x^2-3*x)/exp(x^3)/lo g(3-x)^2,x, algorithm="fricas")
Output:
-(e^(x^3)*log(x)*log(2*log(2)) + 4*log(x))*e^(-x^3)/log(-x + 3)
Time = 0.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=\frac {- \log {\left (2 \right )} \log {\left (x \right )} - \log {\left (x \right )} \log {\left (\log {\left (2 \right )} \right )}}{\log {\left (3 - x \right )}} - \frac {4 e^{- x^{3}} \log {\left (x \right )}}{\log {\left (3 - x \right )}} \] Input:
integrate(((x*exp(x**3)*ln(x)+(3-x)*exp(x**3)*ln(3-x))*ln(2*ln(2))+((12*x* *4-36*x**3)*ln(3-x)+4*x)*ln(x)+(-4*x+12)*ln(3-x))/(x**2-3*x)/exp(x**3)/ln( 3-x)**2,x)
Output:
(-log(2)*log(x) - log(x)*log(log(2)))/log(3 - x) - 4*exp(-x**3)*log(x)/log (3 - x)
Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-\frac {{\left ({\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )} e^{\left (x^{3}\right )} \log \left (x\right ) + 4 \, \log \left (x\right )\right )} e^{\left (-x^{3}\right )}}{\log \left (-x + 3\right )} \] Input:
integrate(((x*exp(x^3)*log(x)+(3-x)*exp(x^3)*log(3-x))*log(2*log(2))+((12* x^4-36*x^3)*log(3-x)+4*x)*log(x)+(-4*x+12)*log(3-x))/(x^2-3*x)/exp(x^3)/lo g(3-x)^2,x, algorithm="maxima")
Output:
-((log(2) + log(log(2)))*e^(x^3)*log(x) + 4*log(x))*e^(-x^3)/log(-x + 3)
Time = 0.14 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-\frac {4 \, e^{\left (-x^{3}\right )} \log \left (x\right ) + \log \left (2\right ) \log \left (x\right ) + \log \left (x\right ) \log \left (\log \left (2\right )\right )}{\log \left (-x + 3\right )} \] Input:
integrate(((x*exp(x^3)*log(x)+(3-x)*exp(x^3)*log(3-x))*log(2*log(2))+((12* x^4-36*x^3)*log(3-x)+4*x)*log(x)+(-4*x+12)*log(3-x))/(x^2-3*x)/exp(x^3)/lo g(3-x)^2,x, algorithm="giac")
Output:
-(4*e^(-x^3)*log(x) + log(2)*log(x) + log(x)*log(log(2)))/log(-x + 3)
Timed out. \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-\int -\frac {{\mathrm {e}}^{-x^3}\,\left (\ln \left (2\,\ln \left (2\right )\right )\,\left ({\mathrm {e}}^{x^3}\,\ln \left (3-x\right )\,\left (x-3\right )-x\,{\mathrm {e}}^{x^3}\,\ln \left (x\right )\right )+\ln \left (3-x\right )\,\left (4\,x-12\right )-\ln \left (x\right )\,\left (4\,x-\ln \left (3-x\right )\,\left (36\,x^3-12\,x^4\right )\right )\right )}{{\ln \left (3-x\right )}^2\,\left (3\,x-x^2\right )} \,d x \] Input:
int((exp(-x^3)*(log(2*log(2))*(exp(x^3)*log(3 - x)*(x - 3) - x*exp(x^3)*lo g(x)) + log(3 - x)*(4*x - 12) - log(x)*(4*x - log(3 - x)*(36*x^3 - 12*x^4) )))/(log(3 - x)^2*(3*x - x^2)),x)
Output:
-int(-(exp(-x^3)*(log(2*log(2))*(exp(x^3)*log(3 - x)*(x - 3) - x*exp(x^3)* log(x)) + log(3 - x)*(4*x - 12) - log(x)*(4*x - log(3 - x)*(36*x^3 - 12*x^ 4))))/(log(3 - x)^2*(3*x - x^2)), x)
Time = 0.41 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=\frac {\mathrm {log}\left (x \right ) \left (-e^{x^{3}} \mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right )-4\right )}{e^{x^{3}} \mathrm {log}\left (-x +3\right )} \] Input:
int(((x*exp(x^3)*log(x)+(3-x)*exp(x^3)*log(3-x))*log(2*log(2))+((12*x^4-36 *x^3)*log(3-x)+4*x)*log(x)+(-4*x+12)*log(3-x))/(x^2-3*x)/exp(x^3)/log(3-x) ^2,x)
Output:
(log(x)*( - e**(x**3)*log(2*log(2)) - 4))/(e**(x**3)*log( - x + 3))