Integrand size = 94, antiderivative size = 28 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-4+\frac {\log (x) \left (-4 e^{-x^3}-\log (\log (4))\right )}{\log (3-x)} \]
Time = 5.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-\frac {\log (x) \left (4 e^{-x^3}+\log (\log (4))\right )}{\log (3-x)} \]
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]
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 |
\(\displaystyle \log (\log (4)) \text {Subst}\left (\int \frac {\log (3-x)}{x \log ^2(x)}dx,x,3-x\right )-\log (\log (4)) \int \frac {1}{x \log (3-x)}dx-\frac {4 e^{-x^3} \left (3 x^3 \log (3-x) \log (x)-x^4 \log (3-x) \log (x)\right )}{(3-x) x^3 \log ^2(3-x)}\) |
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]
3.11.79.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 13.40 (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 | \(-\frac {\ln \left (x \right ) \left (\ln \left (2\right ) {\mathrm e}^{x^{3}}+\ln \left (\ln \left (2\right )\right ) {\mathrm e}^{x^{3}}+4\right ) {\mathrm e}^{-x^{3}}}{\ln \left (-x +3\right )}\) | \(36\) |
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)
Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-\frac {{\left (e^{\left (x^{3}\right )} \log \left (x\right ) \log \left (2 \, \log \left (2\right )\right ) + 4 \, \log \left (x\right )\right )} e^{\left (-x^{3}\right )}}{\log \left (-x + 3\right )} \]
integrate(((x*exp(x^3)*log(x)+(-x+3)*exp(x^3)*log(-x+3))*log(2*log(2))+((1 2*x^4-36*x^3)*log(-x+3)+4*x)*log(x)+(-4*x+12)*log(-x+3))/(x^2-3*x)/exp(x^3 )/log(-x+3)^2,x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=\frac {- \log {\left (2 \right )} \log {\left (x \right )} - \log {\left (x \right )} \log {\left (\log {\left (2 \right )} \right )}}{\log {\left (3 - x \right )}} - \frac {4 e^{- x^{3}} \log {\left (x \right )}}{\log {\left (3 - x \right )}} \]
integrate(((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)
Time = 0.33 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-\frac {{\left ({\left (\log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )} e^{\left (x^{3}\right )} \log \left (x\right ) + 4 \, \log \left (x\right )\right )} e^{\left (-x^{3}\right )}}{\log \left (-x + 3\right )} \]
integrate(((x*exp(x^3)*log(x)+(-x+3)*exp(x^3)*log(-x+3))*log(2*log(2))+((1 2*x^4-36*x^3)*log(-x+3)+4*x)*log(x)+(-4*x+12)*log(-x+3))/(x^2-3*x)/exp(x^3 )/log(-x+3)^2,x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-\frac {4 \, e^{\left (-x^{3}\right )} \log \left (x\right ) + \log \left (2\right ) \log \left (x\right ) + \log \left (x\right ) \log \left (\log \left (2\right )\right )}{\log \left (-x + 3\right )} \]
integrate(((x*exp(x^3)*log(x)+(-x+3)*exp(x^3)*log(-x+3))*log(2*log(2))+((1 2*x^4-36*x^3)*log(-x+3)+4*x)*log(x)+(-4*x+12)*log(-x+3))/(x^2-3*x)/exp(x^3 )/log(-x+3)^2,x, algorithm=\
Timed out. \[ \int \frac {e^{-x^3} \left ((12-4 x) \log (3-x)+\left (4 x+\left (-36 x^3+12 x^4\right ) \log (3-x)\right ) \log (x)+\left (e^{x^3} (3-x) \log (3-x)+e^{x^3} x \log (x)\right ) \log (\log (4))\right )}{\left (-3 x+x^2\right ) \log ^2(3-x)} \, dx=-\int -\frac {{\mathrm {e}}^{-x^3}\,\left (\ln \left (2\,\ln \left (2\right )\right )\,\left ({\mathrm {e}}^{x^3}\,\ln \left (3-x\right )\,\left (x-3\right )-x\,{\mathrm {e}}^{x^3}\,\ln \left (x\right )\right )+\ln \left (3-x\right )\,\left (4\,x-12\right )-\ln \left (x\right )\,\left (4\,x-\ln \left (3-x\right )\,\left (36\,x^3-12\,x^4\right )\right )\right )}{{\ln \left (3-x\right )}^2\,\left (3\,x-x^2\right )} \,d x \]
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)