Integrand size = 130, antiderivative size = 28 \[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=e^{-x+\frac {e^9}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \]
Time = 0.26 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=e^{-x+\frac {e^9}{e^2+\log \left (\frac {\log ^2(-4+x)}{x^2}\right )}} \]
Integrate[(E^((E^9 - x*Log[(E^E^2*Log[-4 + x]^2)/x^2])/Log[(E^E^2*Log[-4 + x]^2)/x^2])*(-2*E^9*x + E^9*(-8 + 2*x)*Log[-4 + x] + (4*x - x^2)*Log[-4 + x]*Log[(E^E^2*Log[-4 + x]^2)/x^2]^2))/((-4*x + x^2)*Log[-4 + x]*Log[(E^E^ 2*Log[-4 + x]^2)/x^2]^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.
| \(\displaystyle \int \frac {\left (\left (4 x-x^2\right ) \log (x-4) \log ^2\left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )-2 e^9 x+e^9 (2 x-8) \log (x-4)\right ) \exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}\right )}{\left (x^2-4 x\right ) \log (x-4) \log ^2\left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (\left (4 x-x^2\right ) \log (x-4) \log ^2\left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )-2 e^9 x+e^9 (2 x-8) \log (x-4)\right ) \exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}\right )}{(x-4) x \log (x-4) \log ^2\left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 (-x+x \log (x-4)-4 \log (x-4)) \exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}+9\right )}{(x-4) x \log (x-4) \left (\log \left (\frac {\log ^2(x-4)}{x^2}\right )+e^2\right )^2}-\exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\int \exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}\right )dx+2 \int \frac {\exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}+9\right )}{x \left (\log \left (\frac {\log ^2(x-4)}{x^2}\right )+e^2\right )^2}dx-2 \int \frac {\exp \left (\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(x-4)}{x^2}\right )}+9\right )}{(x-4) \log (x-4) \left (\log \left (\frac {\log ^2(x-4)}{x^2}\right )+e^2\right )^2}dx\) |
Int[(E^((E^9 - x*Log[(E^E^2*Log[-4 + x]^2)/x^2])/Log[(E^E^2*Log[-4 + x]^2) /x^2])*(-2*E^9*x + E^9*(-8 + 2*x)*Log[-4 + x] + (4*x - x^2)*Log[-4 + x]*Lo g[(E^E^2*Log[-4 + x]^2)/x^2]^2))/((-4*x + x^2)*Log[-4 + x]*Log[(E^E^2*Log[ -4 + x]^2)/x^2]^2),x]
3.19.60.3.1 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 = 265.96 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39
| method | result | size |
| parallelrisch | \({\mathrm e}^{\frac {-x \ln \left (\frac {{\mathrm e}^{{\mathrm e}^{2}} \ln \left (x -4\right )^{2}}{x^{2}}\right )+{\mathrm e}^{9}}{\ln \left (\frac {{\mathrm e}^{{\mathrm e}^{2}} \ln \left (x -4\right )^{2}}{x^{2}}\right )}}\) | \(39\) |
| risch | \({\mathrm e}^{-\frac {-i x \pi \operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{3}+2 i x \pi \operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{2} \operatorname {csgn}\left (i \ln \left (x -4\right )\right )-i x \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x -4\right )\right )^{2}+i x \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2}-i x \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )+i x \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i x \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+i x \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-i x \pi \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{3}+i x \pi \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )+2 \,{\mathrm e}^{2} x -4 x \ln \left (x \right )+4 x \ln \left (\ln \left (x -4\right )\right )-2 \,{\mathrm e}^{9}}{-i \pi \operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{3}+2 i \pi \operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right )^{2} \operatorname {csgn}\left (i \ln \left (x -4\right )\right )-i \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (i \ln \left (x -4\right )\right )^{2}+i \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2}-i \pi \,\operatorname {csgn}\left (i \ln \left (x -4\right )^{2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right ) \operatorname {csgn}\left (\frac {i}{x^{2}}\right )+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-i \pi \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{3}+i \pi \operatorname {csgn}\left (\frac {i \ln \left (x -4\right )^{2}}{x^{2}}\right )^{2} \operatorname {csgn}\left (\frac {i}{x^{2}}\right )+2 \,{\mathrm e}^{2}-4 \ln \left (x \right )+4 \ln \left (\ln \left (x -4\right )\right )}}\) | \(497\) |
int(((-x^2+4*x)*ln(x-4)*ln(exp(exp(2))*ln(x-4)^2/x^2)^2+(2*x-8)*exp(9)*ln( x-4)-2*x*exp(9))*exp((-x*ln(exp(exp(2))*ln(x-4)^2/x^2)+exp(9))/ln(exp(exp( 2))*ln(x-4)^2/x^2))/(x^2-4*x)/ln(x-4)/ln(exp(exp(2))*ln(x-4)^2/x^2)^2,x,me thod=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=e^{\left (-\frac {x \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right ) - e^{9}}{\log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )}\right )} \]
integrate(((-x^2+4*x)*log(x-4)*log(exp(exp(2))*log(x-4)^2/x^2)^2+(2*x-8)*e xp(9)*log(x-4)-2*x*exp(9))*exp((-x*log(exp(exp(2))*log(x-4)^2/x^2)+exp(9)) /log(exp(exp(2))*log(x-4)^2/x^2))/(x^2-4*x)/log(x-4)/log(exp(exp(2))*log(x -4)^2/x^2)^2,x, algorithm=\
Time = 0.66 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=e^{\frac {- x \log {\left (\frac {e^{e^{2}} \log {\left (x - 4 \right )}^{2}}{x^{2}} \right )} + e^{9}}{\log {\left (\frac {e^{e^{2}} \log {\left (x - 4 \right )}^{2}}{x^{2}} \right )}}} \]
integrate(((-x**2+4*x)*ln(x-4)*ln(exp(exp(2))*ln(x-4)**2/x**2)**2+(2*x-8)* exp(9)*ln(x-4)-2*x*exp(9))*exp((-x*ln(exp(exp(2))*ln(x-4)**2/x**2)+exp(9)) /ln(exp(exp(2))*ln(x-4)**2/x**2))/(x**2-4*x)/ln(x-4)/ln(exp(exp(2))*ln(x-4 )**2/x**2)**2,x)
\[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=\int { -\frac {{\left ({\left (x^{2} - 4 \, x\right )} \log \left (x - 4\right ) \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )^{2} - 2 \, {\left (x - 4\right )} e^{9} \log \left (x - 4\right ) + 2 \, x e^{9}\right )} e^{\left (-\frac {x \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right ) - e^{9}}{\log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )}\right )}}{{\left (x^{2} - 4 \, x\right )} \log \left (x - 4\right ) \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )^{2}} \,d x } \]
integrate(((-x^2+4*x)*log(x-4)*log(exp(exp(2))*log(x-4)^2/x^2)^2+(2*x-8)*e xp(9)*log(x-4)-2*x*exp(9))*exp((-x*log(exp(exp(2))*log(x-4)^2/x^2)+exp(9)) /log(exp(exp(2))*log(x-4)^2/x^2))/(x^2-4*x)/log(x-4)/log(exp(exp(2))*log(x -4)^2/x^2)^2,x, algorithm=\
-integrate(((x^2 - 4*x)*log(x - 4)*log(e^(e^2)*log(x - 4)^2/x^2)^2 - 2*(x - 4)*e^9*log(x - 4) + 2*x*e^9)*e^(-(x*log(e^(e^2)*log(x - 4)^2/x^2) - e^9) /log(e^(e^2)*log(x - 4)^2/x^2))/((x^2 - 4*x)*log(x - 4)*log(e^(e^2)*log(x - 4)^2/x^2)^2), x)
\[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=\int { -\frac {{\left ({\left (x^{2} - 4 \, x\right )} \log \left (x - 4\right ) \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )^{2} - 2 \, {\left (x - 4\right )} e^{9} \log \left (x - 4\right ) + 2 \, x e^{9}\right )} e^{\left (-\frac {x \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right ) - e^{9}}{\log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )}\right )}}{{\left (x^{2} - 4 \, x\right )} \log \left (x - 4\right ) \log \left (\frac {e^{\left (e^{2}\right )} \log \left (x - 4\right )^{2}}{x^{2}}\right )^{2}} \,d x } \]
integrate(((-x^2+4*x)*log(x-4)*log(exp(exp(2))*log(x-4)^2/x^2)^2+(2*x-8)*e xp(9)*log(x-4)-2*x*exp(9))*exp((-x*log(exp(exp(2))*log(x-4)^2/x^2)+exp(9)) /log(exp(exp(2))*log(x-4)^2/x^2))/(x^2-4*x)/log(x-4)/log(exp(exp(2))*log(x -4)^2/x^2)^2,x, algorithm=\
Time = 9.62 (sec) , antiderivative size = 92, normalized size of antiderivative = 3.29 \[ \int \frac {e^{\frac {e^9-x \log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}{\log \left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )}} \left (-2 e^9 x+e^9 (-8+2 x) \log (-4+x)+\left (4 x-x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )\right )}{\left (-4 x+x^2\right ) \log (-4+x) \log ^2\left (\frac {e^{e^2} \log ^2(-4+x)}{x^2}\right )} \, dx=\frac {{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^2}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^9}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}\,{\left (x^2\right )}^{\frac {x}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}}{{\left ({\ln \left (x-4\right )}^2\right )}^{\frac {x}{\ln \left ({\ln \left (x-4\right )}^2\right )+\ln \left (\frac {1}{x^2}\right )+{\mathrm {e}}^2}}} \]
int(-(exp((exp(9) - x*log((log(x - 4)^2*exp(exp(2)))/x^2))/log((log(x - 4) ^2*exp(exp(2)))/x^2))*(log(x - 4)*exp(9)*(2*x - 8) - 2*x*exp(9) + log((log (x - 4)^2*exp(exp(2)))/x^2)^2*log(x - 4)*(4*x - x^2)))/(log((log(x - 4)^2* exp(exp(2)))/x^2)^2*log(x - 4)*(4*x - x^2)),x)