Integrand size = 101, antiderivative size = 24 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=5 e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \log (4) \]
Time = 2.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=5 e^{\frac {3 e^{x^2}}{\log (-((-4+x) \log (x)))}} \log (4) \]
Integrate[(E^((3*E^x^2)/Log[(4 - x)*Log[x]])*(E^x^2*(60 - 15*x)*Log[4] - 1 5*E^x^2*x*Log[4]*Log[x] + E^x^2*(-120*x^2 + 30*x^3)*Log[4]*Log[x]*Log[(4 - x)*Log[x]]))/((-4*x + x^2)*Log[x]*Log[(4 - x)*Log[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.
| \(\displaystyle \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (30 x^3-120 x^2\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (x^2-4 x\right ) \log (x) \log ^2((4-x) \log (x))} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (30 x^3-120 x^2\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{(x-4) x \log (x) \log ^2((4-x) \log (x))}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {15 \log (4) e^{x^2+\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (-2 x^3 \log (x) \log (-((x-4) \log (x)))+8 x^2 \log (x) \log (-((x-4) \log (x)))+x+x \log (x)-4\right )}{(4-x) x \log (x) \log ^2((4-x) \log (x))}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 15 \log (4) \int -\frac {e^{x^2+\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (2 \log (x) \log ((4-x) \log (x)) x^3-8 \log (x) \log ((4-x) \log (x)) x^2-\log (x) x-x+4\right )}{(4-x) x \log (x) \log ^2((4-x) \log (x))}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -15 \log (4) \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (2 \log (x) \log ((4-x) \log (x)) x^3-8 \log (x) \log ((4-x) \log (x)) x^2-\log (x) x-x+4\right )}{(4-x) x \log (x) \log ^2((4-x) \log (x))}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -15 \log (4) \int \left (\frac {e^{x^2+\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} (\log (x) x+x-4)}{(x-4) x \log (x) \log ^2(-((x-4) \log (x)))}-\frac {2 e^{x^2+\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} x}{\log (-((x-4) \log (x)))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -15 \log (4) \left (\int \frac {e^{x^2+\frac {3 e^{x^2}}{\log ((4-x) \log (x))}}}{(x-4) \log ^2(-((x-4) \log (x)))}dx+\int \frac {e^{x^2+\frac {3 e^{x^2}}{\log ((4-x) \log (x))}}}{x \log (x) \log ^2(-((x-4) \log (x)))}dx-2 \int \frac {e^{x^2+\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} x}{\log (-((x-4) \log (x)))}dx\right )\) |
Int[(E^((3*E^x^2)/Log[(4 - x)*Log[x]])*(E^x^2*(60 - 15*x)*Log[4] - 15*E^x^ 2*x*Log[4]*Log[x] + E^x^2*(-120*x^2 + 30*x^3)*Log[4]*Log[x]*Log[(4 - x)*Lo g[x]]))/((-4*x + x^2)*Log[x]*Log[(4 - x)*Log[x]]^2),x]
3.4.3.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.15 (sec) , antiderivative size = 129, normalized size of antiderivative = 5.38
\[10 \ln \left (2\right ) {\mathrm e}^{\frac {6 \,{\mathrm e}^{x^{2}}}{i \pi \operatorname {csgn}\left (i \ln \left (x \right ) \left (x -4\right )\right )^{3}+i \operatorname {csgn}\left (i \ln \left (x \right ) \left (x -4\right )\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right ) \pi +i \operatorname {csgn}\left (i \ln \left (x \right ) \left (x -4\right )\right )^{2} \operatorname {csgn}\left (i \left (x -4\right )\right ) \pi -i \operatorname {csgn}\left (i \ln \left (x \right ) \left (x -4\right )\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \left (x -4\right )\right ) \pi -2 i \pi \operatorname {csgn}\left (i \ln \left (x \right ) \left (x -4\right )\right )^{2}+2 i \pi +2 \ln \left (x -4\right )+2 \ln \left (\ln \left (x \right )\right )}}\]
int((2*(30*x^3-120*x^2)*ln(2)*exp(x^2)*ln(x)*ln((-x+4)*ln(x))-30*x*ln(2)*e xp(x^2)*ln(x)+2*(-15*x+60)*ln(2)*exp(x^2))*exp(3*exp(x^2)/ln((-x+4)*ln(x)) )/(x^2-4*x)/ln(x)/ln((-x+4)*ln(x))^2,x)
10*ln(2)*exp(6*exp(x^2)/(I*Pi*csgn(I*ln(x)*(x-4))^3+I*csgn(I*ln(x)*(x-4))^ 2*csgn(I*ln(x))*Pi+I*csgn(I*ln(x)*(x-4))^2*csgn(I*(x-4))*Pi-I*csgn(I*ln(x) *(x-4))*csgn(I*ln(x))*csgn(I*(x-4))*Pi-2*I*Pi*csgn(I*ln(x)*(x-4))^2+2*I*Pi +2*ln(x-4)+2*ln(ln(x))))
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=10 \, e^{\left (\frac {3 \, e^{\left (x^{2}\right )}}{\log \left (-{\left (x - 4\right )} \log \left (x\right )\right )}\right )} \log \left (2\right ) \]
integrate((2*(30*x^3-120*x^2)*log(2)*exp(x^2)*log(x)*log((-x+4)*log(x))-30 *x*log(2)*exp(x^2)*log(x)+2*(-15*x+60)*log(2)*exp(x^2))*exp(3*exp(x^2)/log ((-x+4)*log(x)))/(x^2-4*x)/log(x)/log((-x+4)*log(x))^2,x, algorithm=\
Time = 5.64 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=10 e^{\frac {3 e^{x^{2}}}{\log {\left (\left (4 - x\right ) \log {\left (x \right )} \right )}}} \log {\left (2 \right )} \]
integrate((2*(30*x**3-120*x**2)*ln(2)*exp(x**2)*ln(x)*ln((-x+4)*ln(x))-30* x*ln(2)*exp(x**2)*ln(x)+2*(-15*x+60)*ln(2)*exp(x**2))*exp(3*exp(x**2)/ln(( -x+4)*ln(x)))/(x**2-4*x)/ln(x)/ln((-x+4)*ln(x))**2,x)
Exception generated. \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=\text {Exception raised: RuntimeError} \]
integrate((2*(30*x^3-120*x^2)*log(2)*exp(x^2)*log(x)*log((-x+4)*log(x))-30 *x*log(2)*exp(x^2)*log(x)+2*(-15*x+60)*log(2)*exp(x^2))*exp(3*exp(x^2)/log ((-x+4)*log(x)))/(x^2-4*x)/log(x)/log((-x+4)*log(x))^2,x, algorithm=\
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not of the expected type LIST
Time = 0.92 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=10 \, e^{\left (\frac {3 \, e^{\left (x^{2}\right )}}{\log \left (-x \log \left (x\right ) + 4 \, \log \left (x\right )\right )}\right )} \log \left (2\right ) \]
integrate((2*(30*x^3-120*x^2)*log(2)*exp(x^2)*log(x)*log((-x+4)*log(x))-30 *x*log(2)*exp(x^2)*log(x)+2*(-15*x+60)*log(2)*exp(x^2))*exp(3*exp(x^2)/log ((-x+4)*log(x)))/(x^2-4*x)/log(x)/log((-x+4)*log(x))^2,x, algorithm=\
Time = 12.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {3 e^{x^2}}{\log ((4-x) \log (x))}} \left (e^{x^2} (60-15 x) \log (4)-15 e^{x^2} x \log (4) \log (x)+e^{x^2} \left (-120 x^2+30 x^3\right ) \log (4) \log (x) \log ((4-x) \log (x))\right )}{\left (-4 x+x^2\right ) \log (x) \log ^2((4-x) \log (x))} \, dx=10\,{\mathrm {e}}^{\frac {3\,{\mathrm {e}}^{x^2}}{\ln \left (4\,\ln \left (x\right )-x\,\ln \left (x\right )\right )}}\,\ln \left (2\right ) \]