Integrand size = 127, antiderivative size = 34 \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=e^{4+\frac {x+\frac {-4+\frac {x}{2}-\log (x)}{\log (4-2 x-\log (5))}}{x}} \]
Time = 0.11 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=e^{5+\frac {-8+x-2 \log (x)}{2 x \log (4-2 x-\log (5))}} \]
Integrate[(E^((-8 + x - 2*Log[x] + 10*x*Log[4 - 2*x - Log[5]])/(2*x*Log[4 - 2*x - Log[5]]))*(8*x - x^2 + (-12 + 6*x + 3*Log[5])*Log[4 - 2*x - Log[5] ] + Log[x]*(2*x + (-4 + 2*x + Log[5])*Log[4 - 2*x - Log[5]])))/((-4*x^2 + 2*x^3 + x^2*Log[5])*Log[4 - 2*x - Log[5]]^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.
| \(\displaystyle \int \frac {\left (-x^2+8 x+(6 x-12+3 \log (5)) \log (-2 x+4-\log (5))+\log (x) (2 x+(2 x-4+\log (5)) \log (-2 x+4-\log (5)))\right ) \exp \left (\frac {x+10 x \log (-2 x+4-\log (5))-2 \log (x)-8}{2 x \log (-2 x+4-\log (5))}\right )}{\left (2 x^3-4 x^2+x^2 \log (5)\right ) \log ^2(-2 x+4-\log (5))} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {\left (-x^2+8 x+(6 x-12+3 \log (5)) \log (-2 x+4-\log (5))+\log (x) (2 x+(2 x-4+\log (5)) \log (-2 x+4-\log (5)))\right ) \exp \left (\frac {x+10 x \log (-2 x+4-\log (5))-2 \log (x)-8}{2 x \log (-2 x+4-\log (5))}\right )}{\left (2 x^3+x^2 (\log (5)-4)\right ) \log ^2(-2 x+4-\log (5))}dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (-x^2+8 x+(6 x-12+3 \log (5)) \log (-2 x+4-\log (5))+\log (x) (2 x+(2 x-4+\log (5)) \log (-2 x+4-\log (5)))\right ) \exp \left (\frac {x+10 x \log (-2 x+4-\log (5))-2 \log (x)-8}{2 x \log (-2 x+4-\log (5))}\right )}{x^2 (2 x-4+\log (5)) \log ^2(-2 x+4-\log (5))}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {(\log (x)+3) \exp \left (\frac {x+10 x \log (-2 x+4-\log (5))-2 \log (x)-8}{2 x \log (-2 x+4-\log (5))}\right )}{x^2 \log (-2 x+4-\log (5))}+\frac {(-x+2 \log (x)+8) \exp \left (\frac {x+10 x \log (-2 x+4-\log (5))-2 \log (x)-8}{2 x \log (-2 x+4-\log (5))}\right )}{x (2 x-4+\log (5)) \log ^2(-2 x+4-\log (5))}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 3 \int \frac {\exp \left (\frac {10 \log (-2 x-\log (5)+4) x+x-2 \log (x)-8}{2 x \log (-2 x-\log (5)+4)}\right )}{x^2 \log (-2 x-\log (5)+4)}dx+\int \frac {\exp \left (\frac {10 \log (-2 x-\log (5)+4) x+x-2 \log (x)-8}{2 x \log (-2 x-\log (5)+4)}\right ) \log (x)}{x^2 \log (-2 x-\log (5)+4)}dx-\frac {8 \int \frac {\exp \left (\frac {10 \log (-2 x-\log (5)+4) x+x-2 \log (x)-8}{2 x \log (-2 x-\log (5)+4)}\right )}{x \log ^2(-2 x-\log (5)+4)}dx}{4-\log (5)}+\frac {16 \int \frac {\exp \left (\frac {10 \log (-2 x-\log (5)+4) x+x-2 \log (x)-8}{2 x \log (-2 x-\log (5)+4)}\right )}{(2 x+\log (5)-4) \log ^2(-2 x-\log (5)+4)}dx}{4-\log (5)}-\int \frac {\exp \left (\frac {10 \log (-2 x-\log (5)+4) x+x-2 \log (x)-8}{2 x \log (-2 x-\log (5)+4)}\right )}{(2 x+\log (5)-4) \log ^2(-2 x-\log (5)+4)}dx-\frac {2 \int \frac {\exp \left (\frac {10 \log (-2 x-\log (5)+4) x+x-2 \log (x)-8}{2 x \log (-2 x-\log (5)+4)}\right ) \log (x)}{x \log ^2(-2 x-\log (5)+4)}dx}{4-\log (5)}+\frac {4 \int \frac {\exp \left (\frac {10 \log (-2 x-\log (5)+4) x+x-2 \log (x)-8}{2 x \log (-2 x-\log (5)+4)}\right ) \log (x)}{(2 x+\log (5)-4) \log ^2(-2 x-\log (5)+4)}dx}{4-\log (5)}\) |
Int[(E^((-8 + x - 2*Log[x] + 10*x*Log[4 - 2*x - Log[5]])/(2*x*Log[4 - 2*x - Log[5]]))*(8*x - x^2 + (-12 + 6*x + 3*Log[5])*Log[4 - 2*x - Log[5]] + Lo g[x]*(2*x + (-4 + 2*x + Log[5])*Log[4 - 2*x - Log[5]])))/((-4*x^2 + 2*x^3 + x^2*Log[5])*Log[4 - 2*x - Log[5]]^2),x]
3.7.34.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, 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])
Time = 42.62 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15
| method | result | size |
| parallelrisch | \({\mathrm e}^{\frac {-2 \ln \left (x \right )+10 x \ln \left (-\ln \left (5\right )+4-2 x \right )-8+x}{2 x \ln \left (-\ln \left (5\right )+4-2 x \right )}}\) | \(39\) |
| risch | \({\mathrm e}^{-\frac {2 \ln \left (x \right )-10 x \ln \left (-\ln \left (5\right )+4-2 x \right )+8-x}{2 x \ln \left (-\ln \left (5\right )+4-2 x \right )}}\) | \(41\) |
int((((ln(5)+2*x-4)*ln(-ln(5)+4-2*x)+2*x)*ln(x)+(3*ln(5)+6*x-12)*ln(-ln(5) +4-2*x)-x^2+8*x)*exp(1/2*(-2*ln(x)+10*x*ln(-ln(5)+4-2*x)-8+x)/x/ln(-ln(5)+ 4-2*x))/(x^2*ln(5)+2*x^3-4*x^2)/ln(-ln(5)+4-2*x)^2,x,method=_RETURNVERBOSE )
Time = 0.48 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=e^{\left (\frac {10 \, x \log \left (-2 \, x - \log \left (5\right ) + 4\right ) + x - 2 \, \log \left (x\right ) - 8}{2 \, x \log \left (-2 \, x - \log \left (5\right ) + 4\right )}\right )} \]
integrate((((log(5)+2*x-4)*log(-log(5)+4-2*x)+2*x)*log(x)+(3*log(5)+6*x-12 )*log(-log(5)+4-2*x)-x^2+8*x)*exp(1/2*(-2*log(x)+10*x*log(-log(5)+4-2*x)-8 +x)/x/log(-log(5)+4-2*x))/(x^2*log(5)+2*x^3-4*x^2)/log(-log(5)+4-2*x)^2,x, algorithm=\
Exception generated. \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=\text {Exception raised: TypeError} \]
integrate((((ln(5)+2*x-4)*ln(-ln(5)+4-2*x)+2*x)*ln(x)+(3*ln(5)+6*x-12)*ln( -ln(5)+4-2*x)-x**2+8*x)*exp(1/2*(-2*ln(x)+10*x*ln(-ln(5)+4-2*x)-8+x)/x/ln( -ln(5)+4-2*x))/(x**2*ln(5)+2*x**3-4*x**2)/ln(-ln(5)+4-2*x)**2,x)
Exception generated. \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=\text {Exception raised: RuntimeError} \]
integrate((((log(5)+2*x-4)*log(-log(5)+4-2*x)+2*x)*log(x)+(3*log(5)+6*x-12 )*log(-log(5)+4-2*x)-x^2+8*x)*exp(1/2*(-2*log(x)+10*x*log(-log(5)+4-2*x)-8 +x)/x/log(-log(5)+4-2*x))/(x^2*log(5)+2*x^3-4*x^2)/log(-log(5)+4-2*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 = 0.76 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.56 \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=e^{\left (-\frac {\log \left (x\right )}{x \log \left (-2 \, x - \log \left (5\right ) + 4\right )} + \frac {1}{2 \, \log \left (-2 \, x - \log \left (5\right ) + 4\right )} - \frac {4}{x \log \left (-2 \, x - \log \left (5\right ) + 4\right )} + 5\right )} \]
integrate((((log(5)+2*x-4)*log(-log(5)+4-2*x)+2*x)*log(x)+(3*log(5)+6*x-12 )*log(-log(5)+4-2*x)-x^2+8*x)*exp(1/2*(-2*log(x)+10*x*log(-log(5)+4-2*x)-8 +x)/x/log(-log(5)+4-2*x))/(x^2*log(5)+2*x^3-4*x^2)/log(-log(5)+4-2*x)^2,x, algorithm=\
e^(-log(x)/(x*log(-2*x - log(5) + 4)) + 1/2/log(-2*x - log(5) + 4) - 4/(x* log(-2*x - log(5) + 4)) + 5)
Time = 11.78 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.65 \[ \int \frac {e^{\frac {-8+x-2 \log (x)+10 x \log (4-2 x-\log (5))}{2 x \log (4-2 x-\log (5))}} \left (8 x-x^2+(-12+6 x+3 \log (5)) \log (4-2 x-\log (5))+\log (x) (2 x+(-4+2 x+\log (5)) \log (4-2 x-\log (5)))\right )}{\left (-4 x^2+2 x^3+x^2 \log (5)\right ) \log ^2(4-2 x-\log (5))} \, dx=\frac {{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {4}{x\,\ln \left (4-\ln \left (5\right )-2\,x\right )}}\,{\mathrm {e}}^{\frac {1}{2\,\ln \left (4-\ln \left (5\right )-2\,x\right )}}}{x^{\frac {1}{x\,\ln \left (4-\ln \left (5\right )-2\,x\right )}}} \]
int((exp((x/2 - log(x) + 5*x*log(4 - log(5) - 2*x) - 4)/(x*log(4 - log(5) - 2*x)))*(8*x + log(4 - log(5) - 2*x)*(6*x + 3*log(5) - 12) - x^2 + log(x) *(2*x + log(4 - log(5) - 2*x)*(2*x + log(5) - 4))))/(log(4 - log(5) - 2*x) ^2*(x^2*log(5) - 4*x^2 + 2*x^3)),x)