Integrand size = 93, antiderivative size = 27 \[ \int \frac {(-3+x) \log (3-x)+\frac {e^{e^{e^5}+\frac {e^{e^{e^5}} \left (e^3+x\right )}{\log (3-x)}} \left (-e^3 x-x^2+\left (-3 x+x^2\right ) \log (3-x)\right )}{\log (3-x)}}{\left (-3 x+x^2\right ) \log (3-x)} \, dx=2+e^{\frac {e^{e^{e^5}} \left (e^3+x\right )}{\log (3-x)}}+\log (x) \]
Time = 1.91 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {(-3+x) \log (3-x)+\frac {e^{e^{e^5}+\frac {e^{e^{e^5}} \left (e^3+x\right )}{\log (3-x)}} \left (-e^3 x-x^2+\left (-3 x+x^2\right ) \log (3-x)\right )}{\log (3-x)}}{\left (-3 x+x^2\right ) \log (3-x)} \, dx=e^{\frac {e^{e^{e^5}} \left (e^3+x\right )}{\log (3-x)}}+\log (x) \]
Integrate[((-3 + x)*Log[3 - x] + (E^(E^E^5 + (E^E^E^5*(E^3 + x))/Log[3 - x ])*(-(E^3*x) - x^2 + (-3*x + x^2)*Log[3 - x]))/Log[3 - x])/((-3*x + x^2)*L og[3 - x]),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.
\(\displaystyle \int \frac {\frac {e^{\frac {e^{e^{e^5}} \left (x+e^3\right )}{\log (3-x)}+e^{e^5}} \left (-x^2+\left (x^2-3 x\right ) \log (3-x)-e^3 x\right )}{\log (3-x)}+(x-3) \log (3-x)}{\left (x^2-3 x\right ) \log (3-x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\frac {e^{\frac {e^{e^{e^5}} \left (x+e^3\right )}{\log (3-x)}+e^{e^5}} \left (-x^2+\left (x^2-3 x\right ) \log (3-x)-e^3 x\right )}{\log (3-x)}+(x-3) \log (3-x)}{(x-3) x \log (3-x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {1}{x}+\frac {e^{\frac {e^{e^{e^5}} \left (x+e^3\right )}{\log (3-x)}+e^{e^5}} \left (-x+x \log (3-x)-3 \log (3-x)-e^3\right )}{(x-3) \log ^2(3-x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {e^{\frac {e^{e^{e^5}} \left (x+e^3\right )}{\log (3-x)}+e^{e^5}}}{\log ^2(3-x)}dx-\left (3+e^3\right ) \int \frac {e^{\frac {e^{e^{e^5}} \left (x+e^3\right )}{\log (3-x)}+e^{e^5}}}{(x-3) \log ^2(3-x)}dx+\int \frac {e^{\frac {e^{e^{e^5}} \left (x+e^3\right )}{\log (3-x)}+e^{e^5}}}{\log (3-x)}dx+\log (x)\) |
Int[((-3 + x)*Log[3 - x] + (E^(E^E^5 + (E^E^E^5*(E^3 + x))/Log[3 - x])*(-( E^3*x) - x^2 + (-3*x + x^2)*Log[3 - x]))/Log[3 - x])/((-3*x + x^2)*Log[3 - x]),x]
3.8.72.3.1 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 = 23.72 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81
method | result | size |
default | \(\ln \left (x \right )+{\mathrm e}^{\frac {\left ({\mathrm e}^{3}+x \right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{5}}}}{\ln \left (-x +3\right )}}\) | \(22\) |
norman | \(\ln \left (x \right )+{\mathrm e}^{\frac {\left ({\mathrm e}^{3}+x \right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{5}}}}{\ln \left (-x +3\right )}}\) | \(22\) |
risch | \(\ln \left (x \right )+{\mathrm e}^{\frac {\left ({\mathrm e}^{3}+x \right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{5}}}}{\ln \left (-x +3\right )}}\) | \(22\) |
parts | \(\ln \left (x \right )+{\mathrm e}^{\frac {\left ({\mathrm e}^{3}+x \right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{5}}}}{\ln \left (-x +3\right )}}\) | \(22\) |
parallelrisch | \(\ln \left (x \right )+{\mathrm e}^{\left ({\mathrm e}^{3}+x \right ) {\mathrm e}^{-\ln \left (\ln \left (-x +3\right )\right )+{\mathrm e}^{{\mathrm e}^{5}}}}\) | \(24\) |
int((((x^2-3*x)*ln(-x+3)-x*exp(3)-x^2)*exp(-ln(ln(-x+3))+exp(exp(5)))*exp( (exp(3)+x)*exp(-ln(ln(-x+3))+exp(exp(5))))+(-3+x)*ln(-x+3))/(x^2-3*x)/ln(- x+3),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (24) = 48\).
Time = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.48 \[ \int \frac {(-3+x) \log (3-x)+\frac {e^{e^{e^5}+\frac {e^{e^{e^5}} \left (e^3+x\right )}{\log (3-x)}} \left (-e^3 x-x^2+\left (-3 x+x^2\right ) \log (3-x)\right )}{\log (3-x)}}{\left (-3 x+x^2\right ) \log (3-x)} \, dx={\left (e^{\left (e^{\left (e^{5}\right )}\right )} \log \left (x\right ) + e^{\left (\frac {{\left (x + e^{3}\right )} e^{\left (e^{\left (e^{5}\right )}\right )} + e^{\left (e^{5}\right )} \log \left (-x + 3\right ) - \log \left (-x + 3\right ) \log \left (\log \left (-x + 3\right )\right )}{\log \left (-x + 3\right )}\right )} \log \left (-x + 3\right )\right )} e^{\left (-e^{\left (e^{5}\right )}\right )} \]
integrate((((x^2-3*x)*log(-x+3)-x*exp(3)-x^2)*exp(-log(log(-x+3))+exp(exp( 5)))*exp((exp(3)+x)*exp(-log(log(-x+3))+exp(exp(5))))+(-3+x)*log(-x+3))/(x ^2-3*x)/log(-x+3),x, algorithm=\
(e^(e^(e^5))*log(x) + e^(((x + e^3)*e^(e^(e^5)) + e^(e^5)*log(-x + 3) - lo g(-x + 3)*log(log(-x + 3)))/log(-x + 3))*log(-x + 3))*e^(-e^(e^5))
Exception generated. \[ \int \frac {(-3+x) \log (3-x)+\frac {e^{e^{e^5}+\frac {e^{e^{e^5}} \left (e^3+x\right )}{\log (3-x)}} \left (-e^3 x-x^2+\left (-3 x+x^2\right ) \log (3-x)\right )}{\log (3-x)}}{\left (-3 x+x^2\right ) \log (3-x)} \, dx=\text {Exception raised: TypeError} \]
integrate((((x**2-3*x)*ln(-x+3)-x*exp(3)-x**2)*exp(-ln(ln(-x+3))+exp(exp(5 )))*exp((exp(3)+x)*exp(-ln(ln(-x+3))+exp(exp(5))))+(-3+x)*ln(-x+3))/(x**2- 3*x)/ln(-x+3),x)
Exception generated. \[ \int \frac {(-3+x) \log (3-x)+\frac {e^{e^{e^5}+\frac {e^{e^{e^5}} \left (e^3+x\right )}{\log (3-x)}} \left (-e^3 x-x^2+\left (-3 x+x^2\right ) \log (3-x)\right )}{\log (3-x)}}{\left (-3 x+x^2\right ) \log (3-x)} \, dx=\text {Exception raised: RuntimeError} \]
integrate((((x^2-3*x)*log(-x+3)-x*exp(3)-x^2)*exp(-log(log(-x+3))+exp(exp( 5)))*exp((exp(3)+x)*exp(-log(log(-x+3))+exp(exp(5))))+(-3+x)*log(-x+3))/(x ^2-3*x)/log(-x+3),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
\[ \int \frac {(-3+x) \log (3-x)+\frac {e^{e^{e^5}+\frac {e^{e^{e^5}} \left (e^3+x\right )}{\log (3-x)}} \left (-e^3 x-x^2+\left (-3 x+x^2\right ) \log (3-x)\right )}{\log (3-x)}}{\left (-3 x+x^2\right ) \log (3-x)} \, dx=\int { -\frac {{\left (x^{2} + x e^{3} - {\left (x^{2} - 3 \, x\right )} \log \left (-x + 3\right )\right )} e^{\left ({\left (x + e^{3}\right )} e^{\left (e^{\left (e^{5}\right )} - \log \left (\log \left (-x + 3\right )\right )\right )} + e^{\left (e^{5}\right )} - \log \left (\log \left (-x + 3\right )\right )\right )} - {\left (x - 3\right )} \log \left (-x + 3\right )}{{\left (x^{2} - 3 \, x\right )} \log \left (-x + 3\right )} \,d x } \]
integrate((((x^2-3*x)*log(-x+3)-x*exp(3)-x^2)*exp(-log(log(-x+3))+exp(exp( 5)))*exp((exp(3)+x)*exp(-log(log(-x+3))+exp(exp(5))))+(-3+x)*log(-x+3))/(x ^2-3*x)/log(-x+3),x, algorithm=\
integrate(-((x^2 + x*e^3 - (x^2 - 3*x)*log(-x + 3))*e^((x + e^3)*e^(e^(e^5 ) - log(log(-x + 3))) + e^(e^5) - log(log(-x + 3))) - (x - 3)*log(-x + 3)) /((x^2 - 3*x)*log(-x + 3)), x)
Time = 13.74 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {(-3+x) \log (3-x)+\frac {e^{e^{e^5}+\frac {e^{e^{e^5}} \left (e^3+x\right )}{\log (3-x)}} \left (-e^3 x-x^2+\left (-3 x+x^2\right ) \log (3-x)\right )}{\log (3-x)}}{\left (-3 x+x^2\right ) \log (3-x)} \, dx=\ln \left (x\right )+{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^5}}\,{\mathrm {e}}^3}{\ln \left (3-x\right )}}\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^5}}}{\ln \left (3-x\right )}} \]