Integrand size = 101, antiderivative size = 24 \[ \int \frac {-e^5-x+e^{e^2} \left (2 e^5 x+2 x^2\right )+\left (-x+e^{e^2} x^2\right ) \log \left (x-e^{e^2} x^2\right )}{\left (-e^5 x-x^2+e^{e^2} \left (e^5 x^2+x^3\right )\right ) \log \left (x-e^{e^2} x^2\right )} \, dx=\log (5)+\log \left (4 \left (e^5+x\right ) \log \left (x-e^{e^2} x^2\right )\right ) \]
Time = 0.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {-e^5-x+e^{e^2} \left (2 e^5 x+2 x^2\right )+\left (-x+e^{e^2} x^2\right ) \log \left (x-e^{e^2} x^2\right )}{\left (-e^5 x-x^2+e^{e^2} \left (e^5 x^2+x^3\right )\right ) \log \left (x-e^{e^2} x^2\right )} \, dx=\log \left (e^5+x\right )+\log \left (\log \left (x-e^{e^2} x^2\right )\right ) \]
Integrate[(-E^5 - x + E^E^2*(2*E^5*x + 2*x^2) + (-x + E^E^2*x^2)*Log[x - E ^E^2*x^2])/((-(E^5*x) - x^2 + E^E^2*(E^5*x^2 + x^3))*Log[x - E^E^2*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.
\(\displaystyle \int \frac {e^{e^2} \left (2 x^2+2 e^5 x\right )+\left (e^{e^2} x^2-x\right ) \log \left (x-e^{e^2} x^2\right )-x-e^5}{\left (-x^2+e^{e^2} \left (x^3+e^5 x^2\right )-e^5 x\right ) \log \left (x-e^{e^2} x^2\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{e^2} \left (2 x^2+2 e^5 x\right )+\left (e^{e^2} x^2-x\right ) \log \left (x-e^{e^2} x^2\right )-x-e^5}{x \left (e^{e^2} x^2-\left (1-e^{5+e^2}\right ) x-e^5\right ) \log \left (x-e^{e^2} x^2\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {1}{x+e^5}+\frac {2 e^{e^2} x-1}{x \left (e^{e^2} x-1\right ) \log \left (x \left (1-e^{e^2} x\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {2 e^{e^2} x-1}{x \left (e^{e^2} x-1\right ) \log \left (x \left (1-e^{e^2} x\right )\right )}dx+\log \left (x+e^5\right )\) |
Int[(-E^5 - x + E^E^2*(2*E^5*x + 2*x^2) + (-x + E^E^2*x^2)*Log[x - E^E^2*x ^2])/((-(E^5*x) - x^2 + E^E^2*(E^5*x^2 + x^3))*Log[x - E^E^2*x^2]),x]
3.14.25.3.1 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])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.36 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79
method | result | size |
default | \(\ln \left ({\mathrm e}^{5}+x \right )+\ln \left (\ln \left (-x^{2} {\mathrm e}^{{\mathrm e}^{2}}+x \right )\right )\) | \(19\) |
norman | \(\ln \left ({\mathrm e}^{5}+x \right )+\ln \left (\ln \left (-x^{2} {\mathrm e}^{{\mathrm e}^{2}}+x \right )\right )\) | \(19\) |
risch | \(\ln \left ({\mathrm e}^{5}+x \right )+\ln \left (\ln \left (-x^{2} {\mathrm e}^{{\mathrm e}^{2}}+x \right )\right )\) | \(19\) |
parallelrisch | \(\ln \left ({\mathrm e}^{5}+x \right )+\ln \left (\ln \left (-x^{2} {\mathrm e}^{{\mathrm e}^{2}}+x \right )\right )\) | \(19\) |
parts | \(\ln \left ({\mathrm e}^{5}+x \right )+\ln \left (\ln \left (-x^{2} {\mathrm e}^{{\mathrm e}^{2}}+x \right )\right )\) | \(19\) |
int(((x^2*exp(exp(2))-x)*ln(-x^2*exp(exp(2))+x)+(2*x*exp(5)+2*x^2)*exp(exp (2))-exp(5)-x)/((x^2*exp(5)+x^3)*exp(exp(2))-x*exp(5)-x^2)/ln(-x^2*exp(exp (2))+x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-e^5-x+e^{e^2} \left (2 e^5 x+2 x^2\right )+\left (-x+e^{e^2} x^2\right ) \log \left (x-e^{e^2} x^2\right )}{\left (-e^5 x-x^2+e^{e^2} \left (e^5 x^2+x^3\right )\right ) \log \left (x-e^{e^2} x^2\right )} \, dx=\log \left (x + e^{5}\right ) + \log \left (\log \left (-x^{2} e^{\left (e^{2}\right )} + x\right )\right ) \]
integrate(((x^2*exp(exp(2))-x)*log(-x^2*exp(exp(2))+x)+(2*x*exp(5)+2*x^2)* exp(exp(2))-exp(5)-x)/((x^2*exp(5)+x^3)*exp(exp(2))-x*exp(5)-x^2)/log(-x^2 *exp(exp(2))+x),x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-e^5-x+e^{e^2} \left (2 e^5 x+2 x^2\right )+\left (-x+e^{e^2} x^2\right ) \log \left (x-e^{e^2} x^2\right )}{\left (-e^5 x-x^2+e^{e^2} \left (e^5 x^2+x^3\right )\right ) \log \left (x-e^{e^2} x^2\right )} \, dx=\log {\left (x + e^{5} \right )} + \log {\left (\log {\left (- x^{2} e^{e^{2}} + x \right )} \right )} \]
integrate(((x**2*exp(exp(2))-x)*ln(-x**2*exp(exp(2))+x)+(2*x*exp(5)+2*x**2 )*exp(exp(2))-exp(5)-x)/((x**2*exp(5)+x**3)*exp(exp(2))-x*exp(5)-x**2)/ln( -x**2*exp(exp(2))+x),x)
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-e^5-x+e^{e^2} \left (2 e^5 x+2 x^2\right )+\left (-x+e^{e^2} x^2\right ) \log \left (x-e^{e^2} x^2\right )}{\left (-e^5 x-x^2+e^{e^2} \left (e^5 x^2+x^3\right )\right ) \log \left (x-e^{e^2} x^2\right )} \, dx=\log \left (x + e^{5}\right ) + \log \left (\log \left (-x e^{\left (e^{2}\right )} + 1\right ) + \log \left (x\right )\right ) \]
integrate(((x^2*exp(exp(2))-x)*log(-x^2*exp(exp(2))+x)+(2*x*exp(5)+2*x^2)* exp(exp(2))-exp(5)-x)/((x^2*exp(5)+x^3)*exp(exp(2))-x*exp(5)-x^2)/log(-x^2 *exp(exp(2))+x),x, algorithm=\
\[ \int \frac {-e^5-x+e^{e^2} \left (2 e^5 x+2 x^2\right )+\left (-x+e^{e^2} x^2\right ) \log \left (x-e^{e^2} x^2\right )}{\left (-e^5 x-x^2+e^{e^2} \left (e^5 x^2+x^3\right )\right ) \log \left (x-e^{e^2} x^2\right )} \, dx=\int { -\frac {2 \, {\left (x^{2} + x e^{5}\right )} e^{\left (e^{2}\right )} + {\left (x^{2} e^{\left (e^{2}\right )} - x\right )} \log \left (-x^{2} e^{\left (e^{2}\right )} + x\right ) - x - e^{5}}{{\left (x^{2} + x e^{5} - {\left (x^{3} + x^{2} e^{5}\right )} e^{\left (e^{2}\right )}\right )} \log \left (-x^{2} e^{\left (e^{2}\right )} + x\right )} \,d x } \]
integrate(((x^2*exp(exp(2))-x)*log(-x^2*exp(exp(2))+x)+(2*x*exp(5)+2*x^2)* exp(exp(2))-exp(5)-x)/((x^2*exp(5)+x^3)*exp(exp(2))-x*exp(5)-x^2)/log(-x^2 *exp(exp(2))+x),x, algorithm=\
integrate(-(2*(x^2 + x*e^5)*e^(e^2) + (x^2*e^(e^2) - x)*log(-x^2*e^(e^2) + x) - x - e^5)/((x^2 + x*e^5 - (x^3 + x^2*e^5)*e^(e^2))*log(-x^2*e^(e^2) + x)), x)
Timed out. \[ \int \frac {-e^5-x+e^{e^2} \left (2 e^5 x+2 x^2\right )+\left (-x+e^{e^2} x^2\right ) \log \left (x-e^{e^2} x^2\right )}{\left (-e^5 x-x^2+e^{e^2} \left (e^5 x^2+x^3\right )\right ) \log \left (x-e^{e^2} x^2\right )} \, dx=\int \frac {x+{\mathrm {e}}^5+\ln \left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^2}\right )\,\left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^2}\right )-{\mathrm {e}}^{{\mathrm {e}}^2}\,\left (2\,x^2+2\,{\mathrm {e}}^5\,x\right )}{\ln \left (x-x^2\,{\mathrm {e}}^{{\mathrm {e}}^2}\right )\,\left (x\,{\mathrm {e}}^5+x^2-{\mathrm {e}}^{{\mathrm {e}}^2}\,\left (x^3+{\mathrm {e}}^5\,x^2\right )\right )} \,d x \]
int((x + exp(5) + log(x - x^2*exp(exp(2)))*(x - x^2*exp(exp(2))) - exp(exp (2))*(2*x*exp(5) + 2*x^2))/(log(x - x^2*exp(exp(2)))*(x*exp(5) + x^2 - exp (exp(2))*(x^2*exp(5) + x^3))),x)