Integrand size = 153, antiderivative size = 33 \[ \int \frac {-x+x^2+\left (x^2-x^3\right ) \log (5)+\left (x^2-x^3 \log (5)\right ) \log (x)+e^{-e^{e^5 x}+x} (1-x-x \log (x))+\left (e^{-e^{e^5 x}+x} \left (-x+x^2+e^{5+e^5 x} \left (x-x^2\right )\right ) \log (x)+\left (x-x^2+\left (-2 x^2+2 x^3\right ) \log (5)\right ) \log (x)\right ) \log ((-1+x) \log (x))}{\left (-x+x^2\right ) \log (x) \log ^2((-1+x) \log (x))} \, dx=\frac {e^{-e^{e^5 x}+x}-x+x^2 \log (5)}{\log ((-1+x) \log (x))} \]
Time = 0.35 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {-x+x^2+\left (x^2-x^3\right ) \log (5)+\left (x^2-x^3 \log (5)\right ) \log (x)+e^{-e^{e^5 x}+x} (1-x-x \log (x))+\left (e^{-e^{e^5 x}+x} \left (-x+x^2+e^{5+e^5 x} \left (x-x^2\right )\right ) \log (x)+\left (x-x^2+\left (-2 x^2+2 x^3\right ) \log (5)\right ) \log (x)\right ) \log ((-1+x) \log (x))}{\left (-x+x^2\right ) \log (x) \log ^2((-1+x) \log (x))} \, dx=\frac {e^{-e^{e^5 x}+x}-x+x^2 \log (5)}{\log ((-1+x) \log (x))} \]
Integrate[(-x + x^2 + (x^2 - x^3)*Log[5] + (x^2 - x^3*Log[5])*Log[x] + E^( -E^(E^5*x) + x)*(1 - x - x*Log[x]) + (E^(-E^(E^5*x) + x)*(-x + x^2 + E^(5 + E^5*x)*(x - x^2))*Log[x] + (x - x^2 + (-2*x^2 + 2*x^3)*Log[5])*Log[x])*L og[(-1 + x)*Log[x]])/((-x + x^2)*Log[x]*Log[(-1 + 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.
\(\displaystyle \int \frac {x^2+\left (x^2-x^3 \log (5)\right ) \log (x)+\left (e^{x-e^{e^5 x}} \left (x^2+e^{e^5 x+5} \left (x-x^2\right )-x\right ) \log (x)+\left (-x^2+\left (2 x^3-2 x^2\right ) \log (5)+x\right ) \log (x)\right ) \log ((x-1) \log (x))+\left (x^2-x^3\right ) \log (5)-x+e^{x-e^{e^5 x}} (-x+x (-\log (x))+1)}{\left (x^2-x\right ) \log (x) \log ^2((x-1) \log (x))} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x^2+\left (x^2-x^3 \log (5)\right ) \log (x)+\left (e^{x-e^{e^5 x}} \left (x^2+e^{e^5 x+5} \left (x-x^2\right )-x\right ) \log (x)+\left (-x^2+\left (2 x^3-2 x^2\right ) \log (5)+x\right ) \log (x)\right ) \log ((x-1) \log (x))+\left (x^2-x^3\right ) \log (5)-x+e^{x-e^{e^5 x}} (-x+x (-\log (x))+1)}{(x-1) x \log (x) \log ^2((x-1) \log (x))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {x (x \log (5)-1)}{(x-1) \log ^2((x-1) \log (x))}+\frac {x}{(x-1) \log (x) \log ^2((x-1) \log (x))}-\frac {x \log (5)}{\log (x) \log ^2((x-1) \log (x))}-\frac {1}{(x-1) \log (x) \log ^2((x-1) \log (x))}-\frac {e^{x-e^{e^5 x}} (x+x \log (x)-1)}{(x-1) x \log (x) \log ^2((x-1) \log (x))}+\frac {e^{x-e^{e^5 x}} x}{(x-1) \log ((x-1) \log (x))}-\frac {e^{\left (1+e^5\right ) x-e^{e^5 x}+5}}{\log ((x-1) \log (x))}+\frac {x \log (25)-1}{\log ((x-1) \log (x))}-\frac {e^{x-e^{e^5 x}}}{(x-1) \log ((x-1) \log (x))}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle (1-\log (5)) \int \frac {1}{\log ^2((x-1) \log (x))}dx+(1-\log (5)) \int \frac {1}{(x-1) \log ^2((x-1) \log (x))}dx-\int \frac {e^{x-e^{e^5 x}}}{(x-1) \log ^2((x-1) \log (x))}dx-\log (5) \int \frac {x}{\log ^2((x-1) \log (x))}dx+\int \frac {1}{\log (x) \log ^2((x-1) \log (x))}dx-\int \frac {e^{x-e^{e^5 x}}}{x \log (x) \log ^2((x-1) \log (x))}dx-\log (5) \int \frac {x}{\log (x) \log ^2((x-1) \log (x))}dx-\int \frac {1}{\log ((x-1) \log (x))}dx+\int \frac {e^{x-e^{e^5 x}}}{\log ((x-1) \log (x))}dx-\int \frac {e^{\left (1+e^5\right ) x-e^{e^5 x}+5}}{\log ((x-1) \log (x))}dx+\log (25) \int \frac {x}{\log ((x-1) \log (x))}dx\) |
Int[(-x + x^2 + (x^2 - x^3)*Log[5] + (x^2 - x^3*Log[5])*Log[x] + E^(-E^(E^ 5*x) + x)*(1 - x - x*Log[x]) + (E^(-E^(E^5*x) + x)*(-x + x^2 + E^(5 + E^5* x)*(x - x^2))*Log[x] + (x - x^2 + (-2*x^2 + 2*x^3)*Log[5])*Log[x])*Log[(-1 + x)*Log[x]])/((-x + x^2)*Log[x]*Log[(-1 + x)*Log[x]]^2),x]
3.5.85.3.1 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 = 186.65 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94
method | result | size |
parallelrisch | \(\frac {x^{2} \ln \left (5\right )+{\mathrm e}^{x -{\mathrm e}^{x \,{\mathrm e}^{5}}}-x}{\ln \left (\left (-1+x \right ) \ln \left (x \right )\right )}\) | \(31\) |
risch | \(\frac {2 i \left (x^{2} \ln \left (5\right )+{\mathrm e}^{x -{\mathrm e}^{x \,{\mathrm e}^{5}}}-x \right )}{\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right ) \left (-1+x \right )\right )-\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right ) \left (-1+x \right )\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (i \ln \left (x \right ) \left (-1+x \right )\right )^{2}+\pi \operatorname {csgn}\left (i \ln \left (x \right ) \left (-1+x \right )\right )^{3}+2 i \ln \left (\ln \left (x \right )\right )+2 i \ln \left (-1+x \right )}\) | \(118\) |
int(((((-x^2+x)*exp(5)*exp(x*exp(5))+x^2-x)*ln(x)*exp(x-exp(x*exp(5)))+((2 *x^3-2*x^2)*ln(5)-x^2+x)*ln(x))*ln((-1+x)*ln(x))+(-x*ln(x)-x+1)*exp(x-exp( x*exp(5)))+(-x^3*ln(5)+x^2)*ln(x)+(-x^3+x^2)*ln(5)+x^2-x)/(x^2-x)/ln(x)/ln ((-1+x)*ln(x))^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.15 \[ \int \frac {-x+x^2+\left (x^2-x^3\right ) \log (5)+\left (x^2-x^3 \log (5)\right ) \log (x)+e^{-e^{e^5 x}+x} (1-x-x \log (x))+\left (e^{-e^{e^5 x}+x} \left (-x+x^2+e^{5+e^5 x} \left (x-x^2\right )\right ) \log (x)+\left (x-x^2+\left (-2 x^2+2 x^3\right ) \log (5)\right ) \log (x)\right ) \log ((-1+x) \log (x))}{\left (-x+x^2\right ) \log (x) \log ^2((-1+x) \log (x))} \, dx=\frac {x^{2} \log \left (5\right ) - x + e^{\left ({\left (x e^{5} - e^{\left (x e^{5} + 5\right )}\right )} e^{\left (-5\right )}\right )}}{\log \left ({\left (x - 1\right )} \log \left (x\right )\right )} \]
integrate(((((-x^2+x)*exp(5)*exp(x*exp(5))+x^2-x)*log(x)*exp(x-exp(x*exp(5 )))+((2*x^3-2*x^2)*log(5)-x^2+x)*log(x))*log((-1+x)*log(x))+(-x*log(x)-x+1 )*exp(x-exp(x*exp(5)))+(-x^3*log(5)+x^2)*log(x)+(-x^3+x^2)*log(5)+x^2-x)/( x^2-x)/log(x)/log((-1+x)*log(x))^2,x, algorithm=\
Time = 1.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {-x+x^2+\left (x^2-x^3\right ) \log (5)+\left (x^2-x^3 \log (5)\right ) \log (x)+e^{-e^{e^5 x}+x} (1-x-x \log (x))+\left (e^{-e^{e^5 x}+x} \left (-x+x^2+e^{5+e^5 x} \left (x-x^2\right )\right ) \log (x)+\left (x-x^2+\left (-2 x^2+2 x^3\right ) \log (5)\right ) \log (x)\right ) \log ((-1+x) \log (x))}{\left (-x+x^2\right ) \log (x) \log ^2((-1+x) \log (x))} \, dx=\frac {x^{2} \log {\left (5 \right )} - x}{\log {\left (\left (x - 1\right ) \log {\left (x \right )} \right )}} + \frac {e^{x - e^{x e^{5}}}}{\log {\left (\left (x - 1\right ) \log {\left (x \right )} \right )}} \]
integrate(((((-x**2+x)*exp(5)*exp(x*exp(5))+x**2-x)*ln(x)*exp(x-exp(x*exp( 5)))+((2*x**3-2*x**2)*ln(5)-x**2+x)*ln(x))*ln((-1+x)*ln(x))+(-x*ln(x)-x+1) *exp(x-exp(x*exp(5)))+(-x**3*ln(5)+x**2)*ln(x)+(-x**3+x**2)*ln(5)+x**2-x)/ (x**2-x)/ln(x)/ln((-1+x)*ln(x))**2,x)
Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.94 \[ \int \frac {-x+x^2+\left (x^2-x^3\right ) \log (5)+\left (x^2-x^3 \log (5)\right ) \log (x)+e^{-e^{e^5 x}+x} (1-x-x \log (x))+\left (e^{-e^{e^5 x}+x} \left (-x+x^2+e^{5+e^5 x} \left (x-x^2\right )\right ) \log (x)+\left (x-x^2+\left (-2 x^2+2 x^3\right ) \log (5)\right ) \log (x)\right ) \log ((-1+x) \log (x))}{\left (-x+x^2\right ) \log (x) \log ^2((-1+x) \log (x))} \, dx=\frac {x^{2} \log \left (5\right ) - x + e^{\left (x - e^{\left (x e^{5}\right )}\right )}}{\log \left (x - 1\right ) + \log \left (\log \left (x\right )\right )} \]
integrate(((((-x^2+x)*exp(5)*exp(x*exp(5))+x^2-x)*log(x)*exp(x-exp(x*exp(5 )))+((2*x^3-2*x^2)*log(5)-x^2+x)*log(x))*log((-1+x)*log(x))+(-x*log(x)-x+1 )*exp(x-exp(x*exp(5)))+(-x^3*log(5)+x^2)*log(x)+(-x^3+x^2)*log(5)+x^2-x)/( x^2-x)/log(x)/log((-1+x)*log(x))^2,x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int \frac {-x+x^2+\left (x^2-x^3\right ) \log (5)+\left (x^2-x^3 \log (5)\right ) \log (x)+e^{-e^{e^5 x}+x} (1-x-x \log (x))+\left (e^{-e^{e^5 x}+x} \left (-x+x^2+e^{5+e^5 x} \left (x-x^2\right )\right ) \log (x)+\left (x-x^2+\left (-2 x^2+2 x^3\right ) \log (5)\right ) \log (x)\right ) \log ((-1+x) \log (x))}{\left (-x+x^2\right ) \log (x) \log ^2((-1+x) \log (x))} \, dx=\frac {x^{2} \log \left (5\right ) - x + e^{\left (x - e^{\left (x e^{5}\right )}\right )}}{\log \left (x \log \left (x\right ) - \log \left (x\right )\right )} \]
integrate(((((-x^2+x)*exp(5)*exp(x*exp(5))+x^2-x)*log(x)*exp(x-exp(x*exp(5 )))+((2*x^3-2*x^2)*log(5)-x^2+x)*log(x))*log((-1+x)*log(x))+(-x*log(x)-x+1 )*exp(x-exp(x*exp(5)))+(-x^3*log(5)+x^2)*log(x)+(-x^3+x^2)*log(5)+x^2-x)/( x^2-x)/log(x)/log((-1+x)*log(x))^2,x, algorithm=\
Time = 13.04 (sec) , antiderivative size = 138, normalized size of antiderivative = 4.18 \[ \int \frac {-x+x^2+\left (x^2-x^3\right ) \log (5)+\left (x^2-x^3 \log (5)\right ) \log (x)+e^{-e^{e^5 x}+x} (1-x-x \log (x))+\left (e^{-e^{e^5 x}+x} \left (-x+x^2+e^{5+e^5 x} \left (x-x^2\right )\right ) \log (x)+\left (x-x^2+\left (-2 x^2+2 x^3\right ) \log (5)\right ) \log (x)\right ) \log ((-1+x) \log (x))}{\left (-x+x^2\right ) \log (x) \log ^2((-1+x) \log (x))} \, dx=\frac {{\mathrm {e}}^{x-{\mathrm {e}}^{x\,{\mathrm {e}}^5}}}{\ln \left (\ln \left (x\right )\,\left (x-1\right )\right )}-x\,\left (\ln \left (25\right )+1\right )+2\,x^2\,\ln \left (5\right )+\frac {x\,\left (x\,\ln \left (5\right )-1\right )-\frac {x\,\ln \left (\ln \left (x\right )\,\left (x-1\right )\right )\,\ln \left (x\right )\,\left (2\,x\,\ln \left (5\right )-1\right )\,\left (x-1\right )}{x+x\,\ln \left (x\right )-1}}{\ln \left (\ln \left (x\right )\,\left (x-1\right )\right )}-\frac {x+2\,x\,\ln \left (5\right )-2\,x^2\,\ln \left (5\right )-2\,x^3\,\ln \left (5\right )+2\,x^4\,\ln \left (5\right )+x^2-x^3-1}{\left (x+1\right )\,\left (x+x\,\ln \left (x\right )-1\right )} \]
int((x + exp(x - exp(x*exp(5)))*(x + x*log(x) - 1) + log(log(x)*(x - 1))*( log(x)*(log(5)*(2*x^2 - 2*x^3) - x + x^2) - exp(x - exp(x*exp(5)))*log(x)* (x^2 - x + exp(5)*exp(x*exp(5))*(x - x^2))) - log(5)*(x^2 - x^3) + log(x)* (x^3*log(5) - x^2) - x^2)/(log(log(x)*(x - 1))^2*log(x)*(x - x^2)),x)
exp(x - exp(x*exp(5)))/log(log(x)*(x - 1)) - x*(log(25) + 1) + 2*x^2*log(5 ) + (x*(x*log(5) - 1) - (x*log(log(x)*(x - 1))*log(x)*(2*x*log(5) - 1)*(x - 1))/(x + x*log(x) - 1))/log(log(x)*(x - 1)) - (x + 2*x*log(5) - 2*x^2*lo g(5) - 2*x^3*log(5) + 2*x^4*log(5) + x^2 - x^3 - 1)/((x + 1)*(x + x*log(x) - 1))