Integrand size = 115, antiderivative size = 25 \[ \int \frac {e^{-x} \left (e^x (1-x) \log (-1+x)+e^x (-1+x) \log (-1+x) \log (x)+4^{4 e^{-x}} \left (e^x (-1+x)+e^x (1-x) \log (x)\right )+\left (e^x x \log (x)+4^{4 e^{-x}} \left (-4 x+4 x^2\right ) \log (4) \log (x)\right ) \log \left (\frac {2 x}{\log (x)}\right )\right )}{\left (-x+x^2\right ) \log (x)} \, dx=\left (-4^{4 e^{-x}}+\log (-1+x)\right ) \log \left (\frac {2 x}{\log (x)}\right ) \] Output:
(ln(-1+x)-exp(8*ln(2)/exp(x)))*ln(2*x/ln(x))
Time = 5.42 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x} \left (e^x (1-x) \log (-1+x)+e^x (-1+x) \log (-1+x) \log (x)+4^{4 e^{-x}} \left (e^x (-1+x)+e^x (1-x) \log (x)\right )+\left (e^x x \log (x)+4^{4 e^{-x}} \left (-4 x+4 x^2\right ) \log (4) \log (x)\right ) \log \left (\frac {2 x}{\log (x)}\right )\right )}{\left (-x+x^2\right ) \log (x)} \, dx=-\left (\left (256^{e^{-x}}-\log (-1+x)\right ) \log \left (\frac {2 x}{\log (x)}\right )\right ) \] Input:
Integrate[(E^x*(1 - x)*Log[-1 + x] + E^x*(-1 + x)*Log[-1 + x]*Log[x] + 4^( 4/E^x)*(E^x*(-1 + x) + E^x*(1 - x)*Log[x]) + (E^x*x*Log[x] + 4^(4/E^x)*(-4 *x + 4*x^2)*Log[4]*Log[x])*Log[(2*x)/Log[x]])/(E^x*(-x + x^2)*Log[x]),x]
Output:
-((256^E^(-x) - Log[-1 + x])*Log[(2*x)/Log[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.
| \(\displaystyle \int \frac {e^{-x} \left (\left (4^{4 e^{-x}} \left (4 x^2-4 x\right ) \log (4) \log (x)+e^x x \log (x)\right ) \log \left (\frac {2 x}{\log (x)}\right )+e^x (1-x) \log (x-1)+e^x (x-1) \log (x) \log (x-1)+4^{4 e^{-x}} \left (e^x (x-1)+e^x (1-x) \log (x)\right )\right )}{\left (x^2-x\right ) \log (x)} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {e^{-x} \left (\left (4^{4 e^{-x}} \left (4 x^2-4 x\right ) \log (4) \log (x)+e^x x \log (x)\right ) \log \left (\frac {2 x}{\log (x)}\right )+e^x (1-x) \log (x-1)+e^x (x-1) \log (x) \log (x-1)+4^{4 e^{-x}} \left (e^x (x-1)+e^x (1-x) \log (x)\right )\right )}{(x-1) x \log (x)}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (4^{e^{-x} \left (e^x+4\right )} e^{-x} \log (4) \log \left (\frac {2 x}{\log (x)}\right )-\frac {-256^{e^{-x}} x+256^{e^{-x}}+x \log (x-1)+256^{e^{-x}} x \log (x)-x \log (x-1) \log (x)-x \log (x) \log \left (\frac {2 x}{\log (x)}\right )-\log (x-1)-256^{e^{-x}} \log (x)+\log (x-1) \log (x)}{(x-1) x \log (x)}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\int \frac {\log (x-1)}{x \log (x)}dx+\int \frac {\log \left (\frac {2 x}{\log (x)}\right )}{x-1}dx+\operatorname {PolyLog}(2,1-x)+\log (x-1) \log (x)-4^{4 e^{-x}} \log \left (\frac {2 x}{\log (x)}\right )\) |
Input:
Int[(E^x*(1 - x)*Log[-1 + x] + E^x*(-1 + x)*Log[-1 + x]*Log[x] + 4^(4/E^x) *(E^x*(-1 + x) + E^x*(1 - x)*Log[x]) + (E^x*x*Log[x] + 4^(4/E^x)*(-4*x + 4 *x^2)*Log[4]*Log[x])*Log[(2*x)/Log[x]])/(E^x*(-x + x^2)*Log[x]),x]
Output:
$Aborted
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.70 (sec) , antiderivative size = 261, normalized size of antiderivative = 10.44
\[\left (-\ln \left (-1+x \right )+256^{{\mathrm e}^{-x}}\right ) \ln \left (\ln \left (x \right )\right )+\ln \left (x \right ) \ln \left (-1+x \right )+\frac {i \pi \ln \left (-1+x \right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )}{2}-\frac {i \pi \ln \left (-1+x \right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i x \right )}{2}-\frac {i 256^{{\mathrm e}^{-x}} \pi \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (i x \right )}{2}+\frac {i 256^{{\mathrm e}^{-x}} \pi \,\operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i x \right )}{2}-\frac {i \pi \ln \left (-1+x \right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right )^{3}}{2}-\frac {i 256^{{\mathrm e}^{-x}} \pi \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right )}{2}+\frac {i \pi \ln \left (-1+x \right ) \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (i x \right )}{2}+\frac {i 256^{{\mathrm e}^{-x}} \pi \operatorname {csgn}\left (\frac {i x}{\ln \left (x \right )}\right )^{3}}{2}+\ln \left (2\right ) \ln \left (-1+x \right )-\ln \left (2\right ) 256^{{\mathrm e}^{-x}}-256^{{\mathrm e}^{-x}} \ln \left (x \right )\]
Input:
int(((2*(4*x^2-4*x)*ln(2)*ln(x)*exp(8*ln(2)/exp(x))+x*exp(x)*ln(x))*ln(2*x /ln(x))+((1-x)*exp(x)*ln(x)+(-1+x)*exp(x))*exp(8*ln(2)/exp(x))+(-1+x)*exp( x)*ln(-1+x)*ln(x)+(1-x)*exp(x)*ln(-1+x))/(x^2-x)/exp(x)/ln(x),x)
Output:
(-ln(-1+x)+256^exp(-x))*ln(ln(x))+ln(x)*ln(-1+x)+1/2*I*Pi*ln(-1+x)*csgn(I* x/ln(x))^2*csgn(I/ln(x))-1/2*I*Pi*ln(-1+x)*csgn(I*x/ln(x))*csgn(I/ln(x))*c sgn(I*x)-1/2*I*256^exp(-x)*Pi*csgn(I*x/ln(x))^2*csgn(I*x)+1/2*I*256^exp(-x )*Pi*csgn(I*x/ln(x))*csgn(I/ln(x))*csgn(I*x)-1/2*I*Pi*ln(-1+x)*csgn(I*x/ln (x))^3-1/2*I*256^exp(-x)*Pi*csgn(I*x/ln(x))^2*csgn(I/ln(x))+1/2*I*Pi*ln(-1 +x)*csgn(I*x/ln(x))^2*csgn(I*x)+1/2*I*256^exp(-x)*Pi*csgn(I*x/ln(x))^3+ln( 2)*ln(-1+x)-ln(2)*256^exp(-x)-256^exp(-x)*ln(x)
Time = 0.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (e^x (1-x) \log (-1+x)+e^x (-1+x) \log (-1+x) \log (x)+4^{4 e^{-x}} \left (e^x (-1+x)+e^x (1-x) \log (x)\right )+\left (e^x x \log (x)+4^{4 e^{-x}} \left (-4 x+4 x^2\right ) \log (4) \log (x)\right ) \log \left (\frac {2 x}{\log (x)}\right )\right )}{\left (-x+x^2\right ) \log (x)} \, dx=-{\left (2^{8 \, e^{\left (-x\right )}} - \log \left (x - 1\right )\right )} \log \left (\frac {2 \, x}{\log \left (x\right )}\right ) \] Input:
integrate(((2*(4*x^2-4*x)*log(2)*log(x)*exp(8*log(2)/exp(x))+x*exp(x)*log( x))*log(2*x/log(x))+((1-x)*exp(x)*log(x)+(-1+x)*exp(x))*exp(8*log(2)/exp(x ))+(-1+x)*exp(x)*log(-1+x)*log(x)+(1-x)*exp(x)*log(-1+x))/(x^2-x)/exp(x)/l og(x),x, algorithm="fricas")
Output:
-(2^(8*e^(-x)) - log(x - 1))*log(2*x/log(x))
Timed out. \[ \int \frac {e^{-x} \left (e^x (1-x) \log (-1+x)+e^x (-1+x) \log (-1+x) \log (x)+4^{4 e^{-x}} \left (e^x (-1+x)+e^x (1-x) \log (x)\right )+\left (e^x x \log (x)+4^{4 e^{-x}} \left (-4 x+4 x^2\right ) \log (4) \log (x)\right ) \log \left (\frac {2 x}{\log (x)}\right )\right )}{\left (-x+x^2\right ) \log (x)} \, dx=\text {Timed out} \] Input:
integrate(((2*(4*x**2-4*x)*ln(2)*ln(x)*exp(8*ln(2)/exp(x))+x*exp(x)*ln(x)) *ln(2*x/ln(x))+((1-x)*exp(x)*ln(x)+(-1+x)*exp(x))*exp(8*ln(2)/exp(x))+(-1+ x)*exp(x)*ln(-1+x)*ln(x)+(1-x)*exp(x)*ln(-1+x))/(x**2-x)/exp(x)/ln(x),x)
Output:
Timed out
\[ \int \frac {e^{-x} \left (e^x (1-x) \log (-1+x)+e^x (-1+x) \log (-1+x) \log (x)+4^{4 e^{-x}} \left (e^x (-1+x)+e^x (1-x) \log (x)\right )+\left (e^x x \log (x)+4^{4 e^{-x}} \left (-4 x+4 x^2\right ) \log (4) \log (x)\right ) \log \left (\frac {2 x}{\log (x)}\right )\right )}{\left (-x+x^2\right ) \log (x)} \, dx=\int { \frac {{\left ({\left (x - 1\right )} e^{x} \log \left (x - 1\right ) \log \left (x\right ) - {\left (x - 1\right )} e^{x} \log \left (x - 1\right ) - {\left ({\left (x - 1\right )} e^{x} \log \left (x\right ) - {\left (x - 1\right )} e^{x}\right )} 2^{8 \, e^{\left (-x\right )}} + {\left (8 \, {\left (x^{2} - x\right )} 2^{8 \, e^{\left (-x\right )}} \log \left (2\right ) \log \left (x\right ) + x e^{x} \log \left (x\right )\right )} \log \left (\frac {2 \, x}{\log \left (x\right )}\right )\right )} e^{\left (-x\right )}}{{\left (x^{2} - x\right )} \log \left (x\right )} \,d x } \] Input:
integrate(((2*(4*x^2-4*x)*log(2)*log(x)*exp(8*log(2)/exp(x))+x*exp(x)*log( x))*log(2*x/log(x))+((1-x)*exp(x)*log(x)+(-1+x)*exp(x))*exp(8*log(2)/exp(x ))+(-1+x)*exp(x)*log(-1+x)*log(x)+(1-x)*exp(x)*log(-1+x))/(x^2-x)/exp(x)/l og(x),x, algorithm="maxima")
Output:
(log(2) + log(x) - log(log(x)))*log(x - 1) - integrate(-(8*x*log(2)^2*log( x) + 8*x*log(2)*log(x)^2 - 8*x*log(2)*log(x)*log(log(x)) - (log(x) - 1)*e^ x)*e^(8*e^(-x)*log(2) - x)/(x*log(x)), x)
\[ \int \frac {e^{-x} \left (e^x (1-x) \log (-1+x)+e^x (-1+x) \log (-1+x) \log (x)+4^{4 e^{-x}} \left (e^x (-1+x)+e^x (1-x) \log (x)\right )+\left (e^x x \log (x)+4^{4 e^{-x}} \left (-4 x+4 x^2\right ) \log (4) \log (x)\right ) \log \left (\frac {2 x}{\log (x)}\right )\right )}{\left (-x+x^2\right ) \log (x)} \, dx=\int { \frac {{\left ({\left (x - 1\right )} e^{x} \log \left (x - 1\right ) \log \left (x\right ) - {\left (x - 1\right )} e^{x} \log \left (x - 1\right ) - {\left ({\left (x - 1\right )} e^{x} \log \left (x\right ) - {\left (x - 1\right )} e^{x}\right )} 2^{8 \, e^{\left (-x\right )}} + {\left (8 \, {\left (x^{2} - x\right )} 2^{8 \, e^{\left (-x\right )}} \log \left (2\right ) \log \left (x\right ) + x e^{x} \log \left (x\right )\right )} \log \left (\frac {2 \, x}{\log \left (x\right )}\right )\right )} e^{\left (-x\right )}}{{\left (x^{2} - x\right )} \log \left (x\right )} \,d x } \] Input:
integrate(((2*(4*x^2-4*x)*log(2)*log(x)*exp(8*log(2)/exp(x))+x*exp(x)*log( x))*log(2*x/log(x))+((1-x)*exp(x)*log(x)+(-1+x)*exp(x))*exp(8*log(2)/exp(x ))+(-1+x)*exp(x)*log(-1+x)*log(x)+(1-x)*exp(x)*log(-1+x))/(x^2-x)/exp(x)/l og(x),x, algorithm="giac")
Output:
integrate(((x - 1)*e^x*log(x - 1)*log(x) - (x - 1)*e^x*log(x - 1) - ((x - 1)*e^x*log(x) - (x - 1)*e^x)*2^(8*e^(-x)) + (8*(x^2 - x)*2^(8*e^(-x))*log( 2)*log(x) + x*e^x*log(x))*log(2*x/log(x)))*e^(-x)/((x^2 - x)*log(x)), x)
Timed out. \[ \int \frac {e^{-x} \left (e^x (1-x) \log (-1+x)+e^x (-1+x) \log (-1+x) \log (x)+4^{4 e^{-x}} \left (e^x (-1+x)+e^x (1-x) \log (x)\right )+\left (e^x x \log (x)+4^{4 e^{-x}} \left (-4 x+4 x^2\right ) \log (4) \log (x)\right ) \log \left (\frac {2 x}{\log (x)}\right )\right )}{\left (-x+x^2\right ) \log (x)} \, dx=\int -\frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{8\,{\mathrm {e}}^{-x}\,\ln \left (2\right )}\,\left ({\mathrm {e}}^x\,\left (x-1\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (x-1\right )\right )+\ln \left (\frac {2\,x}{\ln \left (x\right )}\right )\,\left (x\,{\mathrm {e}}^x\,\ln \left (x\right )-2\,{\mathrm {e}}^{8\,{\mathrm {e}}^{-x}\,\ln \left (2\right )}\,\ln \left (2\right )\,\ln \left (x\right )\,\left (4\,x-4\,x^2\right )\right )-\ln \left (x-1\right )\,{\mathrm {e}}^x\,\left (x-1\right )+\ln \left (x-1\right )\,{\mathrm {e}}^x\,\ln \left (x\right )\,\left (x-1\right )\right )}{\ln \left (x\right )\,\left (x-x^2\right )} \,d x \] Input:
int(-(exp(-x)*(exp(8*exp(-x)*log(2))*(exp(x)*(x - 1) - exp(x)*log(x)*(x - 1)) + log((2*x)/log(x))*(x*exp(x)*log(x) - 2*exp(8*exp(-x)*log(2))*log(2)* log(x)*(4*x - 4*x^2)) - log(x - 1)*exp(x)*(x - 1) + log(x - 1)*exp(x)*log( x)*(x - 1)))/(log(x)*(x - x^2)),x)
Output:
int(-(exp(-x)*(exp(8*exp(-x)*log(2))*(exp(x)*(x - 1) - exp(x)*log(x)*(x - 1)) + log((2*x)/log(x))*(x*exp(x)*log(x) - 2*exp(8*exp(-x)*log(2))*log(2)* log(x)*(4*x - 4*x^2)) - log(x - 1)*exp(x)*(x - 1) + log(x - 1)*exp(x)*log( x)*(x - 1)))/(log(x)*(x - x^2)), x)
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-x} \left (e^x (1-x) \log (-1+x)+e^x (-1+x) \log (-1+x) \log (x)+4^{4 e^{-x}} \left (e^x (-1+x)+e^x (1-x) \log (x)\right )+\left (e^x x \log (x)+4^{4 e^{-x}} \left (-4 x+4 x^2\right ) \log (4) \log (x)\right ) \log \left (\frac {2 x}{\log (x)}\right )\right )}{\left (-x+x^2\right ) \log (x)} \, dx=\mathrm {log}\left (\frac {2 x}{\mathrm {log}\left (x \right )}\right ) \left (-e^{\frac {8 \,\mathrm {log}\left (2\right )}{e^{x}}}+\mathrm {log}\left (x -1\right )\right ) \] Input:
int(((2*(4*x^2-4*x)*log(2)*log(x)*exp(8*log(2)/exp(x))+x*exp(x)*log(x))*lo g(2*x/log(x))+((1-x)*exp(x)*log(x)+(-1+x)*exp(x))*exp(8*log(2)/exp(x))+(-1 +x)*exp(x)*log(-1+x)*log(x)+(1-x)*exp(x)*log(-1+x))/(x^2-x)/exp(x)/log(x), x)
Output:
log((2*x)/log(x))*( - e**((8*log(2))/e**x) + log(x - 1))