Integrand size = 88, antiderivative size = 25 \[ \int \frac {256 x-4 e^{4+4 x} x+\left (e^{4+4 x} (1-x)-256 x+256 x^2\right ) \log \left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )}{\left (e^{4+5 x}-256 e^x x\right ) \log ^2\left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )} \, dx=\frac {e^{-x} x}{\log \left (\frac {1}{256} e^{4+4 x}-x\right )} \] Output:
x/ln(1/256*exp(1+x)^4-x)/exp(x)
Time = 0.62 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {256 x-4 e^{4+4 x} x+\left (e^{4+4 x} (1-x)-256 x+256 x^2\right ) \log \left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )}{\left (e^{4+5 x}-256 e^x x\right ) \log ^2\left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )} \, dx=\frac {e^{-x} x}{\log \left (\frac {1}{256} e^{4+4 x}-x\right )} \] Input:
Integrate[(256*x - 4*E^(4 + 4*x)*x + (E^(4 + 4*x)*(1 - x) - 256*x + 256*x^ 2)*Log[(E^(4 + 4*x) - 256*x)/256])/((E^(4 + 5*x) - 256*E^x*x)*Log[(E^(4 + 4*x) - 256*x)/256]^2),x]
Output:
x/(E^x*Log[E^(4 + 4*x)/256 - 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.
\(\displaystyle \int \frac {\left (256 x^2-256 x+e^{4 x+4} (1-x)\right ) \log \left (\frac {1}{256} \left (e^{4 x+4}-256 x\right )\right )-4 e^{4 x+4} x+256 x}{\left (e^{5 x+4}-256 e^x x\right ) \log ^2\left (\frac {1}{256} \left (e^{4 x+4}-256 x\right )\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x} \left (\left (256 x^2-256 x+e^{4 x+4} (1-x)\right ) \log \left (\frac {1}{256} \left (e^{4 x+4}-256 x\right )\right )-4 e^{4 x+4} x+256 x\right )}{\left (e^{4 x+4}-256 x\right ) \log ^2\left (\frac {1}{256} \left (e^{4 x+4}-256 x\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {256 e^{-x} (1-4 x) x}{\left (e^{4 x+4}-256 x\right ) \log ^2\left (\frac {1}{256} \left (e^{4 x+4}-256 x\right )\right )}+\frac {e^{-x} \left (-4 x+x \left (-\log \left (\frac {1}{256} e^{4 x+4}-x\right )\right )+\log \left (\frac {1}{256} e^{4 x+4}-x\right )\right )}{\log ^2\left (\frac {1}{256} \left (e^{4 x+4}-256 x\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 1024 \int \frac {e^{-x} x^2}{\left (256 x-e^{4 x+4}\right ) \log ^2\left (\frac {1}{256} \left (e^{4 x+4}-256 x\right )\right )}dx-4 \int \frac {e^{-x} x}{\log ^2\left (\frac {1}{256} \left (e^{4 x+4}-256 x\right )\right )}dx+256 \int \frac {e^{-x} x}{\left (e^{4 x+4}-256 x\right ) \log ^2\left (\frac {1}{256} \left (e^{4 x+4}-256 x\right )\right )}dx+\int \frac {e^{-x}}{\log \left (\frac {1}{256} \left (e^{4 x+4}-256 x\right )\right )}dx-\int \frac {e^{-x} x}{\log \left (\frac {1}{256} \left (e^{4 x+4}-256 x\right )\right )}dx\) |
Input:
Int[(256*x - 4*E^(4 + 4*x)*x + (E^(4 + 4*x)*(1 - x) - 256*x + 256*x^2)*Log [(E^(4 + 4*x) - 256*x)/256])/((E^(4 + 5*x) - 256*E^x*x)*Log[(E^(4 + 4*x) - 256*x)/256]^2),x]
Output:
$Aborted
Time = 3.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
risch | \(\frac {x \,{\mathrm e}^{-x}}{\ln \left (\frac {{\mathrm e}^{4+4 x}}{256}-x \right )}\) | \(22\) |
parallelrisch | \(\frac {x \,{\mathrm e}^{-x}}{\ln \left (\frac {{\mathrm e}^{4+4 x}}{256}-x \right )}\) | \(22\) |
Input:
int((((1-x)*exp(1+x)^4+256*x^2-256*x)*ln(1/256*exp(1+x)^4-x)-4*x*exp(1+x)^ 4+256*x)/(exp(x)*exp(1+x)^4-256*exp(x)*x)/ln(1/256*exp(1+x)^4-x)^2,x,metho d=_RETURNVERBOSE)
Output:
x*exp(-x)/ln(1/256*exp(4+4*x)-x)
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {256 x-4 e^{4+4 x} x+\left (e^{4+4 x} (1-x)-256 x+256 x^2\right ) \log \left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )}{\left (e^{4+5 x}-256 e^x x\right ) \log ^2\left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )} \, dx=\frac {x e^{\left (-x\right )}}{\log \left (-x + \frac {1}{256} \, e^{\left (4 \, x + 4\right )}\right )} \] Input:
integrate((((1-x)*exp(1+x)^4+256*x^2-256*x)*log(1/256*exp(1+x)^4-x)-4*x*ex p(1+x)^4+256*x)/(exp(x)*exp(1+x)^4-256*exp(x)*x)/log(1/256*exp(1+x)^4-x)^2 ,x, algorithm="fricas")
Output:
x*e^(-x)/log(-x + 1/256*e^(4*x + 4))
Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {256 x-4 e^{4+4 x} x+\left (e^{4+4 x} (1-x)-256 x+256 x^2\right ) \log \left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )}{\left (e^{4+5 x}-256 e^x x\right ) \log ^2\left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )} \, dx=\frac {x e^{- x}}{\log {\left (- x + \frac {e^{4} e^{4 x}}{256} \right )}} \] Input:
integrate((((1-x)*exp(1+x)**4+256*x**2-256*x)*ln(1/256*exp(1+x)**4-x)-4*x* exp(1+x)**4+256*x)/(exp(x)*exp(1+x)**4-256*exp(x)*x)/ln(1/256*exp(1+x)**4- x)**2,x)
Output:
x*exp(-x)/log(-x + exp(4)*exp(4*x)/256)
Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {256 x-4 e^{4+4 x} x+\left (e^{4+4 x} (1-x)-256 x+256 x^2\right ) \log \left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )}{\left (e^{4+5 x}-256 e^x x\right ) \log ^2\left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )} \, dx=-\frac {x}{8 \, e^{x} \log \left (2\right ) - e^{x} \log \left (-256 \, x + e^{\left (4 \, x + 4\right )}\right )} \] Input:
integrate((((1-x)*exp(1+x)^4+256*x^2-256*x)*log(1/256*exp(1+x)^4-x)-4*x*ex p(1+x)^4+256*x)/(exp(x)*exp(1+x)^4-256*exp(x)*x)/log(1/256*exp(1+x)^4-x)^2 ,x, algorithm="maxima")
Output:
-x/(8*e^x*log(2) - e^x*log(-256*x + e^(4*x + 4)))
Time = 0.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {256 x-4 e^{4+4 x} x+\left (e^{4+4 x} (1-x)-256 x+256 x^2\right ) \log \left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )}{\left (e^{4+5 x}-256 e^x x\right ) \log ^2\left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )} \, dx=-\frac {x}{8 \, e^{x} \log \left (2\right ) - e^{x} \log \left (-256 \, x + e^{\left (4 \, x + 4\right )}\right )} \] Input:
integrate((((1-x)*exp(1+x)^4+256*x^2-256*x)*log(1/256*exp(1+x)^4-x)-4*x*ex p(1+x)^4+256*x)/(exp(x)*exp(1+x)^4-256*exp(x)*x)/log(1/256*exp(1+x)^4-x)^2 ,x, algorithm="giac")
Output:
-x/(8*e^x*log(2) - e^x*log(-256*x + e^(4*x + 4)))
Time = 0.83 (sec) , antiderivative size = 106, normalized size of antiderivative = 4.24 \[ \int \frac {256 x-4 e^{4+4 x} x+\left (e^{4+4 x} (1-x)-256 x+256 x^2\right ) \log \left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )}{\left (e^{4+5 x}-256 e^x x\right ) \log ^2\left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )} \, dx=\frac {x\,{\mathrm {e}}^{-x}-\frac {{\mathrm {e}}^{-x}\,\ln \left (\frac {{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^4}{256}-x\right )\,\left (256\,x-{\mathrm {e}}^{4\,x+4}\right )\,\left (x-1\right )}{4\,\left ({\mathrm {e}}^{4\,x+4}-64\right )}}{\ln \left (\frac {{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^4}{256}-x\right )}+{\mathrm {e}}^{-x}\,\left (x-x^2\right )+\frac {{\mathrm {e}}^{3\,x}\,\left (x^2-\frac {5\,x}{4}+\frac {1}{4}\right )}{{\mathrm {e}}^{4\,x}-64\,{\mathrm {e}}^{-4}} \] Input:
int((log(exp(4*x + 4)/256 - x)*(256*x + exp(4*x + 4)*(x - 1) - 256*x^2) - 256*x + 4*x*exp(4*x + 4))/(log(exp(4*x + 4)/256 - x)^2*(256*x*exp(x) - exp (4*x + 4)*exp(x))),x)
Output:
(x*exp(-x) - (exp(-x)*log((exp(4*x)*exp(4))/256 - x)*(256*x - exp(4*x + 4) )*(x - 1))/(4*(exp(4*x + 4) - 64)))/log((exp(4*x)*exp(4))/256 - x) + exp(- x)*(x - x^2) + (exp(3*x)*(x^2 - (5*x)/4 + 1/4))/(exp(4*x) - 64*exp(-4))
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {256 x-4 e^{4+4 x} x+\left (e^{4+4 x} (1-x)-256 x+256 x^2\right ) \log \left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )}{\left (e^{4+5 x}-256 e^x x\right ) \log ^2\left (\frac {1}{256} \left (e^{4+4 x}-256 x\right )\right )} \, dx=\frac {x}{e^{x} \mathrm {log}\left (\frac {e^{4 x} e^{4}}{256}-x \right )} \] Input:
int((((1-x)*exp(1+x)^4+256*x^2-256*x)*log(1/256*exp(1+x)^4-x)-4*x*exp(1+x) ^4+256*x)/(exp(x)*exp(1+x)^4-256*exp(x)*x)/log(1/256*exp(1+x)^4-x)^2,x)
Output:
x/(e**x*log((e**(4*x)*e**4 - 256*x)/256))