Integrand size = 93, antiderivative size = 24 \[ \int \frac {e^{\frac {1}{64} \left (16 e^x x+4 x^2+\left (4 e^x x+x^2\right ) \log (1-x)\right )} \left (-8 x+9 x^2+e^x \left (-16+4 x+16 x^2\right )+\left (-2 x+2 x^2+e^x \left (-4+4 x^2\right )\right ) \log (1-x)\right )}{-64+64 x} \, dx=e^{\frac {1}{16} \left (e^x+\frac {x}{4}\right ) x (4+\log (1-x))} \] Output:
exp(1/16*x*(4+ln(1-x))*(1/4*x+exp(x)))
Time = 1.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{\frac {1}{64} \left (16 e^x x+4 x^2+\left (4 e^x x+x^2\right ) \log (1-x)\right )} \left (-8 x+9 x^2+e^x \left (-16+4 x+16 x^2\right )+\left (-2 x+2 x^2+e^x \left (-4+4 x^2\right )\right ) \log (1-x)\right )}{-64+64 x} \, dx=e^{\frac {1}{16} x \left (4 e^x+x\right )} (1-x)^{\frac {1}{64} x \left (4 e^x+x\right )} \] Input:
Integrate[(E^((16*E^x*x + 4*x^2 + (4*E^x*x + x^2)*Log[1 - x])/64)*(-8*x + 9*x^2 + E^x*(-16 + 4*x + 16*x^2) + (-2*x + 2*x^2 + E^x*(-4 + 4*x^2))*Log[1 - x]))/(-64 + 64*x),x]
Output:
E^((x*(4*E^x + x))/16)*(1 - x)^((x*(4*E^x + x))/64)
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.
\(\displaystyle \int \frac {\left (9 x^2+e^x \left (16 x^2+4 x-16\right )+\left (2 x^2+e^x \left (4 x^2-4\right )-2 x\right ) \log (1-x)-8 x\right ) \exp \left (\frac {1}{64} \left (4 x^2+\left (x^2+4 e^x x\right ) \log (1-x)+16 e^x x\right )\right )}{64 x-64} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {1}{64} x \left (x+4 e^x\right ) (\log (1-x)+4)} \left (-9 x^2-e^x \left (16 x^2+4 x-16\right )-\left (2 x^2+e^x \left (4 x^2-4\right )-2 x\right ) \log (1-x)+8 x\right )}{64-64 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x+\frac {1}{64} \left (x+4 e^x\right ) x (\log (1-x)+4)} \left (4 x^2+x^2 \log (1-x)+x-\log (1-x)-4\right )}{16 (x-1)}+\frac {x e^{\frac {1}{64} x \left (x+4 e^x\right ) (\log (1-x)+4)} (9 x+2 x \log (1-x)-2 \log (1-x)-8)}{64 (x-1)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{64} \int e^{\frac {1}{64} x \left (x+4 e^x\right ) (\log (1-x)+4)}dx+\frac {5}{16} \int e^{\frac {1}{64} \left (x+4 e^x\right ) (\log (1-x)+4) x+x}dx+\frac {1}{64} \int \frac {e^{\frac {1}{64} x \left (x+4 e^x\right ) (\log (1-x)+4)}}{x-1}dx+\frac {1}{16} \int \frac {e^{\frac {1}{64} \left (x+4 e^x\right ) (\log (1-x)+4) x+x}}{x-1}dx+\frac {9}{64} \int e^{\frac {1}{64} x \left (x+4 e^x\right ) (\log (1-x)+4)} xdx+\frac {1}{4} \int e^{\frac {1}{64} \left (x+4 e^x\right ) (\log (1-x)+4) x+x} xdx+\frac {1}{16} \int e^{\frac {1}{64} \left (x+4 e^x\right ) (\log (1-x)+4) x+x} \log (1-x)dx+\frac {1}{32} \int e^{\frac {1}{64} x \left (x+4 e^x\right ) (\log (1-x)+4)} x \log (1-x)dx+\frac {1}{16} \int e^{\frac {1}{64} \left (x+4 e^x\right ) (\log (1-x)+4) x+x} x \log (1-x)dx\) |
Input:
Int[(E^((16*E^x*x + 4*x^2 + (4*E^x*x + x^2)*Log[1 - x])/64)*(-8*x + 9*x^2 + E^x*(-16 + 4*x + 16*x^2) + (-2*x + 2*x^2 + E^x*(-4 + 4*x^2))*Log[1 - x]) )/(-64 + 64*x),x]
Output:
$Aborted
Time = 2.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12
method | result | size |
risch | \(\left (1-x \right )^{\frac {\left (4 \,{\mathrm e}^{x}+x \right ) x}{64}} {\mathrm e}^{\frac {\left (4 \,{\mathrm e}^{x}+x \right ) x}{16}}\) | \(27\) |
parallelrisch | \({\mathrm e}^{\frac {\left (4 \,{\mathrm e}^{x} x +x^{2}\right ) \ln \left (1-x \right )}{64}+\frac {{\mathrm e}^{x} x}{4}+\frac {x^{2}}{16}}\) | \(30\) |
Input:
int((((4*x^2-4)*exp(x)+2*x^2-2*x)*ln(1-x)+(16*x^2+4*x-16)*exp(x)+9*x^2-8*x )*exp(1/64*(4*exp(x)*x+x^2)*ln(1-x)+1/4*exp(x)*x+1/16*x^2)/(64*x-64),x,met hod=_RETURNVERBOSE)
Output:
(1-x)^(1/64*(4*exp(x)+x)*x)*exp(1/16*(4*exp(x)+x)*x)
Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {1}{64} \left (16 e^x x+4 x^2+\left (4 e^x x+x^2\right ) \log (1-x)\right )} \left (-8 x+9 x^2+e^x \left (-16+4 x+16 x^2\right )+\left (-2 x+2 x^2+e^x \left (-4+4 x^2\right )\right ) \log (1-x)\right )}{-64+64 x} \, dx=e^{\left (\frac {1}{16} \, x^{2} + \frac {1}{4} \, x e^{x} + \frac {1}{64} \, {\left (x^{2} + 4 \, x e^{x}\right )} \log \left (-x + 1\right )\right )} \] Input:
integrate((((4*x^2-4)*exp(x)+2*x^2-2*x)*log(1-x)+(16*x^2+4*x-16)*exp(x)+9* x^2-8*x)*exp(1/64*(4*exp(x)*x+x^2)*log(1-x)+1/4*exp(x)*x+1/16*x^2)/(64*x-6 4),x, algorithm="fricas")
Output:
e^(1/16*x^2 + 1/4*x*e^x + 1/64*(x^2 + 4*x*e^x)*log(-x + 1))
Time = 0.35 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {1}{64} \left (16 e^x x+4 x^2+\left (4 e^x x+x^2\right ) \log (1-x)\right )} \left (-8 x+9 x^2+e^x \left (-16+4 x+16 x^2\right )+\left (-2 x+2 x^2+e^x \left (-4+4 x^2\right )\right ) \log (1-x)\right )}{-64+64 x} \, dx=e^{\frac {x^{2}}{16} + \frac {x e^{x}}{4} + \left (\frac {x^{2}}{64} + \frac {x e^{x}}{16}\right ) \log {\left (1 - x \right )}} \] Input:
integrate((((4*x**2-4)*exp(x)+2*x**2-2*x)*ln(1-x)+(16*x**2+4*x-16)*exp(x)+ 9*x**2-8*x)*exp(1/64*(4*exp(x)*x+x**2)*ln(1-x)+1/4*exp(x)*x+1/16*x**2)/(64 *x-64),x)
Output:
exp(x**2/16 + x*exp(x)/4 + (x**2/64 + x*exp(x)/16)*log(1 - x))
Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {1}{64} \left (16 e^x x+4 x^2+\left (4 e^x x+x^2\right ) \log (1-x)\right )} \left (-8 x+9 x^2+e^x \left (-16+4 x+16 x^2\right )+\left (-2 x+2 x^2+e^x \left (-4+4 x^2\right )\right ) \log (1-x)\right )}{-64+64 x} \, dx=e^{\left (\frac {1}{64} \, x^{2} \log \left (-x + 1\right ) + \frac {1}{16} \, x e^{x} \log \left (-x + 1\right ) + \frac {1}{16} \, x^{2} + \frac {1}{4} \, x e^{x}\right )} \] Input:
integrate((((4*x^2-4)*exp(x)+2*x^2-2*x)*log(1-x)+(16*x^2+4*x-16)*exp(x)+9* x^2-8*x)*exp(1/64*(4*exp(x)*x+x^2)*log(1-x)+1/4*exp(x)*x+1/16*x^2)/(64*x-6 4),x, algorithm="maxima")
Output:
e^(1/64*x^2*log(-x + 1) + 1/16*x*e^x*log(-x + 1) + 1/16*x^2 + 1/4*x*e^x)
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{\frac {1}{64} \left (16 e^x x+4 x^2+\left (4 e^x x+x^2\right ) \log (1-x)\right )} \left (-8 x+9 x^2+e^x \left (-16+4 x+16 x^2\right )+\left (-2 x+2 x^2+e^x \left (-4+4 x^2\right )\right ) \log (1-x)\right )}{-64+64 x} \, dx=e^{\left (\frac {1}{64} \, x^{2} \log \left (-x + 1\right ) + \frac {1}{16} \, x e^{x} \log \left (-x + 1\right ) + \frac {1}{16} \, x^{2} + \frac {1}{4} \, x e^{x}\right )} \] Input:
integrate((((4*x^2-4)*exp(x)+2*x^2-2*x)*log(1-x)+(16*x^2+4*x-16)*exp(x)+9* x^2-8*x)*exp(1/64*(4*exp(x)*x+x^2)*log(1-x)+1/4*exp(x)*x+1/16*x^2)/(64*x-6 4),x, algorithm="giac")
Output:
e^(1/64*x^2*log(-x + 1) + 1/16*x*e^x*log(-x + 1) + 1/16*x^2 + 1/4*x*e^x)
Time = 3.78 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {e^{\frac {1}{64} \left (16 e^x x+4 x^2+\left (4 e^x x+x^2\right ) \log (1-x)\right )} \left (-8 x+9 x^2+e^x \left (-16+4 x+16 x^2\right )+\left (-2 x+2 x^2+e^x \left (-4+4 x^2\right )\right ) \log (1-x)\right )}{-64+64 x} \, dx={\mathrm {e}}^{\frac {x\,{\mathrm {e}}^x}{4}+\frac {x^2}{16}}\,{\left (1-x\right )}^{\frac {x\,{\mathrm {e}}^x}{16}+\frac {x^2}{64}} \] Input:
int((exp((x*exp(x))/4 + x^2/16 + (log(1 - x)*(4*x*exp(x) + x^2))/64)*(log( 1 - x)*(exp(x)*(4*x^2 - 4) - 2*x + 2*x^2) - 8*x + exp(x)*(4*x + 16*x^2 - 1 6) + 9*x^2))/(64*x - 64),x)
Output:
exp((x*exp(x))/4 + x^2/16)*(1 - x)^((x*exp(x))/16 + x^2/64)
Time = 3.22 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {e^{\frac {1}{64} \left (16 e^x x+4 x^2+\left (4 e^x x+x^2\right ) \log (1-x)\right )} \left (-8 x+9 x^2+e^x \left (-16+4 x+16 x^2\right )+\left (-2 x+2 x^2+e^x \left (-4+4 x^2\right )\right ) \log (1-x)\right )}{-64+64 x} \, dx=e^{\frac {e^{x} \mathrm {log}\left (1-x \right ) x}{16}+\frac {e^{x} x}{4}+\frac {x^{2}}{16}} \left (1-x \right )^{\frac {x^{2}}{64}} \] Input:
int((((4*x^2-4)*exp(x)+2*x^2-2*x)*log(1-x)+(16*x^2+4*x-16)*exp(x)+9*x^2-8* x)*exp(1/64*(4*exp(x)*x+x^2)*log(1-x)+1/4*exp(x)*x+1/16*x^2)/(64*x-64),x)
Output:
e**((e**x*log( - x + 1)*x + 4*e**x*x + x**2)/16)*( - x + 1)**(x**2/64)