Integrand size = 99, antiderivative size = 21 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=\left (-2 x+\log \left (-2+e^x\right )\right ) \log \left (-x+4 x^2\right ) \] Output:
(ln(exp(x)-2)-2*x)*ln(4*x^2-x)
Time = 0.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=-2 x \log (x (-1+4 x))+\log \left (-2+e^x\right ) \log (x (-1+4 x)) \] Input:
Integrate[(-4*x + 32*x^2 + E^x*(2*x - 16*x^2) + (2 - 16*x + E^x*(-1 + 8*x) )*Log[-2 + E^x] + (-4*x + 16*x^2 + E^x*(x - 4*x^2))*Log[-x + 4*x^2])/(2*x - 8*x^2 + E^x*(-x + 4*x^2)),x]
Output:
-2*x*Log[x*(-1 + 4*x)] + Log[-2 + E^x]*Log[x*(-1 + 4*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.
\(\displaystyle \int \frac {32 x^2+e^x \left (2 x-16 x^2\right )+\left (16 x^2+e^x \left (x-4 x^2\right )-4 x\right ) \log \left (4 x^2-x\right )-4 x+\left (-16 x+e^x (8 x-1)+2\right ) \log \left (e^x-2\right )}{-8 x^2+e^x \left (4 x^2-x\right )+2 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {32 x^2+e^x \left (2 x-16 x^2\right )+\left (16 x^2+e^x \left (x-4 x^2\right )-4 x\right ) \log \left (4 x^2-x\right )-4 x+\left (-16 x+e^x (8 x-1)+2\right ) \log \left (e^x-2\right )}{\left (2-e^x\right ) (1-4 x) x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-16 x^2-4 x^2 \log (x (4 x-1))+2 x+8 x \log \left (e^x-2\right )+x \log (x (4 x-1))-\log \left (e^x-2\right )}{x (4 x-1)}+\frac {2 \log (x (4 x-1))}{e^x-2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {\log \left (-2+e^x\right )}{x}dx+4 \int \frac {\log \left (-2+e^x\right )}{4 x-1}dx+2 \int \frac {\log (x (4 x-1))}{-2+e^x}dx-2 x+x (-\log (-((1-4 x) x)))-\frac {1}{4} \log (1-4 x)\) |
Input:
Int[(-4*x + 32*x^2 + E^x*(2*x - 16*x^2) + (2 - 16*x + E^x*(-1 + 8*x))*Log[ -2 + E^x] + (-4*x + 16*x^2 + E^x*(x - 4*x^2))*Log[-x + 4*x^2])/(2*x - 8*x^ 2 + E^x*(-x + 4*x^2)),x]
Output:
$Aborted
Time = 1.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.48
method | result | size |
parallelrisch | \(-2 \ln \left (4 x^{2}-x \right ) x +\ln \left ({\mathrm e}^{x}-2\right ) \ln \left (4 x^{2}-x \right )\) | \(31\) |
Input:
int((((8*x-1)*exp(x)-16*x+2)*ln(exp(x)-2)+((-4*x^2+x)*exp(x)+16*x^2-4*x)*l n(4*x^2-x)+(-16*x^2+2*x)*exp(x)+32*x^2-4*x)/((4*x^2-x)*exp(x)-8*x^2+2*x),x ,method=_RETURNVERBOSE)
Output:
-2*ln(4*x^2-x)*x+ln(exp(x)-2)*ln(4*x^2-x)
Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=-2 \, x \log \left (4 \, x^{2} - x\right ) + \log \left (4 \, x^{2} - x\right ) \log \left (e^{x} - 2\right ) \] Input:
integrate((((8*x-1)*exp(x)-16*x+2)*log(exp(x)-2)+((-4*x^2+x)*exp(x)+16*x^2 -4*x)*log(4*x^2-x)+(-16*x^2+2*x)*exp(x)+32*x^2-4*x)/((4*x^2-x)*exp(x)-8*x^ 2+2*x),x, algorithm="fricas")
Output:
-2*x*log(4*x^2 - x) + log(4*x^2 - x)*log(e^x - 2)
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=- 2 x \log {\left (4 x^{2} - x \right )} + \log {\left (4 x^{2} - x \right )} \log {\left (e^{x} - 2 \right )} \] Input:
integrate((((8*x-1)*exp(x)-16*x+2)*ln(exp(x)-2)+((-4*x**2+x)*exp(x)+16*x** 2-4*x)*ln(4*x**2-x)+(-16*x**2+2*x)*exp(x)+32*x**2-4*x)/((4*x**2-x)*exp(x)- 8*x**2+2*x),x)
Output:
-2*x*log(4*x**2 - x) + log(4*x**2 - x)*log(exp(x) - 2)
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=-2 \, x \log \left (4 \, x - 1\right ) - 2 \, x \log \left (x\right ) + {\left (\log \left (4 \, x - 1\right ) + \log \left (x\right )\right )} \log \left (e^{x} - 2\right ) \] Input:
integrate((((8*x-1)*exp(x)-16*x+2)*log(exp(x)-2)+((-4*x^2+x)*exp(x)+16*x^2 -4*x)*log(4*x^2-x)+(-16*x^2+2*x)*exp(x)+32*x^2-4*x)/((4*x^2-x)*exp(x)-8*x^ 2+2*x),x, algorithm="maxima")
Output:
-2*x*log(4*x - 1) - 2*x*log(x) + (log(4*x - 1) + log(x))*log(e^x - 2)
Time = 0.14 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=-2 \, x \log \left (4 \, x^{2} - x\right ) + \log \left (4 \, x^{2} - x\right ) \log \left (e^{x} - 2\right ) \] Input:
integrate((((8*x-1)*exp(x)-16*x+2)*log(exp(x)-2)+((-4*x^2+x)*exp(x)+16*x^2 -4*x)*log(4*x^2-x)+(-16*x^2+2*x)*exp(x)+32*x^2-4*x)/((4*x^2-x)*exp(x)-8*x^ 2+2*x),x, algorithm="giac")
Output:
-2*x*log(4*x^2 - x) + log(4*x^2 - x)*log(e^x - 2)
Time = 3.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=-\ln \left (4\,x^2-x\right )\,\left (2\,x-\ln \left ({\mathrm {e}}^x-2\right )\right ) \] Input:
int(-(exp(x)*(2*x - 16*x^2) - 4*x + log(exp(x) - 2)*(exp(x)*(8*x - 1) - 16 *x + 2) + 32*x^2 + log(4*x^2 - x)*(exp(x)*(x - 4*x^2) - 4*x + 16*x^2))/(ex p(x)*(x - 4*x^2) - 2*x + 8*x^2),x)
Output:
-log(4*x^2 - x)*(2*x - log(exp(x) - 2))
Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-4 x+32 x^2+e^x \left (2 x-16 x^2\right )+\left (2-16 x+e^x (-1+8 x)\right ) \log \left (-2+e^x\right )+\left (-4 x+16 x^2+e^x \left (x-4 x^2\right )\right ) \log \left (-x+4 x^2\right )}{2 x-8 x^2+e^x \left (-x+4 x^2\right )} \, dx=\mathrm {log}\left (4 x^{2}-x \right ) \left (\mathrm {log}\left (e^{x}-2\right )-2 x \right ) \] Input:
int((((8*x-1)*exp(x)-16*x+2)*log(exp(x)-2)+((-4*x^2+x)*exp(x)+16*x^2-4*x)* log(4*x^2-x)+(-16*x^2+2*x)*exp(x)+32*x^2-4*x)/((4*x^2-x)*exp(x)-8*x^2+2*x) ,x)
Output:
log(4*x**2 - x)*(log(e**x - 2) - 2*x)