Integrand size = 57, antiderivative size = 18 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=-\frac {3 e^{-1+x}}{x \log \left (-1+\frac {1}{x}\right )} \] Output:
-3*exp(x)/x/ln(-1+1/x)/exp(1)
Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=-\frac {3 e^{-1+x}}{x \log \left (-1+\frac {1}{x}\right )} \] Input:
Integrate[(3*E^x + E^x*(-3 + 6*x - 3*x^2)*Log[(1 - x)/x])/(E*(-x^2 + x^3)* Log[(1 - x)/x]^2),x]
Output:
(-3*E^(-1 + x))/(x*Log[-1 + x^(-1)])
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.
\(\displaystyle \int \frac {e^x \left (-3 x^2+6 x-3\right ) \log \left (\frac {1-x}{x}\right )+3 e^x}{e \left (x^3-x^2\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {3 \left (e^x-e^x \left (x^2-2 x+1\right ) \log \left (\frac {1-x}{x}\right )\right )}{\left (x^2-x^3\right ) \log ^2\left (\frac {1-x}{x}\right )}dx}{e}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3 \int \frac {e^x-e^x \left (x^2-2 x+1\right ) \log \left (\frac {1-x}{x}\right )}{\left (x^2-x^3\right ) \log ^2\left (\frac {1-x}{x}\right )}dx}{e}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle -\frac {3 \int \frac {e^x-e^x \left (x^2-2 x+1\right ) \log \left (\frac {1-x}{x}\right )}{(1-x) x^2 \log ^2\left (\frac {1-x}{x}\right )}dx}{e}\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {3 \int \frac {e^x-e^x \left (x^2-2 x+1\right ) \log \left (\frac {1-x}{x}\right )}{(1-x) x^2 \log ^2\left (\frac {1}{x}-1\right )}dx}{e}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {3 \int \left (\frac {e^x \left (\log \left (\frac {1}{x}-1\right ) x^2-2 \log \left (\frac {1}{x}-1\right ) x+\log \left (\frac {1}{x}-1\right )-1\right )}{(x-1) \log ^2\left (\frac {1}{x}-1\right )}-\frac {e^x \left (\log \left (\frac {1}{x}-1\right ) x^2-2 \log \left (\frac {1}{x}-1\right ) x+\log \left (\frac {1}{x}-1\right )-1\right )}{x \log ^2\left (\frac {1}{x}-1\right )}-\frac {e^x \left (\log \left (\frac {1}{x}-1\right ) x^2-2 \log \left (\frac {1}{x}-1\right ) x+\log \left (\frac {1}{x}-1\right )-1\right )}{x^2 \log ^2\left (\frac {1}{x}-1\right )}\right )dx}{e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 \left (\int \frac {e^x}{x^2 \log ^2\left (\frac {1}{x}-1\right )}dx-\int \frac {e^x}{x^2 \log \left (\frac {1}{x}-1\right )}dx-\int \frac {e^x}{(x-1) \log ^2\left (\frac {1}{x}-1\right )}dx+\int \frac {e^x}{x \log ^2\left (\frac {1}{x}-1\right )}dx+\int \frac {e^x}{x \log \left (\frac {1}{x}-1\right )}dx\right )}{e}\) |
Input:
Int[(3*E^x + E^x*(-3 + 6*x - 3*x^2)*Log[(1 - x)/x])/(E*(-x^2 + x^3)*Log[(1 - x)/x]^2),x]
Output:
$Aborted
Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28
method | result | size |
parallelrisch | \(-\frac {3 \,{\mathrm e}^{-1} {\mathrm e}^{x}}{x \ln \left (-\frac {-1+x}{x}\right )}\) | \(23\) |
risch | \(\frac {6 i {\mathrm e}^{-1+x}}{\left (\pi \,\operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2}-\pi \,\operatorname {csgn}\left (i \left (-1+x \right )\right ) \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )+\pi \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{3}-2 \pi \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2}+\pi \operatorname {csgn}\left (\frac {i \left (-1+x \right )}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 \pi +2 i \ln \left (x \right )-2 i \ln \left (-1+x \right )\right ) x}\) | \(127\) |
Input:
int(((-3*x^2+6*x-3)*exp(x)*ln((1-x)/x)+3*exp(x))/(x^3-x^2)/exp(1)/ln((1-x) /x)^2,x,method=_RETURNVERBOSE)
Output:
-3/exp(1)*exp(x)/x/ln(-(-1+x)/x)
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=-\frac {3 \, e^{\left (x - 1\right )}}{x \log \left (-\frac {x - 1}{x}\right )} \] Input:
integrate(((-3*x^2+6*x-3)*exp(x)*log((1-x)/x)+3*exp(x))/(x^3-x^2)/exp(1)/l og((1-x)/x)^2,x, algorithm="fricas")
Output:
-3*e^(x - 1)/(x*log(-(x - 1)/x))
Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=- \frac {3 e^{x}}{e x \log {\left (\frac {1 - x}{x} \right )}} \] Input:
integrate(((-3*x**2+6*x-3)*exp(x)*ln((1-x)/x)+3*exp(x))/(x**3-x**2)/exp(1) /ln((1-x)/x)**2,x)
Output:
-3*exp(-1)*exp(x)/(x*log((1 - x)/x))
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=\frac {3 \, e^{\left (x - 1\right )}}{x \log \left (x\right ) - x \log \left (-x + 1\right )} \] Input:
integrate(((-3*x^2+6*x-3)*exp(x)*log((1-x)/x)+3*exp(x))/(x^3-x^2)/exp(1)/l og((1-x)/x)^2,x, algorithm="maxima")
Output:
3*e^(x - 1)/(x*log(x) - x*log(-x + 1))
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=-\frac {3 \, e^{\left (x - 1\right )}}{x \log \left (-\frac {x - 1}{x}\right )} \] Input:
integrate(((-3*x^2+6*x-3)*exp(x)*log((1-x)/x)+3*exp(x))/(x^3-x^2)/exp(1)/l og((1-x)/x)^2,x, algorithm="giac")
Output:
-3*e^(x - 1)/(x*log(-(x - 1)/x))
Time = 7.70 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=-\frac {3\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^x}{x\,\ln \left (-\frac {x-1}{x}\right )} \] Input:
int(-(exp(-1)*(3*exp(x) - exp(x)*log(-(x - 1)/x)*(3*x^2 - 6*x + 3)))/(log( -(x - 1)/x)^2*(x^2 - x^3)),x)
Output:
-(3*exp(-1)*exp(x))/(x*log(-(x - 1)/x))
Time = 0.20 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28 \[ \int \frac {3 e^x+e^x \left (-3+6 x-3 x^2\right ) \log \left (\frac {1-x}{x}\right )}{e \left (-x^2+x^3\right ) \log ^2\left (\frac {1-x}{x}\right )} \, dx=-\frac {3 e^{x}}{\mathrm {log}\left (\frac {1-x}{x}\right ) e x} \] Input:
int(((-3*x^2+6*x-3)*exp(x)*log((1-x)/x)+3*exp(x))/(x^3-x^2)/exp(1)/log((1- x)/x)^2,x)
Output:
( - 3*e**x)/(log(( - x + 1)/x)*e*x)