Integrand size = 87, antiderivative size = 27 \[ \int \frac {e^{-3+e^{4+x+e^5 (4+x)}} \left (2 x^2 \log (1-x)+\left (-2 x+2 x^2+e^{4+x+e^5 (4+x)} \left (-x^2+x^3+e^5 \left (-x^2+x^3\right )\right )\right ) \log ^2(1-x)\right )}{-1+x} \, dx=e^{-3+e^{\left (1+e^5\right ) (4+x)}} x^2 \log ^2(1-x) \] Output:
ln(1-x)^2*x^2*exp(exp((4+x)*(exp(5)+1))-3)
Time = 2.55 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3+e^{4+x+e^5 (4+x)}} \left (2 x^2 \log (1-x)+\left (-2 x+2 x^2+e^{4+x+e^5 (4+x)} \left (-x^2+x^3+e^5 \left (-x^2+x^3\right )\right )\right ) \log ^2(1-x)\right )}{-1+x} \, dx=e^{-3+e^{\left (1+e^5\right ) (4+x)}} x^2 \log ^2(1-x) \] Input:
Integrate[(E^(-3 + E^(4 + x + E^5*(4 + x)))*(2*x^2*Log[1 - x] + (-2*x + 2* x^2 + E^(4 + x + E^5*(4 + x))*(-x^2 + x^3 + E^5*(-x^2 + x^3)))*Log[1 - x]^ 2))/(-1 + x),x]
Output:
E^(-3 + E^((1 + E^5)*(4 + x)))*x^2*Log[1 - x]^2
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.
\(\displaystyle \int \frac {e^{e^{x+e^5 (x+4)+4}-3} \left (2 x^2 \log (1-x)+\left (2 x^2+e^{x+e^5 (x+4)+4} \left (x^3-x^2+e^5 \left (x^3-x^2\right )\right )-2 x\right ) \log ^2(1-x)\right )}{x-1} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left ((1+e) \left (1-e+e^2-e^3+e^4\right ) x^2 \exp \left (\left (1+e^5\right ) x+e^{x+e^5 (x+4)+4}+4 \left (1+e^5\right )-3\right ) \log ^2(1-x)+\frac {2 e^{e^{x+e^5 (x+4)+4}-3} x (x+x \log (1-x)-\log (1-x)) \log (1-x)}{x-1}\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \left ((1+e) \left (1-e+e^2-e^3+e^4\right ) x^2 \exp \left (\left (1+e^5\right ) x+e^{x+e^5 (x+4)+4}-3 \left (1-\frac {4}{3} \left (1+e^5\right )\right )\right ) \log ^2(1-x)+\frac {2 e^{e^{x+e^5 (x+4)+4}-3} x (x+x \log (1-x)-\log (1-x)) \log (1-x)}{x-1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \left (1+e^5\right ) \int \exp \left (\left (1+e^5\right ) x+e^{x+e^5 (x+4)+4}-3 \left (1-\frac {4}{3} \left (1+e^5\right )\right )\right ) \log ^2(1-x)dx-2 \left (1+e^5\right ) \int \exp \left (\left (1+e^5\right ) x+e^{x+e^5 (x+4)+4}-3 \left (1-\frac {4}{3} \left (1+e^5\right )\right )\right ) (1-x) \log ^2(1-x)dx+\left (1+e^5\right ) \int \exp \left (\left (1+e^5\right ) x+e^{x+e^5 (x+4)+4}-3 \left (1-\frac {4}{3} \left (1+e^5\right )\right )\right ) (1-x)^2 \log ^2(1-x)dx-\frac {2 \int \frac {\operatorname {ExpIntegralEi}\left (e^{\left (1+e^5\right ) x+4 \left (1+e^5\right )}\right )}{x-1}dx}{e^3 \left (1+e^5\right )}-2 \int \frac {\int \frac {e^{-3+e^{\left (1+e^5\right ) (x+4)}}}{x-1}dx}{x-1}dx-2 \int \frac {\int e^{-3+e^{\left (1+e^5\right ) (x+4)}} xdx}{x-1}dx+2 \int e^{-3+e^{x+e^5 (x+4)+4}} \log ^2(1-x)dx-2 \int e^{-3+e^{x+e^5 (x+4)+4}} (1-x) \log ^2(1-x)dx+2 \log (1-x) \int \frac {e^{-3+e^{x+e^5 (x+4)+4}}}{x-1}dx+2 \log (1-x) \int e^{-3+e^{x+e^5 (x+4)+4}} xdx+\frac {2 \operatorname {ExpIntegralEi}\left (e^{x+e^5 (x+4)+4}\right ) \log (1-x)}{e^3 \left (1+e^5\right )}\) |
Input:
Int[(E^(-3 + E^(4 + x + E^5*(4 + x)))*(2*x^2*Log[1 - x] + (-2*x + 2*x^2 + E^(4 + x + E^5*(4 + x))*(-x^2 + x^3 + E^5*(-x^2 + x^3)))*Log[1 - x]^2))/(- 1 + x),x]
Output:
$Aborted
Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
\[\ln \left (1-x \right )^{2} x^{2} {\mathrm e}^{{\mathrm e}^{\left (4+x \right ) \left ({\mathrm e}^{5}+1\right )}-3}\]
Input:
int(((((x^3-x^2)*exp(5)+x^3-x^2)*exp((4+x)*exp(5)+4+x)+2*x^2-2*x)*ln(1-x)^ 2+2*x^2*ln(1-x))*exp(exp((4+x)*exp(5)+4+x)-3)/(-1+x),x)
Output:
ln(1-x)^2*x^2*exp(exp((4+x)*(exp(5)+1))-3)
Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{-3+e^{4+x+e^5 (4+x)}} \left (2 x^2 \log (1-x)+\left (-2 x+2 x^2+e^{4+x+e^5 (4+x)} \left (-x^2+x^3+e^5 \left (-x^2+x^3\right )\right )\right ) \log ^2(1-x)\right )}{-1+x} \, dx=x^{2} e^{\left (e^{\left ({\left (x + 4\right )} e^{5} + x + 4\right )} - 3\right )} \log \left (-x + 1\right )^{2} \] Input:
integrate(((((x^3-x^2)*exp(5)+x^3-x^2)*exp((4+x)*exp(5)+4+x)+2*x^2-2*x)*lo g(1-x)^2+2*x^2*log(1-x))*exp(exp((4+x)*exp(5)+4+x)-3)/(-1+x),x, algorithm= "fricas")
Output:
x^2*e^(e^((x + 4)*e^5 + x + 4) - 3)*log(-x + 1)^2
Timed out. \[ \int \frac {e^{-3+e^{4+x+e^5 (4+x)}} \left (2 x^2 \log (1-x)+\left (-2 x+2 x^2+e^{4+x+e^5 (4+x)} \left (-x^2+x^3+e^5 \left (-x^2+x^3\right )\right )\right ) \log ^2(1-x)\right )}{-1+x} \, dx=\text {Timed out} \] Input:
integrate(((((x**3-x**2)*exp(5)+x**3-x**2)*exp((4+x)*exp(5)+4+x)+2*x**2-2* x)*ln(1-x)**2+2*x**2*ln(1-x))*exp(exp((4+x)*exp(5)+4+x)-3)/(-1+x),x)
Output:
Timed out
Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3+e^{4+x+e^5 (4+x)}} \left (2 x^2 \log (1-x)+\left (-2 x+2 x^2+e^{4+x+e^5 (4+x)} \left (-x^2+x^3+e^5 \left (-x^2+x^3\right )\right )\right ) \log ^2(1-x)\right )}{-1+x} \, dx=x^{2} e^{\left (e^{\left (x e^{5} + x + 4 \, e^{5} + 4\right )} - 3\right )} \log \left (-x + 1\right )^{2} \] Input:
integrate(((((x^3-x^2)*exp(5)+x^3-x^2)*exp((4+x)*exp(5)+4+x)+2*x^2-2*x)*lo g(1-x)^2+2*x^2*log(1-x))*exp(exp((4+x)*exp(5)+4+x)-3)/(-1+x),x, algorithm= "maxima")
Output:
x^2*e^(e^(x*e^5 + x + 4*e^5 + 4) - 3)*log(-x + 1)^2
\[ \int \frac {e^{-3+e^{4+x+e^5 (4+x)}} \left (2 x^2 \log (1-x)+\left (-2 x+2 x^2+e^{4+x+e^5 (4+x)} \left (-x^2+x^3+e^5 \left (-x^2+x^3\right )\right )\right ) \log ^2(1-x)\right )}{-1+x} \, dx=\int { \frac {{\left (2 \, x^{2} \log \left (-x + 1\right ) + {\left (2 \, x^{2} + {\left (x^{3} - x^{2} + {\left (x^{3} - x^{2}\right )} e^{5}\right )} e^{\left ({\left (x + 4\right )} e^{5} + x + 4\right )} - 2 \, x\right )} \log \left (-x + 1\right )^{2}\right )} e^{\left (e^{\left ({\left (x + 4\right )} e^{5} + x + 4\right )} - 3\right )}}{x - 1} \,d x } \] Input:
integrate(((((x^3-x^2)*exp(5)+x^3-x^2)*exp((4+x)*exp(5)+4+x)+2*x^2-2*x)*lo g(1-x)^2+2*x^2*log(1-x))*exp(exp((4+x)*exp(5)+4+x)-3)/(-1+x),x, algorithm= "giac")
Output:
integrate((2*x^2*log(-x + 1) + (2*x^2 + (x^3 - x^2 + (x^3 - x^2)*e^5)*e^(( x + 4)*e^5 + x + 4) - 2*x)*log(-x + 1)^2)*e^(e^((x + 4)*e^5 + x + 4) - 3)/ (x - 1), x)
Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-3+e^{4+x+e^5 (4+x)}} \left (2 x^2 \log (1-x)+\left (-2 x+2 x^2+e^{4+x+e^5 (4+x)} \left (-x^2+x^3+e^5 \left (-x^2+x^3\right )\right )\right ) \log ^2(1-x)\right )}{-1+x} \, dx=x^2\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{{\mathrm {e}}^{4\,{\mathrm {e}}^5}\,{\mathrm {e}}^4\,{\mathrm {e}}^{x\,{\mathrm {e}}^5}\,{\mathrm {e}}^x}\,{\ln \left (1-x\right )}^2 \] Input:
int(-(exp(exp(x + exp(5)*(x + 4) + 4) - 3)*(log(1 - x)^2*(2*x - 2*x^2 + ex p(x + exp(5)*(x + 4) + 4)*(exp(5)*(x^2 - x^3) + x^2 - x^3)) - 2*x^2*log(1 - x)))/(x - 1),x)
Output:
x^2*exp(-3)*exp(exp(4*exp(5))*exp(4)*exp(x*exp(5))*exp(x))*log(1 - x)^2
Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.30 \[ \int \frac {e^{-3+e^{4+x+e^5 (4+x)}} \left (2 x^2 \log (1-x)+\left (-2 x+2 x^2+e^{4+x+e^5 (4+x)} \left (-x^2+x^3+e^5 \left (-x^2+x^3\right )\right )\right ) \log ^2(1-x)\right )}{-1+x} \, dx=\frac {e^{e^{e^{5} x +4 e^{5}+x} e^{4}} \mathrm {log}\left (1-x \right )^{2} x^{2}}{e^{3}} \] Input:
int(((((x^3-x^2)*exp(5)+x^3-x^2)*exp((4+x)*exp(5)+4+x)+2*x^2-2*x)*log(1-x) ^2+2*x^2*log(1-x))*exp(exp((4+x)*exp(5)+4+x)-3)/(-1+x),x)
Output:
(e**(e**(e**5*x + 4*e**5 + x)*e**4)*log( - x + 1)**2*x**2)/e**3