\(\int \frac {e^{-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x} (-3 x^2+x^3+2 x^4+e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)} (4 x^5+4 x^4 \log (1-x)))}{-1+x} \, dx\) [1695]

Optimal result

Integrand size = 107, antiderivative size = 31 \[ \int \frac {e^{-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x} \left (-3 x^2+x^3+2 x^4+e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)} \left (4 x^5+4 x^4 \log (1-x)\right )\right )}{-1+x} \, dx=e^{-4-2 e^{e^9-(x+\log (1-x))^2}+2 x} x^3 \] Output:

x^3/exp(exp(exp(9)-(ln(1-x)+x)^2)+2-x)^2
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {e^{-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x} \left (-3 x^2+x^3+2 x^4+e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)} \left (4 x^5+4 x^4 \log (1-x)\right )\right )}{-1+x} \, dx=e^{-4-2 e^{e^9-x^2-\log ^2(1-x)} (1-x)^{-2 x}+2 x} x^3 \] Input:

Integrate[(E^(-4 - 2*E^(E^9 - x^2 - 2*x*Log[1 - x] - Log[1 - x]^2) + 2*x)* 
(-3*x^2 + x^3 + 2*x^4 + E^(E^9 - x^2 - 2*x*Log[1 - x] - Log[1 - x]^2)*(4*x 
^5 + 4*x^4*Log[1 - x])))/(-1 + x),x]
 

Output:

E^(-4 - (2*E^(E^9 - x^2 - Log[1 - x]^2))/(1 - x)^(2*x) + 2*x)*x^3
 

Rubi [F]

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.

\(\displaystyle \int \frac {\left (2 x^4+x^3-3 x^2+e^{-x^2-\log ^2(1-x)-2 x \log (1-x)+e^9} \left (4 x^5+4 x^4 \log (1-x)\right )\right ) \exp \left (-2 e^{-x^2-\log ^2(1-x)-2 x \log (1-x)+e^9}+2 x-4\right )}{x-1} \, dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (x^2 (2 x+3) \exp \left (-2 e^{-x^2-\log ^2(1-x)-2 x \log (1-x)+e^9}+2 x-4\right )-4 (1-x)^{-2 x-1} x^4 (x+\log (1-x)) \exp \left (-x^2-2 e^{-x^2-\log ^2(1-x)-2 x \log (1-x)+e^9}+2 x-\log ^2(1-x)+e^9-4\right )\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 \int \exp \left (2 x-2 e^{-x^2-2 \log (1-x) x-\log ^2(1-x)+e^9}-4\right ) x^2dx-4 \int \exp \left (-x^2+2 x-2 e^{-x^2-2 \log (1-x) x-\log ^2(1-x)+e^9}-\log ^2(1-x)-4 \left (1-\frac {e^9}{4}\right )\right ) (1-x)^{-2 x-1} x^5dx-4 \int \exp \left (-x^2+2 x-2 e^{-x^2-2 \log (1-x) x-\log ^2(1-x)+e^9}-\log ^2(1-x)-4 \left (1-\frac {e^9}{4}\right )\right ) (1-x)^{-2 x-1} x^4 \log (1-x)dx+2 \int \exp \left (2 x-2 e^{-x^2-2 \log (1-x) x-\log ^2(1-x)+e^9}-4\right ) x^3dx\)

Input:

Int[(E^(-4 - 2*E^(E^9 - x^2 - 2*x*Log[1 - x] - Log[1 - x]^2) + 2*x)*(-3*x^ 
2 + x^3 + 2*x^4 + E^(E^9 - x^2 - 2*x*Log[1 - x] - Log[1 - x]^2)*(4*x^5 + 4 
*x^4*Log[1 - x])))/(-1 + x),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32

method	result	size
risch	\(x^{3} {\mathrm e}^{-2 \left (1-x \right )^{-2 x} {\mathrm e}^{-\ln \left (1-x \right )^{2}+{\mathrm e}^{9}-x^{2}}-4+2 x}\)	\(41\)
parallelrisch	\(x^{3} {\mathrm e}^{-2 \,{\mathrm e}^{-\ln \left (1-x \right )^{2}-2 x \ln \left (1-x \right )+{\mathrm e}^{9}-x^{2}}-4+2 x}\)	\(41\)

Input:

int(((4*x^4*ln(1-x)+4*x^5)*exp(-ln(1-x)^2-2*x*ln(1-x)+exp(9)-x^2)+2*x^4+x^ 
3-3*x^2)/(-1+x)/exp(exp(-ln(1-x)^2-2*x*ln(1-x)+exp(9)-x^2)+2-x)^2,x,method 
=_RETURNVERBOSE)
 

Output:

x^3*exp(-2*(1-x)^(-2*x)*exp(-ln(1-x)^2+exp(9)-x^2)-4+2*x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.29 \[ \int \frac {e^{-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x} \left (-3 x^2+x^3+2 x^4+e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)} \left (4 x^5+4 x^4 \log (1-x)\right )\right )}{-1+x} \, dx=x^{3} e^{\left (2 \, x - 2 \, e^{\left (-x^{2} - 2 \, x \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} + e^{9}\right )} - 4\right )} \] Input:

integrate(((4*x^4*log(1-x)+4*x^5)*exp(-log(1-x)^2-2*x*log(1-x)+exp(9)-x^2) 
+2*x^4+x^3-3*x^2)/(-1+x)/exp(exp(-log(1-x)^2-2*x*log(1-x)+exp(9)-x^2)+2-x) 
^2,x, algorithm="fricas")
 

Output:

x^3*e^(2*x - 2*e^(-x^2 - 2*x*log(-x + 1) - log(-x + 1)^2 + e^9) - 4)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x} \left (-3 x^2+x^3+2 x^4+e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)} \left (4 x^5+4 x^4 \log (1-x)\right )\right )}{-1+x} \, dx=\text {Timed out} \] Input:

integrate(((4*x**4*ln(1-x)+4*x**5)*exp(-ln(1-x)**2-2*x*ln(1-x)+exp(9)-x**2 
)+2*x**4+x**3-3*x**2)/(-1+x)/exp(exp(-ln(1-x)**2-2*x*ln(1-x)+exp(9)-x**2)+ 
2-x)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {e^{-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x} \left (-3 x^2+x^3+2 x^4+e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)} \left (4 x^5+4 x^4 \log (1-x)\right )\right )}{-1+x} \, dx=\int { \frac {{\left (2 \, x^{4} + x^{3} - 3 \, x^{2} + 4 \, {\left (x^{5} + x^{4} \log \left (-x + 1\right )\right )} e^{\left (-x^{2} - 2 \, x \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} + e^{9}\right )}\right )} e^{\left (2 \, x - 2 \, e^{\left (-x^{2} - 2 \, x \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} + e^{9}\right )} - 4\right )}}{x - 1} \,d x } \] Input:

integrate(((4*x^4*log(1-x)+4*x^5)*exp(-log(1-x)^2-2*x*log(1-x)+exp(9)-x^2) 
+2*x^4+x^3-3*x^2)/(-1+x)/exp(exp(-log(1-x)^2-2*x*log(1-x)+exp(9)-x^2)+2-x) 
^2,x, algorithm="maxima")
 

Output:

integrate((2*x^4 + x^3 - 3*x^2 + 4*(x^5 + x^4*log(-x + 1))*e^(-x^2 - 2*x*l 
og(-x + 1) - log(-x + 1)^2 + e^9))*e^(2*x - 2*e^(-x^2 - 2*x*log(-x + 1) - 
log(-x + 1)^2 + e^9) - 4)/(x - 1), x)
 

Giac [F]

\[ \int \frac {e^{-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x} \left (-3 x^2+x^3+2 x^4+e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)} \left (4 x^5+4 x^4 \log (1-x)\right )\right )}{-1+x} \, dx=\int { \frac {{\left (2 \, x^{4} + x^{3} - 3 \, x^{2} + 4 \, {\left (x^{5} + x^{4} \log \left (-x + 1\right )\right )} e^{\left (-x^{2} - 2 \, x \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} + e^{9}\right )}\right )} e^{\left (2 \, x - 2 \, e^{\left (-x^{2} - 2 \, x \log \left (-x + 1\right ) - \log \left (-x + 1\right )^{2} + e^{9}\right )} - 4\right )}}{x - 1} \,d x } \] Input:

integrate(((4*x^4*log(1-x)+4*x^5)*exp(-log(1-x)^2-2*x*log(1-x)+exp(9)-x^2) 
+2*x^4+x^3-3*x^2)/(-1+x)/exp(exp(-log(1-x)^2-2*x*log(1-x)+exp(9)-x^2)+2-x) 
^2,x, algorithm="giac")
 

Output:

integrate((2*x^4 + x^3 - 3*x^2 + 4*(x^5 + x^4*log(-x + 1))*e^(-x^2 - 2*x*l 
og(-x + 1) - log(-x + 1)^2 + e^9))*e^(2*x - 2*e^(-x^2 - 2*x*log(-x + 1) - 
log(-x + 1)^2 + e^9) - 4)/(x - 1), x)
 

Mupad [B] (verification not implemented)

Time = 1.53 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x} \left (-3 x^2+x^3+2 x^4+e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)} \left (4 x^5+4 x^4 \log (1-x)\right )\right )}{-1+x} \, dx=x^3\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-\frac {2\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{-{\ln \left (1-x\right )}^2}\,{\mathrm {e}}^{{\mathrm {e}}^9}}{{\left (1-x\right )}^{2\,x}}}\,{\mathrm {e}}^{-4} \] Input:

int((exp(2*x - 2*exp(exp(9) - log(1 - x)^2 - 2*x*log(1 - x) - x^2) - 4)*(e 
xp(exp(9) - log(1 - x)^2 - 2*x*log(1 - x) - x^2)*(4*x^5 + 4*x^4*log(1 - x) 
) - 3*x^2 + x^3 + 2*x^4))/(x - 1),x)
 

Output:

x^3*exp(2*x)*exp(-(2*exp(-x^2)*exp(-log(1 - x)^2)*exp(exp(9)))/(1 - x)^(2* 
x))*exp(-4)
 

Reduce [B] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61 \[ \int \frac {e^{-4-2 e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)}+2 x} \left (-3 x^2+x^3+2 x^4+e^{e^9-x^2-2 x \log (1-x)-\log ^2(1-x)} \left (4 x^5+4 x^4 \log (1-x)\right )\right )}{-1+x} \, dx=\frac {e^{2 x} x^{3}}{e^{\frac {2 e^{e^{9}}}{e^{\mathrm {log}\left (1-x \right )^{2}+x^{2}} \left (1-x \right )^{2 x}}} e^{4}} \] Input:

int(((4*x^4*log(1-x)+4*x^5)*exp(-log(1-x)^2-2*x*log(1-x)+exp(9)-x^2)+2*x^4 
+x^3-3*x^2)/(-1+x)/exp(exp(-log(1-x)^2-2*x*log(1-x)+exp(9)-x^2)+2-x)^2,x)
 

Output:

(e**(2*x)*x**3)/(e**((2*e**(e**9))/(e**(log( - x + 1)**2 + x**2)*( - x + 1 
)**(2*x)))*e**4)