Integrand size = 194, antiderivative size = 29 \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=e^{-2 e^{e^x}} x \left (e^{\frac {4}{-1+x}}-\log (-1+x)\right )^2 \] Output:
x*(exp(4/(-1+x))-ln(-1+x))^2/exp(exp(exp(x)))^2
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=e^{-2 e^{e^x}} x \left (e^{\frac {4}{-1+x}}-\log (-1+x)\right )^2 \] Input:
Integrate[(E^(4/(-1 + x))*(2*x - 2*x^2) + E^(8/(-1 + x))*(1 - 10*x + x^2) + (-2*x + 2*x^2 + E^(4/(-1 + x))*(-2 + 12*x - 2*x^2))*Log[-1 + x] + (1 - 2 *x + x^2)*Log[-1 + x]^2 + E^E^x*(E^(8/(-1 + x) + x)*(-2*x + 4*x^2 - 2*x^3) + E^(4/(-1 + x) + x)*(4*x - 8*x^2 + 4*x^3)*Log[-1 + x] + E^x*(-2*x + 4*x^ 2 - 2*x^3)*Log[-1 + x]^2))/(E^(2*E^E^x)*(1 - 2*x + x^2)),x]
Output:
(x*(E^(4/(-1 + x)) - Log[-1 + x])^2)/E^(2*E^E^x)
Leaf count is larger than twice the leaf count of optimal. \(98\) vs. \(2(29)=58\).
Time = 1.22 (sec) , antiderivative size = 98, normalized size of antiderivative = 3.38, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.005, Rules used = {2726}
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.
| \(\displaystyle \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{x-1}} \left (2 x-2 x^2\right )+e^{\frac {8}{x-1}} \left (x^2-10 x+1\right )+\left (x^2-2 x+1\right ) \log ^2(x-1)+\left (2 x^2+e^{\frac {4}{x-1}} \left (-2 x^2+12 x-2\right )-2 x\right ) \log (x-1)+e^{e^x} \left (e^{x+\frac {8}{x-1}} \left (-2 x^3+4 x^2-2 x\right )+e^x \left (-2 x^3+4 x^2-2 x\right ) \log ^2(x-1)+e^{x+\frac {4}{x-1}} \left (4 x^3-8 x^2+4 x\right ) \log (x-1)\right )\right )}{x^2-2 x+1} \, dx\) |
| \(\Big \downarrow \) 2726 |
| \(\displaystyle \frac {e^{-x-2 e^{e^x}} \left (e^{x-\frac {8}{1-x}} \left (x^3-2 x^2+x\right )+e^x \left (x^3-2 x^2+x\right ) \log ^2(x-1)-2 e^{x-\frac {4}{1-x}} \left (x^3-2 x^2+x\right ) \log (x-1)\right )}{x^2-2 x+1}\) |
Input:
Int[(E^(4/(-1 + x))*(2*x - 2*x^2) + E^(8/(-1 + x))*(1 - 10*x + x^2) + (-2* x + 2*x^2 + E^(4/(-1 + x))*(-2 + 12*x - 2*x^2))*Log[-1 + x] + (1 - 2*x + x ^2)*Log[-1 + x]^2 + E^E^x*(E^(8/(-1 + x) + x)*(-2*x + 4*x^2 - 2*x^3) + E^( 4/(-1 + x) + x)*(4*x - 8*x^2 + 4*x^3)*Log[-1 + x] + E^x*(-2*x + 4*x^2 - 2* x^3)*Log[-1 + x]^2))/(E^(2*E^E^x)*(1 - 2*x + x^2)),x]
Output:
(E^(-2*E^E^x - x)*(E^(-8/(1 - x) + x)*(x - 2*x^2 + x^3) - 2*E^(-4/(1 - x) + x)*(x - 2*x^2 + x^3)*Log[-1 + x] + E^x*(x - 2*x^2 + x^3)*Log[-1 + x]^2)) /(1 - 2*x + x^2)
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 164.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31
| method | result | size |
| risch | \(\left ({\mathrm e}^{\frac {8}{-1+x}}-2 \,{\mathrm e}^{\frac {4}{-1+x}} \ln \left (-1+x \right )+\ln \left (-1+x \right )^{2}\right ) x \,{\mathrm e}^{-2 \,{\mathrm e}^{{\mathrm e}^{x}}}\) | \(38\) |
| parallelrisch | \(\frac {\left (4 x \ln \left (-1+x \right )^{2}-8 x \,{\mathrm e}^{\frac {4}{-1+x}} \ln \left (-1+x \right )+4 \,{\mathrm e}^{\frac {8}{-1+x}} x \right ) {\mathrm e}^{-2 \,{\mathrm e}^{{\mathrm e}^{x}}}}{4}\) | \(47\) |
Input:
int((((-2*x^3+4*x^2-2*x)*exp(x)*ln(-1+x)^2+(4*x^3-8*x^2+4*x)*exp(4/(-1+x)) *exp(x)*ln(-1+x)+(-2*x^3+4*x^2-2*x)*exp(4/(-1+x))^2*exp(x))*exp(exp(x))+(x ^2-2*x+1)*ln(-1+x)^2+((-2*x^2+12*x-2)*exp(4/(-1+x))+2*x^2-2*x)*ln(-1+x)+(x ^2-10*x+1)*exp(4/(-1+x))^2+(-2*x^2+2*x)*exp(4/(-1+x)))/(x^2-2*x+1)/exp(exp (exp(x)))^2,x,method=_RETURNVERBOSE)
Output:
(exp(8/(-1+x))-2*exp(4/(-1+x))*ln(-1+x)+ln(-1+x)^2)*x*exp(-2*exp(exp(x)))
Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (25) = 50\).
Time = 0.11 (sec) , antiderivative size = 133, normalized size of antiderivative = 4.59 \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx={\left (x e^{\left (\frac {2 \, {\left (x^{2} - x + 4\right )}}{x - 1}\right )} \log \left (x - 1\right )^{2} - 2 \, x e^{\left (\frac {x^{2} - x + 8}{x - 1} + \frac {x^{2} - x + 4}{x - 1}\right )} \log \left (x - 1\right ) + x e^{\left (\frac {2 \, {\left (x^{2} - x + 8\right )}}{x - 1}\right )}\right )} e^{\left (-\frac {2 \, {\left (x^{2} - x + 4\right )}}{x - 1} - 2 \, e^{\left (e^{\left (-\frac {x^{2} - x + 8}{x - 1} + \frac {2 \, {\left (x^{2} - x + 4\right )}}{x - 1}\right )}\right )}\right )} \] Input:
integrate((((-2*x^3+4*x^2-2*x)*exp(x)*log(-1+x)^2+(4*x^3-8*x^2+4*x)*exp(4/ (-1+x))*exp(x)*log(-1+x)+(-2*x^3+4*x^2-2*x)*exp(4/(-1+x))^2*exp(x))*exp(ex p(x))+(x^2-2*x+1)*log(-1+x)^2+((-2*x^2+12*x-2)*exp(4/(-1+x))+2*x^2-2*x)*lo g(-1+x)+(x^2-10*x+1)*exp(4/(-1+x))^2+(-2*x^2+2*x)*exp(4/(-1+x)))/(x^2-2*x+ 1)/exp(exp(exp(x)))^2,x, algorithm="fricas")
Output:
(x*e^(2*(x^2 - x + 4)/(x - 1))*log(x - 1)^2 - 2*x*e^((x^2 - x + 8)/(x - 1) + (x^2 - x + 4)/(x - 1))*log(x - 1) + x*e^(2*(x^2 - x + 8)/(x - 1)))*e^(- 2*(x^2 - x + 4)/(x - 1) - 2*e^(e^(-(x^2 - x + 8)/(x - 1) + 2*(x^2 - x + 4) /(x - 1))))
Timed out. \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=\text {Timed out} \] Input:
integrate((((-2*x**3+4*x**2-2*x)*exp(x)*ln(-1+x)**2+(4*x**3-8*x**2+4*x)*ex p(4/(-1+x))*exp(x)*ln(-1+x)+(-2*x**3+4*x**2-2*x)*exp(4/(-1+x))**2*exp(x))* exp(exp(x))+(x**2-2*x+1)*ln(-1+x)**2+((-2*x**2+12*x-2)*exp(4/(-1+x))+2*x** 2-2*x)*ln(-1+x)+(x**2-10*x+1)*exp(4/(-1+x))**2+(-2*x**2+2*x)*exp(4/(-1+x)) )/(x**2-2*x+1)/exp(exp(exp(x)))**2,x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=-{\left (2 \, x e^{\left (\frac {4}{x - 1}\right )} \log \left (x - 1\right ) - x \log \left (x - 1\right )^{2} - x e^{\left (\frac {8}{x - 1}\right )}\right )} e^{\left (-2 \, e^{\left (e^{x}\right )}\right )} \] Input:
integrate((((-2*x^3+4*x^2-2*x)*exp(x)*log(-1+x)^2+(4*x^3-8*x^2+4*x)*exp(4/ (-1+x))*exp(x)*log(-1+x)+(-2*x^3+4*x^2-2*x)*exp(4/(-1+x))^2*exp(x))*exp(ex p(x))+(x^2-2*x+1)*log(-1+x)^2+((-2*x^2+12*x-2)*exp(4/(-1+x))+2*x^2-2*x)*lo g(-1+x)+(x^2-10*x+1)*exp(4/(-1+x))^2+(-2*x^2+2*x)*exp(4/(-1+x)))/(x^2-2*x+ 1)/exp(exp(exp(x)))^2,x, algorithm="maxima")
Output:
-(2*x*e^(4/(x - 1))*log(x - 1) - x*log(x - 1)^2 - x*e^(8/(x - 1)))*e^(-2*e ^(e^x))
\[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=\int { \frac {{\left ({\left (x^{2} - 2 \, x + 1\right )} \log \left (x - 1\right )^{2} + {\left (x^{2} - 10 \, x + 1\right )} e^{\left (\frac {8}{x - 1}\right )} - 2 \, {\left (x^{2} - x\right )} e^{\left (\frac {4}{x - 1}\right )} - 2 \, {\left ({\left (x^{3} - 2 \, x^{2} + x\right )} e^{x} \log \left (x - 1\right )^{2} - 2 \, {\left (x^{3} - 2 \, x^{2} + x\right )} e^{\left (x + \frac {4}{x - 1}\right )} \log \left (x - 1\right ) + {\left (x^{3} - 2 \, x^{2} + x\right )} e^{\left (x + \frac {8}{x - 1}\right )}\right )} e^{\left (e^{x}\right )} + 2 \, {\left (x^{2} - {\left (x^{2} - 6 \, x + 1\right )} e^{\left (\frac {4}{x - 1}\right )} - x\right )} \log \left (x - 1\right )\right )} e^{\left (-2 \, e^{\left (e^{x}\right )}\right )}}{x^{2} - 2 \, x + 1} \,d x } \] Input:
integrate((((-2*x^3+4*x^2-2*x)*exp(x)*log(-1+x)^2+(4*x^3-8*x^2+4*x)*exp(4/ (-1+x))*exp(x)*log(-1+x)+(-2*x^3+4*x^2-2*x)*exp(4/(-1+x))^2*exp(x))*exp(ex p(x))+(x^2-2*x+1)*log(-1+x)^2+((-2*x^2+12*x-2)*exp(4/(-1+x))+2*x^2-2*x)*lo g(-1+x)+(x^2-10*x+1)*exp(4/(-1+x))^2+(-2*x^2+2*x)*exp(4/(-1+x)))/(x^2-2*x+ 1)/exp(exp(exp(x)))^2,x, algorithm="giac")
Output:
integrate(((x^2 - 2*x + 1)*log(x - 1)^2 + (x^2 - 10*x + 1)*e^(8/(x - 1)) - 2*(x^2 - x)*e^(4/(x - 1)) - 2*((x^3 - 2*x^2 + x)*e^x*log(x - 1)^2 - 2*(x^ 3 - 2*x^2 + x)*e^(x + 4/(x - 1))*log(x - 1) + (x^3 - 2*x^2 + x)*e^(x + 8/( x - 1)))*e^(e^x) + 2*(x^2 - (x^2 - 6*x + 1)*e^(4/(x - 1)) - x)*log(x - 1)) *e^(-2*e^(e^x))/(x^2 - 2*x + 1), x)
Time = 3.57 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx={\mathrm {e}}^{-2\,{\mathrm {e}}^{{\mathrm {e}}^x}}\,\left (x\,{\ln \left (x-1\right )}^2-2\,x\,{\mathrm {e}}^{\frac {4}{x-1}}\,\ln \left (x-1\right )+x\,{\mathrm {e}}^{\frac {8}{x-1}}\right ) \] Input:
int((exp(-2*exp(exp(x)))*(log(x - 1)^2*(x^2 - 2*x + 1) + exp(8/(x - 1))*(x ^2 - 10*x + 1) - exp(exp(x))*(log(x - 1)^2*exp(x)*(2*x - 4*x^2 + 2*x^3) + exp(x)*exp(8/(x - 1))*(2*x - 4*x^2 + 2*x^3) - log(x - 1)*exp(x)*exp(4/(x - 1))*(4*x - 8*x^2 + 4*x^3)) + exp(4/(x - 1))*(2*x - 2*x^2) - log(x - 1)*(2 *x + exp(4/(x - 1))*(2*x^2 - 12*x + 2) - 2*x^2)))/(x^2 - 2*x + 1),x)
Output:
exp(-2*exp(exp(x)))*(x*log(x - 1)^2 + x*exp(8/(x - 1)) - 2*x*log(x - 1)*ex p(4/(x - 1)))
\[ \int \frac {e^{-2 e^{e^x}} \left (e^{\frac {4}{-1+x}} \left (2 x-2 x^2\right )+e^{\frac {8}{-1+x}} \left (1-10 x+x^2\right )+\left (-2 x+2 x^2+e^{\frac {4}{-1+x}} \left (-2+12 x-2 x^2\right )\right ) \log (-1+x)+\left (1-2 x+x^2\right ) \log ^2(-1+x)+e^{e^x} \left (e^{\frac {8}{-1+x}+x} \left (-2 x+4 x^2-2 x^3\right )+e^{\frac {4}{-1+x}+x} \left (4 x-8 x^2+4 x^3\right ) \log (-1+x)+e^x \left (-2 x+4 x^2-2 x^3\right ) \log ^2(-1+x)\right )\right )}{1-2 x+x^2} \, dx=\int \frac {\left (\left (-2 x^{3}+4 x^{2}-2 x \right ) {\mathrm e}^{x} \mathrm {log}\left (x -1\right )^{2}+\left (4 x^{3}-8 x^{2}+4 x \right ) {\mathrm e}^{\frac {4}{x -1}} {\mathrm e}^{x} \mathrm {log}\left (x -1\right )+\left (-2 x^{3}+4 x^{2}-2 x \right ) \left ({\mathrm e}^{\frac {4}{x -1}}\right )^{2} {\mathrm e}^{x}\right ) {\mathrm e}^{{\mathrm e}^{x}}+\left (x^{2}-2 x +1\right ) \mathrm {log}\left (x -1\right )^{2}+\left (\left (-2 x^{2}+12 x -2\right ) {\mathrm e}^{\frac {4}{x -1}}+2 x^{2}-2 x \right ) \mathrm {log}\left (x -1\right )+\left (x^{2}-10 x +1\right ) \left ({\mathrm e}^{\frac {4}{x -1}}\right )^{2}+\left (-2 x^{2}+2 x \right ) {\mathrm e}^{\frac {4}{x -1}}}{\left (x^{2}-2 x +1\right ) \left ({\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}}\right )^{2}}d x \] Input:
int((((-2*x^3+4*x^2-2*x)*exp(x)*log(-1+x)^2+(4*x^3-8*x^2+4*x)*exp(4/(-1+x) )*exp(x)*log(-1+x)+(-2*x^3+4*x^2-2*x)*exp(4/(-1+x))^2*exp(x))*exp(exp(x))+ (x^2-2*x+1)*log(-1+x)^2+((-2*x^2+12*x-2)*exp(4/(-1+x))+2*x^2-2*x)*log(-1+x )+(x^2-10*x+1)*exp(4/(-1+x))^2+(-2*x^2+2*x)*exp(4/(-1+x)))/(x^2-2*x+1)/exp (exp(exp(x)))^2,x)
Output:
int((((-2*x^3+4*x^2-2*x)*exp(x)*log(-1+x)^2+(4*x^3-8*x^2+4*x)*exp(4/(-1+x) )*exp(x)*log(-1+x)+(-2*x^3+4*x^2-2*x)*exp(4/(-1+x))^2*exp(x))*exp(exp(x))+ (x^2-2*x+1)*log(-1+x)^2+((-2*x^2+12*x-2)*exp(4/(-1+x))+2*x^2-2*x)*log(-1+x )+(x^2-10*x+1)*exp(4/(-1+x))^2+(-2*x^2+2*x)*exp(4/(-1+x)))/(x^2-2*x+1)/exp (exp(exp(x)))^2,x)