Integrand size = 184, antiderivative size = 24 \[ \int \frac {e^{-\frac {8 x^2}{e^{2 x}-2 e^x x+x^2}} \left (e^{3 x} (1-x)-x^3+x^4+e^{2 x} \left (-3 x+3 x^2\right )+e^x \left (3 x^2-3 x^3\right )+\left (e^x \left (16 x^3-16 x^4\right )+e^x \left (-16 x^2+16 x^3\right ) \log (x)\right ) \log (-x+\log (x))\right )}{-e^{3 x} x^2+3 e^{2 x} x^3-3 e^x x^4+x^5+\left (e^{3 x} x-3 e^{2 x} x^2+3 e^x x^3-x^4\right ) \log (x)} \, dx=e^{-\frac {8 x^2}{\left (e^x-x\right )^2}} \log (-x+\log (x)) \] Output:
ln(ln(x)-x)/exp(8*x^2/(exp(x)-x)^2)
Time = 0.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {8 x^2}{e^{2 x}-2 e^x x+x^2}} \left (e^{3 x} (1-x)-x^3+x^4+e^{2 x} \left (-3 x+3 x^2\right )+e^x \left (3 x^2-3 x^3\right )+\left (e^x \left (16 x^3-16 x^4\right )+e^x \left (-16 x^2+16 x^3\right ) \log (x)\right ) \log (-x+\log (x))\right )}{-e^{3 x} x^2+3 e^{2 x} x^3-3 e^x x^4+x^5+\left (e^{3 x} x-3 e^{2 x} x^2+3 e^x x^3-x^4\right ) \log (x)} \, dx=e^{-\frac {8 x^2}{\left (e^x-x\right )^2}} \log (-x+\log (x)) \] Input:
Integrate[(E^(3*x)*(1 - x) - x^3 + x^4 + E^(2*x)*(-3*x + 3*x^2) + E^x*(3*x ^2 - 3*x^3) + (E^x*(16*x^3 - 16*x^4) + E^x*(-16*x^2 + 16*x^3)*Log[x])*Log[ -x + Log[x]])/(E^((8*x^2)/(E^(2*x) - 2*E^x*x + x^2))*(-(E^(3*x)*x^2) + 3*E ^(2*x)*x^3 - 3*E^x*x^4 + x^5 + (E^(3*x)*x - 3*E^(2*x)*x^2 + 3*E^x*x^3 - x^ 4)*Log[x])),x]
Output:
Log[-x + Log[x]]/E^((8*x^2)/(E^x - x)^2)
Leaf count is larger than twice the leaf count of optimal. \(194\) vs. \(2(24)=48\).
Time = 0.74 (sec) , antiderivative size = 194, normalized size of antiderivative = 8.08, 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.
\(\displaystyle \int \frac {e^{-\frac {8 x^2}{x^2-2 e^x x+e^{2 x}}} \left (x^4-x^3+e^{2 x} \left (3 x^2-3 x\right )+e^x \left (3 x^2-3 x^3\right )+\left (e^x \left (16 x^3-16 x^4\right )+e^x \left (16 x^3-16 x^2\right ) \log (x)\right ) \log (\log (x)-x)+e^{3 x} (1-x)\right )}{x^5-3 e^x x^4+3 e^{2 x} x^3-e^{3 x} x^2+\left (-x^4+3 e^x x^3-3 e^{2 x} x^2+e^{3 x} x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {e^{-\frac {8 x^2}{x^2-2 e^x x+e^{2 x}}} \left (e^x \left (x^3-x^4\right )-e^x \left (x^2-x^3\right ) \log (x)\right ) \log (\log (x)-x)}{\left (\frac {\left (e^x x-x+e^x-e^{2 x}\right ) x^2}{\left (x^2-2 e^x x+e^{2 x}\right )^2}+\frac {x}{x^2-2 e^x x+e^{2 x}}\right ) \left (-x^5+3 e^x x^4-3 e^{2 x} x^3+e^{3 x} x^2-\left (-x^4+3 e^x x^3-3 e^{2 x} x^2+e^{3 x} x\right ) \log (x)\right )}\) |
Input:
Int[(E^(3*x)*(1 - x) - x^3 + x^4 + E^(2*x)*(-3*x + 3*x^2) + E^x*(3*x^2 - 3 *x^3) + (E^x*(16*x^3 - 16*x^4) + E^x*(-16*x^2 + 16*x^3)*Log[x])*Log[-x + L og[x]])/(E^((8*x^2)/(E^(2*x) - 2*E^x*x + x^2))*(-(E^(3*x)*x^2) + 3*E^(2*x) *x^3 - 3*E^x*x^4 + x^5 + (E^(3*x)*x - 3*E^(2*x)*x^2 + 3*E^x*x^3 - x^4)*Log [x])),x]
Output:
((E^x*(x^3 - x^4) - E^x*(x^2 - x^3)*Log[x])*Log[-x + Log[x]])/(E^((8*x^2)/ (E^(2*x) - 2*E^x*x + x^2))*((x^2*(E^x - E^(2*x) - x + E^x*x))/(E^(2*x) - 2 *E^x*x + x^2)^2 + x/(E^(2*x) - 2*E^x*x + x^2))*(E^(3*x)*x^2 - 3*E^(2*x)*x^ 3 + 3*E^x*x^4 - x^5 - (E^(3*x)*x - 3*E^(2*x)*x^2 + 3*E^x*x^3 - x^4)*Log[x] ))
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 = 0.06 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42
\[{\mathrm e}^{\frac {8 x^{2}}{2 \,{\mathrm e}^{x} x -x^{2}-{\mathrm e}^{2 x}}} \ln \left (\ln \left (x \right )-x \right )\]
Input:
int((((16*x^3-16*x^2)*exp(x)*ln(x)+(-16*x^4+16*x^3)*exp(x))*ln(ln(x)-x)+(1 -x)*exp(x)^3+(3*x^2-3*x)*exp(x)^2+(-3*x^3+3*x^2)*exp(x)+x^4-x^3)/((x*exp(x )^3-3*exp(x)^2*x^2+3*exp(x)*x^3-x^4)*ln(x)-x^2*exp(x)^3+3*exp(x)^2*x^3-3*e xp(x)*x^4+x^5)/exp(8*x^2/(exp(x)^2-2*exp(x)*x+x^2)),x)
Output:
exp(8*x^2/(2*exp(x)*x-x^2-exp(2*x)))*ln(ln(x)-x)
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{-\frac {8 x^2}{e^{2 x}-2 e^x x+x^2}} \left (e^{3 x} (1-x)-x^3+x^4+e^{2 x} \left (-3 x+3 x^2\right )+e^x \left (3 x^2-3 x^3\right )+\left (e^x \left (16 x^3-16 x^4\right )+e^x \left (-16 x^2+16 x^3\right ) \log (x)\right ) \log (-x+\log (x))\right )}{-e^{3 x} x^2+3 e^{2 x} x^3-3 e^x x^4+x^5+\left (e^{3 x} x-3 e^{2 x} x^2+3 e^x x^3-x^4\right ) \log (x)} \, dx=e^{\left (-\frac {8 \, x^{2}}{x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}}\right )} \log \left (-x + \log \left (x\right )\right ) \] Input:
integrate((((16*x^3-16*x^2)*exp(x)*log(x)+(-16*x^4+16*x^3)*exp(x))*log(log (x)-x)+(1-x)*exp(x)^3+(3*x^2-3*x)*exp(x)^2+(-3*x^3+3*x^2)*exp(x)+x^4-x^3)/ ((x*exp(x)^3-3*exp(x)^2*x^2+3*exp(x)*x^3-x^4)*log(x)-x^2*exp(x)^3+3*exp(x) ^2*x^3-3*exp(x)*x^4+x^5)/exp(8*x^2/(exp(x)^2-2*exp(x)*x+x^2)),x, algorithm ="fricas")
Output:
e^(-8*x^2/(x^2 - 2*x*e^x + e^(2*x)))*log(-x + log(x))
Time = 56.57 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-\frac {8 x^2}{e^{2 x}-2 e^x x+x^2}} \left (e^{3 x} (1-x)-x^3+x^4+e^{2 x} \left (-3 x+3 x^2\right )+e^x \left (3 x^2-3 x^3\right )+\left (e^x \left (16 x^3-16 x^4\right )+e^x \left (-16 x^2+16 x^3\right ) \log (x)\right ) \log (-x+\log (x))\right )}{-e^{3 x} x^2+3 e^{2 x} x^3-3 e^x x^4+x^5+\left (e^{3 x} x-3 e^{2 x} x^2+3 e^x x^3-x^4\right ) \log (x)} \, dx=e^{- \frac {8 x^{2}}{x^{2} - 2 x e^{x} + e^{2 x}}} \log {\left (- x + \log {\left (x \right )} \right )} \] Input:
integrate((((16*x**3-16*x**2)*exp(x)*ln(x)+(-16*x**4+16*x**3)*exp(x))*ln(l n(x)-x)+(1-x)*exp(x)**3+(3*x**2-3*x)*exp(x)**2+(-3*x**3+3*x**2)*exp(x)+x** 4-x**3)/((x*exp(x)**3-3*exp(x)**2*x**2+3*exp(x)*x**3-x**4)*ln(x)-x**2*exp( x)**3+3*exp(x)**2*x**3-3*exp(x)*x**4+x**5)/exp(8*x**2/(exp(x)**2-2*exp(x)* x+x**2)),x)
Output:
exp(-8*x**2/(x**2 - 2*x*exp(x) + exp(2*x)))*log(-x + log(x))
Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.83 \[ \int \frac {e^{-\frac {8 x^2}{e^{2 x}-2 e^x x+x^2}} \left (e^{3 x} (1-x)-x^3+x^4+e^{2 x} \left (-3 x+3 x^2\right )+e^x \left (3 x^2-3 x^3\right )+\left (e^x \left (16 x^3-16 x^4\right )+e^x \left (-16 x^2+16 x^3\right ) \log (x)\right ) \log (-x+\log (x))\right )}{-e^{3 x} x^2+3 e^{2 x} x^3-3 e^x x^4+x^5+\left (e^{3 x} x-3 e^{2 x} x^2+3 e^x x^3-x^4\right ) \log (x)} \, dx=e^{\left (-\frac {8 \, e^{\left (2 \, x\right )}}{x^{2} - 2 \, x e^{x} + e^{\left (2 \, x\right )}} - \frac {16 \, e^{x}}{x - e^{x}} - 8\right )} \log \left (-x + \log \left (x\right )\right ) \] Input:
integrate((((16*x^3-16*x^2)*exp(x)*log(x)+(-16*x^4+16*x^3)*exp(x))*log(log (x)-x)+(1-x)*exp(x)^3+(3*x^2-3*x)*exp(x)^2+(-3*x^3+3*x^2)*exp(x)+x^4-x^3)/ ((x*exp(x)^3-3*exp(x)^2*x^2+3*exp(x)*x^3-x^4)*log(x)-x^2*exp(x)^3+3*exp(x) ^2*x^3-3*exp(x)*x^4+x^5)/exp(8*x^2/(exp(x)^2-2*exp(x)*x+x^2)),x, algorithm ="maxima")
Output:
e^(-8*e^(2*x)/(x^2 - 2*x*e^x + e^(2*x)) - 16*e^x/(x - e^x) - 8)*log(-x + l og(x))
Exception generated. \[ \int \frac {e^{-\frac {8 x^2}{e^{2 x}-2 e^x x+x^2}} \left (e^{3 x} (1-x)-x^3+x^4+e^{2 x} \left (-3 x+3 x^2\right )+e^x \left (3 x^2-3 x^3\right )+\left (e^x \left (16 x^3-16 x^4\right )+e^x \left (-16 x^2+16 x^3\right ) \log (x)\right ) \log (-x+\log (x))\right )}{-e^{3 x} x^2+3 e^{2 x} x^3-3 e^x x^4+x^5+\left (e^{3 x} x-3 e^{2 x} x^2+3 e^x x^3-x^4\right ) \log (x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((((16*x^3-16*x^2)*exp(x)*log(x)+(-16*x^4+16*x^3)*exp(x))*log(log (x)-x)+(1-x)*exp(x)^3+(3*x^2-3*x)*exp(x)^2+(-3*x^3+3*x^2)*exp(x)+x^4-x^3)/ ((x*exp(x)^3-3*exp(x)^2*x^2+3*exp(x)*x^3-x^4)*log(x)-x^2*exp(x)^3+3*exp(x) ^2*x^3-3*exp(x)*x^4+x^5)/exp(8*x^2/(exp(x)^2-2*exp(x)*x+x^2)),x, algorithm ="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{-16,[1,23,7]%%%}+%%%{64,[1,23,6]%%%}+%%%{-96,[1,23,5]%%%}+ %%%{64,[1
Timed out. \[ \int \frac {e^{-\frac {8 x^2}{e^{2 x}-2 e^x x+x^2}} \left (e^{3 x} (1-x)-x^3+x^4+e^{2 x} \left (-3 x+3 x^2\right )+e^x \left (3 x^2-3 x^3\right )+\left (e^x \left (16 x^3-16 x^4\right )+e^x \left (-16 x^2+16 x^3\right ) \log (x)\right ) \log (-x+\log (x))\right )}{-e^{3 x} x^2+3 e^{2 x} x^3-3 e^x x^4+x^5+\left (e^{3 x} x-3 e^{2 x} x^2+3 e^x x^3-x^4\right ) \log (x)} \, dx=\int -\frac {{\mathrm {e}}^{-\frac {8\,x^2}{{\mathrm {e}}^{2\,x}-2\,x\,{\mathrm {e}}^x+x^2}}\,\left ({\mathrm {e}}^{2\,x}\,\left (3\,x-3\,x^2\right )-{\mathrm {e}}^x\,\left (3\,x^2-3\,x^3\right )-\ln \left (\ln \left (x\right )-x\right )\,\left ({\mathrm {e}}^x\,\left (16\,x^3-16\,x^4\right )-{\mathrm {e}}^x\,\ln \left (x\right )\,\left (16\,x^2-16\,x^3\right )\right )+{\mathrm {e}}^{3\,x}\,\left (x-1\right )+x^3-x^4\right )}{3\,x^3\,{\mathrm {e}}^{2\,x}-x^2\,{\mathrm {e}}^{3\,x}-3\,x^4\,{\mathrm {e}}^x+\ln \left (x\right )\,\left (x\,{\mathrm {e}}^{3\,x}+3\,x^3\,{\mathrm {e}}^x-3\,x^2\,{\mathrm {e}}^{2\,x}-x^4\right )+x^5} \,d x \] Input:
int(-(exp(-(8*x^2)/(exp(2*x) - 2*x*exp(x) + x^2))*(exp(2*x)*(3*x - 3*x^2) - exp(x)*(3*x^2 - 3*x^3) - log(log(x) - x)*(exp(x)*(16*x^3 - 16*x^4) - exp (x)*log(x)*(16*x^2 - 16*x^3)) + exp(3*x)*(x - 1) + x^3 - x^4))/(3*x^3*exp( 2*x) - x^2*exp(3*x) - 3*x^4*exp(x) + log(x)*(x*exp(3*x) + 3*x^3*exp(x) - 3 *x^2*exp(2*x) - x^4) + x^5),x)
Output:
int(-(exp(-(8*x^2)/(exp(2*x) - 2*x*exp(x) + x^2))*(exp(2*x)*(3*x - 3*x^2) - exp(x)*(3*x^2 - 3*x^3) - log(log(x) - x)*(exp(x)*(16*x^3 - 16*x^4) - exp (x)*log(x)*(16*x^2 - 16*x^3)) + exp(3*x)*(x - 1) + x^3 - x^4))/(3*x^3*exp( 2*x) - x^2*exp(3*x) - 3*x^4*exp(x) + log(x)*(x*exp(3*x) + 3*x^3*exp(x) - 3 *x^2*exp(2*x) - x^4) + x^5), x)
Time = 142.79 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-\frac {8 x^2}{e^{2 x}-2 e^x x+x^2}} \left (e^{3 x} (1-x)-x^3+x^4+e^{2 x} \left (-3 x+3 x^2\right )+e^x \left (3 x^2-3 x^3\right )+\left (e^x \left (16 x^3-16 x^4\right )+e^x \left (-16 x^2+16 x^3\right ) \log (x)\right ) \log (-x+\log (x))\right )}{-e^{3 x} x^2+3 e^{2 x} x^3-3 e^x x^4+x^5+\left (e^{3 x} x-3 e^{2 x} x^2+3 e^x x^3-x^4\right ) \log (x)} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (x \right )-x \right )}{e^{\frac {8 x^{2}}{e^{2 x}-2 e^{x} x +x^{2}}}} \] Input:
int((((16*x^3-16*x^2)*exp(x)*log(x)+(-16*x^4+16*x^3)*exp(x))*log(log(x)-x) +(1-x)*exp(x)^3+(3*x^2-3*x)*exp(x)^2+(-3*x^3+3*x^2)*exp(x)+x^4-x^3)/((x*ex p(x)^3-3*exp(x)^2*x^2+3*exp(x)*x^3-x^4)*log(x)-x^2*exp(x)^3+3*exp(x)^2*x^3 -3*exp(x)*x^4+x^5)/exp(8*x^2/(exp(x)^2-2*exp(x)*x+x^2)),x)
Output:
log(log(x) - x)/e**((8*x**2)/(e**(2*x) - 2*e**x*x + x**2))