Integrand size = 79, antiderivative size = 27 \[ \int \frac {e^{e^{\frac {-e^{2-4 x-x^2}-4 x}{x}}} \left (-10 x+e^{2+\frac {-e^{2-4 x-x^2}-4 x}{x}-4 x-x^2} \left (10+40 x+20 x^2\right )\right )}{x^3} \, dx=\frac {10 e^{e^{-4-\frac {e^{2+(-4-x) x}}{x}}}}{x} \] Output:
10*exp(exp(-4-exp(2+x*(-4-x))/x)-ln(x))
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {e^{e^{\frac {-e^{2-4 x-x^2}-4 x}{x}}} \left (-10 x+e^{2+\frac {-e^{2-4 x-x^2}-4 x}{x}-4 x-x^2} \left (10+40 x+20 x^2\right )\right )}{x^3} \, dx=\frac {10 e^{e^{-4-\frac {e^{2-4 x-x^2}}{x}}}}{x} \] Input:
Integrate[(E^E^((-E^(2 - 4*x - x^2) - 4*x)/x)*(-10*x + E^(2 + (-E^(2 - 4*x - x^2) - 4*x)/x - 4*x - x^2)*(10 + 40*x + 20*x^2)))/x^3,x]
Output:
(10*E^E^(-4 - E^(2 - 4*x - x^2)/x))/x
Leaf count is larger than twice the leaf count of optimal. \(97\) vs. \(2(27)=54\).
Time = 0.43 (sec) , antiderivative size = 97, normalized size of antiderivative = 3.59, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, 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.
\(\displaystyle \int \frac {e^{e^{\frac {-e^{-x^2-4 x+2}-4 x}{x}}} \left (\left (20 x^2+40 x+10\right ) \exp \left (-x^2+\frac {-e^{-x^2-4 x+2}-4 x}{x}-4 x+2\right )-10 x\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {10 \left (2 x^2+4 x+1\right ) \exp \left (-x^2+e^{-\frac {e^{-x^2-4 x+2}+4 x}{x}}-4 x+2\right )}{x^3 \left (\frac {e^{-x^2-4 x+2}+4 x}{x^2}-\frac {2 \left (2-e^{-x^2-4 x+2} (x+2)\right )}{x}\right )}\) |
Input:
Int[(E^E^((-E^(2 - 4*x - x^2) - 4*x)/x)*(-10*x + E^(2 + (-E^(2 - 4*x - x^2 ) - 4*x)/x - 4*x - x^2)*(10 + 40*x + 20*x^2)))/x^3,x]
Output:
(10*E^(2 + E^(-((E^(2 - 4*x - x^2) + 4*x)/x)) - 4*x - x^2)*(1 + 4*x + 2*x^ 2))/(x^3*((E^(2 - 4*x - x^2) + 4*x)/x^2 - (2*(2 - E^(2 - 4*x - x^2)*(2 + x )))/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 = 0.87 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\frac {10 \,{\mathrm e}^{{\mathrm e}^{-\frac {{\mathrm e}^{-x^{2}-4 x +2}+4 x}{x}}}}{x}\) | \(28\) |
parallelrisch | \(10 \,{\mathrm e}^{{\mathrm e}^{-\frac {{\mathrm e}^{-x^{2}-4 x +2}+4 x}{x}}-\ln \left (x \right )}\) | \(30\) |
Input:
int(((20*x^2+40*x+10)*exp(-x^2-4*x+2)*exp((-exp(-x^2-4*x+2)-4*x)/x)-10*x)* exp(exp((-exp(-x^2-4*x+2)-4*x)/x)-ln(x))/x^2,x,method=_RETURNVERBOSE)
Output:
10/x*exp(exp(-(exp(-x^2-4*x+2)+4*x)/x))
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{e^{\frac {-e^{2-4 x-x^2}-4 x}{x}}} \left (-10 x+e^{2+\frac {-e^{2-4 x-x^2}-4 x}{x}-4 x-x^2} \left (10+40 x+20 x^2\right )\right )}{x^3} \, dx=10 \, e^{\left (e^{\left (-\frac {4 \, x + e^{\left (-x^{2} - 4 \, x + 2\right )}}{x}\right )} - \log \left (x\right )\right )} \] Input:
integrate(((20*x^2+40*x+10)*exp(-x^2-4*x+2)*exp((-exp(-x^2-4*x+2)-4*x)/x)- 10*x)*exp(exp((-exp(-x^2-4*x+2)-4*x)/x)-log(x))/x^2,x, algorithm="fricas")
Output:
10*e^(e^(-(4*x + e^(-x^2 - 4*x + 2))/x) - log(x))
Time = 0.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{e^{\frac {-e^{2-4 x-x^2}-4 x}{x}}} \left (-10 x+e^{2+\frac {-e^{2-4 x-x^2}-4 x}{x}-4 x-x^2} \left (10+40 x+20 x^2\right )\right )}{x^3} \, dx=\frac {10 e^{e^{\frac {- 4 x - e^{- x^{2} - 4 x + 2}}{x}}}}{x} \] Input:
integrate(((20*x**2+40*x+10)*exp(-x**2-4*x+2)*exp((-exp(-x**2-4*x+2)-4*x)/ x)-10*x)*exp(exp((-exp(-x**2-4*x+2)-4*x)/x)-ln(x))/x**2,x)
Output:
10*exp(exp((-4*x - exp(-x**2 - 4*x + 2))/x))/x
\[ \int \frac {e^{e^{\frac {-e^{2-4 x-x^2}-4 x}{x}}} \left (-10 x+e^{2+\frac {-e^{2-4 x-x^2}-4 x}{x}-4 x-x^2} \left (10+40 x+20 x^2\right )\right )}{x^3} \, dx=\int { \frac {10 \, {\left ({\left (2 \, x^{2} + 4 \, x + 1\right )} e^{\left (-x^{2} - 4 \, x - \frac {4 \, x + e^{\left (-x^{2} - 4 \, x + 2\right )}}{x} + 2\right )} - x\right )} e^{\left (e^{\left (-\frac {4 \, x + e^{\left (-x^{2} - 4 \, x + 2\right )}}{x}\right )} - \log \left (x\right )\right )}}{x^{2}} \,d x } \] Input:
integrate(((20*x^2+40*x+10)*exp(-x^2-4*x+2)*exp((-exp(-x^2-4*x+2)-4*x)/x)- 10*x)*exp(exp((-exp(-x^2-4*x+2)-4*x)/x)-log(x))/x^2,x, algorithm="maxima")
Output:
10*integrate(((2*x^2 + 4*x + 1)*e^(-x^2 - 4*x - (4*x + e^(-x^2 - 4*x + 2)) /x + 2) - x)*e^(e^(-(4*x + e^(-x^2 - 4*x + 2))/x))/x^3, x)
\[ \int \frac {e^{e^{\frac {-e^{2-4 x-x^2}-4 x}{x}}} \left (-10 x+e^{2+\frac {-e^{2-4 x-x^2}-4 x}{x}-4 x-x^2} \left (10+40 x+20 x^2\right )\right )}{x^3} \, dx=\int { \frac {10 \, {\left ({\left (2 \, x^{2} + 4 \, x + 1\right )} e^{\left (-x^{2} - 4 \, x - \frac {4 \, x + e^{\left (-x^{2} - 4 \, x + 2\right )}}{x} + 2\right )} - x\right )} e^{\left (e^{\left (-\frac {4 \, x + e^{\left (-x^{2} - 4 \, x + 2\right )}}{x}\right )} - \log \left (x\right )\right )}}{x^{2}} \,d x } \] Input:
integrate(((20*x^2+40*x+10)*exp(-x^2-4*x+2)*exp((-exp(-x^2-4*x+2)-4*x)/x)- 10*x)*exp(exp((-exp(-x^2-4*x+2)-4*x)/x)-log(x))/x^2,x, algorithm="giac")
Output:
integrate(10*((2*x^2 + 4*x + 1)*e^(-x^2 - 4*x - (4*x + e^(-x^2 - 4*x + 2)) /x + 2) - x)*e^(e^(-(4*x + e^(-x^2 - 4*x + 2))/x) - log(x))/x^2, x)
Time = 3.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{\frac {-e^{2-4 x-x^2}-4 x}{x}}} \left (-10 x+e^{2+\frac {-e^{2-4 x-x^2}-4 x}{x}-4 x-x^2} \left (10+40 x+20 x^2\right )\right )}{x^3} \, dx=\frac {10\,{\mathrm {e}}^{{\mathrm {e}}^{-4}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-4\,x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{-x^2}}{x}}}}{x} \] Input:
int(-(exp(exp(-(4*x + exp(2 - x^2 - 4*x))/x) - log(x))*(10*x - exp(-(4*x + exp(2 - x^2 - 4*x))/x)*exp(2 - x^2 - 4*x)*(40*x + 20*x^2 + 10)))/x^2,x)
Output:
(10*exp(exp(-4)*exp(-(exp(-4*x)*exp(2)*exp(-x^2))/x)))/x
\[ \int \frac {e^{e^{\frac {-e^{2-4 x-x^2}-4 x}{x}}} \left (-10 x+e^{2+\frac {-e^{2-4 x-x^2}-4 x}{x}-4 x-x^2} \left (10+40 x+20 x^2\right )\right )}{x^3} \, dx=\frac {-10 \left (\int \frac {e^{\frac {1}{e^{\frac {e^{2}}{e^{x^{2}+4 x} x}} e^{4}}}}{x^{2}}d x \right ) e^{2}+10 \left (\int \frac {e^{\frac {1}{e^{\frac {e^{2}}{e^{x^{2}+4 x} x}} e^{4}}}}{e^{\frac {e^{x^{2}+4 x} x^{3}+4 e^{x^{2}+4 x} x^{2}+e^{2}}{e^{x^{2}+4 x} x}} x^{3}}d x \right )+40 \left (\int \frac {e^{\frac {1}{e^{\frac {e^{2}}{e^{x^{2}+4 x} x}} e^{4}}}}{e^{\frac {e^{x^{2}+4 x} x^{3}+4 e^{x^{2}+4 x} x^{2}+e^{2}}{e^{x^{2}+4 x} x}} x^{2}}d x \right )+20 \left (\int \frac {e^{\frac {1}{e^{\frac {e^{2}}{e^{x^{2}+4 x} x}} e^{4}}}}{e^{\frac {e^{x^{2}+4 x} x^{3}+4 e^{x^{2}+4 x} x^{2}+e^{2}}{e^{x^{2}+4 x} x}} x}d x \right )}{e^{2}} \] Input:
int(((20*x^2+40*x+10)*exp(-x^2-4*x+2)*exp((-exp(-x^2-4*x+2)-4*x)/x)-10*x)* exp(exp((-exp(-x^2-4*x+2)-4*x)/x)-log(x))/x^2,x)
Output:
(10*( - int(e**(1/(e**(e**2/(e**(x**2 + 4*x)*x))*e**4))/x**2,x)*e**2 + int (e**(1/(e**(e**2/(e**(x**2 + 4*x)*x))*e**4))/(e**((e**(x**2 + 4*x)*x**3 + 4*e**(x**2 + 4*x)*x**2 + e**2)/(e**(x**2 + 4*x)*x))*x**3),x) + 4*int(e**(1 /(e**(e**2/(e**(x**2 + 4*x)*x))*e**4))/(e**((e**(x**2 + 4*x)*x**3 + 4*e**( x**2 + 4*x)*x**2 + e**2)/(e**(x**2 + 4*x)*x))*x**2),x) + 2*int(e**(1/(e**( e**2/(e**(x**2 + 4*x)*x))*e**4))/(e**((e**(x**2 + 4*x)*x**3 + 4*e**(x**2 + 4*x)*x**2 + e**2)/(e**(x**2 + 4*x)*x))*x),x)))/e**2