Integrand size = 126, antiderivative size = 31 \[ \int \frac {e^{-x} \left (-e^x x+e^{-24-e^{\frac {16 e^{-x}}{x}}+10 x-x^2+e^{\frac {8 e^{-x}}{x}} (-10+2 x)} \left (e^{\frac {16 e^{-x}}{x}} (16+16 x)+e^{\frac {8 e^{-x}}{x}} \left (80+64 x-16 x^2+2 e^x x^2\right )+e^x \left (10 x^2-2 x^3\right )\right )\right )}{x^2} \, dx=3+e^{1-\left (5+e^{\frac {8 e^{-x}}{x}}-x\right )^2}-\log (x) \] Output:
3-ln(x)+exp(1-(exp(4/exp(x)/x)^2-x+5)^2)
Time = 0.14 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.55 \[ \int \frac {e^{-x} \left (-e^x x+e^{-24-e^{\frac {16 e^{-x}}{x}}+10 x-x^2+e^{\frac {8 e^{-x}}{x}} (-10+2 x)} \left (e^{\frac {16 e^{-x}}{x}} (16+16 x)+e^{\frac {8 e^{-x}}{x}} \left (80+64 x-16 x^2+2 e^x x^2\right )+e^x \left (10 x^2-2 x^3\right )\right )\right )}{x^2} \, dx=e^{-24-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (-5+x)+10 x-x^2}-\log (x) \] Input:
Integrate[(-(E^x*x) + E^(-24 - E^(16/(E^x*x)) + 10*x - x^2 + E^(8/(E^x*x)) *(-10 + 2*x))*(E^(16/(E^x*x))*(16 + 16*x) + E^(8/(E^x*x))*(80 + 64*x - 16* x^2 + 2*E^x*x^2) + E^x*(10*x^2 - 2*x^3)))/(E^x*x^2),x]
Output:
E^(-24 - E^(16/(E^x*x)) + 2*E^(8/(E^x*x))*(-5 + x) + 10*x - x^2) - Log[x]
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.
| \(\displaystyle \int \frac {e^{-x} \left (\left (e^{\frac {8 e^{-x}}{x}} \left (2 e^x x^2-16 x^2+64 x+80\right )+e^x \left (10 x^2-2 x^3\right )+e^{\frac {16 e^{-x}}{x}} (16 x+16)\right ) \exp \left (-x^2+10 x-e^{\frac {16 e^{-x}}{x}}+e^{\frac {8 e^{-x}}{x}} (2 x-10)-24\right )-e^x x\right )}{x^2} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {2 \left (-x+e^{\frac {8 e^{-x}}{x}}+5\right ) \left (e^x x^2+8 e^{\frac {8 e^{-x}}{x}} x+8 e^{\frac {8 e^{-x}}{x}}\right ) \exp \left (-x^2+9 x-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (x-5)-24\right )}{x^2}-\frac {1}{x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -16 \int \exp \left (-x^2+9 x-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (x-5)-24+\frac {8 e^{-x}}{x}\right )dx+10 \int \exp \left (-x^2+10 x-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (x-5)-24\right )dx+2 \int \exp \left (-x^2+10 x-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (x-5)-24+\frac {8 e^{-x}}{x}\right )dx+80 \int \frac {\exp \left (-x^2+9 x-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (x-5)-24+\frac {8 e^{-x}}{x}\right )}{x^2}dx+16 \int \frac {\exp \left (-x^2+9 x-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (x-5)-24+\frac {16 e^{-x}}{x}\right )}{x^2}dx+64 \int \frac {\exp \left (-x^2+9 x-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (x-5)-24+\frac {8 e^{-x}}{x}\right )}{x}dx+16 \int \frac {\exp \left (-x^2+9 x-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (x-5)-24+\frac {16 e^{-x}}{x}\right )}{x}dx-2 \int \exp \left (-x^2+10 x-e^{\frac {16 e^{-x}}{x}}+2 e^{\frac {8 e^{-x}}{x}} (x-5)-24\right ) xdx-\log (x)\) |
Input:
Int[(-(E^x*x) + E^(-24 - E^(16/(E^x*x)) + 10*x - x^2 + E^(8/(E^x*x))*(-10 + 2*x))*(E^(16/(E^x*x))*(16 + 16*x) + E^(8/(E^x*x))*(80 + 64*x - 16*x^2 + 2*E^x*x^2) + E^x*(10*x^2 - 2*x^3)))/(E^x*x^2),x]
Output:
$Aborted
Time = 1.81 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58
| method | result | size |
| parallelrisch | \(-\ln \left (x \right )+{\mathrm e}^{-{\mathrm e}^{\frac {16 \,{\mathrm e}^{-x}}{x}}+\left (2 x -10\right ) {\mathrm e}^{\frac {8 \,{\mathrm e}^{-x}}{x}}-x^{2}+10 x -24}\) | \(49\) |
| risch | \(-\ln \left (x \right )+{\mathrm e}^{2 \,{\mathrm e}^{\frac {8 \,{\mathrm e}^{-x}}{x}} x -x^{2}-{\mathrm e}^{\frac {16 \,{\mathrm e}^{-x}}{x}}-10 \,{\mathrm e}^{\frac {8 \,{\mathrm e}^{-x}}{x}}+10 x -24}\) | \(54\) |
Input:
int((((16*x+16)*exp(4/exp(x)/x)^4+(2*exp(x)*x^2-16*x^2+64*x+80)*exp(4/exp( x)/x)^2+(-2*x^3+10*x^2)*exp(x))*exp(-exp(4/exp(x)/x)^4+(2*x-10)*exp(4/exp( x)/x)^2-x^2+10*x-24)-exp(x)*x)/exp(x)/x^2,x,method=_RETURNVERBOSE)
Output:
-ln(x)+exp(-exp(4/exp(x)/x)^4+(2*x-10)*exp(4/exp(x)/x)^2-x^2+10*x-24)
Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {e^{-x} \left (-e^x x+e^{-24-e^{\frac {16 e^{-x}}{x}}+10 x-x^2+e^{\frac {8 e^{-x}}{x}} (-10+2 x)} \left (e^{\frac {16 e^{-x}}{x}} (16+16 x)+e^{\frac {8 e^{-x}}{x}} \left (80+64 x-16 x^2+2 e^x x^2\right )+e^x \left (10 x^2-2 x^3\right )\right )\right )}{x^2} \, dx=e^{\left (-x^{2} + 2 \, {\left (x - 5\right )} e^{\left (\frac {8 \, e^{\left (-x\right )}}{x}\right )} + 10 \, x - e^{\left (\frac {16 \, e^{\left (-x\right )}}{x}\right )} - 24\right )} - \log \left (x\right ) \] Input:
integrate((((16*x+16)*exp(4/exp(x)/x)^4+(2*exp(x)*x^2-16*x^2+64*x+80)*exp( 4/exp(x)/x)^2+(-2*x^3+10*x^2)*exp(x))*exp(-exp(4/exp(x)/x)^4+(2*x-10)*exp( 4/exp(x)/x)^2-x^2+10*x-24)-exp(x)*x)/exp(x)/x^2,x, algorithm="fricas")
Output:
e^(-x^2 + 2*(x - 5)*e^(8*e^(-x)/x) + 10*x - e^(16*e^(-x)/x) - 24) - log(x)
Time = 1.90 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-x} \left (-e^x x+e^{-24-e^{\frac {16 e^{-x}}{x}}+10 x-x^2+e^{\frac {8 e^{-x}}{x}} (-10+2 x)} \left (e^{\frac {16 e^{-x}}{x}} (16+16 x)+e^{\frac {8 e^{-x}}{x}} \left (80+64 x-16 x^2+2 e^x x^2\right )+e^x \left (10 x^2-2 x^3\right )\right )\right )}{x^2} \, dx=e^{- x^{2} + 10 x + \left (2 x - 10\right ) e^{\frac {8 e^{- x}}{x}} - e^{\frac {16 e^{- x}}{x}} - 24} - \log {\left (x \right )} \] Input:
integrate((((16*x+16)*exp(4/exp(x)/x)**4+(2*exp(x)*x**2-16*x**2+64*x+80)*e xp(4/exp(x)/x)**2+(-2*x**3+10*x**2)*exp(x))*exp(-exp(4/exp(x)/x)**4+(2*x-1 0)*exp(4/exp(x)/x)**2-x**2+10*x-24)-exp(x)*x)/exp(x)/x**2,x)
Output:
exp(-x**2 + 10*x + (2*x - 10)*exp(8*exp(-x)/x) - exp(16*exp(-x)/x) - 24) - log(x)
Time = 0.28 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.71 \[ \int \frac {e^{-x} \left (-e^x x+e^{-24-e^{\frac {16 e^{-x}}{x}}+10 x-x^2+e^{\frac {8 e^{-x}}{x}} (-10+2 x)} \left (e^{\frac {16 e^{-x}}{x}} (16+16 x)+e^{\frac {8 e^{-x}}{x}} \left (80+64 x-16 x^2+2 e^x x^2\right )+e^x \left (10 x^2-2 x^3\right )\right )\right )}{x^2} \, dx=e^{\left (-x^{2} + 2 \, x e^{\left (\frac {8 \, e^{\left (-x\right )}}{x}\right )} + 10 \, x - e^{\left (\frac {16 \, e^{\left (-x\right )}}{x}\right )} - 10 \, e^{\left (\frac {8 \, e^{\left (-x\right )}}{x}\right )} - 24\right )} - \log \left (x\right ) \] Input:
integrate((((16*x+16)*exp(4/exp(x)/x)^4+(2*exp(x)*x^2-16*x^2+64*x+80)*exp( 4/exp(x)/x)^2+(-2*x^3+10*x^2)*exp(x))*exp(-exp(4/exp(x)/x)^4+(2*x-10)*exp( 4/exp(x)/x)^2-x^2+10*x-24)-exp(x)*x)/exp(x)/x^2,x, algorithm="maxima")
Output:
e^(-x^2 + 2*x*e^(8*e^(-x)/x) + 10*x - e^(16*e^(-x)/x) - 10*e^(8*e^(-x)/x) - 24) - log(x)
\[ \int \frac {e^{-x} \left (-e^x x+e^{-24-e^{\frac {16 e^{-x}}{x}}+10 x-x^2+e^{\frac {8 e^{-x}}{x}} (-10+2 x)} \left (e^{\frac {16 e^{-x}}{x}} (16+16 x)+e^{\frac {8 e^{-x}}{x}} \left (80+64 x-16 x^2+2 e^x x^2\right )+e^x \left (10 x^2-2 x^3\right )\right )\right )}{x^2} \, dx=\int { -\frac {{\left (2 \, {\left ({\left (x^{3} - 5 \, x^{2}\right )} e^{x} - 8 \, {\left (x + 1\right )} e^{\left (\frac {16 \, e^{\left (-x\right )}}{x}\right )} - {\left (x^{2} e^{x} - 8 \, x^{2} + 32 \, x + 40\right )} e^{\left (\frac {8 \, e^{\left (-x\right )}}{x}\right )}\right )} e^{\left (-x^{2} + 2 \, {\left (x - 5\right )} e^{\left (\frac {8 \, e^{\left (-x\right )}}{x}\right )} + 10 \, x - e^{\left (\frac {16 \, e^{\left (-x\right )}}{x}\right )} - 24\right )} + x e^{x}\right )} e^{\left (-x\right )}}{x^{2}} \,d x } \] Input:
integrate((((16*x+16)*exp(4/exp(x)/x)^4+(2*exp(x)*x^2-16*x^2+64*x+80)*exp( 4/exp(x)/x)^2+(-2*x^3+10*x^2)*exp(x))*exp(-exp(4/exp(x)/x)^4+(2*x-10)*exp( 4/exp(x)/x)^2-x^2+10*x-24)-exp(x)*x)/exp(x)/x^2,x, algorithm="giac")
Output:
integrate(-(2*((x^3 - 5*x^2)*e^x - 8*(x + 1)*e^(16*e^(-x)/x) - (x^2*e^x - 8*x^2 + 32*x + 40)*e^(8*e^(-x)/x))*e^(-x^2 + 2*(x - 5)*e^(8*e^(-x)/x) + 10 *x - e^(16*e^(-x)/x) - 24) + x*e^x)*e^(-x)/x^2, x)
Time = 4.44 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.87 \[ \int \frac {e^{-x} \left (-e^x x+e^{-24-e^{\frac {16 e^{-x}}{x}}+10 x-x^2+e^{\frac {8 e^{-x}}{x}} (-10+2 x)} \left (e^{\frac {16 e^{-x}}{x}} (16+16 x)+e^{\frac {8 e^{-x}}{x}} \left (80+64 x-16 x^2+2 e^x x^2\right )+e^x \left (10 x^2-2 x^3\right )\right )\right )}{x^2} \, dx={\mathrm {e}}^{2\,x\,{\mathrm {e}}^{\frac {8\,{\mathrm {e}}^{-x}}{x}}}\,{\mathrm {e}}^{10\,x}\,{\mathrm {e}}^{-24}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{-{\mathrm {e}}^{\frac {16\,{\mathrm {e}}^{-x}}{x}}}\,{\mathrm {e}}^{-10\,{\mathrm {e}}^{\frac {8\,{\mathrm {e}}^{-x}}{x}}}-\ln \left (x\right ) \] Input:
int((exp(-x)*(exp(10*x - exp((16*exp(-x))/x) + exp((8*exp(-x))/x)*(2*x - 1 0) - x^2 - 24)*(exp(x)*(10*x^2 - 2*x^3) + exp((16*exp(-x))/x)*(16*x + 16) + exp((8*exp(-x))/x)*(64*x + 2*x^2*exp(x) - 16*x^2 + 80)) - x*exp(x)))/x^2 ,x)
Output:
exp(2*x*exp((8*exp(-x))/x))*exp(10*x)*exp(-24)*exp(-x^2)*exp(-exp((16*exp( -x))/x))*exp(-10*exp((8*exp(-x))/x)) - log(x)
Time = 0.55 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.19 \[ \int \frac {e^{-x} \left (-e^x x+e^{-24-e^{\frac {16 e^{-x}}{x}}+10 x-x^2+e^{\frac {8 e^{-x}}{x}} (-10+2 x)} \left (e^{\frac {16 e^{-x}}{x}} (16+16 x)+e^{\frac {8 e^{-x}}{x}} \left (80+64 x-16 x^2+2 e^x x^2\right )+e^x \left (10 x^2-2 x^3\right )\right )\right )}{x^2} \, dx=\frac {-e^{e^{\frac {16}{e^{x} x}}+10 e^{\frac {8}{e^{x} x}}+x^{2}} \mathrm {log}\left (x \right ) e^{24}+e^{2 e^{\frac {8}{e^{x} x}} x +10 x}}{e^{e^{\frac {16}{e^{x} x}}+10 e^{\frac {8}{e^{x} x}}+x^{2}} e^{24}} \] Input:
int((((16*x+16)*exp(4/exp(x)/x)^4+(2*exp(x)*x^2-16*x^2+64*x+80)*exp(4/exp( x)/x)^2+(-2*x^3+10*x^2)*exp(x))*exp(-exp(4/exp(x)/x)^4+(2*x-10)*exp(4/exp( x)/x)^2-x^2+10*x-24)-exp(x)*x)/exp(x)/x^2,x)
Output:
( - e**(e**(16/(e**x*x)) + 10*e**(8/(e**x*x)) + x**2)*log(x)*e**24 + e**(2 *e**(8/(e**x*x))*x + 10*x))/(e**(e**(16/(e**x*x)) + 10*e**(8/(e**x*x)) + x **2)*e**24)