Integrand size = 117, antiderivative size = 32 \[ \int \frac {16 x+4 e^4 x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} \left (-52 x+2 x^2+e^4 \left (-4 x+4 e^2 x\right )+e^2 \left (16 x-4 x^2\right )\right )}{e^{2 x-2 e^2 x} x-2 e^{x-e^2 x} x^2+x^3} \, dx=x-\frac {4 \left (4+e^4-x\right )}{-e^{x-e^2 x}+x}+16 \log (x) \] Output:
16*ln(x)+x-4/(x-exp(-exp(2)*x+x))*(4-x+exp(4))
Time = 10.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {16 x+4 e^4 x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} \left (-52 x+2 x^2+e^4 \left (-4 x+4 e^2 x\right )+e^2 \left (16 x-4 x^2\right )\right )}{e^{2 x-2 e^2 x} x-2 e^{x-e^2 x} x^2+x^3} \, dx=-\frac {4 \left (4+e^4\right )}{x}+x+\frac {4 e^x \left (-4-e^4+x\right )}{x \left (-e^x+e^{e^2 x} x\right )}+16 \log (x) \] Input:
Integrate[(16*x + 4*E^4*x + 16*x^2 + x^3 + E^(2*x - 2*E^2*x)*(16 + x) + E^ (x - E^2*x)*(-52*x + 2*x^2 + E^4*(-4*x + 4*E^2*x) + E^2*(16*x - 4*x^2)))/( E^(2*x - 2*E^2*x)*x - 2*E^(x - E^2*x)*x^2 + x^3),x]
Output:
(-4*(4 + E^4))/x + x + (4*E^x*(-4 - E^4 + x))/(x*(-E^x + E^(E^2*x)*x)) + 1 6*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.
| \(\displaystyle \int \frac {x^3+16 x^2+e^{x-e^2 x} \left (2 x^2+e^2 \left (16 x-4 x^2\right )-52 x+e^4 \left (4 e^2 x-4 x\right )\right )+4 e^4 x+16 x+e^{2 x-2 e^2 x} (x+16)}{x^3-2 e^{x-e^2 x} x^2+e^{2 x-2 e^2 x} x} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {x^3+16 x^2+e^{x-e^2 x} \left (2 x^2+e^2 \left (16 x-4 x^2\right )-52 x+e^4 \left (4 e^2 x-4 x\right )\right )+\left (16+4 e^4\right ) x+e^{2 x-2 e^2 x} (x+16)}{x^3-2 e^{x-e^2 x} x^2+e^{2 x-2 e^2 x} x}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {e^{2 e^2 x} \left (x^3+16 x^2+e^{x-e^2 x} \left (2 x^2+e^2 \left (16 x-4 x^2\right )-52 x+e^4 \left (4 e^2 x-4 x\right )\right )+\left (16+4 e^4\right ) x+e^{2 x-2 e^2 x} (x+16)\right )}{x \left (e^x-e^{e^2 x} x\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {4 e^{2 e^2 x} \left (\left (1-e^2\right ) x^2-\left (5-4 e^2+e^4-e^6\right ) x+e^4+4\right )}{\left (e^x-e^{e^2 x} x\right )^2}+\frac {x+16}{x}+4 e^{2 e^2 x-\left (1+e^2\right ) x} \left (\left (1-e^2\right ) x+e^6-e^4+4 e^2-5\right )+\frac {4 e^{2 e^2 x-x} x \left (\left (1-e^2\right ) x+e^6-e^4+4 e^2-5\right )}{e^x-e^{e^2 x} x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 4 (1-e) (1+e) \int \frac {e^{2 e^2 x} x^2}{\left (e^{e^2 x} x-e^x\right )^2}dx-4 (1-e) (1+e) \int \frac {e^{-\left (\left (1-2 e^2\right ) x\right )} x^2}{e^{e^2 x} x-e^x}dx+4 \left (4+e^4\right ) \int \frac {e^{2 e^2 x}}{\left (e^x-e^{e^2 x} x\right )^2}dx-4 \left (5-4 e^2+e^4-e^6\right ) \int \frac {e^{2 e^2 x} x}{\left (e^{e^2 x} x-e^x\right )^2}dx+4 \left (5-4 e^2+e^4-e^6\right ) \int \frac {e^{-\left (\left (1-2 e^2\right ) x\right )} x}{e^{e^2 x} x-e^x}dx+x+\frac {4 e^{-\left (\left (1-e^2\right ) x\right )} \left (-\left (\left (1-e^2\right ) x\right )-e^6+e^4-4 e^2+5\right )}{1-e^2}-\frac {4 e^{-\left (\left (1-e^2\right ) x\right )}}{1-e^2}+16 \log (x)\) |
Input:
Int[(16*x + 4*E^4*x + 16*x^2 + x^3 + E^(2*x - 2*E^2*x)*(16 + x) + E^(x - E ^2*x)*(-52*x + 2*x^2 + E^4*(-4*x + 4*E^2*x) + E^2*(16*x - 4*x^2)))/(E^(2*x - 2*E^2*x)*x - 2*E^(x - E^2*x)*x^2 + x^3),x]
Output:
$Aborted
Time = 18.87 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94
| method | result | size |
| risch | \(x +16 \ln \left (x \right )-\frac {4 \left (4-x +{\mathrm e}^{4}\right )}{x -{\mathrm e}^{-\left ({\mathrm e}^{2}-1\right ) x}}\) | \(30\) |
| norman | \(\frac {x^{2}+4 \,{\mathrm e}^{-{\mathrm e}^{2} x +x}-x \,{\mathrm e}^{-{\mathrm e}^{2} x +x}-16-4 \,{\mathrm e}^{4}}{x -{\mathrm e}^{-{\mathrm e}^{2} x +x}}+16 \ln \left (x \right )\) | \(51\) |
| parallelrisch | \(-\frac {-16 x \ln \left (x \right )+16 \ln \left (x \right ) {\mathrm e}^{-{\mathrm e}^{2} x +x}-x^{2}+x \,{\mathrm e}^{-{\mathrm e}^{2} x +x}+16+4 \,{\mathrm e}^{4}-4 x}{x -{\mathrm e}^{-{\mathrm e}^{2} x +x}}\) | \(58\) |
Input:
int(((x+16)*exp(-exp(2)*x+x)^2+((4*exp(2)*x-4*x)*exp(4)+(-4*x^2+16*x)*exp( 2)+2*x^2-52*x)*exp(-exp(2)*x+x)+4*x*exp(4)+x^3+16*x^2+16*x)/(x*exp(-exp(2) *x+x)^2-2*x^2*exp(-exp(2)*x+x)+x^3),x,method=_RETURNVERBOSE)
Output:
x+16*ln(x)-4*(4-x+exp(4))/(x-exp(-(exp(2)-1)*x))
Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.69 \[ \int \frac {16 x+4 e^4 x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} \left (-52 x+2 x^2+e^4 \left (-4 x+4 e^2 x\right )+e^2 \left (16 x-4 x^2\right )\right )}{e^{2 x-2 e^2 x} x-2 e^{x-e^2 x} x^2+x^3} \, dx=\frac {x^{2} - x e^{\left (-x e^{2} + x\right )} + 16 \, {\left (x - e^{\left (-x e^{2} + x\right )}\right )} \log \left (x\right ) + 4 \, x - 4 \, e^{4} - 16}{x - e^{\left (-x e^{2} + x\right )}} \] Input:
integrate(((x+16)*exp(-exp(2)*x+x)^2+((4*exp(2)*x-4*x)*exp(4)+(-4*x^2+16*x )*exp(2)+2*x^2-52*x)*exp(-exp(2)*x+x)+4*x*exp(4)+x^3+16*x^2+16*x)/(x*exp(- exp(2)*x+x)^2-2*x^2*exp(-exp(2)*x+x)+x^3),x, algorithm="fricas")
Output:
(x^2 - x*e^(-x*e^2 + x) + 16*(x - e^(-x*e^2 + x))*log(x) + 4*x - 4*e^4 - 1 6)/(x - e^(-x*e^2 + x))
Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {16 x+4 e^4 x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} \left (-52 x+2 x^2+e^4 \left (-4 x+4 e^2 x\right )+e^2 \left (16 x-4 x^2\right )\right )}{e^{2 x-2 e^2 x} x-2 e^{x-e^2 x} x^2+x^3} \, dx=x + 16 \log {\left (x \right )} + \frac {- 4 x + 16 + 4 e^{4}}{- x + e^{- x e^{2} + x}} \] Input:
integrate(((x+16)*exp(-exp(2)*x+x)**2+((4*exp(2)*x-4*x)*exp(4)+(-4*x**2+16 *x)*exp(2)+2*x**2-52*x)*exp(-exp(2)*x+x)+4*x*exp(4)+x**3+16*x**2+16*x)/(x* exp(-exp(2)*x+x)**2-2*x**2*exp(-exp(2)*x+x)+x**3),x)
Output:
x + 16*log(x) + (-4*x + 16 + 4*exp(4))/(-x + exp(-x*exp(2) + x))
Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.34 \[ \int \frac {16 x+4 e^4 x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} \left (-52 x+2 x^2+e^4 \left (-4 x+4 e^2 x\right )+e^2 \left (16 x-4 x^2\right )\right )}{e^{2 x-2 e^2 x} x-2 e^{x-e^2 x} x^2+x^3} \, dx=\frac {{\left (x^{2} - 4 \, e^{4} - 16\right )} e^{\left (x e^{2}\right )} - {\left (x - 4\right )} e^{x}}{x e^{\left (x e^{2}\right )} - e^{x}} + 16 \, \log \left (x\right ) \] Input:
integrate(((x+16)*exp(-exp(2)*x+x)^2+((4*exp(2)*x-4*x)*exp(4)+(-4*x^2+16*x )*exp(2)+2*x^2-52*x)*exp(-exp(2)*x+x)+4*x*exp(4)+x^3+16*x^2+16*x)/(x*exp(- exp(2)*x+x)^2-2*x^2*exp(-exp(2)*x+x)+x^3),x, algorithm="maxima")
Output:
((x^2 - 4*e^4 - 16)*e^(x*e^2) - (x - 4)*e^x)/(x*e^(x*e^2) - e^x) + 16*log( x)
Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (29) = 58\).
Time = 2.37 (sec) , antiderivative size = 198, normalized size of antiderivative = 6.19 \[ \int \frac {16 x+4 e^4 x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} \left (-52 x+2 x^2+e^4 \left (-4 x+4 e^2 x\right )+e^2 \left (16 x-4 x^2\right )\right )}{e^{2 x-2 e^2 x} x-2 e^{x-e^2 x} x^2+x^3} \, dx=\frac {x^{2} {\left (e^{2} - 1\right )}^{2} e^{\left (x {\left (e^{2} - 1\right )}\right )} + 16 \, x {\left (e^{2} - 1\right )} e^{\left (x {\left (e^{2} - 1\right )} + 2\right )} \log \left (x {\left (e^{2} - 1\right )}\right ) - 16 \, x {\left (e^{2} - 1\right )} e^{\left (x {\left (e^{2} - 1\right )}\right )} \log \left (x {\left (e^{2} - 1\right )}\right ) - x {\left (e^{2} - 1\right )} e^{2} + x {\left (e^{2} - 1\right )} - 16 \, e^{4} \log \left (x {\left (e^{2} - 1\right )}\right ) + 32 \, e^{2} \log \left (x {\left (e^{2} - 1\right )}\right ) + 4 \, e^{4} - 8 \, e^{2} - 4 \, e^{\left (x {\left (e^{2} - 1\right )} + 8\right )} + 8 \, e^{\left (x {\left (e^{2} - 1\right )} + 6\right )} - 20 \, e^{\left (x {\left (e^{2} - 1\right )} + 4\right )} + 32 \, e^{\left (x {\left (e^{2} - 1\right )} + 2\right )} - 16 \, e^{\left (x {\left (e^{2} - 1\right )}\right )} - 16 \, \log \left (x {\left (e^{2} - 1\right )}\right ) + 4}{{\left (x {\left (e^{2} - 1\right )} e^{\left (x {\left (e^{2} - 1\right )}\right )} - e^{2} + 1\right )} {\left (e^{2} - 1\right )}} \] Input:
integrate(((x+16)*exp(-exp(2)*x+x)^2+((4*exp(2)*x-4*x)*exp(4)+(-4*x^2+16*x )*exp(2)+2*x^2-52*x)*exp(-exp(2)*x+x)+4*x*exp(4)+x^3+16*x^2+16*x)/(x*exp(- exp(2)*x+x)^2-2*x^2*exp(-exp(2)*x+x)+x^3),x, algorithm="giac")
Output:
(x^2*(e^2 - 1)^2*e^(x*(e^2 - 1)) + 16*x*(e^2 - 1)*e^(x*(e^2 - 1) + 2)*log( x*(e^2 - 1)) - 16*x*(e^2 - 1)*e^(x*(e^2 - 1))*log(x*(e^2 - 1)) - x*(e^2 - 1)*e^2 + x*(e^2 - 1) - 16*e^4*log(x*(e^2 - 1)) + 32*e^2*log(x*(e^2 - 1)) + 4*e^4 - 8*e^2 - 4*e^(x*(e^2 - 1) + 8) + 8*e^(x*(e^2 - 1) + 6) - 20*e^(x*( e^2 - 1) + 4) + 32*e^(x*(e^2 - 1) + 2) - 16*e^(x*(e^2 - 1)) - 16*log(x*(e^ 2 - 1)) + 4)/((x*(e^2 - 1)*e^(x*(e^2 - 1)) - e^2 + 1)*(e^2 - 1))
Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.41 \[ \int \frac {16 x+4 e^4 x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} \left (-52 x+2 x^2+e^4 \left (-4 x+4 e^2 x\right )+e^2 \left (16 x-4 x^2\right )\right )}{e^{2 x-2 e^2 x} x-2 e^{x-e^2 x} x^2+x^3} \, dx=16\,\ln \left (x\right )-\frac {4\,{\mathrm {e}}^4-4\,x+x\,{\mathrm {e}}^{x-x\,{\mathrm {e}}^2}-x^2+16}{x-{\mathrm {e}}^{x-x\,{\mathrm {e}}^2}} \] Input:
int((16*x + exp(2*x - 2*x*exp(2))*(x + 16) + 4*x*exp(4) - exp(x - x*exp(2) )*(52*x - exp(2)*(16*x - 4*x^2) + exp(4)*(4*x - 4*x*exp(2)) - 2*x^2) + 16* x^2 + x^3)/(x*exp(2*x - 2*x*exp(2)) - 2*x^2*exp(x - x*exp(2)) + x^3),x)
Output:
16*log(x) - (4*exp(4) - 4*x + x*exp(x - x*exp(2)) - x^2 + 16)/(x - exp(x - x*exp(2)))
Time = 0.19 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.69 \[ \int \frac {16 x+4 e^4 x+16 x^2+x^3+e^{2 x-2 e^2 x} (16+x)+e^{x-e^2 x} \left (-52 x+2 x^2+e^4 \left (-4 x+4 e^2 x\right )+e^2 \left (16 x-4 x^2\right )\right )}{e^{2 x-2 e^2 x} x-2 e^{x-e^2 x} x^2+x^3} \, dx=\frac {16 e^{e^{2} x} \mathrm {log}\left (x \right ) x -4 e^{e^{2} x} e^{4}+e^{e^{2} x} x^{2}+4 e^{e^{2} x} x -16 e^{e^{2} x}-16 e^{x} \mathrm {log}\left (x \right )-e^{x} x}{e^{e^{2} x} x -e^{x}} \] Input:
int(((x+16)*exp(-exp(2)*x+x)^2+((4*exp(2)*x-4*x)*exp(4)+(-4*x^2+16*x)*exp( 2)+2*x^2-52*x)*exp(-exp(2)*x+x)+4*x*exp(4)+x^3+16*x^2+16*x)/(x*exp(-exp(2) *x+x)^2-2*x^2*exp(-exp(2)*x+x)+x^3),x)
Output:
(16*e**(e**2*x)*log(x)*x - 4*e**(e**2*x)*e**4 + e**(e**2*x)*x**2 + 4*e**(e **2*x)*x - 16*e**(e**2*x) - 16*e**x*log(x) - e**x*x)/(e**(e**2*x)*x - e**x )