Integrand size = 140, antiderivative size = 30 \[ \int \frac {e^{e^{e^x}-e^{2+x^2}+e^{e^{e^x}-e^{2+x^2}} \left (16+x-8 \log (x)+\log ^2(x)\right )} \left (-8+x+e^{2+x^2} \left (-32 x^2-2 x^3\right )+\left (2+16 e^{2+x^2} x^2\right ) \log (x)-2 e^{2+x^2} x^2 \log ^2(x)+e^{e^x} \left (e^x \left (16 x+x^2\right )-8 e^x x \log (x)+e^x x \log ^2(x)\right )\right )}{x} \, dx=e^{e^{e^{e^x}-e^{2+x^2}} \left (x+(4-\log (x))^2\right )} \]
Time = 0.29 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {e^{e^{e^x}-e^{2+x^2}+e^{e^{e^x}-e^{2+x^2}} \left (16+x-8 \log (x)+\log ^2(x)\right )} \left (-8+x+e^{2+x^2} \left (-32 x^2-2 x^3\right )+\left (2+16 e^{2+x^2} x^2\right ) \log (x)-2 e^{2+x^2} x^2 \log ^2(x)+e^{e^x} \left (e^x \left (16 x+x^2\right )-8 e^x x \log (x)+e^x x \log ^2(x)\right )\right )}{x} \, dx=e^{e^{e^{e^x}-e^{2+x^2}} \left (16+x-8 \log (x)+\log ^2(x)\right )} \]
Integrate[(E^(E^E^x - E^(2 + x^2) + E^(E^E^x - E^(2 + x^2))*(16 + x - 8*Lo g[x] + Log[x]^2))*(-8 + x + E^(2 + x^2)*(-32*x^2 - 2*x^3) + (2 + 16*E^(2 + x^2)*x^2)*Log[x] - 2*E^(2 + x^2)*x^2*Log[x]^2 + E^E^x*(E^x*(16*x + x^2) - 8*E^x*x*Log[x] + E^x*x*Log[x]^2)))/x,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.
| \(\displaystyle \int \frac {\left (-2 e^{x^2+2} x^2 \log ^2(x)+e^{e^x} \left (e^x \left (x^2+16 x\right )+e^x x \log ^2(x)-8 e^x x \log (x)\right )+\left (16 e^{x^2+2} x^2+2\right ) \log (x)+e^{x^2+2} \left (-2 x^3-32 x^2\right )+x-8\right ) \exp \left (-e^{x^2+2}+e^{e^{e^x}-e^{x^2+2}} \left (x+\log ^2(x)-8 \log (x)+16\right )+e^{e^x}\right )}{x} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {\left (e^{x+e^x} x^2+16 e^{x+e^x} x+x+e^{x+e^x} x \log ^2(x)-8 e^{x+e^x} x \log (x)+2 \log (x)-8\right ) \exp \left (-e^{x^2+2}+e^{e^{e^x}-e^{x^2+2}} \left (x+\log ^2(x)-8 \log (x)+16\right )+e^{e^x}\right )}{x}-2 x \left (x+\log ^2(x)-8 \log (x)+16\right ) \exp \left (x^2-e^{x^2+2}+e^{e^{e^x}-e^{x^2+2}} \left (x+\log ^2(x)-8 \log (x)+16\right )+e^{e^x}+2\right )\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (\frac {\left (e^{x+e^x} x^2+16 e^{x+e^x} x+x+e^{x+e^x} x \log ^2(x)-8 e^{x+e^x} x \log (x)+2 \log (x)-8\right ) \exp \left (-e^{x^2+2}+e^{e^{e^x}-e^{x^2+2}} \left (x+\log ^2(x)-8 \log (x)+16\right )+e^{e^x}\right )}{x}-2 x \left (x+\log ^2(x)-8 \log (x)+16\right ) \exp \left (x^2-e^{x^2+2}+e^{e^{e^x}-e^{x^2+2}} \left (x+\log ^2(x)-8 \log (x)+16\right )+e^{e^x}+2\right )\right )dx\) |
Int[(E^(E^E^x - E^(2 + x^2) + E^(E^E^x - E^(2 + x^2))*(16 + x - 8*Log[x] + Log[x]^2))*(-8 + x + E^(2 + x^2)*(-32*x^2 - 2*x^3) + (2 + 16*E^(2 + x^2)* x^2)*Log[x] - 2*E^(2 + x^2)*x^2*Log[x]^2 + E^E^x*(E^x*(16*x + x^2) - 8*E^x *x*Log[x] + E^x*x*Log[x]^2)))/x,x]
3.26.61.3.1 Defintions of rubi rules used
Time = 62.06 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90
| method | result | size |
| risch | \({\mathrm e}^{\left (\ln \left (x \right )^{2}-8 \ln \left (x \right )+x +16\right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}-{\mathrm e}^{x^{2}+2}}}\) | \(27\) |
| parallelrisch | \({\mathrm e}^{\left (\ln \left (x \right )^{2}-8 \ln \left (x \right )+x +16\right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}-{\mathrm e}^{x^{2}+2}}}\) | \(27\) |
int(((x*exp(x)*ln(x)^2-8*x*exp(x)*ln(x)+(x^2+16*x)*exp(x))*exp(exp(x))-2*x ^2*exp(x^2+2)*ln(x)^2+(16*x^2*exp(x^2+2)+2)*ln(x)+(-2*x^3-32*x^2)*exp(x^2+ 2)-8+x)*exp(exp(exp(x))-exp(x^2+2))*exp((ln(x)^2-8*ln(x)+x+16)*exp(exp(exp (x))-exp(x^2+2)))/x,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {e^{e^{e^x}-e^{2+x^2}+e^{e^{e^x}-e^{2+x^2}} \left (16+x-8 \log (x)+\log ^2(x)\right )} \left (-8+x+e^{2+x^2} \left (-32 x^2-2 x^3\right )+\left (2+16 e^{2+x^2} x^2\right ) \log (x)-2 e^{2+x^2} x^2 \log ^2(x)+e^{e^x} \left (e^x \left (16 x+x^2\right )-8 e^x x \log (x)+e^x x \log ^2(x)\right )\right )}{x} \, dx=e^{\left ({\left (\log \left (x\right )^{2} + x - 8 \, \log \left (x\right ) + 16\right )} e^{\left (-e^{\left (x^{2} + 2\right )} + e^{\left (e^{x}\right )}\right )}\right )} \]
integrate(((x*exp(x)*log(x)^2-8*x*exp(x)*log(x)+(x^2+16*x)*exp(x))*exp(exp (x))-2*x^2*exp(x^2+2)*log(x)^2+(16*x^2*exp(x^2+2)+2)*log(x)+(-2*x^3-32*x^2 )*exp(x^2+2)-8+x)*exp(exp(exp(x))-exp(x^2+2))*exp((log(x)^2-8*log(x)+x+16) *exp(exp(exp(x))-exp(x^2+2)))/x,x, algorithm=\
Timed out. \[ \int \frac {e^{e^{e^x}-e^{2+x^2}+e^{e^{e^x}-e^{2+x^2}} \left (16+x-8 \log (x)+\log ^2(x)\right )} \left (-8+x+e^{2+x^2} \left (-32 x^2-2 x^3\right )+\left (2+16 e^{2+x^2} x^2\right ) \log (x)-2 e^{2+x^2} x^2 \log ^2(x)+e^{e^x} \left (e^x \left (16 x+x^2\right )-8 e^x x \log (x)+e^x x \log ^2(x)\right )\right )}{x} \, dx=\text {Timed out} \]
integrate(((x*exp(x)*ln(x)**2-8*x*exp(x)*ln(x)+(x**2+16*x)*exp(x))*exp(exp (x))-2*x**2*exp(x**2+2)*ln(x)**2+(16*x**2*exp(x**2+2)+2)*ln(x)+(-2*x**3-32 *x**2)*exp(x**2+2)-8+x)*exp(exp(exp(x))-exp(x**2+2))*exp((ln(x)**2-8*ln(x) +x+16)*exp(exp(exp(x))-exp(x**2+2)))/x,x)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (23) = 46\).
Time = 1.01 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.23 \[ \int \frac {e^{e^{e^x}-e^{2+x^2}+e^{e^{e^x}-e^{2+x^2}} \left (16+x-8 \log (x)+\log ^2(x)\right )} \left (-8+x+e^{2+x^2} \left (-32 x^2-2 x^3\right )+\left (2+16 e^{2+x^2} x^2\right ) \log (x)-2 e^{2+x^2} x^2 \log ^2(x)+e^{e^x} \left (e^x \left (16 x+x^2\right )-8 e^x x \log (x)+e^x x \log ^2(x)\right )\right )}{x} \, dx=e^{\left (e^{\left (-e^{\left (x^{2} + 2\right )} + e^{\left (e^{x}\right )}\right )} \log \left (x\right )^{2} + x e^{\left (-e^{\left (x^{2} + 2\right )} + e^{\left (e^{x}\right )}\right )} - 8 \, e^{\left (-e^{\left (x^{2} + 2\right )} + e^{\left (e^{x}\right )}\right )} \log \left (x\right ) + 16 \, e^{\left (-e^{\left (x^{2} + 2\right )} + e^{\left (e^{x}\right )}\right )}\right )} \]
integrate(((x*exp(x)*log(x)^2-8*x*exp(x)*log(x)+(x^2+16*x)*exp(x))*exp(exp (x))-2*x^2*exp(x^2+2)*log(x)^2+(16*x^2*exp(x^2+2)+2)*log(x)+(-2*x^3-32*x^2 )*exp(x^2+2)-8+x)*exp(exp(exp(x))-exp(x^2+2))*exp((log(x)^2-8*log(x)+x+16) *exp(exp(exp(x))-exp(x^2+2)))/x,x, algorithm=\
e^(e^(-e^(x^2 + 2) + e^(e^x))*log(x)^2 + x*e^(-e^(x^2 + 2) + e^(e^x)) - 8* e^(-e^(x^2 + 2) + e^(e^x))*log(x) + 16*e^(-e^(x^2 + 2) + e^(e^x)))
\[ \int \frac {e^{e^{e^x}-e^{2+x^2}+e^{e^{e^x}-e^{2+x^2}} \left (16+x-8 \log (x)+\log ^2(x)\right )} \left (-8+x+e^{2+x^2} \left (-32 x^2-2 x^3\right )+\left (2+16 e^{2+x^2} x^2\right ) \log (x)-2 e^{2+x^2} x^2 \log ^2(x)+e^{e^x} \left (e^x \left (16 x+x^2\right )-8 e^x x \log (x)+e^x x \log ^2(x)\right )\right )}{x} \, dx=\int { -\frac {{\left (2 \, x^{2} e^{\left (x^{2} + 2\right )} \log \left (x\right )^{2} + 2 \, {\left (x^{3} + 16 \, x^{2}\right )} e^{\left (x^{2} + 2\right )} - {\left (x e^{x} \log \left (x\right )^{2} - 8 \, x e^{x} \log \left (x\right ) + {\left (x^{2} + 16 \, x\right )} e^{x}\right )} e^{\left (e^{x}\right )} - 2 \, {\left (8 \, x^{2} e^{\left (x^{2} + 2\right )} + 1\right )} \log \left (x\right ) - x + 8\right )} e^{\left ({\left (\log \left (x\right )^{2} + x - 8 \, \log \left (x\right ) + 16\right )} e^{\left (-e^{\left (x^{2} + 2\right )} + e^{\left (e^{x}\right )}\right )} - e^{\left (x^{2} + 2\right )} + e^{\left (e^{x}\right )}\right )}}{x} \,d x } \]
integrate(((x*exp(x)*log(x)^2-8*x*exp(x)*log(x)+(x^2+16*x)*exp(x))*exp(exp (x))-2*x^2*exp(x^2+2)*log(x)^2+(16*x^2*exp(x^2+2)+2)*log(x)+(-2*x^3-32*x^2 )*exp(x^2+2)-8+x)*exp(exp(exp(x))-exp(x^2+2))*exp((log(x)^2-8*log(x)+x+16) *exp(exp(exp(x))-exp(x^2+2)))/x,x, algorithm=\
integrate(-(2*x^2*e^(x^2 + 2)*log(x)^2 + 2*(x^3 + 16*x^2)*e^(x^2 + 2) - (x *e^x*log(x)^2 - 8*x*e^x*log(x) + (x^2 + 16*x)*e^x)*e^(e^x) - 2*(8*x^2*e^(x ^2 + 2) + 1)*log(x) - x + 8)*e^((log(x)^2 + x - 8*log(x) + 16)*e^(-e^(x^2 + 2) + e^(e^x)) - e^(x^2 + 2) + e^(e^x))/x, x)
Time = 14.48 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.37 \[ \int \frac {e^{e^{e^x}-e^{2+x^2}+e^{e^{e^x}-e^{2+x^2}} \left (16+x-8 \log (x)+\log ^2(x)\right )} \left (-8+x+e^{2+x^2} \left (-32 x^2-2 x^3\right )+\left (2+16 e^{2+x^2} x^2\right ) \log (x)-2 e^{2+x^2} x^2 \log ^2(x)+e^{e^x} \left (e^x \left (16 x+x^2\right )-8 e^x x \log (x)+e^x x \log ^2(x)\right )\right )}{x} \, dx=\frac {{\mathrm {e}}^{{\mathrm {e}}^{-{\mathrm {e}}^{x^2}\,{\mathrm {e}}^2}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}\,{\ln \left (x\right )}^2}\,{\mathrm {e}}^{x\,{\mathrm {e}}^{-{\mathrm {e}}^{x^2}\,{\mathrm {e}}^2}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}}\,{\mathrm {e}}^{16\,{\mathrm {e}}^{-{\mathrm {e}}^{x^2}\,{\mathrm {e}}^2}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}}}{x^{8\,{\mathrm {e}}^{-{\mathrm {e}}^{x^2}\,{\mathrm {e}}^2}\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}}} \]
int((exp(exp(exp(x)) - exp(x^2 + 2))*exp(exp(exp(exp(x)) - exp(x^2 + 2))*( x - 8*log(x) + log(x)^2 + 16))*(x + log(x)*(16*x^2*exp(x^2 + 2) + 2) - exp (x^2 + 2)*(32*x^2 + 2*x^3) + exp(exp(x))*(exp(x)*(16*x + x^2) - 8*x*exp(x) *log(x) + x*exp(x)*log(x)^2) - 2*x^2*exp(x^2 + 2)*log(x)^2 - 8))/x,x)