Integrand size = 95, antiderivative size = 25 \[ \int \frac {e^{e^{\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}}+\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}} \left (-2 x^2-2 x^3+3 x^4+\left (2-2 x^2\right ) \log (x)-\log ^2(x)\right )}{x^2} \, dx=e^{e^{11-x \left (x-\left (-x+\frac {\log (x)}{x}\right )^2\right )}} \]
\[ \int \frac {e^{e^{\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}}+\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}} \left (-2 x^2-2 x^3+3 x^4+\left (2-2 x^2\right ) \log (x)-\log ^2(x)\right )}{x^2} \, dx=\int \frac {e^{e^{\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}}+\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}} \left (-2 x^2-2 x^3+3 x^4+\left (2-2 x^2\right ) \log (x)-\log ^2(x)\right )}{x^2} \, dx \]
Integrate[(E^(E^((11*x - x^3 + x^4 - 2*x^2*Log[x] + Log[x]^2)/x) + (11*x - x^3 + x^4 - 2*x^2*Log[x] + Log[x]^2)/x)*(-2*x^2 - 2*x^3 + 3*x^4 + (2 - 2* x^2)*Log[x] - Log[x]^2))/x^2,x]
Integrate[(E^(E^((11*x - x^3 + x^4 - 2*x^2*Log[x] + Log[x]^2)/x) + (11*x - x^3 + x^4 - 2*x^2*Log[x] + Log[x]^2)/x)*(-2*x^2 - 2*x^3 + 3*x^4 + (2 - 2* x^2)*Log[x] - Log[x]^2))/x^2, 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.
\(\displaystyle \int \frac {\left (3 x^4-2 x^3-2 x^2+\left (2-2 x^2\right ) \log (x)-\log ^2(x)\right ) \exp \left (\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}}\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (3 x^2 \exp \left (\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}}\right )-2 x \exp \left (\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}}\right )-2 \exp \left (\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}}\right )-\frac {\log ^2(x) \exp \left (\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}}\right )}{x^2}-\frac {2 \left (x^2-1\right ) \log (x) \exp \left (\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 x^2 \log (x)+11 x+\log ^2(x)}{x}}\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \int \exp \left (\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}}\right )dx-2 \int \exp \left (\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}}\right ) xdx+3 \int \exp \left (\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}}\right ) x^2dx-2 \int \exp \left (\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}}\right ) \log (x)dx+2 \int \frac {\exp \left (\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}}\right ) \log (x)}{x^2}dx-\int \frac {\exp \left (\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}+e^{\frac {x^4-x^3-2 \log (x) x^2+11 x+\log ^2(x)}{x}}\right ) \log ^2(x)}{x^2}dx\) |
Int[(E^(E^((11*x - x^3 + x^4 - 2*x^2*Log[x] + Log[x]^2)/x) + (11*x - x^3 + x^4 - 2*x^2*Log[x] + Log[x]^2)/x)*(-2*x^2 - 2*x^3 + 3*x^4 + (2 - 2*x^2)*L og[x] - Log[x]^2))/x^2,x]
3.10.92.3.1 Defintions of rubi rules used
Time = 0.38 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16
method | result | size |
risch | \({\mathrm e}^{x^{-2 x} {\mathrm e}^{\frac {x^{4}-x^{3}+\ln \left (x \right )^{2}+11 x}{x}}}\) | \(29\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\frac {\ln \left (x \right )^{2}-2 x^{2} \ln \left (x \right )+x^{4}-x^{3}+11 x}{x}}}\) | \(30\) |
int((-ln(x)^2+(-2*x^2+2)*ln(x)+3*x^4-2*x^3-2*x^2)*exp((ln(x)^2-2*x^2*ln(x) +x^4-x^3+11*x)/x)*exp(exp((ln(x)^2-2*x^2*ln(x)+x^4-x^3+11*x)/x))/x^2,x,met hod=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (21) = 42\).
Time = 0.25 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.48 \[ \int \frac {e^{e^{\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}}+\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}} \left (-2 x^2-2 x^3+3 x^4+\left (2-2 x^2\right ) \log (x)-\log ^2(x)\right )}{x^2} \, dx=e^{\left (\frac {x^{4} - x^{3} - 2 \, x^{2} \log \left (x\right ) + x e^{\left (\frac {x^{4} - x^{3} - 2 \, x^{2} \log \left (x\right ) + \log \left (x\right )^{2} + 11 \, x}{x}\right )} + \log \left (x\right )^{2} + 11 \, x}{x} - \frac {x^{4} - x^{3} - 2 \, x^{2} \log \left (x\right ) + \log \left (x\right )^{2} + 11 \, x}{x}\right )} \]
integrate((-log(x)^2+(-2*x^2+2)*log(x)+3*x^4-2*x^3-2*x^2)*exp((log(x)^2-2* x^2*log(x)+x^4-x^3+11*x)/x)*exp(exp((log(x)^2-2*x^2*log(x)+x^4-x^3+11*x)/x ))/x^2,x, algorithm=\
e^((x^4 - x^3 - 2*x^2*log(x) + x*e^((x^4 - x^3 - 2*x^2*log(x) + log(x)^2 + 11*x)/x) + log(x)^2 + 11*x)/x - (x^4 - x^3 - 2*x^2*log(x) + log(x)^2 + 11 *x)/x)
Time = 0.52 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {e^{e^{\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}}+\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}} \left (-2 x^2-2 x^3+3 x^4+\left (2-2 x^2\right ) \log (x)-\log ^2(x)\right )}{x^2} \, dx=e^{e^{\frac {x^{4} - x^{3} - 2 x^{2} \log {\left (x \right )} + 11 x + \log {\left (x \right )}^{2}}{x}}} \]
integrate((-ln(x)**2+(-2*x**2+2)*ln(x)+3*x**4-2*x**3-2*x**2)*exp((ln(x)**2 -2*x**2*ln(x)+x**4-x**3+11*x)/x)*exp(exp((ln(x)**2-2*x**2*ln(x)+x**4-x**3+ 11*x)/x))/x**2,x)
Time = 0.47 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}}+\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}} \left (-2 x^2-2 x^3+3 x^4+\left (2-2 x^2\right ) \log (x)-\log ^2(x)\right )}{x^2} \, dx=e^{\left (e^{\left (x^{3} - x^{2} - 2 \, x \log \left (x\right ) + \frac {\log \left (x\right )^{2}}{x} + 11\right )}\right )} \]
integrate((-log(x)^2+(-2*x^2+2)*log(x)+3*x^4-2*x^3-2*x^2)*exp((log(x)^2-2* x^2*log(x)+x^4-x^3+11*x)/x)*exp(exp((log(x)^2-2*x^2*log(x)+x^4-x^3+11*x)/x ))/x^2,x, algorithm=\
\[ \int \frac {e^{e^{\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}}+\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}} \left (-2 x^2-2 x^3+3 x^4+\left (2-2 x^2\right ) \log (x)-\log ^2(x)\right )}{x^2} \, dx=\int { \frac {{\left (3 \, x^{4} - 2 \, x^{3} - 2 \, x^{2} - 2 \, {\left (x^{2} - 1\right )} \log \left (x\right ) - \log \left (x\right )^{2}\right )} e^{\left (\frac {x^{4} - x^{3} - 2 \, x^{2} \log \left (x\right ) + \log \left (x\right )^{2} + 11 \, x}{x} + e^{\left (\frac {x^{4} - x^{3} - 2 \, x^{2} \log \left (x\right ) + \log \left (x\right )^{2} + 11 \, x}{x}\right )}\right )}}{x^{2}} \,d x } \]
integrate((-log(x)^2+(-2*x^2+2)*log(x)+3*x^4-2*x^3-2*x^2)*exp((log(x)^2-2* x^2*log(x)+x^4-x^3+11*x)/x)*exp(exp((log(x)^2-2*x^2*log(x)+x^4-x^3+11*x)/x ))/x^2,x, algorithm=\
integrate((3*x^4 - 2*x^3 - 2*x^2 - 2*(x^2 - 1)*log(x) - log(x)^2)*e^((x^4 - x^3 - 2*x^2*log(x) + log(x)^2 + 11*x)/x + e^((x^4 - x^3 - 2*x^2*log(x) + log(x)^2 + 11*x)/x))/x^2, x)
Time = 9.34 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{e^{\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}}+\frac {11 x-x^3+x^4-2 x^2 \log (x)+\log ^2(x)}{x}} \left (-2 x^2-2 x^3+3 x^4+\left (2-2 x^2\right ) \log (x)-\log ^2(x)\right )}{x^2} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{11}\,{\mathrm {e}}^{-x^2}\,{\mathrm {e}}^{\frac {{\ln \left (x\right )}^2}{x}}}{x^{2\,x}}} \]