Integrand size = 81, antiderivative size = 22 \begin {dmath*} \int e^{e^{-\frac {e^{25+e^x+x} (-4+4 x)}{x}} x-\frac {e^{25+e^x+x} (-4+4 x)}{x}} \left (1+\frac {e^{25+e^x+x} \left (-4+4 x-4 x^2+e^x \left (4 x-4 x^2\right )\right )}{x}\right ) \, dx=e^{e^{-\frac {4 e^{25+e^x+x} (-1+x)}{x}} x} \end {dmath*}
Time = 5.24 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \begin {dmath*} \int e^{e^{-\frac {e^{25+e^x+x} (-4+4 x)}{x}} x-\frac {e^{25+e^x+x} (-4+4 x)}{x}} \left (1+\frac {e^{25+e^x+x} \left (-4+4 x-4 x^2+e^x \left (4 x-4 x^2\right )\right )}{x}\right ) \, dx=e^{e^{-\frac {4 e^{25+e^x+x} (-1+x)}{x}} x} \end {dmath*}
Integrate[E^(x/E^((E^(25 + E^x + x)*(-4 + 4*x))/x) - (E^(25 + E^x + x)*(-4 + 4*x))/x)*(1 + (E^(25 + E^x + x)*(-4 + 4*x - 4*x^2 + E^x*(4*x - 4*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.
\(\displaystyle \int \left (\frac {e^{x+e^x+25} \left (-4 x^2+e^x \left (4 x-4 x^2\right )+4 x-4\right )}{x}+1\right ) \exp \left (e^{-\frac {e^{x+e^x+25} (4 x-4)}{x}} x-\frac {e^{x+e^x+25} (4 x-4)}{x}\right ) \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\exp \left (e^{-\frac {e^{x+e^x+25} (4 x-4)}{x}} x-\frac {e^{x+e^x+25} (4 x-4)}{x}\right )-\frac {4 \left (e^x x^2+x^2-e^x x-x+1\right ) \exp \left (e^{-\frac {e^{x+e^x+25} (4 x-4)}{x}} x+x+e^x-\frac {e^{x+e^x+25} (4 x-4)}{x}+25\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \exp \left (e^{-\frac {e^{x+e^x+25} (4 x-4)}{x}} x-\frac {e^{x+e^x+25} (4 x-4)}{x}\right )dx+4 \int \exp \left (e^{-\frac {e^{x+e^x+25} (4 x-4)}{x}} x+x+e^x+25-\frac {e^{x+e^x+25} (4 x-4)}{x}\right )dx+4 \int \exp \left (e^{-\frac {e^{x+e^x+25} (4 x-4)}{x}} x+2 x+e^x+25-\frac {e^{x+e^x+25} (4 x-4)}{x}\right )dx-4 \int \frac {\exp \left (e^{-\frac {e^{x+e^x+25} (4 x-4)}{x}} x+x+e^x+25-\frac {e^{x+e^x+25} (4 x-4)}{x}\right )}{x}dx-4 \int \exp \left (e^{-\frac {e^{x+e^x+25} (4 x-4)}{x}} x+x+e^x+25-\frac {e^{x+e^x+25} (4 x-4)}{x}\right ) xdx-4 \int \exp \left (e^{-\frac {e^{x+e^x+25} (4 x-4)}{x}} x+2 x+e^x+25-\frac {e^{x+e^x+25} (4 x-4)}{x}\right ) xdx\) |
Int[E^(x/E^((E^(25 + E^x + x)*(-4 + 4*x))/x) - (E^(25 + E^x + x)*(-4 + 4*x ))/x)*(1 + (E^(25 + E^x + x)*(-4 + 4*x - 4*x^2 + E^x*(4*x - 4*x^2)))/x),x]
3.6.96.3.1 Defintions of rubi rules used
Time = 5.31 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86
method | result | size |
risch | \({\mathrm e}^{x \,{\mathrm e}^{-\frac {4 \left (-1+x \right ) {\mathrm e}^{25+{\mathrm e}^{x}+x}}{x}}}\) | \(19\) |
parallelrisch | \({\mathrm e}^{x \,{\mathrm e}^{-\frac {\left (-4+4 x \right ) {\mathrm e}^{25+{\mathrm e}^{x}+x}}{x}}}\) | \(25\) |
int((((-4*x^2+4*x)*exp(x)-4*x^2+4*x-4)*exp(ln(exp(25)/x)+exp(x)+x)+1)*exp( x/exp((-4+4*x)*exp(ln(exp(25)/x)+exp(x)+x)))/exp((-4+4*x)*exp(ln(exp(25)/x )+exp(x)+x)),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \begin {dmath*} \int e^{e^{-\frac {e^{25+e^x+x} (-4+4 x)}{x}} x-\frac {e^{25+e^x+x} (-4+4 x)}{x}} \left (1+\frac {e^{25+e^x+x} \left (-4+4 x-4 x^2+e^x \left (4 x-4 x^2\right )\right )}{x}\right ) \, dx=e^{\left (x e^{\left (-4 \, {\left (x - 1\right )} e^{\left (x + e^{x} + \log \left (\frac {e^{25}}{x}\right )\right )}\right )}\right )} \end {dmath*}
integrate((((-4*x^2+4*x)*exp(x)-4*x^2+4*x-4)*exp(log(exp(25)/x)+exp(x)+x)+ 1)*exp(x/exp((-4+4*x)*exp(log(exp(25)/x)+exp(x)+x)))/exp((-4+4*x)*exp(log( exp(25)/x)+exp(x)+x)),x, algorithm=\
Time = 24.51 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \begin {dmath*} \int e^{e^{-\frac {e^{25+e^x+x} (-4+4 x)}{x}} x-\frac {e^{25+e^x+x} (-4+4 x)}{x}} \left (1+\frac {e^{25+e^x+x} \left (-4+4 x-4 x^2+e^x \left (4 x-4 x^2\right )\right )}{x}\right ) \, dx=e^{x e^{- \frac {\left (4 x - 4\right ) e^{25} e^{x + e^{x}}}{x}}} \end {dmath*}
integrate((((-4*x**2+4*x)*exp(x)-4*x**2+4*x-4)*exp(ln(exp(25)/x)+exp(x)+x) +1)*exp(x/exp((-4+4*x)*exp(ln(exp(25)/x)+exp(x)+x)))/exp((-4+4*x)*exp(ln(e xp(25)/x)+exp(x)+x)),x)
Time = 0.55 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \begin {dmath*} \int e^{e^{-\frac {e^{25+e^x+x} (-4+4 x)}{x}} x-\frac {e^{25+e^x+x} (-4+4 x)}{x}} \left (1+\frac {e^{25+e^x+x} \left (-4+4 x-4 x^2+e^x \left (4 x-4 x^2\right )\right )}{x}\right ) \, dx=e^{\left (x e^{\left (\frac {4 \, e^{\left (x + e^{x} + 25\right )}}{x} - 4 \, e^{\left (x + e^{x} + 25\right )}\right )}\right )} \end {dmath*}
integrate((((-4*x^2+4*x)*exp(x)-4*x^2+4*x-4)*exp(log(exp(25)/x)+exp(x)+x)+ 1)*exp(x/exp((-4+4*x)*exp(log(exp(25)/x)+exp(x)+x)))/exp((-4+4*x)*exp(log( exp(25)/x)+exp(x)+x)),x, algorithm=\
\begin {dmath*} \int e^{e^{-\frac {e^{25+e^x+x} (-4+4 x)}{x}} x-\frac {e^{25+e^x+x} (-4+4 x)}{x}} \left (1+\frac {e^{25+e^x+x} \left (-4+4 x-4 x^2+e^x \left (4 x-4 x^2\right )\right )}{x}\right ) \, dx=\int { -{\left (4 \, {\left (x^{2} + {\left (x^{2} - x\right )} e^{x} - x + 1\right )} e^{\left (x + e^{x} + \log \left (\frac {e^{25}}{x}\right )\right )} - 1\right )} e^{\left (x e^{\left (-4 \, {\left (x - 1\right )} e^{\left (x + e^{x} + \log \left (\frac {e^{25}}{x}\right )\right )}\right )} - 4 \, {\left (x - 1\right )} e^{\left (x + e^{x} + \log \left (\frac {e^{25}}{x}\right )\right )}\right )} \,d x } \end {dmath*}
integrate((((-4*x^2+4*x)*exp(x)-4*x^2+4*x-4)*exp(log(exp(25)/x)+exp(x)+x)+ 1)*exp(x/exp((-4+4*x)*exp(log(exp(25)/x)+exp(x)+x)))/exp((-4+4*x)*exp(log( exp(25)/x)+exp(x)+x)),x, algorithm=\
integrate(-(4*(x^2 + (x^2 - x)*e^x - x + 1)*e^(x + e^x + log(e^25/x)) - 1) *e^(x*e^(-4*(x - 1)*e^(x + e^x + log(e^25/x))) - 4*(x - 1)*e^(x + e^x + lo g(e^25/x))), x)
Time = 14.39 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \begin {dmath*} \int e^{e^{-\frac {e^{25+e^x+x} (-4+4 x)}{x}} x-\frac {e^{25+e^x+x} (-4+4 x)}{x}} \left (1+\frac {e^{25+e^x+x} \left (-4+4 x-4 x^2+e^x \left (4 x-4 x^2\right )\right )}{x}\right ) \, dx={\mathrm {e}}^{x\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{25}\,{\mathrm {e}}^x}{x}}\,{\mathrm {e}}^{-4\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^{25}\,{\mathrm {e}}^x}} \end {dmath*}