Integrand size = 68, antiderivative size = 26 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=e^{\frac {25}{9} \left (-8+\frac {1}{3} e^{\frac {\left (2+e^x\right ) (1+x)}{x}}\right )} \]
Time = 3.38 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=e^{\frac {25}{27} \left (-24+e^{\frac {\left (2+e^x\right ) (1+x)}{x}}\right )} \]
Integrate[(E^((-600 + 25*E^((2 + 2*x + E^x*(1 + x))/x))/27 + (2 + 2*x + E^ x*(1 + x))/x)*(-50 + E^x*(-25 + 25*x + 25*x^2)))/(27*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.
\(\displaystyle \int \frac {\left (e^x \left (25 x^2+25 x-25\right )-50\right ) \exp \left (\frac {1}{27} \left (25 e^{\frac {2 x+e^x (x+1)+2}{x}}-600\right )+\frac {2 x+e^x (x+1)+2}{x}\right )}{27 x^2} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{27} \int -\frac {25 \exp \left (\frac {2 x+e^x (x+1)+2}{x}-\frac {25}{27} \left (24-e^{\frac {2 x+e^x (x+1)+2}{x}}\right )\right ) \left (e^x \left (-x^2-x+1\right )+2\right )}{x^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {25}{27} \int \frac {\exp \left (\frac {2 x+e^x (x+1)+2}{x}-\frac {25}{27} \left (24-e^{\frac {2 x+e^x (x+1)+2}{x}}\right )\right ) \left (e^x \left (-x^2-x+1\right )+2\right )}{x^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle -\frac {25}{27} \int \frac {\exp \left (\frac {27 e^x x+25 e^{\frac {\left (2+e^x\right ) (x+1)}{x}} x-546 x+27 e^x+54}{27 x}\right ) \left (e^x \left (-x^2-x+1\right )+2\right )}{x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {25}{27} \int \left (\frac {2 \exp \left (\frac {27 e^x x+25 e^{\frac {\left (2+e^x\right ) (x+1)}{x}} x-546 x+27 e^x+54}{27 x}\right )}{x^2}-\frac {\exp \left (x+\frac {27 e^x x+25 e^{\frac {\left (2+e^x\right ) (x+1)}{x}} x-546 x+27 e^x+54}{27 x}\right ) \left (x^2+x-1\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {25}{27} \left (2 \int \frac {\exp \left (\frac {27 e^x x+25 e^{\frac {\left (2+e^x\right ) (x+1)}{x}} x-546 x+27 e^x+54}{27 x}\right )}{x^2}dx+\int \frac {\exp \left (x+\frac {27 e^x x+25 e^{\frac {\left (2+e^x\right ) (x+1)}{x}} x-546 x+27 e^x+54}{27 x}\right )}{x^2}dx-\int \exp \left (x+\frac {27 e^x x+25 e^{\frac {\left (2+e^x\right ) (x+1)}{x}} x-546 x+27 e^x+54}{27 x}\right )dx-\int \frac {\exp \left (x+\frac {27 e^x x+25 e^{\frac {\left (2+e^x\right ) (x+1)}{x}} x-546 x+27 e^x+54}{27 x}\right )}{x}dx\right )\) |
Int[(E^((-600 + 25*E^((2 + 2*x + E^x*(1 + x))/x))/27 + (2 + 2*x + E^x*(1 + x))/x)*(-50 + E^x*(-25 + 25*x + 25*x^2)))/(27*x^2),x]
3.18.6.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.64 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69
method | result | size |
risch | \({\mathrm e}^{\frac {25 \,{\mathrm e}^{\frac {\left ({\mathrm e}^{x}+2\right ) \left (1+x \right )}{x}}}{27}-\frac {200}{9}}\) | \(18\) |
norman | \({\mathrm e}^{\frac {25 \,{\mathrm e}^{\frac {\left (1+x \right ) {\mathrm e}^{x}+2 x +2}{x}}}{27}-\frac {200}{9}}\) | \(22\) |
parallelrisch | \({\mathrm e}^{\frac {25 \,{\mathrm e}^{\frac {\left (1+x \right ) {\mathrm e}^{x}+2 x +2}{x}}}{27}-\frac {200}{9}}\) | \(22\) |
int(1/27*((25*x^2+25*x-25)*exp(x)-50)*exp(((1+x)*exp(x)+2*x+2)/x)*exp(25/2 7*exp(((1+x)*exp(x)+2*x+2)/x)-200/9)/x^2,x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (17) = 34\).
Time = 0.34 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.08 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=e^{\left (\frac {27 \, {\left (x + 1\right )} e^{x} + 25 \, x e^{\left (\frac {{\left (x + 1\right )} e^{x} + 2 \, x + 2}{x}\right )} - 546 \, x + 54}{27 \, x} - \frac {{\left (x + 1\right )} e^{x} + 2 \, x + 2}{x}\right )} \]
integrate(1/27*((25*x^2+25*x-25)*exp(x)-50)*exp(((1+x)*exp(x)+2*x+2)/x)*ex p(25/27*exp(((1+x)*exp(x)+2*x+2)/x)-200/9)/x^2,x, algorithm=\
e^(1/27*(27*(x + 1)*e^x + 25*x*e^(((x + 1)*e^x + 2*x + 2)/x) - 546*x + 54) /x - ((x + 1)*e^x + 2*x + 2)/x)
Time = 0.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=e^{\frac {25 e^{\frac {2 x + \left (x + 1\right ) e^{x} + 2}{x}}}{27} - \frac {200}{9}} \]
integrate(1/27*((25*x**2+25*x-25)*exp(x)-50)*exp(((1+x)*exp(x)+2*x+2)/x)*e xp(25/27*exp(((1+x)*exp(x)+2*x+2)/x)-200/9)/x**2,x)
Time = 0.33 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=e^{\left (\frac {25}{27} \, e^{\left (\frac {e^{x}}{x} + \frac {2}{x} + e^{x} + 2\right )} - \frac {200}{9}\right )} \]
integrate(1/27*((25*x^2+25*x-25)*exp(x)-50)*exp(((1+x)*exp(x)+2*x+2)/x)*ex p(25/27*exp(((1+x)*exp(x)+2*x+2)/x)-200/9)/x^2,x, algorithm=\
\[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx=\int { \frac {25 \, {\left ({\left (x^{2} + x - 1\right )} e^{x} - 2\right )} e^{\left (\frac {{\left (x + 1\right )} e^{x} + 2 \, x + 2}{x} + \frac {25}{27} \, e^{\left (\frac {{\left (x + 1\right )} e^{x} + 2 \, x + 2}{x}\right )} - \frac {200}{9}\right )}}{27 \, x^{2}} \,d x } \]
integrate(1/27*((25*x^2+25*x-25)*exp(x)-50)*exp(((1+x)*exp(x)+2*x+2)/x)*ex p(25/27*exp(((1+x)*exp(x)+2*x+2)/x)-200/9)/x^2,x, algorithm=\
integrate(25/27*((x^2 + x - 1)*e^x - 2)*e^(((x + 1)*e^x + 2*x + 2)/x + 25/ 27*e^(((x + 1)*e^x + 2*x + 2)/x) - 200/9)/x^2, x)
Time = 13.40 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {1}{27} \left (-600+25 e^{\frac {2+2 x+e^x (1+x)}{x}}\right )+\frac {2+2 x+e^x (1+x)}{x}} \left (-50+e^x \left (-25+25 x+25 x^2\right )\right )}{27 x^2} \, dx={\mathrm {e}}^{-\frac {200}{9}}\,{\mathrm {e}}^{\frac {25\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x}}\,{\mathrm {e}}^{2/x}}{27}} \]