Integrand size = 78, antiderivative size = 24 \[ \int \frac {e^{-\frac {e^{2 x}}{-25+5 x}} \left (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} \left (250 x-100 x^2+10 x^3\right )\right )}{125-50 x+5 x^2} \, dx=e^{-\frac {e^{2 x}}{5 (-5+x)}} (13+e)+x^2 \] Output:
(exp(1)+13)/exp(1/5*exp(x)^2/(-5+x))+x^2
Time = 1.45 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-\frac {e^{2 x}}{-25+5 x}} \left (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} \left (250 x-100 x^2+10 x^3\right )\right )}{125-50 x+5 x^2} \, dx=-25+e^{1+\frac {e^{2 x}}{25-5 x}}+13 e^{\frac {e^{2 x}}{25-5 x}}+x^2 \] Input:
Integrate[(E^(2*x)*(143 + E*(11 - 2*x) - 26*x) + E^(E^(2*x)/(-25 + 5*x))*( 250*x - 100*x^2 + 10*x^3))/(E^(E^(2*x)/(-25 + 5*x))*(125 - 50*x + 5*x^2)), x]
Output:
-25 + E^(1 + E^(2*x)/(25 - 5*x)) + 13*E^(E^(2*x)/(25 - 5*x)) + x^2
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.
\(\displaystyle \int \frac {e^{-\frac {e^{2 x}}{5 x-25}} \left (e^{\frac {e^{2 x}}{5 x-25}} \left (10 x^3-100 x^2+250 x\right )+e^{2 x} (e (11-2 x)-26 x+143)\right )}{5 x^2-50 x+125} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 20 \int \frac {e^{\frac {e^{2 x}}{5 (5-x)}} \left (e^{2 x} (e (11-2 x)-26 x+143)+10 e^{-\frac {e^{2 x}}{5 (5-x)}} \left (x^3-10 x^2+25 x\right )\right )}{100 (5-x)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {e^{\frac {e^{2 x}}{5 (5-x)}} \left (e^{2 x} (e (11-2 x)-26 x+143)+10 e^{-\frac {e^{2 x}}{5 (5-x)}} \left (x^3-10 x^2+25 x\right )\right )}{(5-x)^2}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \frac {1}{5} \int \frac {10 x (x-5)^2+\left (1+\frac {e}{13}\right ) e^{2 x+\frac {e^{2 x}}{25-5 x}} (143-26 x)}{(5-x)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (10 x-\frac {e^{2 x+\frac {e^{2 x}}{25-5 x}} (13+e) (2 x-11)}{(x-5)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left ((13+e) \int \frac {e^{2 x+\frac {e^{2 x}}{25-5 x}}}{(x-5)^2}dx-2 (13+e) \int \frac {e^{2 x+\frac {e^{2 x}}{25-5 x}}}{x-5}dx+5 x^2\right )\) |
Input:
Int[(E^(2*x)*(143 + E*(11 - 2*x) - 26*x) + E^(E^(2*x)/(-25 + 5*x))*(250*x - 100*x^2 + 10*x^3))/(E^(E^(2*x)/(-25 + 5*x))*(125 - 50*x + 5*x^2)),x]
Output:
$Aborted
Time = 0.56 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42
method | result | size |
risch | \({\mathrm e}^{-\frac {{\mathrm e}^{2 x}}{5 \left (-5+x \right )}} {\mathrm e}+x^{2}+13 \,{\mathrm e}^{-\frac {{\mathrm e}^{2 x}}{5 \left (-5+x \right )}}\) | \(34\) |
parts | \(x^{2}+\frac {\left (\left ({\mathrm e}+13\right ) x -65-5 \,{\mathrm e}\right ) {\mathrm e}^{-\frac {{\mathrm e}^{2 x}}{5 x -25}}}{-5+x}\) | \(38\) |
norman | \(\frac {\left (\left ({\mathrm e}+13\right ) x +{\mathrm e}^{\frac {{\mathrm e}^{2 x}}{5 x -25}} x^{3}-65-5 \,{\mathrm e}^{\frac {{\mathrm e}^{2 x}}{5 x -25}} x^{2}-5 \,{\mathrm e}\right ) {\mathrm e}^{-\frac {{\mathrm e}^{2 x}}{5 x -25}}}{-5+x}\) | \(69\) |
parallelrisch | \(\frac {\left (-325+5 \,{\mathrm e}^{\frac {{\mathrm e}^{2 x}}{5 x -25}} x^{3}-25 \,{\mathrm e}^{\frac {{\mathrm e}^{2 x}}{5 x -25}} x^{2}+5 x \,{\mathrm e}-750 \,{\mathrm e}^{\frac {{\mathrm e}^{2 x}}{5 x -25}} x -25 \,{\mathrm e}+65 x +3750 \,{\mathrm e}^{\frac {{\mathrm e}^{2 x}}{5 x -25}}\right ) {\mathrm e}^{-\frac {{\mathrm e}^{2 x}}{5 \left (-5+x \right )}}}{5 x -25}\) | \(99\) |
Input:
int(((10*x^3-100*x^2+250*x)*exp(exp(x)^2/(5*x-25))+((-2*x+11)*exp(1)-26*x+ 143)*exp(x)^2)/(5*x^2-50*x+125)/exp(exp(x)^2/(5*x-25)),x,method=_RETURNVER BOSE)
Output:
exp(-1/5*exp(2*x)/(-5+x))*exp(1)+x^2+13*exp(-1/5*exp(2*x)/(-5+x))
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-\frac {e^{2 x}}{-25+5 x}} \left (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} \left (250 x-100 x^2+10 x^3\right )\right )}{125-50 x+5 x^2} \, dx={\left (x^{2} e^{\left (\frac {e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}}\right )} + e + 13\right )} e^{\left (-\frac {e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}}\right )} \] Input:
integrate(((10*x^3-100*x^2+250*x)*exp(exp(x)^2/(5*x-25))+((-2*x+11)*exp(1) -26*x+143)*exp(x)^2)/(5*x^2-50*x+125)/exp(exp(x)^2/(5*x-25)),x, algorithm= "fricas")
Output:
(x^2*e^(1/5*e^(2*x)/(x - 5)) + e + 13)*e^(-1/5*e^(2*x)/(x - 5))
Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-\frac {e^{2 x}}{-25+5 x}} \left (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} \left (250 x-100 x^2+10 x^3\right )\right )}{125-50 x+5 x^2} \, dx=x^{2} + \left (e + 13\right ) e^{- \frac {e^{2 x}}{5 x - 25}} \] Input:
integrate(((10*x**3-100*x**2+250*x)*exp(exp(x)**2/(5*x-25))+((-2*x+11)*exp (1)-26*x+143)*exp(x)**2)/(5*x**2-50*x+125)/exp(exp(x)**2/(5*x-25)),x)
Output:
x**2 + (E + 13)*exp(-exp(2*x)/(5*x - 25))
Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {e^{-\frac {e^{2 x}}{-25+5 x}} \left (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} \left (250 x-100 x^2+10 x^3\right )\right )}{125-50 x+5 x^2} \, dx={\left (x^{2} e^{\left (\frac {e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}}\right )} + e + 13\right )} e^{\left (-\frac {e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}}\right )} \] Input:
integrate(((10*x^3-100*x^2+250*x)*exp(exp(x)^2/(5*x-25))+((-2*x+11)*exp(1) -26*x+143)*exp(x)^2)/(5*x^2-50*x+125)/exp(exp(x)^2/(5*x-25)),x, algorithm= "maxima")
Output:
(x^2*e^(1/5*e^(2*x)/(x - 5)) + e + 13)*e^(-1/5*e^(2*x)/(x - 5))
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (21) = 42\).
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.67 \[ \int \frac {e^{-\frac {e^{2 x}}{-25+5 x}} \left (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} \left (250 x-100 x^2+10 x^3\right )\right )}{125-50 x+5 x^2} \, dx={\left (x^{2} e^{\left (2 \, x\right )} + e^{\left (\frac {10 \, x^{2} - 50 \, x - e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}} + 1\right )} + 13 \, e^{\left (\frac {10 \, x^{2} - 50 \, x - e^{\left (2 \, x\right )}}{5 \, {\left (x - 5\right )}}\right )}\right )} e^{\left (-2 \, x\right )} \] Input:
integrate(((10*x^3-100*x^2+250*x)*exp(exp(x)^2/(5*x-25))+((-2*x+11)*exp(1) -26*x+143)*exp(x)^2)/(5*x^2-50*x+125)/exp(exp(x)^2/(5*x-25)),x, algorithm= "giac")
Output:
(x^2*e^(2*x) + e^(1/5*(10*x^2 - 50*x - e^(2*x))/(x - 5) + 1) + 13*e^(1/5*( 10*x^2 - 50*x - e^(2*x))/(x - 5)))*e^(-2*x)
Time = 2.50 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-\frac {e^{2 x}}{-25+5 x}} \left (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} \left (250 x-100 x^2+10 x^3\right )\right )}{125-50 x+5 x^2} \, dx={\mathrm {e}}^{-\frac {{\mathrm {e}}^{2\,x}}{5\,x-25}}\,\left (\mathrm {e}+13\right )+x^2 \] Input:
int((exp(-exp(2*x)/(5*x - 25))*(exp(exp(2*x)/(5*x - 25))*(250*x - 100*x^2 + 10*x^3) - exp(2*x)*(26*x + exp(1)*(2*x - 11) - 143)))/(5*x^2 - 50*x + 12 5),x)
Output:
exp(-exp(2*x)/(5*x - 25))*(exp(1) + 13) + x^2
Time = 0.22 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {e^{-\frac {e^{2 x}}{-25+5 x}} \left (e^{2 x} (143+e (11-2 x)-26 x)+e^{\frac {e^{2 x}}{-25+5 x}} \left (250 x-100 x^2+10 x^3\right )\right )}{125-50 x+5 x^2} \, dx=\frac {e^{\frac {e^{2 x}}{5 x -25}} x^{2}+e +13}{e^{\frac {e^{2 x}}{5 x -25}}} \] Input:
int(((10*x^3-100*x^2+250*x)*exp(exp(x)^2/(5*x-25))+((-2*x+11)*exp(1)-26*x+ 143)*exp(x)^2)/(5*x^2-50*x+125)/exp(exp(x)^2/(5*x-25)),x)
Output:
(e**(e**(2*x)/(5*x - 25))*x**2 + e + 13)/e**(e**(2*x)/(5*x - 25))