Integrand size = 96, antiderivative size = 32 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5 e^{\frac {e^x+\frac {1}{4} e^{-e^x x^2} x^2}{x}}}{x} \]
Time = 5.13 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5 e^{\frac {e^x}{x}+\frac {1}{4} e^{-e^x x^2} x}}{x} \]
Integrate[(E^(-(E^x*x^2) + (4*E^(x + E^x*x^2) + x^2)/(4*E^(E^x*x^2)*x))*(5 *x^2 + E^x*(-10*x^4 - 5*x^5) + E^(E^x*x^2)*(-20*x + E^x*(-20 + 20*x))))/(4 *x^3),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.
\(\displaystyle \int \frac {\left (5 x^2+e^{e^x x^2} \left (e^x (20 x-20)-20 x\right )+e^x \left (-5 x^5-10 x^4\right )\right ) \exp \left (\frac {e^{-e^x x^2} \left (x^2+4 e^{e^x x^2+x}\right )}{4 x}-e^x x^2\right )}{4 x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int \frac {5 \exp \left (\frac {e^{-e^x x^2} \left (x^2+4 e^{e^x x^2+x}\right )}{4 x}-e^x x^2\right ) \left (x^2-4 e^{e^x x^2} \left (e^x (1-x)+x\right )-e^x \left (x^5+2 x^4\right )\right )}{x^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{4} \int \frac {\exp \left (\frac {e^{-e^x x^2} \left (x^2+4 e^{e^x x^2+x}\right )}{4 x}-e^x x^2\right ) \left (x^2-4 e^{e^x x^2} \left (e^x (1-x)+x\right )-e^x \left (x^5+2 x^4\right )\right )}{x^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {5}{4} \int \left (\frac {4 \exp \left (\frac {e^{-e^x x^2} \left (x^2+4 e^{e^x x^2+x}\right )}{4 x}\right ) \left (e^x x-x-e^x\right )}{x^3}-\frac {\exp \left (\frac {e^{-e^x x^2} \left (x^2+4 e^{e^x x^2+x}\right )}{4 x}-e^x x^2\right ) \left (e^x x^3+2 e^x x^2-1\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5}{4} \left (-4 \int \frac {\exp \left (\frac {e^{-e^x x^2} \left (x^2+4 e^{e^x x^2+x}\right )}{4 x}\right )}{x^2}dx+4 \int \frac {\exp \left (x+\frac {e^{-e^x x^2} \left (x^2+4 e^{e^x x^2+x}\right )}{4 x}\right )}{x^2}dx+\int \frac {\exp \left (\frac {e^{-e^x x^2} \left (x^2+4 e^{e^x x^2+x}\right )}{4 x}-e^x x^2\right )}{x}dx-2 \int \exp \left (-e^x x^2+x+\frac {e^{-e^x x^2} \left (x^2+4 e^{e^x x^2+x}\right )}{4 x}\right ) xdx-\int \exp \left (-e^x x^2+x+\frac {e^{-e^x x^2} \left (x^2+4 e^{e^x x^2+x}\right )}{4 x}\right ) x^2dx-4 \int \frac {\exp \left (x+\frac {e^{-e^x x^2} \left (x^2+4 e^{e^x x^2+x}\right )}{4 x}\right )}{x^3}dx\right )\) |
Int[(E^(-(E^x*x^2) + (4*E^(x + E^x*x^2) + x^2)/(4*E^(E^x*x^2)*x))*(5*x^2 + E^x*(-10*x^4 - 5*x^5) + E^(E^x*x^2)*(-20*x + E^x*(-20 + 20*x))))/(4*x^3), x]
3.29.52.3.1 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 = 1.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09
method | result | size |
risch | \(\frac {5 \,{\mathrm e}^{\frac {\left (4 \,{\mathrm e}^{x \left ({\mathrm e}^{x} x +1\right )}+x^{2}\right ) {\mathrm e}^{-{\mathrm e}^{x} x^{2}}}{4 x}}}{x}\) | \(35\) |
parallelrisch | \(\frac {5 \,{\mathrm e}^{\frac {\left (4 \,{\mathrm e}^{x} {\mathrm e}^{{\mathrm e}^{x} x^{2}}+x^{2}\right ) {\mathrm e}^{-{\mathrm e}^{x} x^{2}}}{4 x}}}{x}\) | \(36\) |
int(1/4*(((20*x-20)*exp(x)-20*x)*exp(exp(x)*x^2)+(-5*x^5-10*x^4)*exp(x)+5* x^2)*exp(1/4*(4*exp(x)*exp(exp(x)*x^2)+x^2)/x/exp(exp(x)*x^2))/x^3/exp(exp (x)*x^2),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.72 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5 \, e^{\left (x^{2} e^{x} + \frac {{\left (x^{2} e^{x} - 4 \, {\left (x^{3} - 1\right )} e^{\left (x^{2} e^{x} + 2 \, x\right )}\right )} e^{\left (-x^{2} e^{x} - x\right )}}{4 \, x}\right )}}{x} \]
integrate(1/4*(((20*x-20)*exp(x)-20*x)*exp(exp(x)*x^2)+(-5*x^5-10*x^4)*exp (x)+5*x^2)*exp(1/4*(4*exp(x)*exp(exp(x)*x^2)+x^2)/x/exp(exp(x)*x^2))/x^3/e xp(exp(x)*x^2),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5 e^{\frac {\left (\frac {x^{2}}{4} + e^{x} e^{x^{2} e^{x}}\right ) e^{- x^{2} e^{x}}}{x}}}{x} \]
integrate(1/4*(((20*x-20)*exp(x)-20*x)*exp(exp(x)*x**2)+(-5*x**5-10*x**4)* exp(x)+5*x**2)*exp(1/4*(4*exp(x)*exp(exp(x)*x**2)+x**2)/x/exp(exp(x)*x**2) )/x**3/exp(exp(x)*x**2),x)
Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5 \, e^{\left (\frac {1}{4} \, x e^{\left (-x^{2} e^{x}\right )} + \frac {e^{x}}{x}\right )}}{x} \]
integrate(1/4*(((20*x-20)*exp(x)-20*x)*exp(exp(x)*x^2)+(-5*x^5-10*x^4)*exp (x)+5*x^2)*exp(1/4*(4*exp(x)*exp(exp(x)*x^2)+x^2)/x/exp(exp(x)*x^2))/x^3/e xp(exp(x)*x^2),x, algorithm=\
\[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\int { \frac {5 \, {\left (x^{2} + 4 \, {\left ({\left (x - 1\right )} e^{x} - x\right )} e^{\left (x^{2} e^{x}\right )} - {\left (x^{5} + 2 \, x^{4}\right )} e^{x}\right )} e^{\left (-x^{2} e^{x} + \frac {{\left (x^{2} + 4 \, e^{\left (x^{2} e^{x} + x\right )}\right )} e^{\left (-x^{2} e^{x}\right )}}{4 \, x}\right )}}{4 \, x^{3}} \,d x } \]
integrate(1/4*(((20*x-20)*exp(x)-20*x)*exp(exp(x)*x^2)+(-5*x^5-10*x^4)*exp (x)+5*x^2)*exp(1/4*(4*exp(x)*exp(exp(x)*x^2)+x^2)/x/exp(exp(x)*x^2))/x^3/e xp(exp(x)*x^2),x, algorithm=\
integrate(5/4*(x^2 + 4*((x - 1)*e^x - x)*e^(x^2*e^x) - (x^5 + 2*x^4)*e^x)* e^(-x^2*e^x + 1/4*(x^2 + 4*e^(x^2*e^x + x))*e^(-x^2*e^x)/x)/x^3, x)
Time = 12.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-e^x x^2+\frac {e^{-e^x x^2} \left (4 e^{x+e^x x^2}+x^2\right )}{4 x}} \left (5 x^2+e^x \left (-10 x^4-5 x^5\right )+e^{e^x x^2} \left (-20 x+e^x (-20+20 x)\right )\right )}{4 x^3} \, dx=\frac {5\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{x}+\frac {x\,{\mathrm {e}}^{-x^2\,{\mathrm {e}}^x}}{4}}}{x} \]