Integrand size = 82, antiderivative size = 25 \[ \int \frac {1}{5} \left (-2 x+30 x^2+e^x \left (-10 x-5 x^2\right )+e^{e^x x^2} \left (-1+20 x+e^{2 x} \left (-10 x^2-5 x^3\right )+e^x \left (-5-5 x-2 x^2+19 x^3+10 x^4\right )\right )\right ) \, dx=x \left (e^{e^x x^2}+x\right ) \left (-\frac {1}{5}-e^x+2 x\right ) \]
Time = 5.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {1}{5} \left (-2 x+30 x^2+e^x \left (-10 x-5 x^2\right )+e^{e^x x^2} \left (-1+20 x+e^{2 x} \left (-10 x^2-5 x^3\right )+e^x \left (-5-5 x-2 x^2+19 x^3+10 x^4\right )\right )\right ) \, dx=\frac {1}{5} x \left (e^{e^x x^2}+x\right ) \left (-1-5 e^x+10 x\right ) \]
Integrate[(-2*x + 30*x^2 + E^x*(-10*x - 5*x^2) + E^(E^x*x^2)*(-1 + 20*x + E^(2*x)*(-10*x^2 - 5*x^3) + E^x*(-5 - 5*x - 2*x^2 + 19*x^3 + 10*x^4)))/5,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.
\(\displaystyle \int \frac {1}{5} \left (30 x^2+e^x \left (-5 x^2-10 x\right )+e^{e^x x^2} \left (e^{2 x} \left (-5 x^3-10 x^2\right )+e^x \left (10 x^4+19 x^3-2 x^2-5 x-5\right )+20 x-1\right )-2 x\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \left (30 x^2-2 x-5 e^x \left (x^2+2 x\right )-e^{e^x x^2} \left (-20 x+5 e^{2 x} \left (x^3+2 x^2\right )+e^x \left (-10 x^4-19 x^3+2 x^2+5 x+5\right )+1\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (-\int e^{e^x x^2}dx-5 \int e^{e^x x^2+x}dx+20 \int e^{e^x x^2} xdx-5 \int e^{e^x x^2+x} xdx-2 \int e^{e^x x^2+x} x^2dx-10 \int e^{e^x x^2+2 x} x^2dx+10 \int e^{e^x x^2+x} x^4dx+19 \int e^{e^x x^2+x} x^3dx-5 \int e^{e^x x^2+2 x} x^3dx+10 x^3-5 e^x x^2-x^2\right )\) |
Int[(-2*x + 30*x^2 + E^x*(-10*x - 5*x^2) + E^(E^x*x^2)*(-1 + 20*x + E^(2*x )*(-10*x^2 - 5*x^3) + E^x*(-5 - 5*x - 2*x^2 + 19*x^3 + 10*x^4)))/5,x]
3.11.93.3.1 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 = 0.10 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52
method | result | size |
risch | \(\frac {\left (10 x -5 \,{\mathrm e}^{x}-1\right ) x \,{\mathrm e}^{{\mathrm e}^{x} x^{2}}}{5}-{\mathrm e}^{x} x^{2}+2 x^{3}-\frac {x^{2}}{5}\) | \(38\) |
default | \(2 x^{2} {\mathrm e}^{{\mathrm e}^{x} x^{2}}-\frac {{\mathrm e}^{{\mathrm e}^{x} x^{2}} x}{5}-{\mathrm e}^{{\mathrm e}^{x} x^{2}} {\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}-\frac {x^{2}}{5}+2 x^{3}\) | \(53\) |
norman | \(2 x^{2} {\mathrm e}^{{\mathrm e}^{x} x^{2}}-\frac {{\mathrm e}^{{\mathrm e}^{x} x^{2}} x}{5}-{\mathrm e}^{{\mathrm e}^{x} x^{2}} {\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}-\frac {x^{2}}{5}+2 x^{3}\) | \(53\) |
parallelrisch | \(2 x^{2} {\mathrm e}^{{\mathrm e}^{x} x^{2}}-\frac {{\mathrm e}^{{\mathrm e}^{x} x^{2}} x}{5}-{\mathrm e}^{{\mathrm e}^{x} x^{2}} {\mathrm e}^{x} x -{\mathrm e}^{x} x^{2}-\frac {x^{2}}{5}+2 x^{3}\) | \(53\) |
int(1/5*((-5*x^3-10*x^2)*exp(x)^2+(10*x^4+19*x^3-2*x^2-5*x-5)*exp(x)+20*x- 1)*exp(exp(x)*x^2)+1/5*(-5*x^2-10*x)*exp(x)+6*x^2-2/5*x,x,method=_RETURNVE RBOSE)
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {1}{5} \left (-2 x+30 x^2+e^x \left (-10 x-5 x^2\right )+e^{e^x x^2} \left (-1+20 x+e^{2 x} \left (-10 x^2-5 x^3\right )+e^x \left (-5-5 x-2 x^2+19 x^3+10 x^4\right )\right )\right ) \, dx=2 \, x^{3} - x^{2} e^{x} - \frac {1}{5} \, x^{2} + \frac {1}{5} \, {\left (10 \, x^{2} - 5 \, x e^{x} - x\right )} e^{\left (x^{2} e^{x}\right )} \]
integrate(1/5*((-5*x^3-10*x^2)*exp(x)^2+(10*x^4+19*x^3-2*x^2-5*x-5)*exp(x) +20*x-1)*exp(exp(x)*x^2)+1/5*(-5*x^2-10*x)*exp(x)+6*x^2-2/5*x,x, algorithm =\
Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {1}{5} \left (-2 x+30 x^2+e^x \left (-10 x-5 x^2\right )+e^{e^x x^2} \left (-1+20 x+e^{2 x} \left (-10 x^2-5 x^3\right )+e^x \left (-5-5 x-2 x^2+19 x^3+10 x^4\right )\right )\right ) \, dx=2 x^{3} - x^{2} e^{x} - \frac {x^{2}}{5} + \frac {\left (10 x^{2} - 5 x e^{x} - x\right ) e^{x^{2} e^{x}}}{5} \]
integrate(1/5*((-5*x**3-10*x**2)*exp(x)**2+(10*x**4+19*x**3-2*x**2-5*x-5)* exp(x)+20*x-1)*exp(exp(x)*x**2)+1/5*(-5*x**2-10*x)*exp(x)+6*x**2-2/5*x,x)
Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {1}{5} \left (-2 x+30 x^2+e^x \left (-10 x-5 x^2\right )+e^{e^x x^2} \left (-1+20 x+e^{2 x} \left (-10 x^2-5 x^3\right )+e^x \left (-5-5 x-2 x^2+19 x^3+10 x^4\right )\right )\right ) \, dx=2 \, x^{3} - x^{2} e^{x} - \frac {1}{5} \, x^{2} + \frac {1}{5} \, {\left (10 \, x^{2} - 5 \, x e^{x} - x\right )} e^{\left (x^{2} e^{x}\right )} \]
integrate(1/5*((-5*x^3-10*x^2)*exp(x)^2+(10*x^4+19*x^3-2*x^2-5*x-5)*exp(x) +20*x-1)*exp(exp(x)*x^2)+1/5*(-5*x^2-10*x)*exp(x)+6*x^2-2/5*x,x, algorithm =\
\[ \int \frac {1}{5} \left (-2 x+30 x^2+e^x \left (-10 x-5 x^2\right )+e^{e^x x^2} \left (-1+20 x+e^{2 x} \left (-10 x^2-5 x^3\right )+e^x \left (-5-5 x-2 x^2+19 x^3+10 x^4\right )\right )\right ) \, dx=\int { 6 \, x^{2} - \frac {1}{5} \, {\left (5 \, {\left (x^{3} + 2 \, x^{2}\right )} e^{\left (2 \, x\right )} - {\left (10 \, x^{4} + 19 \, x^{3} - 2 \, x^{2} - 5 \, x - 5\right )} e^{x} - 20 \, x + 1\right )} e^{\left (x^{2} e^{x}\right )} - {\left (x^{2} + 2 \, x\right )} e^{x} - \frac {2}{5} \, x \,d x } \]
integrate(1/5*((-5*x^3-10*x^2)*exp(x)^2+(10*x^4+19*x^3-2*x^2-5*x-5)*exp(x) +20*x-1)*exp(exp(x)*x^2)+1/5*(-5*x^2-10*x)*exp(x)+6*x^2-2/5*x,x, algorithm =\
integrate(6*x^2 - 1/5*(5*(x^3 + 2*x^2)*e^(2*x) - (10*x^4 + 19*x^3 - 2*x^2 - 5*x - 5)*e^x - 20*x + 1)*e^(x^2*e^x) - (x^2 + 2*x)*e^x - 2/5*x, x)
Time = 13.91 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {1}{5} \left (-2 x+30 x^2+e^x \left (-10 x-5 x^2\right )+e^{e^x x^2} \left (-1+20 x+e^{2 x} \left (-10 x^2-5 x^3\right )+e^x \left (-5-5 x-2 x^2+19 x^3+10 x^4\right )\right )\right ) \, dx=-\frac {x\,\left (x+{\mathrm {e}}^{x^2\,{\mathrm {e}}^x}\right )\,\left (5\,{\mathrm {e}}^x-10\,x+1\right )}{5} \]