Integrand size = 59, antiderivative size = 23 \[ \int \frac {e^{4+3 x} \left (1280000 x-640000 x^2\right )+e^{4+2 x} \left (-2560000+5120000 x-4480000 x^2+1280000 x^3\right )}{e^{2 x}-2 e^x x+x^2} \, dx=\frac {640000 e^{4+2 x} (-2+x)^2}{-e^x+x} \]
Time = 1.43 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {e^{4+3 x} \left (1280000 x-640000 x^2\right )+e^{4+2 x} \left (-2560000+5120000 x-4480000 x^2+1280000 x^3\right )}{e^{2 x}-2 e^x x+x^2} \, dx=-\frac {640000 e^{4+2 x} (-2+x)^2}{e^x-x} \]
Integrate[(E^(4 + 3*x)*(1280000*x - 640000*x^2) + E^(4 + 2*x)*(-2560000 + 5120000*x - 4480000*x^2 + 1280000*x^3))/(E^(2*x) - 2*E^x*x + 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.
| \(\displaystyle \int \frac {e^{3 x+4} \left (1280000 x-640000 x^2\right )+e^{2 x+4} \left (1280000 x^3-4480000 x^2+5120000 x-2560000\right )}{x^2-2 e^x x+e^{2 x}} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {640000 e^{2 x+4} (2-x) \left (-2 x^2+e^x x+3 x-2\right )}{\left (e^x-x\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 640000 \int -\frac {e^{2 x+4} (2-x) \left (2 x^2-e^x x-3 x+2\right )}{\left (e^x-x\right )^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -640000 \int \frac {e^{2 x+4} (2-x) \left (2 x^2-e^x x-3 x+2\right )}{\left (e^x-x\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -640000 \int \left (\frac {e^{2 x+4} (x-2) x}{e^x-x}-\frac {e^{2 x+4} (x-2)^2 (x-1)}{\left (e^x-x\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -640000 \left (-\int \frac {e^{2 x+4} x^3}{\left (e^x-x\right )^2}dx+5 \int \frac {e^{2 x+4} x^2}{\left (e^x-x\right )^2}dx+\int \frac {e^{2 x+4} x^2}{e^x-x}dx+4 \int \frac {e^{2 x+4}}{\left (e^x-x\right )^2}dx-8 \int \frac {e^{2 x+4} x}{\left (e^x-x\right )^2}dx-2 \int \frac {e^{2 x+4} x}{e^x-x}dx\right )\) |
Int[(E^(4 + 3*x)*(1280000*x - 640000*x^2) + E^(4 + 2*x)*(-2560000 + 512000 0*x - 4480000*x^2 + 1280000*x^3))/(E^(2*x) - 2*E^x*x + x^2),x]
3.19.8.3.1 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 = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
| method | result | size |
| risch | \(-\frac {640000 \left (x^{2}-4 x +4\right ) {\mathrm e}^{4+2 x}}{{\mathrm e}^{x}-x}\) | \(25\) |
| norman | \(\frac {2560000 \,{\mathrm e}^{4} {\mathrm e}^{2 x}-2560000 \,{\mathrm e}^{4} {\mathrm e}^{2 x} x +640000 \,{\mathrm e}^{4} {\mathrm e}^{2 x} x^{2}}{x -{\mathrm e}^{x}}\) | \(45\) |
| parallelrisch | \(\frac {2560000 \,{\mathrm e}^{4} {\mathrm e}^{2 x}-2560000 \,{\mathrm e}^{4} {\mathrm e}^{2 x} x +640000 \,{\mathrm e}^{4} {\mathrm e}^{2 x} x^{2}}{x -{\mathrm e}^{x}}\) | \(45\) |
int(((-640000*x^2+1280000*x)*exp(2)^2*exp(x)^3+(1280000*x^3-4480000*x^2+51 20000*x-2560000)*exp(2)^2*exp(x)^2)/(exp(x)^2-2*exp(x)*x+x^2),x,method=_RE TURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {e^{4+3 x} \left (1280000 x-640000 x^2\right )+e^{4+2 x} \left (-2560000+5120000 x-4480000 x^2+1280000 x^3\right )}{e^{2 x}-2 e^x x+x^2} \, dx=\frac {640000 \, {\left (x^{2} - 4 \, x + 4\right )} e^{\left (2 \, x + 4\right )}}{x - e^{x}} \]
integrate(((-640000*x^2+1280000*x)*exp(2)^2*exp(x)^3+(1280000*x^3-4480000* x^2+5120000*x-2560000)*exp(2)^2*exp(x)^2)/(exp(x)^2-2*exp(x)*x+x^2),x, alg orithm=\
Leaf count of result is larger than twice the leaf count of optimal. 76 vs. \(2 (19) = 38\).
Time = 0.10 (sec) , antiderivative size = 76, normalized size of antiderivative = 3.30 \[ \int \frac {e^{4+3 x} \left (1280000 x-640000 x^2\right )+e^{4+2 x} \left (-2560000+5120000 x-4480000 x^2+1280000 x^3\right )}{e^{2 x}-2 e^x x+x^2} \, dx=- 640000 x^{3} e^{4} + 2560000 x^{2} e^{4} - 2560000 x e^{4} + \left (- 640000 x^{2} e^{4} + 2560000 x e^{4} - 2560000 e^{4}\right ) e^{x} + \frac {- 640000 x^{4} e^{4} + 2560000 x^{3} e^{4} - 2560000 x^{2} e^{4}}{- x + e^{x}} \]
integrate(((-640000*x**2+1280000*x)*exp(2)**2*exp(x)**3+(1280000*x**3-4480 000*x**2+5120000*x-2560000)*exp(2)**2*exp(x)**2)/(exp(x)**2-2*exp(x)*x+x** 2),x)
-640000*x**3*exp(4) + 2560000*x**2*exp(4) - 2560000*x*exp(4) + (-640000*x* *2*exp(4) + 2560000*x*exp(4) - 2560000*exp(4))*exp(x) + (-640000*x**4*exp( 4) + 2560000*x**3*exp(4) - 2560000*x**2*exp(4))/(-x + exp(x))
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {e^{4+3 x} \left (1280000 x-640000 x^2\right )+e^{4+2 x} \left (-2560000+5120000 x-4480000 x^2+1280000 x^3\right )}{e^{2 x}-2 e^x x+x^2} \, dx=\frac {640000 \, {\left (x^{2} e^{4} - 4 \, x e^{4} + 4 \, e^{4}\right )} e^{\left (2 \, x\right )}}{x - e^{x}} \]
integrate(((-640000*x^2+1280000*x)*exp(2)^2*exp(x)^3+(1280000*x^3-4480000* x^2+5120000*x-2560000)*exp(2)^2*exp(x)^2)/(exp(x)^2-2*exp(x)*x+x^2),x, alg orithm=\
Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {e^{4+3 x} \left (1280000 x-640000 x^2\right )+e^{4+2 x} \left (-2560000+5120000 x-4480000 x^2+1280000 x^3\right )}{e^{2 x}-2 e^x x+x^2} \, dx=\frac {640000 \, {\left (x^{2} e^{\left (2 \, x + 4\right )} - 4 \, x e^{\left (2 \, x + 4\right )} + 4 \, e^{\left (2 \, x + 4\right )}\right )}}{x - e^{x}} \]
integrate(((-640000*x^2+1280000*x)*exp(2)^2*exp(x)^3+(1280000*x^3-4480000* x^2+5120000*x-2560000)*exp(2)^2*exp(x)^2)/(exp(x)^2-2*exp(x)*x+x^2),x, alg orithm=\
Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{4+3 x} \left (1280000 x-640000 x^2\right )+e^{4+2 x} \left (-2560000+5120000 x-4480000 x^2+1280000 x^3\right )}{e^{2 x}-2 e^x x+x^2} \, dx=\frac {640000\,{\mathrm {e}}^{2\,x+4}\,{\left (x-2\right )}^2}{x-{\mathrm {e}}^x} \]