Integrand size = 82, antiderivative size = 36 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=\frac {1}{7} \left (-4+\frac {\left (e^{\frac {1}{25} (-2+x)^2 x^2}+\frac {x}{5}\right )^2}{x}-x\right ) x \] Output:
1/7*((exp(1/5*x^2*(-2+x)*(1/5*x-2/5))+1/5*x)^2/x-4-x)*x
Time = 0.87 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.28 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=\frac {1}{175} \left (25 e^{\frac {2}{25} (-2+x)^2 x^2}+10 e^{\frac {1}{25} (-2+x)^2 x^2} x-4 x (25+6 x)\right ) \] Input:
Integrate[(-500 - 240*x + E^((2*(4*x^2 - 4*x^3 + x^4))/25)*(80*x - 120*x^2 + 40*x^3) + E^((4*x^2 - 4*x^3 + x^4)/25)*(50 + 16*x^2 - 24*x^3 + 8*x^4))/ 875,x]
Output:
(25*E^((2*(-2 + x)^2*x^2)/25) + 10*E^(((-2 + x)^2*x^2)/25)*x - 4*x*(25 + 6 *x))/175
Leaf count is larger than twice the leaf count of optimal. \(85\) vs. \(2(36)=72\).
Time = 0.35 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.36, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {27, 2009}
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.
\(\displaystyle \int \frac {1}{875} \left (e^{\frac {2}{25} \left (x^4-4 x^3+4 x^2\right )} \left (40 x^3-120 x^2+80 x\right )+e^{\frac {1}{25} \left (x^4-4 x^3+4 x^2\right )} \left (8 x^4-24 x^3+16 x^2+50\right )-240 x-500\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{875} \int \left (-240 x+40 e^{\frac {2}{25} \left (x^4-4 x^3+4 x^2\right )} \left (x^3-3 x^2+2 x\right )+2 e^{\frac {1}{25} \left (x^4-4 x^3+4 x^2\right )} \left (4 x^4-12 x^3+8 x^2+25\right )-500\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{875} \left (-120 x^2+125 e^{\frac {2}{25} \left (x^4-4 x^3+4 x^2\right )}+\frac {50 e^{\frac {1}{25} \left (x^4-4 x^3+4 x^2\right )} \left (x^4-3 x^3+2 x^2\right )}{x^3-3 x^2+2 x}-500 x\right )\) |
Input:
Int[(-500 - 240*x + E^((2*(4*x^2 - 4*x^3 + x^4))/25)*(80*x - 120*x^2 + 40* x^3) + E^((4*x^2 - 4*x^3 + x^4)/25)*(50 + 16*x^2 - 24*x^3 + 8*x^4))/875,x]
Output:
(125*E^((2*(4*x^2 - 4*x^3 + x^4))/25) - 500*x - 120*x^2 + (50*E^((4*x^2 - 4*x^3 + x^4)/25)*(2*x^2 - 3*x^3 + x^4))/(2*x - 3*x^2 + x^3))/875
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.83 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.03
method | result | size |
risch | \(-\frac {4 x}{7}+\frac {{\mathrm e}^{\frac {2 \left (-2+x \right )^{2} x^{2}}{25}}}{7}+\frac {2 \,{\mathrm e}^{\frac {\left (-2+x \right )^{2} x^{2}}{25}} x}{35}-\frac {24 x^{2}}{175}\) | \(37\) |
parallelrisch | \(\frac {{\mathrm e}^{\frac {2 x^{2} \left (x^{2}-4 x +4\right )}{25}}}{7}-\frac {24 x^{2}}{175}+\frac {2 \,{\mathrm e}^{\frac {x^{2} \left (x^{2}-4 x +4\right )}{25}} x}{35}-\frac {4 x}{7}\) | \(45\) |
default | \(-\frac {4 x}{7}+\frac {{\mathrm e}^{\frac {2}{25} x^{4}-\frac {8}{25} x^{3}+\frac {8}{25} x^{2}}}{7}+\frac {2 \,{\mathrm e}^{\frac {1}{25} x^{4}-\frac {4}{25} x^{3}+\frac {4}{25} x^{2}} x}{35}-\frac {24 x^{2}}{175}\) | \(51\) |
norman | \(-\frac {4 x}{7}+\frac {{\mathrm e}^{\frac {2}{25} x^{4}-\frac {8}{25} x^{3}+\frac {8}{25} x^{2}}}{7}+\frac {2 \,{\mathrm e}^{\frac {1}{25} x^{4}-\frac {4}{25} x^{3}+\frac {4}{25} x^{2}} x}{35}-\frac {24 x^{2}}{175}\) | \(51\) |
parts | \(-\frac {4 x}{7}+\frac {{\mathrm e}^{\frac {2}{25} x^{4}-\frac {8}{25} x^{3}+\frac {8}{25} x^{2}}}{7}+\frac {2 \,{\mathrm e}^{\frac {1}{25} x^{4}-\frac {4}{25} x^{3}+\frac {4}{25} x^{2}} x}{35}-\frac {24 x^{2}}{175}\) | \(51\) |
orering | \(\text {Expression too large to display}\) | \(1072\) |
Input:
int(1/875*(40*x^3-120*x^2+80*x)*exp(1/25*x^4-4/25*x^3+4/25*x^2)^2+1/875*(8 *x^4-24*x^3+16*x^2+50)*exp(1/25*x^4-4/25*x^3+4/25*x^2)-48/175*x-4/7,x,meth od=_RETURNVERBOSE)
Output:
-4/7*x+1/7*exp(2/25*(-2+x)^2*x^2)+2/35*exp(1/25*(-2+x)^2*x^2)*x-24/175*x^2
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=-\frac {24}{175} \, x^{2} + \frac {2}{35} \, x e^{\left (\frac {1}{25} \, x^{4} - \frac {4}{25} \, x^{3} + \frac {4}{25} \, x^{2}\right )} - \frac {4}{7} \, x + \frac {1}{7} \, e^{\left (\frac {2}{25} \, x^{4} - \frac {8}{25} \, x^{3} + \frac {8}{25} \, x^{2}\right )} \] Input:
integrate(1/875*(40*x^3-120*x^2+80*x)*exp(1/25*x^4-4/25*x^3+4/25*x^2)^2+1/ 875*(8*x^4-24*x^3+16*x^2+50)*exp(1/25*x^4-4/25*x^3+4/25*x^2)-48/175*x-4/7, x, algorithm="fricas")
Output:
-24/175*x^2 + 2/35*x*e^(1/25*x^4 - 4/25*x^3 + 4/25*x^2) - 4/7*x + 1/7*e^(2 /25*x^4 - 8/25*x^3 + 8/25*x^2)
Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.67 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=- \frac {24 x^{2}}{175} + \frac {2 x e^{\frac {x^{4}}{25} - \frac {4 x^{3}}{25} + \frac {4 x^{2}}{25}}}{35} - \frac {4 x}{7} + \frac {e^{\frac {2 x^{4}}{25} - \frac {8 x^{3}}{25} + \frac {8 x^{2}}{25}}}{7} \] Input:
integrate(1/875*(40*x**3-120*x**2+80*x)*exp(1/25*x**4-4/25*x**3+4/25*x**2) **2+1/875*(8*x**4-24*x**3+16*x**2+50)*exp(1/25*x**4-4/25*x**3+4/25*x**2)-4 8/175*x-4/7,x)
Output:
-24*x**2/175 + 2*x*exp(x**4/25 - 4*x**3/25 + 4*x**2/25)/35 - 4*x/7 + exp(2 *x**4/25 - 8*x**3/25 + 8*x**2/25)/7
Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=-\frac {24}{175} \, x^{2} + \frac {2}{35} \, x e^{\left (\frac {1}{25} \, x^{4} - \frac {4}{25} \, x^{3} + \frac {4}{25} \, x^{2}\right )} - \frac {4}{7} \, x + \frac {1}{7} \, e^{\left (\frac {2}{25} \, x^{4} - \frac {8}{25} \, x^{3} + \frac {8}{25} \, x^{2}\right )} \] Input:
integrate(1/875*(40*x^3-120*x^2+80*x)*exp(1/25*x^4-4/25*x^3+4/25*x^2)^2+1/ 875*(8*x^4-24*x^3+16*x^2+50)*exp(1/25*x^4-4/25*x^3+4/25*x^2)-48/175*x-4/7, x, algorithm="maxima")
Output:
-24/175*x^2 + 2/35*x*e^(1/25*x^4 - 4/25*x^3 + 4/25*x^2) - 4/7*x + 1/7*e^(2 /25*x^4 - 8/25*x^3 + 8/25*x^2)
Time = 0.13 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=-\frac {24}{175} \, x^{2} + \frac {2}{35} \, x e^{\left (\frac {1}{25} \, x^{4} - \frac {4}{25} \, x^{3} + \frac {4}{25} \, x^{2}\right )} - \frac {4}{7} \, x + \frac {1}{7} \, e^{\left (\frac {2}{25} \, x^{4} - \frac {8}{25} \, x^{3} + \frac {8}{25} \, x^{2}\right )} \] Input:
integrate(1/875*(40*x^3-120*x^2+80*x)*exp(1/25*x^4-4/25*x^3+4/25*x^2)^2+1/ 875*(8*x^4-24*x^3+16*x^2+50)*exp(1/25*x^4-4/25*x^3+4/25*x^2)-48/175*x-4/7, x, algorithm="giac")
Output:
-24/175*x^2 + 2/35*x*e^(1/25*x^4 - 4/25*x^3 + 4/25*x^2) - 4/7*x + 1/7*e^(2 /25*x^4 - 8/25*x^3 + 8/25*x^2)
Time = 3.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=\frac {{\mathrm {e}}^{\frac {2\,x^4}{25}-\frac {8\,x^3}{25}+\frac {8\,x^2}{25}}}{7}-\frac {4\,x}{7}+\frac {2\,x\,{\mathrm {e}}^{\frac {x^4}{25}-\frac {4\,x^3}{25}+\frac {4\,x^2}{25}}}{35}-\frac {24\,x^2}{175} \] Input:
int((exp((8*x^2)/25 - (8*x^3)/25 + (2*x^4)/25)*(80*x - 120*x^2 + 40*x^3))/ 875 - (48*x)/175 + (exp((4*x^2)/25 - (4*x^3)/25 + x^4/25)*(16*x^2 - 24*x^3 + 8*x^4 + 50))/875 - 4/7,x)
Output:
exp((8*x^2)/25 - (8*x^3)/25 + (2*x^4)/25)/7 - (4*x)/7 + (2*x*exp((4*x^2)/2 5 - (4*x^3)/25 + x^4/25))/35 - (24*x^2)/175
Time = 0.18 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.94 \[ \int \frac {1}{875} \left (-500-240 x+e^{\frac {2}{25} \left (4 x^2-4 x^3+x^4\right )} \left (80 x-120 x^2+40 x^3\right )+e^{\frac {1}{25} \left (4 x^2-4 x^3+x^4\right )} \left (50+16 x^2-24 x^3+8 x^4\right )\right ) \, dx=\frac {25 e^{\frac {2}{25} x^{4}+\frac {8}{25} x^{2}}+10 e^{\frac {1}{25} x^{4}+\frac {4}{25} x^{3}+\frac {4}{25} x^{2}} x -24 e^{\frac {8 x^{3}}{25}} x^{2}-100 e^{\frac {8 x^{3}}{25}} x}{175 e^{\frac {8 x^{3}}{25}}} \] Input:
int(1/875*(40*x^3-120*x^2+80*x)*exp(1/25*x^4-4/25*x^3+4/25*x^2)^2+1/875*(8 *x^4-24*x^3+16*x^2+50)*exp(1/25*x^4-4/25*x^3+4/25*x^2)-48/175*x-4/7,x)
Output:
(25*e**((2*x**4 + 8*x**2)/25) + 10*e**((x**4 + 4*x**3 + 4*x**2)/25)*x - 24 *e**((8*x**3)/25)*x**2 - 100*e**((8*x**3)/25)*x)/(175*e**((8*x**3)/25))