Integrand size = 94, antiderivative size = 23 \[ \int \frac {162 x+e^{e \left (20 x-20 x^2+5 x^3\right )} \left (x^2+e \left (-20 x^3+40 x^4-15 x^5\right )\right )}{6561+162 e^{e \left (20 x-20 x^2+5 x^3\right )} x+e^{2 e \left (20 x-20 x^2+5 x^3\right )} x^2} \, dx=\frac {x}{e^{5 e (2-x)^2 x}+\frac {81}{x}} \] Output:
x/(exp(5*exp(1)*(2-x)^2*x)+81/x)
Time = 10.40 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {162 x+e^{e \left (20 x-20 x^2+5 x^3\right )} \left (x^2+e \left (-20 x^3+40 x^4-15 x^5\right )\right )}{6561+162 e^{e \left (20 x-20 x^2+5 x^3\right )} x+e^{2 e \left (20 x-20 x^2+5 x^3\right )} x^2} \, dx=\frac {x^2}{81+e^{5 e (-2+x)^2 x} x} \] Input:
Integrate[(162*x + E^(E*(20*x - 20*x^2 + 5*x^3))*(x^2 + E*(-20*x^3 + 40*x^ 4 - 15*x^5)))/(6561 + 162*E^(E*(20*x - 20*x^2 + 5*x^3))*x + E^(2*E*(20*x - 20*x^2 + 5*x^3))*x^2),x]
Output:
x^2/(81 + E^(5*E*(-2 + x)^2*x)*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.
\(\displaystyle \int \frac {e^{e \left (5 x^3-20 x^2+20 x\right )} \left (x^2+e \left (-15 x^5+40 x^4-20 x^3\right )\right )+162 x}{e^{2 e \left (5 x^3-20 x^2+20 x\right )} x^2+162 e^{e \left (5 x^3-20 x^2+20 x\right )} x+6561} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{e \left (5 x^3-20 x^2+20 x\right )} \left (x^2+e \left (-15 x^5+40 x^4-20 x^3\right )\right )+162 x}{\left (e^{5 e (x-2)^2 x} x+81\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {81 x \left (15 e x^3-40 e x^2+20 e x+1\right )}{\left (e^{5 e (x-2)^2 x} x+81\right )^2}-\frac {x \left (15 e x^3-40 e x^2+20 e x-1\right )}{e^{5 e (x-2)^2 x} x+81}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 1215 e \int \frac {x^4}{\left (e^{5 e (x-2)^2 x} x+81\right )^2}dx-15 e \int \frac {x^4}{e^{5 e (x-2)^2 x} x+81}dx-3240 e \int \frac {x^3}{\left (e^{5 e (x-2)^2 x} x+81\right )^2}dx+40 e \int \frac {x^3}{e^{5 e (x-2)^2 x} x+81}dx+1620 e \int \frac {x^2}{\left (e^{5 e (x-2)^2 x} x+81\right )^2}dx-20 e \int \frac {x^2}{e^{5 e (x-2)^2 x} x+81}dx+81 \int \frac {x}{\left (e^{5 e (x-2)^2 x} x+81\right )^2}dx+\int \frac {x}{e^{5 e (x-2)^2 x} x+81}dx\) |
Input:
Int[(162*x + E^(E*(20*x - 20*x^2 + 5*x^3))*(x^2 + E*(-20*x^3 + 40*x^4 - 15 *x^5)))/(6561 + 162*E^(E*(20*x - 20*x^2 + 5*x^3))*x + E^(2*E*(20*x - 20*x^ 2 + 5*x^3))*x^2),x]
Output:
$Aborted
Time = 0.38 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {x^{2}}{x \,{\mathrm e}^{5 x \left (-2+x \right )^{2} {\mathrm e}}+81}\) | \(22\) |
norman | \(\frac {x^{2}}{x \,{\mathrm e}^{\left (5 x^{3}-20 x^{2}+20 x \right ) {\mathrm e}}+81}\) | \(29\) |
parallelrisch | \(\frac {x^{2}}{x \,{\mathrm e}^{\left (5 x^{3}-20 x^{2}+20 x \right ) {\mathrm e}}+81}\) | \(29\) |
Input:
int((((-15*x^5+40*x^4-20*x^3)*exp(1)+x^2)*exp((5*x^3-20*x^2+20*x)*exp(1))+ 162*x)/(x^2*exp((5*x^3-20*x^2+20*x)*exp(1))^2+162*x*exp((5*x^3-20*x^2+20*x )*exp(1))+6561),x,method=_RETURNVERBOSE)
Output:
x^2/(x*exp(5*x*(-2+x)^2*exp(1))+81)
Time = 0.07 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {162 x+e^{e \left (20 x-20 x^2+5 x^3\right )} \left (x^2+e \left (-20 x^3+40 x^4-15 x^5\right )\right )}{6561+162 e^{e \left (20 x-20 x^2+5 x^3\right )} x+e^{2 e \left (20 x-20 x^2+5 x^3\right )} x^2} \, dx=\frac {x^{2}}{x e^{\left (5 \, {\left (x^{3} - 4 \, x^{2} + 4 \, x\right )} e\right )} + 81} \] Input:
integrate((((-15*x^5+40*x^4-20*x^3)*exp(1)+x^2)*exp((5*x^3-20*x^2+20*x)*ex p(1))+162*x)/(x^2*exp((5*x^3-20*x^2+20*x)*exp(1))^2+162*x*exp((5*x^3-20*x^ 2+20*x)*exp(1))+6561),x, algorithm="fricas")
Output:
x^2/(x*e^(5*(x^3 - 4*x^2 + 4*x)*e) + 81)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {162 x+e^{e \left (20 x-20 x^2+5 x^3\right )} \left (x^2+e \left (-20 x^3+40 x^4-15 x^5\right )\right )}{6561+162 e^{e \left (20 x-20 x^2+5 x^3\right )} x+e^{2 e \left (20 x-20 x^2+5 x^3\right )} x^2} \, dx=\frac {x^{2}}{x e^{e \left (5 x^{3} - 20 x^{2} + 20 x\right )} + 81} \] Input:
integrate((((-15*x**5+40*x**4-20*x**3)*exp(1)+x**2)*exp((5*x**3-20*x**2+20 *x)*exp(1))+162*x)/(x**2*exp((5*x**3-20*x**2+20*x)*exp(1))**2+162*x*exp((5 *x**3-20*x**2+20*x)*exp(1))+6561),x)
Output:
x**2/(x*exp(E*(5*x**3 - 20*x**2 + 20*x)) + 81)
Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {162 x+e^{e \left (20 x-20 x^2+5 x^3\right )} \left (x^2+e \left (-20 x^3+40 x^4-15 x^5\right )\right )}{6561+162 e^{e \left (20 x-20 x^2+5 x^3\right )} x+e^{2 e \left (20 x-20 x^2+5 x^3\right )} x^2} \, dx=\frac {x^{2} e^{\left (20 \, x^{2} e\right )}}{x e^{\left (5 \, x^{3} e + 20 \, x e\right )} + 81 \, e^{\left (20 \, x^{2} e\right )}} \] Input:
integrate((((-15*x^5+40*x^4-20*x^3)*exp(1)+x^2)*exp((5*x^3-20*x^2+20*x)*ex p(1))+162*x)/(x^2*exp((5*x^3-20*x^2+20*x)*exp(1))^2+162*x*exp((5*x^3-20*x^ 2+20*x)*exp(1))+6561),x, algorithm="maxima")
Output:
x^2*e^(20*x^2*e)/(x*e^(5*x^3*e + 20*x*e) + 81*e^(20*x^2*e))
Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {162 x+e^{e \left (20 x-20 x^2+5 x^3\right )} \left (x^2+e \left (-20 x^3+40 x^4-15 x^5\right )\right )}{6561+162 e^{e \left (20 x-20 x^2+5 x^3\right )} x+e^{2 e \left (20 x-20 x^2+5 x^3\right )} x^2} \, dx=\frac {x^{2}}{x e^{\left (5 \, x^{3} e - 20 \, x^{2} e + 20 \, x e\right )} + 81} \] Input:
integrate((((-15*x^5+40*x^4-20*x^3)*exp(1)+x^2)*exp((5*x^3-20*x^2+20*x)*ex p(1))+162*x)/(x^2*exp((5*x^3-20*x^2+20*x)*exp(1))^2+162*x*exp((5*x^3-20*x^ 2+20*x)*exp(1))+6561),x, algorithm="giac")
Output:
x^2/(x*e^(5*x^3*e - 20*x^2*e + 20*x*e) + 81)
Time = 1.70 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {162 x+e^{e \left (20 x-20 x^2+5 x^3\right )} \left (x^2+e \left (-20 x^3+40 x^4-15 x^5\right )\right )}{6561+162 e^{e \left (20 x-20 x^2+5 x^3\right )} x+e^{2 e \left (20 x-20 x^2+5 x^3\right )} x^2} \, dx=\frac {x^2}{x\,{\mathrm {e}}^{5\,x^3\,\mathrm {e}}\,{\mathrm {e}}^{-20\,x^2\,\mathrm {e}}\,{\mathrm {e}}^{20\,x\,\mathrm {e}}+81} \] Input:
int((162*x - exp(exp(1)*(20*x - 20*x^2 + 5*x^3))*(exp(1)*(20*x^3 - 40*x^4 + 15*x^5) - x^2))/(x^2*exp(2*exp(1)*(20*x - 20*x^2 + 5*x^3)) + 162*x*exp(e xp(1)*(20*x - 20*x^2 + 5*x^3)) + 6561),x)
Output:
x^2/(x*exp(5*x^3*exp(1))*exp(-20*x^2*exp(1))*exp(20*x*exp(1)) + 81)
Time = 0.21 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74 \[ \int \frac {162 x+e^{e \left (20 x-20 x^2+5 x^3\right )} \left (x^2+e \left (-20 x^3+40 x^4-15 x^5\right )\right )}{6561+162 e^{e \left (20 x-20 x^2+5 x^3\right )} x+e^{2 e \left (20 x-20 x^2+5 x^3\right )} x^2} \, dx=\frac {e^{20 e \,x^{2}} x^{2}}{e^{5 e \,x^{3}+20 e x} x +81 e^{20 e \,x^{2}}} \] Input:
int((((-15*x^5+40*x^4-20*x^3)*exp(1)+x^2)*exp((5*x^3-20*x^2+20*x)*exp(1))+ 162*x)/(x^2*exp((5*x^3-20*x^2+20*x)*exp(1))^2+162*x*exp((5*x^3-20*x^2+20*x )*exp(1))+6561),x)
Output:
(e**(20*e*x**2)*x**2)/(e**(5*e*x**3 + 20*e*x)*x + 81*e**(20*e*x**2))