Integrand size = 122, antiderivative size = 29 \[ \int \frac {80 e^2+e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+e^{e^{3+4 e^4+x}} \left (80+2 e^2 x^2+e^{3+4 e^4+x} \left (80 x-80 x^2\right )\right )}{e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+2 e^{2+e^{3+4 e^4+x}} x^2} \, dx=\frac {20 \left (4-\frac {4}{x}\right )}{e^2+e^{e^{3+4 e^4+x}}}+x \]
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {80 e^2+e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+e^{e^{3+4 e^4+x}} \left (80+2 e^2 x^2+e^{3+4 e^4+x} \left (80 x-80 x^2\right )\right )}{e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+2 e^{2+e^{3+4 e^4+x}} x^2} \, dx=\frac {80 (-1+x)}{\left (e^2+e^{e^{3+4 e^4+x}}\right ) x}+x \]
Integrate[(80*E^2 + E^4*x^2 + E^(2*E^(3 + 4*E^4 + x))*x^2 + E^E^(3 + 4*E^4 + x)*(80 + 2*E^2*x^2 + E^(3 + 4*E^4 + x)*(80*x - 80*x^2)))/(E^4*x^2 + E^( 2*E^(3 + 4*E^4 + x))*x^2 + 2*E^(2 + E^(3 + 4*E^4 + 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.
\(\displaystyle \int \frac {e^{2 e^{x+4 e^4+3}} x^2+e^4 x^2+e^{e^{x+4 e^4+3}} \left (2 e^2 x^2+e^{x+4 e^4+3} \left (80 x-80 x^2\right )+80\right )+80 e^2}{e^{2 e^{x+4 e^4+3}} x^2+2 e^{e^{x+4 e^4+3}+2} x^2+e^4 x^2} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 e^{x+4 e^4+3}} x^2+e^4 x^2+e^{e^{x+4 e^4+3}} \left (2 e^2 x^2+e^{x+4 e^4+3} \left (80 x-80 x^2\right )+80\right )+80 e^2}{\left (e^{e^{x+4 e^4+3}}+e^2\right )^2 x^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {80 e^{e^{x+4 e^4+3}}}{\left (e^{e^{x+4 e^4+3}}+e^2\right )^2 x^2}+\frac {80 e^2}{\left (e^{e^{x+4 e^4+3}}+e^2\right )^2 x^2}+\frac {80 e^{x+e^{x+4 e^4+3}+3 \left (1+\frac {4 e^4}{3}\right )} (1-x)}{\left (e^{e^{x+4 e^4+3}}+e^2\right )^2 x}+\frac {e^{2 e^{x+4 e^4+3}}}{\left (e^{e^{x+4 e^4+3}}+e^2\right )^2}+\frac {2 e^{e^{x+4 e^4+3}+2}}{\left (e^{e^{x+4 e^4+3}}+e^2\right )^2}+\frac {e^4}{\left (e^{e^{x+4 e^4+3}}+e^2\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 e^2 \text {Subst}\left (\int \frac {1}{\left (e^2+e^x\right ) x^2}dx,x,e^{x+4 e^4+3}\right )+2 e^4 \text {Subst}\left (\int \frac {1}{\left (e^2+e^x\right )^2 x}dx,x,e^{x+4 e^4+3}\right )-2 e^2 \text {Subst}\left (\int \frac {1}{\left (e^2+e^x\right ) x}dx,x,e^{x+4 e^4+3}\right )+80 e^2 \int \frac {1}{\left (e^2+e^{e^{x+4 e^4+3}}\right )^2 x^2}dx+80 \int \frac {e^{e^{x+4 e^4+3}}}{\left (e^2+e^{e^{x+4 e^4+3}}\right )^2 x^2}dx+80 \int \frac {e^{x+e^{x+4 e^4+3}+3 \left (1+\frac {4 e^4}{3}\right )}}{\left (e^2+e^{e^{x+4 e^4+3}}\right )^2 x}dx+x-\frac {2 e^{-x-4 e^4-1}}{e^{e^{x+4 e^4+3}}+e^2}+\frac {80}{e^{e^{x+4 e^4+3}}+e^2}\) |
Int[(80*E^2 + E^4*x^2 + E^(2*E^(3 + 4*E^4 + x))*x^2 + E^E^(3 + 4*E^4 + x)* (80 + 2*E^2*x^2 + E^(3 + 4*E^4 + x)*(80*x - 80*x^2)))/(E^4*x^2 + E^(2*E^(3 + 4*E^4 + x))*x^2 + 2*E^(2 + E^(3 + 4*E^4 + x))*x^2),x]
3.29.1.3.1 Defintions of rubi rules used
Time = 0.77 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86
method | result | size |
risch | \(x +\frac {-80+80 x}{x \left ({\mathrm e}^{2}+{\mathrm e}^{{\mathrm e}^{4 \,{\mathrm e}^{4}+3+x}}\right )}\) | \(25\) |
norman | \(\frac {-80+x^{2} {\mathrm e}^{2}+x^{2} {\mathrm e}^{{\mathrm e}^{4 \,{\mathrm e}^{4}+3+x}}+80 x}{x \left ({\mathrm e}^{2}+{\mathrm e}^{{\mathrm e}^{4 \,{\mathrm e}^{4}+3+x}}\right )}\) | \(43\) |
parallelrisch | \(\frac {-80+x^{2} {\mathrm e}^{2}+x^{2} {\mathrm e}^{{\mathrm e}^{4 \,{\mathrm e}^{4}+3+x}}+80 x}{x \left ({\mathrm e}^{2}+{\mathrm e}^{{\mathrm e}^{4 \,{\mathrm e}^{4}+3+x}}\right )}\) | \(43\) |
int((x^2*exp(exp(4*exp(4)+3+x))^2+((-80*x^2+80*x)*exp(4*exp(4)+3+x)+2*x^2* exp(2)+80)*exp(exp(4*exp(4)+3+x))+x^2*exp(2)^2+80*exp(2))/(x^2*exp(exp(4*e xp(4)+3+x))^2+2*x^2*exp(2)*exp(exp(4*exp(4)+3+x))+x^2*exp(2)^2),x,method=_ RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (23) = 46\).
Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72 \[ \int \frac {80 e^2+e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+e^{e^{3+4 e^4+x}} \left (80+2 e^2 x^2+e^{3+4 e^4+x} \left (80 x-80 x^2\right )\right )}{e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+2 e^{2+e^{3+4 e^4+x}} x^2} \, dx=\frac {x^{2} e^{4} + x^{2} e^{\left (e^{\left (x + 4 \, e^{4} + 3\right )} + 2\right )} + 80 \, {\left (x - 1\right )} e^{2}}{x e^{4} + x e^{\left (e^{\left (x + 4 \, e^{4} + 3\right )} + 2\right )}} \]
integrate((x^2*exp(exp(4*exp(4)+3+x))^2+((-80*x^2+80*x)*exp(4*exp(4)+3+x)+ 2*x^2*exp(2)+80)*exp(exp(4*exp(4)+3+x))+x^2*exp(2)^2+80*exp(2))/(x^2*exp(e xp(4*exp(4)+3+x))^2+2*x^2*exp(2)*exp(exp(4*exp(4)+3+x))+x^2*exp(2)^2),x, a lgorithm=\
(x^2*e^4 + x^2*e^(e^(x + 4*e^4 + 3) + 2) + 80*(x - 1)*e^2)/(x*e^4 + x*e^(e ^(x + 4*e^4 + 3) + 2))
Time = 0.08 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {80 e^2+e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+e^{e^{3+4 e^4+x}} \left (80+2 e^2 x^2+e^{3+4 e^4+x} \left (80 x-80 x^2\right )\right )}{e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+2 e^{2+e^{3+4 e^4+x}} x^2} \, dx=x + \frac {80 x - 80}{x e^{e^{x + 3 + 4 e^{4}}} + x e^{2}} \]
integrate((x**2*exp(exp(4*exp(4)+3+x))**2+((-80*x**2+80*x)*exp(4*exp(4)+3+ x)+2*x**2*exp(2)+80)*exp(exp(4*exp(4)+3+x))+x**2*exp(2)**2+80*exp(2))/(x** 2*exp(exp(4*exp(4)+3+x))**2+2*x**2*exp(2)*exp(exp(4*exp(4)+3+x))+x**2*exp( 2)**2),x)
Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.48 \[ \int \frac {80 e^2+e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+e^{e^{3+4 e^4+x}} \left (80+2 e^2 x^2+e^{3+4 e^4+x} \left (80 x-80 x^2\right )\right )}{e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+2 e^{2+e^{3+4 e^4+x}} x^2} \, dx=\frac {x^{2} e^{2} + x^{2} e^{\left (e^{\left (x + 4 \, e^{4} + 3\right )}\right )} + 80 \, x - 80}{x e^{2} + x e^{\left (e^{\left (x + 4 \, e^{4} + 3\right )}\right )}} \]
integrate((x^2*exp(exp(4*exp(4)+3+x))^2+((-80*x^2+80*x)*exp(4*exp(4)+3+x)+ 2*x^2*exp(2)+80)*exp(exp(4*exp(4)+3+x))+x^2*exp(2)^2+80*exp(2))/(x^2*exp(e xp(4*exp(4)+3+x))^2+2*x^2*exp(2)*exp(exp(4*exp(4)+3+x))+x^2*exp(2)^2),x, a lgorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 86 vs. \(2 (23) = 46\).
Time = 0.27 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.97 \[ \int \frac {80 e^2+e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+e^{e^{3+4 e^4+x}} \left (80+2 e^2 x^2+e^{3+4 e^4+x} \left (80 x-80 x^2\right )\right )}{e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+2 e^{2+e^{3+4 e^4+x}} x^2} \, dx=\frac {x^{2} e^{\left (x + 4 \, e^{4} + e^{\left (x + 4 \, e^{4} + 3\right )} + 3\right )} + x^{2} e^{\left (x + 4 \, e^{4} + 5\right )} + 80 \, x e^{\left (x + 4 \, e^{4} + 3\right )} - 80 \, e^{\left (x + 4 \, e^{4} + 3\right )}}{x e^{\left (x + 4 \, e^{4} + e^{\left (x + 4 \, e^{4} + 3\right )} + 3\right )} + x e^{\left (x + 4 \, e^{4} + 5\right )}} \]
integrate((x^2*exp(exp(4*exp(4)+3+x))^2+((-80*x^2+80*x)*exp(4*exp(4)+3+x)+ 2*x^2*exp(2)+80)*exp(exp(4*exp(4)+3+x))+x^2*exp(2)^2+80*exp(2))/(x^2*exp(e xp(4*exp(4)+3+x))^2+2*x^2*exp(2)*exp(exp(4*exp(4)+3+x))+x^2*exp(2)^2),x, a lgorithm=\
(x^2*e^(x + 4*e^4 + e^(x + 4*e^4 + 3) + 3) + x^2*e^(x + 4*e^4 + 5) + 80*x* e^(x + 4*e^4 + 3) - 80*e^(x + 4*e^4 + 3))/(x*e^(x + 4*e^4 + e^(x + 4*e^4 + 3) + 3) + x*e^(x + 4*e^4 + 5))
Time = 9.58 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.93 \[ \int \frac {80 e^2+e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+e^{e^{3+4 e^4+x}} \left (80+2 e^2 x^2+e^{3+4 e^4+x} \left (80 x-80 x^2\right )\right )}{e^4 x^2+e^{2 e^{3+4 e^4+x}} x^2+2 e^{2+e^{3+4 e^4+x}} x^2} \, dx=x+\frac {80\,x-80}{x\,\left ({\mathrm {e}}^{{\mathrm {e}}^{4\,{\mathrm {e}}^4}\,{\mathrm {e}}^3\,{\mathrm {e}}^x}+{\mathrm {e}}^2\right )} \]