Integrand size = 117, antiderivative size = 24 \[ \int \frac {e^{\frac {2 x^2}{10+7 x+x^2+e^{5+x} (10+2 x)}} \left (40 x+14 x^2+e^{5+x} \left (40 x-16 x^2-4 x^3\right )\right )}{100+140 x+69 x^2+14 x^3+x^4+e^{10+2 x} \left (100+40 x+4 x^2\right )+e^{5+x} \left (200+180 x+48 x^2+4 x^3\right )} \, dx=e^{\frac {2 x^2}{(5+x) \left (2+2 e^{5+x}+x\right )}} \]
Time = 1.19 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {2 x^2}{10+7 x+x^2+e^{5+x} (10+2 x)}} \left (40 x+14 x^2+e^{5+x} \left (40 x-16 x^2-4 x^3\right )\right )}{100+140 x+69 x^2+14 x^3+x^4+e^{10+2 x} \left (100+40 x+4 x^2\right )+e^{5+x} \left (200+180 x+48 x^2+4 x^3\right )} \, dx=e^{\frac {2 x^2}{(5+x) \left (2+2 e^{5+x}+x\right )}} \]
Integrate[(E^((2*x^2)/(10 + 7*x + x^2 + E^(5 + x)*(10 + 2*x)))*(40*x + 14* x^2 + E^(5 + x)*(40*x - 16*x^2 - 4*x^3)))/(100 + 140*x + 69*x^2 + 14*x^3 + x^4 + E^(10 + 2*x)*(100 + 40*x + 4*x^2) + E^(5 + x)*(200 + 180*x + 48*x^2 + 4*x^3)),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.
\(\displaystyle \int \frac {e^{\frac {2 x^2}{x^2+7 x+e^{x+5} (2 x+10)+10}} \left (14 x^2+e^{x+5} \left (-4 x^3-16 x^2+40 x\right )+40 x\right )}{x^4+14 x^3+69 x^2+e^{2 x+10} \left (4 x^2+40 x+100\right )+e^{x+5} \left (4 x^3+48 x^2+180 x+200\right )+140 x+100} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}} x \left (-2 e^{x+5} \left (x^2+4 x-10\right )+7 x+20\right )}{(x+5)^2 \left (x+2 e^{x+5}+2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}} x \left (7 x+2 e^{x+5} \left (-x^2-4 x+10\right )+20\right )}{(x+5)^2 \left (x+2 e^{x+5}+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (\frac {e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}} x^2 (x+1)}{(x+5) \left (x+2 e^{x+5}+2\right )^2}-\frac {e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}} x \left (x^2+4 x-10\right )}{(x+5)^2 \left (x+2 e^{x+5}+2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (20 \int \frac {e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}}}{\left (x+2 e^{x+5}+2\right )^2}dx-4 \int \frac {e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}} x}{\left (x+2 e^{x+5}+2\right )^2}dx+\int \frac {e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}} x^2}{\left (x+2 e^{x+5}+2\right )^2}dx-100 \int \frac {e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}}}{(x+5) \left (x+2 e^{x+5}+2\right )^2}dx+6 \int \frac {e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}}}{x+2 e^{x+5}+2}dx-\int \frac {e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}} x}{x+2 e^{x+5}+2}dx-25 \int \frac {e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}}}{(x+5)^2 \left (x+2 e^{x+5}+2\right )}dx-25 \int \frac {e^{\frac {2 x^2}{(x+5) \left (x+2 e^{x+5}+2\right )}}}{(x+5) \left (x+2 e^{x+5}+2\right )}dx\right )\) |
Int[(E^((2*x^2)/(10 + 7*x + x^2 + E^(5 + x)*(10 + 2*x)))*(40*x + 14*x^2 + E^(5 + x)*(40*x - 16*x^2 - 4*x^3)))/(100 + 140*x + 69*x^2 + 14*x^3 + x^4 + E^(10 + 2*x)*(100 + 40*x + 4*x^2) + E^(5 + x)*(200 + 180*x + 48*x^2 + 4*x ^3)),x]
3.1.7.3.1 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]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.68 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96
method | result | size |
risch | \({\mathrm e}^{\frac {2 x^{2}}{\left (5+x \right ) \left (x +2 \,{\mathrm e}^{5+x}+2\right )}}\) | \(23\) |
parallelrisch | \({\mathrm e}^{\frac {2 x^{2}}{2 x \,{\mathrm e}^{5} {\mathrm e}^{x}+10 \,{\mathrm e}^{5} {\mathrm e}^{x}+x^{2}+7 x +10}}\) | \(30\) |
norman | \(\frac {x^{2} {\mathrm e}^{\frac {2 x^{2}}{\left (2 x +10\right ) {\mathrm e}^{5} {\mathrm e}^{x}+x^{2}+7 x +10}}+7 x \,{\mathrm e}^{\frac {2 x^{2}}{\left (2 x +10\right ) {\mathrm e}^{5} {\mathrm e}^{x}+x^{2}+7 x +10}}+10 \,{\mathrm e}^{5} {\mathrm e}^{x} {\mathrm e}^{\frac {2 x^{2}}{\left (2 x +10\right ) {\mathrm e}^{5} {\mathrm e}^{x}+x^{2}+7 x +10}}+2 x \,{\mathrm e}^{5} {\mathrm e}^{x} {\mathrm e}^{\frac {2 x^{2}}{\left (2 x +10\right ) {\mathrm e}^{5} {\mathrm e}^{x}+x^{2}+7 x +10}}+10 \,{\mathrm e}^{\frac {2 x^{2}}{\left (2 x +10\right ) {\mathrm e}^{5} {\mathrm e}^{x}+x^{2}+7 x +10}}}{\left (5+x \right ) \left (2 \,{\mathrm e}^{5} {\mathrm e}^{x}+2+x \right )}\) | \(171\) |
int(((-4*x^3-16*x^2+40*x)*exp(5)*exp(x)+14*x^2+40*x)*exp(2*x^2/((2*x+10)*e xp(5)*exp(x)+x^2+7*x+10))/((4*x^2+40*x+100)*exp(5)^2*exp(x)^2+(4*x^3+48*x^ 2+180*x+200)*exp(5)*exp(x)+x^4+14*x^3+69*x^2+140*x+100),x,method=_RETURNVE RBOSE)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {e^{\frac {2 x^2}{10+7 x+x^2+e^{5+x} (10+2 x)}} \left (40 x+14 x^2+e^{5+x} \left (40 x-16 x^2-4 x^3\right )\right )}{100+140 x+69 x^2+14 x^3+x^4+e^{10+2 x} \left (100+40 x+4 x^2\right )+e^{5+x} \left (200+180 x+48 x^2+4 x^3\right )} \, dx=e^{\left (\frac {2 \, x^{2}}{x^{2} + 2 \, {\left (x + 5\right )} e^{\left (x + 5\right )} + 7 \, x + 10}\right )} \]
integrate(((-4*x^3-16*x^2+40*x)*exp(5)*exp(x)+14*x^2+40*x)*exp(2*x^2/((2*x +10)*exp(5)*exp(x)+x^2+7*x+10))/((4*x^2+40*x+100)*exp(5)^2*exp(x)^2+(4*x^3 +48*x^2+180*x+200)*exp(5)*exp(x)+x^4+14*x^3+69*x^2+140*x+100),x, algorithm =\
Time = 0.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{\frac {2 x^2}{10+7 x+x^2+e^{5+x} (10+2 x)}} \left (40 x+14 x^2+e^{5+x} \left (40 x-16 x^2-4 x^3\right )\right )}{100+140 x+69 x^2+14 x^3+x^4+e^{10+2 x} \left (100+40 x+4 x^2\right )+e^{5+x} \left (200+180 x+48 x^2+4 x^3\right )} \, dx=e^{\frac {2 x^{2}}{x^{2} + 7 x + \left (2 x + 10\right ) e^{5} e^{x} + 10}} \]
integrate(((-4*x**3-16*x**2+40*x)*exp(5)*exp(x)+14*x**2+40*x)*exp(2*x**2/( (2*x+10)*exp(5)*exp(x)+x**2+7*x+10))/((4*x**2+40*x+100)*exp(5)**2*exp(x)** 2+(4*x**3+48*x**2+180*x+200)*exp(5)*exp(x)+x**4+14*x**3+69*x**2+140*x+100) ,x)
Leaf count of result is larger than twice the leaf count of optimal. 125 vs. \(2 (22) = 44\).
Time = 0.57 (sec) , antiderivative size = 125, normalized size of antiderivative = 5.21 \[ \int \frac {e^{\frac {2 x^2}{10+7 x+x^2+e^{5+x} (10+2 x)}} \left (40 x+14 x^2+e^{5+x} \left (40 x-16 x^2-4 x^3\right )\right )}{100+140 x+69 x^2+14 x^3+x^4+e^{10+2 x} \left (100+40 x+4 x^2\right )+e^{5+x} \left (200+180 x+48 x^2+4 x^3\right )} \, dx=e^{\left (-\frac {8 \, e^{\left (2 \, x + 10\right )}}{2 \, {\left (x e^{5} - e^{5}\right )} e^{x} - 3 \, x + 4 \, e^{\left (2 \, x + 10\right )} - 6} - \frac {16 \, e^{\left (x + 5\right )}}{2 \, {\left (x e^{5} - e^{5}\right )} e^{x} - 3 \, x + 4 \, e^{\left (2 \, x + 10\right )} - 6} + \frac {50}{2 \, {\left (x e^{5} + 5 \, e^{5}\right )} e^{x} - 3 \, x - 15} - \frac {8}{2 \, {\left (x e^{5} - e^{5}\right )} e^{x} - 3 \, x + 4 \, e^{\left (2 \, x + 10\right )} - 6} + 2\right )} \]
integrate(((-4*x^3-16*x^2+40*x)*exp(5)*exp(x)+14*x^2+40*x)*exp(2*x^2/((2*x +10)*exp(5)*exp(x)+x^2+7*x+10))/((4*x^2+40*x+100)*exp(5)^2*exp(x)^2+(4*x^3 +48*x^2+180*x+200)*exp(5)*exp(x)+x^4+14*x^3+69*x^2+140*x+100),x, algorithm =\
e^(-8*e^(2*x + 10)/(2*(x*e^5 - e^5)*e^x - 3*x + 4*e^(2*x + 10) - 6) - 16*e ^(x + 5)/(2*(x*e^5 - e^5)*e^x - 3*x + 4*e^(2*x + 10) - 6) + 50/(2*(x*e^5 + 5*e^5)*e^x - 3*x - 15) - 8/(2*(x*e^5 - e^5)*e^x - 3*x + 4*e^(2*x + 10) - 6) + 2)
Exception generated. \[ \int \frac {e^{\frac {2 x^2}{10+7 x+x^2+e^{5+x} (10+2 x)}} \left (40 x+14 x^2+e^{5+x} \left (40 x-16 x^2-4 x^3\right )\right )}{100+140 x+69 x^2+14 x^3+x^4+e^{10+2 x} \left (100+40 x+4 x^2\right )+e^{5+x} \left (200+180 x+48 x^2+4 x^3\right )} \, dx=\text {Exception raised: TypeError} \]
integrate(((-4*x^3-16*x^2+40*x)*exp(5)*exp(x)+14*x^2+40*x)*exp(2*x^2/((2*x +10)*exp(5)*exp(x)+x^2+7*x+10))/((4*x^2+40*x+100)*exp(5)^2*exp(x)^2+(4*x^3 +48*x^2+180*x+200)*exp(5)*exp(x)+x^4+14*x^3+69*x^2+140*x+100),x, algorithm =\
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{4,[1,16]%%%}+%%%{-176,[1,15]%%%}+%%%{3392,[1,14]%%%}+%%%{- 37408,[1,
Time = 7.50 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {e^{\frac {2 x^2}{10+7 x+x^2+e^{5+x} (10+2 x)}} \left (40 x+14 x^2+e^{5+x} \left (40 x-16 x^2-4 x^3\right )\right )}{100+140 x+69 x^2+14 x^3+x^4+e^{10+2 x} \left (100+40 x+4 x^2\right )+e^{5+x} \left (200+180 x+48 x^2+4 x^3\right )} \, dx={\mathrm {e}}^{\frac {2\,x^2}{7\,x+10\,{\mathrm {e}}^5\,{\mathrm {e}}^x+x^2+2\,x\,{\mathrm {e}}^5\,{\mathrm {e}}^x+10}} \]