Integrand size = 108, antiderivative size = 26 \[ \int \frac {e^{\frac {e^{9+2 x} \left (10+2 x^2\right )+e^{9+x} \left (100 x+20 x^3\right )+e^9 \left (250 x^2+50 x^4\right )}{x^2}} \left (100 e^9 x^4+e^{9+2 x} \left (-20+20 x+4 x^3\right )+e^{9+x} \left (-100 x+100 x^2+20 x^3+20 x^4\right )\right )}{x^3} \, dx=e^{\frac {50 e^9 \left (\frac {e^x}{5}+x\right )^2 \left (5+x^2\right )}{x^2}} \]
Time = 0.12 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {e^{9+2 x} \left (10+2 x^2\right )+e^{9+x} \left (100 x+20 x^3\right )+e^9 \left (250 x^2+50 x^4\right )}{x^2}} \left (100 e^9 x^4+e^{9+2 x} \left (-20+20 x+4 x^3\right )+e^{9+x} \left (-100 x+100 x^2+20 x^3+20 x^4\right )\right )}{x^3} \, dx=e^{\frac {2 e^9 \left (e^x+5 x\right )^2 \left (5+x^2\right )}{x^2}} \]
Integrate[(E^((E^(9 + 2*x)*(10 + 2*x^2) + E^(9 + x)*(100*x + 20*x^3) + E^9 *(250*x^2 + 50*x^4))/x^2)*(100*E^9*x^4 + E^(9 + 2*x)*(-20 + 20*x + 4*x^3) + E^(9 + x)*(-100*x + 100*x^2 + 20*x^3 + 20*x^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.
\(\displaystyle \int \frac {\left (100 e^9 x^4+e^{2 x+9} \left (4 x^3+20 x-20\right )+e^{x+9} \left (20 x^4+20 x^3+100 x^2-100 x\right )\right ) \exp \left (\frac {e^{x+9} \left (20 x^3+100 x\right )+e^{2 x+9} \left (2 x^2+10\right )+e^9 \left (50 x^4+250 x^2\right )}{x^2}\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {2 e^9 \left (5 x+e^x\right )^2 \left (x^2+5\right )}{x^2}} \left (100 e^9 x^4+e^{2 x+9} \left (4 x^3+20 x-20\right )+e^{x+9} \left (20 x^4+20 x^3+100 x^2-100 x\right )\right )}{x^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 \left (x^3+5 x-5\right ) \exp \left (\frac {2 e^9 \left (x^2+5\right ) \left (5 x+e^x\right )^2}{x^2}+2 x+9\right )}{x^3}+100 e^{\frac {2 e^9 \left (x^2+5\right ) \left (5 x+e^x\right )^2}{x^2}+9} x+\frac {20 e^{\frac {2 e^9 \left (x^2+5\right ) \left (5 x+e^x\right )^2}{x^2}+x+9} \left (x^3+x^2+5 x-5\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \exp \left (\frac {2 e^9 \left (x^2+5\right ) \left (5 x+e^x\right )^2}{x^2}+2 x+9\right )dx+20 \int \frac {\exp \left (\frac {2 e^9 \left (x^2+5\right ) \left (5 x+e^x\right )^2}{x^2}+2 x+9\right )}{x^2}dx-20 \int \frac {\exp \left (\frac {2 e^9 \left (x^2+5\right ) \left (5 x+e^x\right )^2}{x^2}+2 x+9\right )}{x^3}dx+20 \int e^{\frac {2 e^9 \left (x^2+5\right ) \left (5 x+e^x\right )^2}{x^2}+x+9}dx-100 \int \frac {e^{\frac {2 e^9 \left (x^2+5\right ) \left (5 x+e^x\right )^2}{x^2}+x+9}}{x^2}dx+100 \int \frac {e^{\frac {2 e^9 \left (x^2+5\right ) \left (5 x+e^x\right )^2}{x^2}+x+9}}{x}dx+100 \int e^{\frac {2 e^9 \left (x^2+5\right ) \left (5 x+e^x\right )^2}{x^2}+9} xdx+20 \int e^{\frac {2 e^9 \left (x^2+5\right ) \left (5 x+e^x\right )^2}{x^2}+x+9} xdx\) |
Int[(E^((E^(9 + 2*x)*(10 + 2*x^2) + E^(9 + x)*(100*x + 20*x^3) + E^9*(250* x^2 + 50*x^4))/x^2)*(100*E^9*x^4 + E^(9 + 2*x)*(-20 + 20*x + 4*x^3) + E^(9 + x)*(-100*x + 100*x^2 + 20*x^3 + 20*x^4)))/x^3,x]
3.16.69.3.1 Defintions of rubi rules used
Time = 1.63 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27
method | result | size |
risch | \({\mathrm e}^{\frac {2 \left (x^{2}+5\right ) \left (25 x^{2} {\mathrm e}^{9}+10 x \,{\mathrm e}^{x +9}+{\mathrm e}^{2 x +9}\right )}{x^{2}}}\) | \(33\) |
parallelrisch | \({\mathrm e}^{\frac {2 \,{\mathrm e}^{9} \left ({\mathrm e}^{2 x} x^{2}+10 \,{\mathrm e}^{x} x^{3}+25 x^{4}+5 \,{\mathrm e}^{2 x}+50 \,{\mathrm e}^{x} x +125 x^{2}\right )}{x^{2}}}\) | \(46\) |
norman | \({\mathrm e}^{\frac {\left (2 x^{2}+10\right ) {\mathrm e}^{9} {\mathrm e}^{2 x}+\left (20 x^{3}+100 x \right ) {\mathrm e}^{9} {\mathrm e}^{x}+\left (50 x^{4}+250 x^{2}\right ) {\mathrm e}^{9}}{x^{2}}}\) | \(49\) |
int(((4*x^3+20*x-20)*exp(9)*exp(x)^2+(20*x^4+20*x^3+100*x^2-100*x)*exp(9)* exp(x)+100*x^4*exp(9))*exp(((2*x^2+10)*exp(9)*exp(x)^2+(20*x^3+100*x)*exp( 9)*exp(x)+(50*x^4+250*x^2)*exp(9))/x^2)/x^3,x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (21) = 42\).
Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \[ \int \frac {e^{\frac {e^{9+2 x} \left (10+2 x^2\right )+e^{9+x} \left (100 x+20 x^3\right )+e^9 \left (250 x^2+50 x^4\right )}{x^2}} \left (100 e^9 x^4+e^{9+2 x} \left (-20+20 x+4 x^3\right )+e^{9+x} \left (-100 x+100 x^2+20 x^3+20 x^4\right )\right )}{x^3} \, dx=e^{\left (\frac {2 \, {\left (25 \, {\left (x^{4} + 5 \, x^{2}\right )} e^{18} + {\left (x^{2} + 5\right )} e^{\left (2 \, x + 18\right )} + 10 \, {\left (x^{3} + 5 \, x\right )} e^{\left (x + 18\right )}\right )} e^{\left (-9\right )}}{x^{2}}\right )} \]
integrate(((4*x^3+20*x-20)*exp(9)*exp(x)^2+(20*x^4+20*x^3+100*x^2-100*x)*e xp(9)*exp(x)+100*x^4*exp(9))*exp(((2*x^2+10)*exp(9)*exp(x)^2+(20*x^3+100*x )*exp(9)*exp(x)+(50*x^4+250*x^2)*exp(9))/x^2)/x^3,x, algorithm=\
e^(2*(25*(x^4 + 5*x^2)*e^18 + (x^2 + 5)*e^(2*x + 18) + 10*(x^3 + 5*x)*e^(x + 18))*e^(-9)/x^2)
Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {e^{\frac {e^{9+2 x} \left (10+2 x^2\right )+e^{9+x} \left (100 x+20 x^3\right )+e^9 \left (250 x^2+50 x^4\right )}{x^2}} \left (100 e^9 x^4+e^{9+2 x} \left (-20+20 x+4 x^3\right )+e^{9+x} \left (-100 x+100 x^2+20 x^3+20 x^4\right )\right )}{x^3} \, dx=e^{\frac {\left (2 x^{2} + 10\right ) e^{9} e^{2 x} + \left (20 x^{3} + 100 x\right ) e^{9} e^{x} + \left (50 x^{4} + 250 x^{2}\right ) e^{9}}{x^{2}}} \]
integrate(((4*x**3+20*x-20)*exp(9)*exp(x)**2+(20*x**4+20*x**3+100*x**2-100 *x)*exp(9)*exp(x)+100*x**4*exp(9))*exp(((2*x**2+10)*exp(9)*exp(x)**2+(20*x **3+100*x)*exp(9)*exp(x)+(50*x**4+250*x**2)*exp(9))/x**2)/x**3,x)
exp(((2*x**2 + 10)*exp(9)*exp(2*x) + (20*x**3 + 100*x)*exp(9)*exp(x) + (50 *x**4 + 250*x**2)*exp(9))/x**2)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (21) = 42\).
Time = 0.39 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \frac {e^{\frac {e^{9+2 x} \left (10+2 x^2\right )+e^{9+x} \left (100 x+20 x^3\right )+e^9 \left (250 x^2+50 x^4\right )}{x^2}} \left (100 e^9 x^4+e^{9+2 x} \left (-20+20 x+4 x^3\right )+e^{9+x} \left (-100 x+100 x^2+20 x^3+20 x^4\right )\right )}{x^3} \, dx=e^{\left (50 \, x^{2} e^{9} + 20 \, x e^{\left (x + 9\right )} + \frac {100 \, e^{\left (x + 9\right )}}{x} + \frac {10 \, e^{\left (2 \, x + 9\right )}}{x^{2}} + 250 \, e^{9} + 2 \, e^{\left (2 \, x + 9\right )}\right )} \]
integrate(((4*x^3+20*x-20)*exp(9)*exp(x)^2+(20*x^4+20*x^3+100*x^2-100*x)*e xp(9)*exp(x)+100*x^4*exp(9))*exp(((2*x^2+10)*exp(9)*exp(x)^2+(20*x^3+100*x )*exp(9)*exp(x)+(50*x^4+250*x^2)*exp(9))/x^2)/x^3,x, algorithm=\
e^(50*x^2*e^9 + 20*x*e^(x + 9) + 100*e^(x + 9)/x + 10*e^(2*x + 9)/x^2 + 25 0*e^9 + 2*e^(2*x + 9))
\[ \int \frac {e^{\frac {e^{9+2 x} \left (10+2 x^2\right )+e^{9+x} \left (100 x+20 x^3\right )+e^9 \left (250 x^2+50 x^4\right )}{x^2}} \left (100 e^9 x^4+e^{9+2 x} \left (-20+20 x+4 x^3\right )+e^{9+x} \left (-100 x+100 x^2+20 x^3+20 x^4\right )\right )}{x^3} \, dx=\int { \frac {4 \, {\left (25 \, x^{4} e^{9} + {\left (x^{3} + 5 \, x - 5\right )} e^{\left (2 \, x + 9\right )} + 5 \, {\left (x^{4} + x^{3} + 5 \, x^{2} - 5 \, x\right )} e^{\left (x + 9\right )}\right )} e^{\left (\frac {2 \, {\left (25 \, {\left (x^{4} + 5 \, x^{2}\right )} e^{9} + {\left (x^{2} + 5\right )} e^{\left (2 \, x + 9\right )} + 10 \, {\left (x^{3} + 5 \, x\right )} e^{\left (x + 9\right )}\right )}}{x^{2}}\right )}}{x^{3}} \,d x } \]
integrate(((4*x^3+20*x-20)*exp(9)*exp(x)^2+(20*x^4+20*x^3+100*x^2-100*x)*e xp(9)*exp(x)+100*x^4*exp(9))*exp(((2*x^2+10)*exp(9)*exp(x)^2+(20*x^3+100*x )*exp(9)*exp(x)+(50*x^4+250*x^2)*exp(9))/x^2)/x^3,x, algorithm=\
integrate(4*(25*x^4*e^9 + (x^3 + 5*x - 5)*e^(2*x + 9) + 5*(x^4 + x^3 + 5*x ^2 - 5*x)*e^(x + 9))*e^(2*(25*(x^4 + 5*x^2)*e^9 + (x^2 + 5)*e^(2*x + 9) + 10*(x^3 + 5*x)*e^(x + 9))/x^2)/x^3, x)
Time = 8.79 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.04 \[ \int \frac {e^{\frac {e^{9+2 x} \left (10+2 x^2\right )+e^{9+x} \left (100 x+20 x^3\right )+e^9 \left (250 x^2+50 x^4\right )}{x^2}} \left (100 e^9 x^4+e^{9+2 x} \left (-20+20 x+4 x^3\right )+e^{9+x} \left (-100 x+100 x^2+20 x^3+20 x^4\right )\right )}{x^3} \, dx={\mathrm {e}}^{50\,x^2\,{\mathrm {e}}^9}\,{\mathrm {e}}^{250\,{\mathrm {e}}^9}\,{\mathrm {e}}^{\frac {10\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^9}{x^2}}\,{\mathrm {e}}^{2\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^9}\,{\mathrm {e}}^{20\,x\,{\mathrm {e}}^9\,{\mathrm {e}}^x}\,{\mathrm {e}}^{\frac {100\,{\mathrm {e}}^9\,{\mathrm {e}}^x}{x}} \]
int((exp((exp(9)*(250*x^2 + 50*x^4) + exp(9)*exp(x)*(100*x + 20*x^3) + exp (2*x)*exp(9)*(2*x^2 + 10))/x^2)*(100*x^4*exp(9) + exp(2*x)*exp(9)*(20*x + 4*x^3 - 20) + exp(9)*exp(x)*(100*x^2 - 100*x + 20*x^3 + 20*x^4)))/x^3,x)