Integrand size = 118, antiderivative size = 28 \[ \int \frac {-20000 e^{2 e^x} x+1600 x^2+288 x^3+\left (100000 x+5500 x^2\right ) \log (3)-500000 x \log ^2(3)+e^{e^x} \left (20000 x+1100 x^2+e^x \left (-10000 x^2-2500 x^3\right )-200000 x \log (3)\right )}{625 e^{2 e^x}+16 x^2+1000 x \log (3)+15625 \log ^2(3)+e^{e^x} (200 x+6250 \log (3))} \, dx=4 x^2 \left (-4+\frac {4+x}{e^{e^x}+\frac {4 x}{25}+5 \log (3)}\right ) \] Output:
x^2*(4*(4+x)/(exp(exp(x))+5*ln(3)+4/25*x)-16)
Leaf count is larger than twice the leaf count of optimal. \(91\) vs. \(2(28)=56\).
Time = 0.08 (sec) , antiderivative size = 91, normalized size of antiderivative = 3.25 \[ \int \frac {-20000 e^{2 e^x} x+1600 x^2+288 x^3+\left (100000 x+5500 x^2\right ) \log (3)-500000 x \log ^2(3)+e^{e^x} \left (20000 x+1100 x^2+e^x \left (-10000 x^2-2500 x^3\right )-200000 x \log (3)\right )}{625 e^{2 e^x}+16 x^2+1000 x \log (3)+15625 \log ^2(3)+e^{e^x} (200 x+6250 \log (3))} \, dx=-\frac {4 x^2 \left (-36 e^x x^2+x \left (36+400 e^{e^x+x}+25 e^x (-16+35 \log (3))\right )+100 \left (-4+125 e^x \log (3)\right ) \left (-1+e^{e^x}+\log (243)\right )\right )}{\left (25 e^{e^x}+4 x+125 \log (3)\right ) \left (-4+e^x (4 x+125 \log (3))\right )} \] Input:
Integrate[(-20000*E^(2*E^x)*x + 1600*x^2 + 288*x^3 + (100000*x + 5500*x^2) *Log[3] - 500000*x*Log[3]^2 + E^E^x*(20000*x + 1100*x^2 + E^x*(-10000*x^2 - 2500*x^3) - 200000*x*Log[3]))/(625*E^(2*E^x) + 16*x^2 + 1000*x*Log[3] + 15625*Log[3]^2 + E^E^x*(200*x + 6250*Log[3])),x]
Output:
(-4*x^2*(-36*E^x*x^2 + x*(36 + 400*E^(E^x + x) + 25*E^x*(-16 + 35*Log[3])) + 100*(-4 + 125*E^x*Log[3])*(-1 + E^E^x + Log[243])))/((25*E^E^x + 4*x + 125*Log[3])*(-4 + E^x*(4*x + 125*Log[3])))
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.
| \(\displaystyle \int \frac {288 x^3+1600 x^2+\left (5500 x^2+100000 x\right ) \log (3)+e^{e^x} \left (1100 x^2+e^x \left (-2500 x^3-10000 x^2\right )+20000 x-200000 x \log (3)\right )-20000 e^{2 e^x} x-500000 x \log ^2(3)}{16 x^2+625 e^{2 e^x}+1000 x \log (3)+e^{e^x} (200 x+6250 \log (3))+15625 \log ^2(3)} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int \frac {288 x^3+1600 x^2+\left (5500 x^2+100000 x\right ) \log (3)+e^{e^x} \left (1100 x^2+e^x \left (-2500 x^3-10000 x^2\right )+20000 x-200000 x \log (3)\right )-20000 e^{2 e^x} x-500000 x \log ^2(3)}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {288 x^3}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}+\frac {1100 e^{e^x} x^2}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}-\frac {2500 e^{x+e^x} (x+4) x^2}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}+\frac {1600 x^2}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}-\frac {500000 x \log ^2(3)}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}-\frac {20000 e^{2 e^x} x}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}+\frac {500 (11 x+200) x \log (3)}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}+\frac {20000 e^{e^x} x (1-10 \log (3))}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 288 \int \frac {x^3}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}dx-2500 \int \frac {e^{x+e^x} x^3}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}dx+5500 \log (3) \int \frac {x^2}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}dx+1600 \int \frac {x^2}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}dx+1100 \int \frac {e^{e^x} x^2}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}dx-10000 \int \frac {e^{x+e^x} x^2}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}dx-500000 \log ^2(3) \int \frac {x}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}dx+100000 \log (3) \int \frac {x}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}dx+20000 (1-10 \log (3)) \int \frac {e^{e^x} x}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}dx-20000 \int \frac {e^{2 e^x} x}{\left (4 x+25 e^{e^x}+125 \log (3)\right )^2}dx\) |
Input:
Int[(-20000*E^(2*E^x)*x + 1600*x^2 + 288*x^3 + (100000*x + 5500*x^2)*Log[3 ] - 500000*x*Log[3]^2 + E^E^x*(20000*x + 1100*x^2 + E^x*(-10000*x^2 - 2500 *x^3) - 200000*x*Log[3]))/(625*E^(2*E^x) + 16*x^2 + 1000*x*Log[3] + 15625* Log[3]^2 + E^E^x*(200*x + 6250*Log[3])),x]
Output:
$Aborted
Time = 0.85 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07
| method | result | size |
| risch | \(-16 x^{2}+\frac {100 x^{2} \left (4+x \right )}{4 x +25 \,{\mathrm e}^{{\mathrm e}^{x}}+125 \ln \left (3\right )}\) | \(30\) |
| norman | \(\frac {\left (-2000 \ln \left (3\right )+400\right ) x^{2}+36 x^{3}-400 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}}{4 x +25 \,{\mathrm e}^{{\mathrm e}^{x}}+125 \ln \left (3\right )}\) | \(41\) |
| parallelrisch | \(-\frac {8000 x^{2} \ln \left (3\right )-144 x^{3}+1600 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-1600 x^{2}}{4 \left (4 x +25 \,{\mathrm e}^{{\mathrm e}^{x}}+125 \ln \left (3\right )\right )}\) | \(44\) |
Input:
int((-20000*x*exp(exp(x))^2+((-2500*x^3-10000*x^2)*exp(x)-200000*x*ln(3)+1 100*x^2+20000*x)*exp(exp(x))-500000*x*ln(3)^2+(5500*x^2+100000*x)*ln(3)+28 8*x^3+1600*x^2)/(625*exp(exp(x))^2+(6250*ln(3)+200*x)*exp(exp(x))+15625*ln (3)^2+1000*x*ln(3)+16*x^2),x,method=_RETURNVERBOSE)
Output:
-16*x^2+100*x^2*(4+x)/(4*x+25*exp(exp(x))+125*ln(3))
Time = 0.07 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {-20000 e^{2 e^x} x+1600 x^2+288 x^3+\left (100000 x+5500 x^2\right ) \log (3)-500000 x \log ^2(3)+e^{e^x} \left (20000 x+1100 x^2+e^x \left (-10000 x^2-2500 x^3\right )-200000 x \log (3)\right )}{625 e^{2 e^x}+16 x^2+1000 x \log (3)+15625 \log ^2(3)+e^{e^x} (200 x+6250 \log (3))} \, dx=\frac {4 \, {\left (9 \, x^{3} - 100 \, x^{2} e^{\left (e^{x}\right )} - 500 \, x^{2} \log \left (3\right ) + 100 \, x^{2}\right )}}{4 \, x + 25 \, e^{\left (e^{x}\right )} + 125 \, \log \left (3\right )} \] Input:
integrate((-20000*x*exp(exp(x))^2+((-2500*x^3-10000*x^2)*exp(x)-200000*x*l og(3)+1100*x^2+20000*x)*exp(exp(x))-500000*x*log(3)^2+(5500*x^2+100000*x)* log(3)+288*x^3+1600*x^2)/(625*exp(exp(x))^2+(6250*log(3)+200*x)*exp(exp(x) )+15625*log(3)^2+1000*x*log(3)+16*x^2),x, algorithm="fricas")
Output:
4*(9*x^3 - 100*x^2*e^(e^x) - 500*x^2*log(3) + 100*x^2)/(4*x + 25*e^(e^x) + 125*log(3))
Time = 0.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {-20000 e^{2 e^x} x+1600 x^2+288 x^3+\left (100000 x+5500 x^2\right ) \log (3)-500000 x \log ^2(3)+e^{e^x} \left (20000 x+1100 x^2+e^x \left (-10000 x^2-2500 x^3\right )-200000 x \log (3)\right )}{625 e^{2 e^x}+16 x^2+1000 x \log (3)+15625 \log ^2(3)+e^{e^x} (200 x+6250 \log (3))} \, dx=- 16 x^{2} + \frac {4 x^{3} + 16 x^{2}}{\frac {4 x}{25} + e^{e^{x}} + 5 \log {\left (3 \right )}} \] Input:
integrate((-20000*x*exp(exp(x))**2+((-2500*x**3-10000*x**2)*exp(x)-200000* x*ln(3)+1100*x**2+20000*x)*exp(exp(x))-500000*x*ln(3)**2+(5500*x**2+100000 *x)*ln(3)+288*x**3+1600*x**2)/(625*exp(exp(x))**2+(6250*ln(3)+200*x)*exp(e xp(x))+15625*ln(3)**2+1000*x*ln(3)+16*x**2),x)
Output:
-16*x**2 + (4*x**3 + 16*x**2)/(4*x/25 + exp(exp(x)) + 5*log(3))
Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {-20000 e^{2 e^x} x+1600 x^2+288 x^3+\left (100000 x+5500 x^2\right ) \log (3)-500000 x \log ^2(3)+e^{e^x} \left (20000 x+1100 x^2+e^x \left (-10000 x^2-2500 x^3\right )-200000 x \log (3)\right )}{625 e^{2 e^x}+16 x^2+1000 x \log (3)+15625 \log ^2(3)+e^{e^x} (200 x+6250 \log (3))} \, dx=\frac {4 \, {\left (9 \, x^{3} - 100 \, x^{2} {\left (5 \, \log \left (3\right ) - 1\right )} - 100 \, x^{2} e^{\left (e^{x}\right )}\right )}}{4 \, x + 25 \, e^{\left (e^{x}\right )} + 125 \, \log \left (3\right )} \] Input:
integrate((-20000*x*exp(exp(x))^2+((-2500*x^3-10000*x^2)*exp(x)-200000*x*l og(3)+1100*x^2+20000*x)*exp(exp(x))-500000*x*log(3)^2+(5500*x^2+100000*x)* log(3)+288*x^3+1600*x^2)/(625*exp(exp(x))^2+(6250*log(3)+200*x)*exp(exp(x) )+15625*log(3)^2+1000*x*log(3)+16*x^2),x, algorithm="maxima")
Output:
4*(9*x^3 - 100*x^2*(5*log(3) - 1) - 100*x^2*e^(e^x))/(4*x + 25*e^(e^x) + 1 25*log(3))
Time = 0.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {-20000 e^{2 e^x} x+1600 x^2+288 x^3+\left (100000 x+5500 x^2\right ) \log (3)-500000 x \log ^2(3)+e^{e^x} \left (20000 x+1100 x^2+e^x \left (-10000 x^2-2500 x^3\right )-200000 x \log (3)\right )}{625 e^{2 e^x}+16 x^2+1000 x \log (3)+15625 \log ^2(3)+e^{e^x} (200 x+6250 \log (3))} \, dx=\frac {4 \, {\left (9 \, x^{3} - 100 \, x^{2} e^{\left (e^{x}\right )} - 500 \, x^{2} \log \left (3\right ) + 100 \, x^{2}\right )}}{4 \, x + 25 \, e^{\left (e^{x}\right )} + 125 \, \log \left (3\right )} \] Input:
integrate((-20000*x*exp(exp(x))^2+((-2500*x^3-10000*x^2)*exp(x)-200000*x*l og(3)+1100*x^2+20000*x)*exp(exp(x))-500000*x*log(3)^2+(5500*x^2+100000*x)* log(3)+288*x^3+1600*x^2)/(625*exp(exp(x))^2+(6250*log(3)+200*x)*exp(exp(x) )+15625*log(3)^2+1000*x*log(3)+16*x^2),x, algorithm="giac")
Output:
4*(9*x^3 - 100*x^2*e^(e^x) - 500*x^2*log(3) + 100*x^2)/(4*x + 25*e^(e^x) + 125*log(3))
Timed out. \[ \int \frac {-20000 e^{2 e^x} x+1600 x^2+288 x^3+\left (100000 x+5500 x^2\right ) \log (3)-500000 x \log ^2(3)+e^{e^x} \left (20000 x+1100 x^2+e^x \left (-10000 x^2-2500 x^3\right )-200000 x \log (3)\right )}{625 e^{2 e^x}+16 x^2+1000 x \log (3)+15625 \log ^2(3)+e^{e^x} (200 x+6250 \log (3))} \, dx=\int \frac {\ln \left (3\right )\,\left (5500\,x^2+100000\,x\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (20000\,x-{\mathrm {e}}^x\,\left (2500\,x^3+10000\,x^2\right )-200000\,x\,\ln \left (3\right )+1100\,x^2\right )-20000\,x\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}-500000\,x\,{\ln \left (3\right )}^2+1600\,x^2+288\,x^3}{625\,{\mathrm {e}}^{2\,{\mathrm {e}}^x}+1000\,x\,\ln \left (3\right )+{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (200\,x+6250\,\ln \left (3\right )\right )+15625\,{\ln \left (3\right )}^2+16\,x^2} \,d x \] Input:
int((log(3)*(100000*x + 5500*x^2) + exp(exp(x))*(20000*x - exp(x)*(10000*x ^2 + 2500*x^3) - 200000*x*log(3) + 1100*x^2) - 20000*x*exp(2*exp(x)) - 500 000*x*log(3)^2 + 1600*x^2 + 288*x^3)/(625*exp(2*exp(x)) + 1000*x*log(3) + exp(exp(x))*(200*x + 6250*log(3)) + 15625*log(3)^2 + 16*x^2),x)
Output:
int((log(3)*(100000*x + 5500*x^2) + exp(exp(x))*(20000*x - exp(x)*(10000*x ^2 + 2500*x^3) - 200000*x*log(3) + 1100*x^2) - 20000*x*exp(2*exp(x)) - 500 000*x*log(3)^2 + 1600*x^2 + 288*x^3)/(625*exp(2*exp(x)) + 1000*x*log(3) + exp(exp(x))*(200*x + 6250*log(3)) + 15625*log(3)^2 + 16*x^2), x)
Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.36 \[ \int \frac {-20000 e^{2 e^x} x+1600 x^2+288 x^3+\left (100000 x+5500 x^2\right ) \log (3)-500000 x \log ^2(3)+e^{e^x} \left (20000 x+1100 x^2+e^x \left (-10000 x^2-2500 x^3\right )-200000 x \log (3)\right )}{625 e^{2 e^x}+16 x^2+1000 x \log (3)+15625 \log ^2(3)+e^{e^x} (200 x+6250 \log (3))} \, dx=\frac {4 x^{2} \left (-100 e^{e^{x}}-500 \,\mathrm {log}\left (3\right )+9 x +100\right )}{25 e^{e^{x}}+125 \,\mathrm {log}\left (3\right )+4 x} \] Input:
int((-20000*x*exp(exp(x))^2+((-2500*x^3-10000*x^2)*exp(x)-200000*x*log(3)+ 1100*x^2+20000*x)*exp(exp(x))-500000*x*log(3)^2+(5500*x^2+100000*x)*log(3) +288*x^3+1600*x^2)/(625*exp(exp(x))^2+(6250*log(3)+200*x)*exp(exp(x))+1562 5*log(3)^2+1000*x*log(3)+16*x^2),x)
Output:
(4*x**2*( - 100*e**(e**x) - 500*log(3) + 9*x + 100))/(25*e**(e**x) + 125*l og(3) + 4*x)