Integrand size = 116, antiderivative size = 26 \[ \int \frac {e^{162-36 x+2 x^2} \left (-17 x^2+180 x^3-20 x^4+e^x \left (-24 x+350 x^2-40 x^3\right )\right )+e^{162-36 x+2 x^2} \left (3 x^2-36 x^3+4 x^4+e^x \left (4 x-70 x^2+8 x^3\right )\right ) \log (x)}{-125+75 \log (x)-15 \log ^2(x)+\log ^3(x)} \, dx=\frac {e^{2 (-9+x)^2} x^2 \left (2 e^x+x\right )}{(-5+\log (x))^2} \]
Time = 0.51 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^{162-36 x+2 x^2} \left (-17 x^2+180 x^3-20 x^4+e^x \left (-24 x+350 x^2-40 x^3\right )\right )+e^{162-36 x+2 x^2} \left (3 x^2-36 x^3+4 x^4+e^x \left (4 x-70 x^2+8 x^3\right )\right ) \log (x)}{-125+75 \log (x)-15 \log ^2(x)+\log ^3(x)} \, dx=\frac {e^{2 (-9+x)^2} x^2 \left (2 e^x+x\right )}{(-5+\log (x))^2} \]
Integrate[(E^(162 - 36*x + 2*x^2)*(-17*x^2 + 180*x^3 - 20*x^4 + E^x*(-24*x + 350*x^2 - 40*x^3)) + E^(162 - 36*x + 2*x^2)*(3*x^2 - 36*x^3 + 4*x^4 + E ^x*(4*x - 70*x^2 + 8*x^3))*Log[x])/(-125 + 75*Log[x] - 15*Log[x]^2 + Log[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.
\(\displaystyle \int \frac {e^{2 x^2-36 x+162} \left (-20 x^4+180 x^3-17 x^2+e^x \left (-40 x^3+350 x^2-24 x\right )\right )+e^{2 x^2-36 x+162} \left (4 x^4-36 x^3+3 x^2+e^x \left (8 x^3-70 x^2+4 x\right )\right ) \log (x)}{\log ^3(x)-15 \log ^2(x)+75 \log (x)-125} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 (x-9)^2} x \left (20 x^3-4 x^3 \log (x)+40 e^x x^2-180 x^2-8 e^x x^2 \log (x)+36 x^2 \log (x)-350 e^x x+17 x+24 e^x+70 e^x x \log (x)-3 x \log (x)-4 e^x \log (x)\right )}{(5-\log (x))^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {4 e^{2 (x-9)^2} x^4 \log (x)}{(\log (x)-5)^3}-\frac {20 e^{2 (x-9)^2} x^4}{(\log (x)-5)^3}-\frac {36 e^{2 (x-9)^2} x^3 \log (x)}{(\log (x)-5)^3}+\frac {180 e^{2 (x-9)^2} x^3}{(\log (x)-5)^3}+\frac {3 e^{2 (x-9)^2} x^2 \log (x)}{(\log (x)-5)^3}-\frac {17 e^{2 (x-9)^2} x^2}{(\log (x)-5)^3}+\frac {2 e^{2 (x-9)^2+x} x \left (-20 x^2+4 x^2 \log (x)+175 x-35 x \log (x)+2 \log (x)-12\right )}{(\log (x)-5)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \int \frac {e^{2 (x-9)^2} x^4}{(\log (x)-5)^2}dx-36 \int \frac {e^{2 (x-9)^2} x^3}{(\log (x)-5)^2}dx-2 \int \frac {e^{2 (x-9)^2} x^2}{(\log (x)-5)^3}dx+3 \int \frac {e^{2 (x-9)^2} x^2}{(\log (x)-5)^2}dx-\frac {2 e^{2 (9-x)^2+x} x \left (-20 x^2+4 x^2 \log (x)+175 x-35 x \log (x)\right )}{(1-4 (9-x)) (5-\log (x))^3}\) |
Int[(E^(162 - 36*x + 2*x^2)*(-17*x^2 + 180*x^3 - 20*x^4 + E^x*(-24*x + 350 *x^2 - 40*x^3)) + E^(162 - 36*x + 2*x^2)*(3*x^2 - 36*x^3 + 4*x^4 + E^x*(4* x - 70*x^2 + 8*x^3))*Log[x])/(-125 + 75*Log[x] - 15*Log[x]^2 + Log[x]^3),x ]
3.23.93.3.1 Defintions of rubi rules used
Time = 1.73 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {x^{2} {\mathrm e}^{2 \left (x -9\right )^{2}} \left (2 \,{\mathrm e}^{x}+x \right )}{\left (\ln \left (x \right )-5\right )^{2}}\) | \(25\) |
parallelrisch | \(-\frac {-10 \,{\mathrm e}^{2 x^{2}-36 x +162} x^{3}-20 \,{\mathrm e}^{2 x^{2}-36 x +162} x^{2} {\mathrm e}^{x}}{10 \left (\ln \left (x \right )^{2}-10 \ln \left (x \right )+25\right )}\) | \(50\) |
int((((8*x^3-70*x^2+4*x)*exp(x)+4*x^4-36*x^3+3*x^2)*exp(x^2-18*x+81)^2*ln( x)+((-40*x^3+350*x^2-24*x)*exp(x)-20*x^4+180*x^3-17*x^2)*exp(x^2-18*x+81)^ 2)/(ln(x)^3-15*ln(x)^2+75*ln(x)-125),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {e^{162-36 x+2 x^2} \left (-17 x^2+180 x^3-20 x^4+e^x \left (-24 x+350 x^2-40 x^3\right )\right )+e^{162-36 x+2 x^2} \left (3 x^2-36 x^3+4 x^4+e^x \left (4 x-70 x^2+8 x^3\right )\right ) \log (x)}{-125+75 \log (x)-15 \log ^2(x)+\log ^3(x)} \, dx=\frac {{\left (x^{3} + 2 \, x^{2} e^{x}\right )} e^{\left (2 \, x^{2} - 36 \, x + 162\right )}}{\log \left (x\right )^{2} - 10 \, \log \left (x\right ) + 25} \]
integrate((((8*x^3-70*x^2+4*x)*exp(x)+4*x^4-36*x^3+3*x^2)*exp(x^2-18*x+81) ^2*log(x)+((-40*x^3+350*x^2-24*x)*exp(x)-20*x^4+180*x^3-17*x^2)*exp(x^2-18 *x+81)^2)/(log(x)^3-15*log(x)^2+75*log(x)-125),x, algorithm=\
Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {e^{162-36 x+2 x^2} \left (-17 x^2+180 x^3-20 x^4+e^x \left (-24 x+350 x^2-40 x^3\right )\right )+e^{162-36 x+2 x^2} \left (3 x^2-36 x^3+4 x^4+e^x \left (4 x-70 x^2+8 x^3\right )\right ) \log (x)}{-125+75 \log (x)-15 \log ^2(x)+\log ^3(x)} \, dx=\frac {\left (x^{3} + 2 x^{2} e^{x}\right ) e^{2 x^{2} - 36 x + 162}}{\log {\left (x \right )}^{2} - 10 \log {\left (x \right )} + 25} \]
integrate((((8*x**3-70*x**2+4*x)*exp(x)+4*x**4-36*x**3+3*x**2)*exp(x**2-18 *x+81)**2*ln(x)+((-40*x**3+350*x**2-24*x)*exp(x)-20*x**4+180*x**3-17*x**2) *exp(x**2-18*x+81)**2)/(ln(x)**3-15*ln(x)**2+75*ln(x)-125),x)
Time = 0.26 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {e^{162-36 x+2 x^2} \left (-17 x^2+180 x^3-20 x^4+e^x \left (-24 x+350 x^2-40 x^3\right )\right )+e^{162-36 x+2 x^2} \left (3 x^2-36 x^3+4 x^4+e^x \left (4 x-70 x^2+8 x^3\right )\right ) \log (x)}{-125+75 \log (x)-15 \log ^2(x)+\log ^3(x)} \, dx=\frac {{\left (x^{3} e^{162} + 2 \, x^{2} e^{\left (x + 162\right )}\right )} e^{\left (2 \, x^{2} - 36 \, x\right )}}{\log \left (x\right )^{2} - 10 \, \log \left (x\right ) + 25} \]
integrate((((8*x^3-70*x^2+4*x)*exp(x)+4*x^4-36*x^3+3*x^2)*exp(x^2-18*x+81) ^2*log(x)+((-40*x^3+350*x^2-24*x)*exp(x)-20*x^4+180*x^3-17*x^2)*exp(x^2-18 *x+81)^2)/(log(x)^3-15*log(x)^2+75*log(x)-125),x, algorithm=\
Time = 0.33 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {e^{162-36 x+2 x^2} \left (-17 x^2+180 x^3-20 x^4+e^x \left (-24 x+350 x^2-40 x^3\right )\right )+e^{162-36 x+2 x^2} \left (3 x^2-36 x^3+4 x^4+e^x \left (4 x-70 x^2+8 x^3\right )\right ) \log (x)}{-125+75 \log (x)-15 \log ^2(x)+\log ^3(x)} \, dx=\frac {x^{3} e^{\left (2 \, x^{2} - 36 \, x + 162\right )} + 2 \, x^{2} e^{\left (2 \, x^{2} - 35 \, x + 162\right )}}{\log \left (x\right )^{2} - 10 \, \log \left (x\right ) + 25} \]
integrate((((8*x^3-70*x^2+4*x)*exp(x)+4*x^4-36*x^3+3*x^2)*exp(x^2-18*x+81) ^2*log(x)+((-40*x^3+350*x^2-24*x)*exp(x)-20*x^4+180*x^3-17*x^2)*exp(x^2-18 *x+81)^2)/(log(x)^3-15*log(x)^2+75*log(x)-125),x, algorithm=\
Time = 14.70 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^{162-36 x+2 x^2} \left (-17 x^2+180 x^3-20 x^4+e^x \left (-24 x+350 x^2-40 x^3\right )\right )+e^{162-36 x+2 x^2} \left (3 x^2-36 x^3+4 x^4+e^x \left (4 x-70 x^2+8 x^3\right )\right ) \log (x)}{-125+75 \log (x)-15 \log ^2(x)+\log ^3(x)} \, dx=\frac {x^2\,{\mathrm {e}}^{2\,x^2-36\,x+162}\,\left (x+2\,{\mathrm {e}}^x\right )}{{\left (\ln \left (x\right )-5\right )}^2} \]