Integrand size = 108, antiderivative size = 22 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 e^x}{(-3+x) \log (x) (-1+x \log (x))} \]
Time = 0.69 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 e^x}{(-3+x) \log (x) (-1+x \log (x))} \]
Integrate[(E^x*(-60 + 20*x) + E^x*(200*x - 60*x^2)*Log[x] + E^x*(60*x - 10 0*x^2 + 20*x^3)*Log[x]^2)/((9*x - 6*x^2 + x^3)*Log[x]^2 + (-18*x^2 + 12*x^ 3 - 2*x^4)*Log[x]^3 + (9*x^3 - 6*x^4 + x^5)*Log[x]^4),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.
\(\displaystyle \int \frac {e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (20 x^3-100 x^2+60 x\right ) \log ^2(x)+e^x (20 x-60)}{\left (x^3-6 x^2+9 x\right ) \log ^2(x)+\left (x^5-6 x^4+9 x^3\right ) \log ^4(x)+\left (-2 x^4+12 x^3-18 x^2\right ) \log ^3(x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {20 e^x \left (x \left (x^2-5 x+3\right ) \log ^2(x)+x+(10-3 x) x \log (x)-3\right )}{(3-x)^2 x \log ^2(x) (1-x \log (x))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 20 \int -\frac {e^x \left (-x \left (x^2-5 x+3\right ) \log ^2(x)-(10-3 x) x \log (x)-x+3\right )}{(3-x)^2 x \log ^2(x) (1-x \log (x))^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -20 \int \frac {e^x \left (-x \left (x^2-5 x+3\right ) \log ^2(x)-(10-3 x) x \log (x)-x+3\right )}{(3-x)^2 x \log ^2(x) (1-x \log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -20 \int \left (\frac {e^x (x-4)}{(x-3)^2 \log (x)}-\frac {e^x x (x-4)}{(x-3)^2 (x \log (x)-1)}-\frac {e^x}{(x-3) x \log ^2(x)}+\frac {e^x (x+1)}{(x-3) (x \log (x)-1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -20 \left (-\frac {1}{3} \int \frac {e^x}{(x-3) \log ^2(x)}dx+\frac {1}{3} \int \frac {e^x}{x \log ^2(x)}dx-\int \frac {e^x}{(x-3)^2 \log (x)}dx+\int \frac {e^x}{(x-3) \log (x)}dx+\int \frac {e^x}{(x \log (x)-1)^2}dx+4 \int \frac {e^x}{(x-3) (x \log (x)-1)^2}dx-\int \frac {e^x}{x \log (x)-1}dx+3 \int \frac {e^x}{(x-3)^2 (x \log (x)-1)}dx-2 \int \frac {e^x}{(x-3) (x \log (x)-1)}dx\right )\) |
Int[(E^x*(-60 + 20*x) + E^x*(200*x - 60*x^2)*Log[x] + E^x*(60*x - 100*x^2 + 20*x^3)*Log[x]^2)/((9*x - 6*x^2 + x^3)*Log[x]^2 + (-18*x^2 + 12*x^3 - 2* x^4)*Log[x]^3 + (9*x^3 - 6*x^4 + x^5)*Log[x]^4),x]
3.6.6.3.1 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 = 1.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00
method | result | size |
risch | \(\frac {20 \,{\mathrm e}^{x}}{\left (-3+x \right ) \ln \left (x \right ) \left (x \ln \left (x \right )-1\right )}\) | \(22\) |
parallelrisch | \(\frac {20 \,{\mathrm e}^{x}}{\ln \left (x \right ) \left (x^{2} \ln \left (x \right )-3 x \ln \left (x \right )-x +3\right )}\) | \(27\) |
int(((20*x^3-100*x^2+60*x)*exp(x)*ln(x)^2+(-60*x^2+200*x)*exp(x)*ln(x)+(20 *x-60)*exp(x))/((x^5-6*x^4+9*x^3)*ln(x)^4+(-2*x^4+12*x^3-18*x^2)*ln(x)^3+( x^3-6*x^2+9*x)*ln(x)^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 \, e^{x}}{{\left (x^{2} - 3 \, x\right )} \log \left (x\right )^{2} - {\left (x - 3\right )} \log \left (x\right )} \]
integrate(((20*x^3-100*x^2+60*x)*exp(x)*log(x)^2+(-60*x^2+200*x)*exp(x)*lo g(x)+(20*x-60)*exp(x))/((x^5-6*x^4+9*x^3)*log(x)^4+(-2*x^4+12*x^3-18*x^2)* log(x)^3+(x^3-6*x^2+9*x)*log(x)^2),x, algorithm=\
Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 e^{x}}{x^{2} \log {\left (x \right )}^{2} - 3 x \log {\left (x \right )}^{2} - x \log {\left (x \right )} + 3 \log {\left (x \right )}} \]
integrate(((20*x**3-100*x**2+60*x)*exp(x)*ln(x)**2+(-60*x**2+200*x)*exp(x) *ln(x)+(20*x-60)*exp(x))/((x**5-6*x**4+9*x**3)*ln(x)**4+(-2*x**4+12*x**3-1 8*x**2)*ln(x)**3+(x**3-6*x**2+9*x)*ln(x)**2),x)
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 \, e^{x}}{{\left (x^{2} - 3 \, x\right )} \log \left (x\right )^{2} - {\left (x - 3\right )} \log \left (x\right )} \]
integrate(((20*x^3-100*x^2+60*x)*exp(x)*log(x)^2+(-60*x^2+200*x)*exp(x)*lo g(x)+(20*x-60)*exp(x))/((x^5-6*x^4+9*x^3)*log(x)^4+(-2*x^4+12*x^3-18*x^2)* log(x)^3+(x^3-6*x^2+9*x)*log(x)^2),x, algorithm=\
Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=\frac {20 \, e^{x}}{x^{2} \log \left (x\right )^{2} - 3 \, x \log \left (x\right )^{2} - x \log \left (x\right ) + 3 \, \log \left (x\right )} \]
integrate(((20*x^3-100*x^2+60*x)*exp(x)*log(x)^2+(-60*x^2+200*x)*exp(x)*lo g(x)+(20*x-60)*exp(x))/((x^5-6*x^4+9*x^3)*log(x)^4+(-2*x^4+12*x^3-18*x^2)* log(x)^3+(x^3-6*x^2+9*x)*log(x)^2),x, algorithm=\
Time = 7.40 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {e^x (-60+20 x)+e^x \left (200 x-60 x^2\right ) \log (x)+e^x \left (60 x-100 x^2+20 x^3\right ) \log ^2(x)}{\left (9 x-6 x^2+x^3\right ) \log ^2(x)+\left (-18 x^2+12 x^3-2 x^4\right ) \log ^3(x)+\left (9 x^3-6 x^4+x^5\right ) \log ^4(x)} \, dx=-\frac {20\,{\mathrm {e}}^x}{\left (\ln \left (x\right )-x\,{\ln \left (x\right )}^2\right )\,\left (x-3\right )} \]