Integrand size = 108, antiderivative size = 27 \[ \int \frac {-30+e^x \left (-200 x-200 x^2\right )+e^x \left (-80 x-80 x^2\right ) \log (x)+e^x \left (-8 x-8 x^2\right ) \log ^2(x)}{225 x-600 e^x x^2+400 e^{2 x} x^3+\left (-120 e^x x^2+160 e^{2 x} x^3\right ) \log (x)+16 e^{2 x} x^3 \log ^2(x)} \, dx=\frac {2}{4 e^x x+\frac {1}{-1+\frac {1}{3} \left (2-\frac {\log (x)}{5}\right )}} \]
Time = 0.35 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-30+e^x \left (-200 x-200 x^2\right )+e^x \left (-80 x-80 x^2\right ) \log (x)+e^x \left (-8 x-8 x^2\right ) \log ^2(x)}{225 x-600 e^x x^2+400 e^{2 x} x^3+\left (-120 e^x x^2+160 e^{2 x} x^3\right ) \log (x)+16 e^{2 x} x^3 \log ^2(x)} \, dx=-\frac {2 (5+\log (x))}{15-20 e^x x-4 e^x x \log (x)} \]
Integrate[(-30 + E^x*(-200*x - 200*x^2) + E^x*(-80*x - 80*x^2)*Log[x] + E^ x*(-8*x - 8*x^2)*Log[x]^2)/(225*x - 600*E^x*x^2 + 400*E^(2*x)*x^3 + (-120* E^x*x^2 + 160*E^(2*x)*x^3)*Log[x] + 16*E^(2*x)*x^3*Log[x]^2),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.
\(\displaystyle \int \frac {e^x \left (-200 x^2-200 x\right )+e^x \left (-8 x^2-8 x\right ) \log ^2(x)+e^x \left (-80 x^2-80 x\right ) \log (x)-30}{400 e^{2 x} x^3+16 e^{2 x} x^3 \log ^2(x)-600 e^x x^2+\left (160 e^{2 x} x^3-120 e^x x^2\right ) \log (x)+225 x} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {2 \left (-100 e^x x (x+1)-4 e^x x (x+1) \log ^2(x)-40 e^x x (x+1) \log (x)-15\right )}{x \left (-20 e^x x-4 e^x x \log (x)+15\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int -\frac {4 e^x x (x+1) \log ^2(x)+40 e^x x (x+1) \log (x)+100 e^x x (x+1)+15}{x \left (-20 e^x x-4 e^x \log (x) x+15\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {4 e^x x (x+1) \log ^2(x)+40 e^x x (x+1) \log (x)+100 e^x x (x+1)+15}{x \left (-20 e^x x-4 e^x \log (x) x+15\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -2 \int \left (\frac {(x+1) (\log (x)+5)}{x \left (20 e^x x+4 e^x \log (x) x-15\right )}+\frac {15 (\log (x) x+5 x+\log (x)+6)}{x \left (20 e^x x+4 e^x \log (x) x-15\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -2 \left (75 \int \frac {1}{\left (20 e^x x+4 e^x \log (x) x-15\right )^2}dx+90 \int \frac {1}{x \left (20 e^x x+4 e^x \log (x) x-15\right )^2}dx+15 \int \frac {\log (x)}{\left (20 e^x x+4 e^x \log (x) x-15\right )^2}dx+15 \int \frac {\log (x)}{x \left (20 e^x x+4 e^x \log (x) x-15\right )^2}dx+5 \int \frac {1}{20 e^x x+4 e^x \log (x) x-15}dx+5 \int \frac {1}{x \left (20 e^x x+4 e^x \log (x) x-15\right )}dx+\int \frac {\log (x)}{20 e^x x+4 e^x \log (x) x-15}dx+\int \frac {\log (x)}{x \left (20 e^x x+4 e^x \log (x) x-15\right )}dx\right )\) |
Int[(-30 + E^x*(-200*x - 200*x^2) + E^x*(-80*x - 80*x^2)*Log[x] + E^x*(-8* x - 8*x^2)*Log[x]^2)/(225*x - 600*E^x*x^2 + 400*E^(2*x)*x^3 + (-120*E^x*x^ 2 + 160*E^(2*x)*x^3)*Log[x] + 16*E^(2*x)*x^3*Log[x]^2),x]
3.7.62.3.1 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 = 0.89 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85
method | result | size |
risch | \(\frac {2 \ln \left (x \right )+10}{4 x \,{\mathrm e}^{x} \ln \left (x \right )+20 \,{\mathrm e}^{x} x -15}\) | \(23\) |
parallelrisch | \(\frac {40 x \,{\mathrm e}^{x} \ln \left (x \right )+200 \,{\mathrm e}^{x} x +30 \ln \left (x \right )}{60 x \,{\mathrm e}^{x} \ln \left (x \right )+300 \,{\mathrm e}^{x} x -225}\) | \(36\) |
int(((-8*x^2-8*x)*exp(x)*ln(x)^2+(-80*x^2-80*x)*exp(x)*ln(x)+(-200*x^2-200 *x)*exp(x)-30)/(16*x^3*exp(x)^2*ln(x)^2+(160*exp(x)^2*x^3-120*exp(x)*x^2)* ln(x)+400*exp(x)^2*x^3-600*exp(x)*x^2+225*x),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-30+e^x \left (-200 x-200 x^2\right )+e^x \left (-80 x-80 x^2\right ) \log (x)+e^x \left (-8 x-8 x^2\right ) \log ^2(x)}{225 x-600 e^x x^2+400 e^{2 x} x^3+\left (-120 e^x x^2+160 e^{2 x} x^3\right ) \log (x)+16 e^{2 x} x^3 \log ^2(x)} \, dx=\frac {2 \, {\left (\log \left (x\right ) + 5\right )}}{4 \, x e^{x} \log \left (x\right ) + 20 \, x e^{x} - 15} \]
integrate(((-8*x^2-8*x)*exp(x)*log(x)^2+(-80*x^2-80*x)*exp(x)*log(x)+(-200 *x^2-200*x)*exp(x)-30)/(16*x^3*exp(x)^2*log(x)^2+(160*exp(x)^2*x^3-120*exp (x)*x^2)*log(x)+400*exp(x)^2*x^3-600*exp(x)*x^2+225*x),x, algorithm=\
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {-30+e^x \left (-200 x-200 x^2\right )+e^x \left (-80 x-80 x^2\right ) \log (x)+e^x \left (-8 x-8 x^2\right ) \log ^2(x)}{225 x-600 e^x x^2+400 e^{2 x} x^3+\left (-120 e^x x^2+160 e^{2 x} x^3\right ) \log (x)+16 e^{2 x} x^3 \log ^2(x)} \, dx=\frac {2 \log {\left (x \right )} + 10}{\left (4 x \log {\left (x \right )} + 20 x\right ) e^{x} - 15} \]
integrate(((-8*x**2-8*x)*exp(x)*ln(x)**2+(-80*x**2-80*x)*exp(x)*ln(x)+(-20 0*x**2-200*x)*exp(x)-30)/(16*x**3*exp(x)**2*ln(x)**2+(160*exp(x)**2*x**3-1 20*exp(x)*x**2)*ln(x)+400*exp(x)**2*x**3-600*exp(x)*x**2+225*x),x)
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-30+e^x \left (-200 x-200 x^2\right )+e^x \left (-80 x-80 x^2\right ) \log (x)+e^x \left (-8 x-8 x^2\right ) \log ^2(x)}{225 x-600 e^x x^2+400 e^{2 x} x^3+\left (-120 e^x x^2+160 e^{2 x} x^3\right ) \log (x)+16 e^{2 x} x^3 \log ^2(x)} \, dx=\frac {2 \, {\left (\log \left (x\right ) + 5\right )}}{4 \, {\left (x \log \left (x\right ) + 5 \, x\right )} e^{x} - 15} \]
integrate(((-8*x^2-8*x)*exp(x)*log(x)^2+(-80*x^2-80*x)*exp(x)*log(x)+(-200 *x^2-200*x)*exp(x)-30)/(16*x^3*exp(x)^2*log(x)^2+(160*exp(x)^2*x^3-120*exp (x)*x^2)*log(x)+400*exp(x)^2*x^3-600*exp(x)*x^2+225*x),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {-30+e^x \left (-200 x-200 x^2\right )+e^x \left (-80 x-80 x^2\right ) \log (x)+e^x \left (-8 x-8 x^2\right ) \log ^2(x)}{225 x-600 e^x x^2+400 e^{2 x} x^3+\left (-120 e^x x^2+160 e^{2 x} x^3\right ) \log (x)+16 e^{2 x} x^3 \log ^2(x)} \, dx=\frac {2 \, {\left (\log \left (x\right ) + 5\right )}}{4 \, x e^{x} \log \left (x\right ) + 20 \, x e^{x} - 15} \]
integrate(((-8*x^2-8*x)*exp(x)*log(x)^2+(-80*x^2-80*x)*exp(x)*log(x)+(-200 *x^2-200*x)*exp(x)-30)/(16*x^3*exp(x)^2*log(x)^2+(160*exp(x)^2*x^3-120*exp (x)*x^2)*log(x)+400*exp(x)^2*x^3-600*exp(x)*x^2+225*x),x, algorithm=\
Timed out. \[ \int \frac {-30+e^x \left (-200 x-200 x^2\right )+e^x \left (-80 x-80 x^2\right ) \log (x)+e^x \left (-8 x-8 x^2\right ) \log ^2(x)}{225 x-600 e^x x^2+400 e^{2 x} x^3+\left (-120 e^x x^2+160 e^{2 x} x^3\right ) \log (x)+16 e^{2 x} x^3 \log ^2(x)} \, dx=\int -\frac {{\mathrm {e}}^x\,\left (8\,x^2+8\,x\right )\,{\ln \left (x\right )}^2+{\mathrm {e}}^x\,\left (80\,x^2+80\,x\right )\,\ln \left (x\right )+{\mathrm {e}}^x\,\left (200\,x^2+200\,x\right )+30}{225\,x-600\,x^2\,{\mathrm {e}}^x-\ln \left (x\right )\,\left (120\,x^2\,{\mathrm {e}}^x-160\,x^3\,{\mathrm {e}}^{2\,x}\right )+400\,x^3\,{\mathrm {e}}^{2\,x}+16\,x^3\,{\mathrm {e}}^{2\,x}\,{\ln \left (x\right )}^2} \,d x \]
int(-(exp(x)*(200*x + 200*x^2) + exp(x)*log(x)*(80*x + 80*x^2) + exp(x)*lo g(x)^2*(8*x + 8*x^2) + 30)/(225*x - 600*x^2*exp(x) - log(x)*(120*x^2*exp(x ) - 160*x^3*exp(2*x)) + 400*x^3*exp(2*x) + 16*x^3*exp(2*x)*log(x)^2),x)