Integrand size = 103, antiderivative size = 28 \[ \int \frac {-140 x+130 x^3+20 x^4+10 x^6+\left (40 x-40 x^3\right ) \log (x)+e^{e^x} \left (-70 x+10 x^4+e^x \left (-30 x^2-10 x^5\right )+\left (20 x+10 e^x x^2\right ) \log (x)\right )}{9+6 x^3+x^6+\left (-6-2 x^3\right ) \log (x)+\log ^2(x)} \, dx=\frac {10 x^2 \left (2+e^{e^x}-x^2\right )}{-3-x^3+\log (x)} \]
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-140 x+130 x^3+20 x^4+10 x^6+\left (40 x-40 x^3\right ) \log (x)+e^{e^x} \left (-70 x+10 x^4+e^x \left (-30 x^2-10 x^5\right )+\left (20 x+10 e^x x^2\right ) \log (x)\right )}{9+6 x^3+x^6+\left (-6-2 x^3\right ) \log (x)+\log ^2(x)} \, dx=\frac {10 x^2 \left (-2-e^{e^x}+x^2\right )}{3+x^3-\log (x)} \]
Integrate[(-140*x + 130*x^3 + 20*x^4 + 10*x^6 + (40*x - 40*x^3)*Log[x] + E ^E^x*(-70*x + 10*x^4 + E^x*(-30*x^2 - 10*x^5) + (20*x + 10*E^x*x^2)*Log[x] ))/(9 + 6*x^3 + x^6 + (-6 - 2*x^3)*Log[x] + 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.
\(\displaystyle \int \frac {10 x^6+20 x^4+130 x^3+\left (40 x-40 x^3\right ) \log (x)+e^{e^x} \left (10 x^4+\left (10 e^x x^2+20 x\right ) \log (x)+e^x \left (-10 x^5-30 x^2\right )-70 x\right )-140 x}{x^6+6 x^3+\left (-2 x^3-6\right ) \log (x)+\log ^2(x)+9} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {10 x^6+20 x^4+130 x^3+\left (40 x-40 x^3\right ) \log (x)+e^{e^x} \left (10 x^4+\left (10 e^x x^2+20 x\right ) \log (x)+e^x \left (-10 x^5-30 x^2\right )-70 x\right )-140 x}{\left (x^3-\log (x)+3\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {130 x^3}{\left (x^3-\log (x)+3\right )^2}+\frac {20 e^{e^x} x \log (x)}{\left (x^3-\log (x)+3\right )^2}-\frac {40 (x-1) (x+1) x \log (x)}{\left (x^3-\log (x)+3\right )^2}-\frac {70 e^{e^x} x}{\left (x^3-\log (x)+3\right )^2}-\frac {140 x}{\left (x^3-\log (x)+3\right )^2}+\frac {10 x^6}{\left (x^3-\log (x)+3\right )^2}+\frac {10 e^{e^x} x^4}{\left (x^3-\log (x)+3\right )^2}+\frac {20 x^4}{\left (x^3-\log (x)+3\right )^2}-\frac {10 e^{x+e^x} x^2}{x^3-\log (x)+3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -20 \int \frac {x}{\left (x^3-\log (x)+3\right )^2}dx-10 \int \frac {e^{e^x} x}{\left (x^3-\log (x)+3\right )^2}dx+10 \int \frac {x^3}{\left (x^3-\log (x)+3\right )^2}dx-40 \int \frac {x}{x^3-\log (x)+3}dx-20 \int \frac {e^{e^x} x}{x^3-\log (x)+3}dx+40 \int \frac {x^3}{x^3-\log (x)+3}dx-30 \int \frac {x^6}{\left (x^3-\log (x)+3\right )^2}dx+60 \int \frac {x^4}{\left (x^3-\log (x)+3\right )^2}dx+30 \int \frac {e^{e^x} x^4}{\left (x^3-\log (x)+3\right )^2}dx-10 \int \frac {e^{x+e^x} x^2}{x^3-\log (x)+3}dx\) |
Int[(-140*x + 130*x^3 + 20*x^4 + 10*x^6 + (40*x - 40*x^3)*Log[x] + E^E^x*( -70*x + 10*x^4 + E^x*(-30*x^2 - 10*x^5) + (20*x + 10*E^x*x^2)*Log[x]))/(9 + 6*x^3 + x^6 + (-6 - 2*x^3)*Log[x] + Log[x]^2),x]
3.28.70.3.1 Defintions of rubi rules used
Time = 6.38 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14
method | result | size |
parallelrisch | \(\frac {10 x^{4}-10 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}-20 x^{2}}{x^{3}-\ln \left (x \right )+3}\) | \(32\) |
int((((10*exp(x)*x^2+20*x)*ln(x)+(-10*x^5-30*x^2)*exp(x)+10*x^4-70*x)*exp( exp(x))+(-40*x^3+40*x)*ln(x)+10*x^6+20*x^4+130*x^3-140*x)/(ln(x)^2+(-2*x^3 -6)*ln(x)+x^6+6*x^3+9),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {-140 x+130 x^3+20 x^4+10 x^6+\left (40 x-40 x^3\right ) \log (x)+e^{e^x} \left (-70 x+10 x^4+e^x \left (-30 x^2-10 x^5\right )+\left (20 x+10 e^x x^2\right ) \log (x)\right )}{9+6 x^3+x^6+\left (-6-2 x^3\right ) \log (x)+\log ^2(x)} \, dx=\frac {10 \, {\left (x^{4} - x^{2} e^{\left (e^{x}\right )} - 2 \, x^{2}\right )}}{x^{3} - \log \left (x\right ) + 3} \]
integrate((((10*exp(x)*x^2+20*x)*log(x)+(-10*x^5-30*x^2)*exp(x)+10*x^4-70* x)*exp(exp(x))+(-40*x^3+40*x)*log(x)+10*x^6+20*x^4+130*x^3-140*x)/(log(x)^ 2+(-2*x^3-6)*log(x)+x^6+6*x^3+9),x, algorithm=\
Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {-140 x+130 x^3+20 x^4+10 x^6+\left (40 x-40 x^3\right ) \log (x)+e^{e^x} \left (-70 x+10 x^4+e^x \left (-30 x^2-10 x^5\right )+\left (20 x+10 e^x x^2\right ) \log (x)\right )}{9+6 x^3+x^6+\left (-6-2 x^3\right ) \log (x)+\log ^2(x)} \, dx=- \frac {10 x^{2} e^{e^{x}}}{x^{3} - \log {\left (x \right )} + 3} + \frac {- 10 x^{4} + 20 x^{2}}{- x^{3} + \log {\left (x \right )} - 3} \]
integrate((((10*exp(x)*x**2+20*x)*ln(x)+(-10*x**5-30*x**2)*exp(x)+10*x**4- 70*x)*exp(exp(x))+(-40*x**3+40*x)*ln(x)+10*x**6+20*x**4+130*x**3-140*x)/(l n(x)**2+(-2*x**3-6)*ln(x)+x**6+6*x**3+9),x)
Time = 0.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {-140 x+130 x^3+20 x^4+10 x^6+\left (40 x-40 x^3\right ) \log (x)+e^{e^x} \left (-70 x+10 x^4+e^x \left (-30 x^2-10 x^5\right )+\left (20 x+10 e^x x^2\right ) \log (x)\right )}{9+6 x^3+x^6+\left (-6-2 x^3\right ) \log (x)+\log ^2(x)} \, dx=\frac {10 \, {\left (x^{4} - x^{2} e^{\left (e^{x}\right )} - 2 \, x^{2}\right )}}{x^{3} - \log \left (x\right ) + 3} \]
integrate((((10*exp(x)*x^2+20*x)*log(x)+(-10*x^5-30*x^2)*exp(x)+10*x^4-70* x)*exp(exp(x))+(-40*x^3+40*x)*log(x)+10*x^6+20*x^4+130*x^3-140*x)/(log(x)^ 2+(-2*x^3-6)*log(x)+x^6+6*x^3+9),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.61 \[ \int \frac {-140 x+130 x^3+20 x^4+10 x^6+\left (40 x-40 x^3\right ) \log (x)+e^{e^x} \left (-70 x+10 x^4+e^x \left (-30 x^2-10 x^5\right )+\left (20 x+10 e^x x^2\right ) \log (x)\right )}{9+6 x^3+x^6+\left (-6-2 x^3\right ) \log (x)+\log ^2(x)} \, dx=\frac {10 \, {\left (x^{4} e^{x} - x^{2} e^{\left (x + e^{x}\right )} - 2 \, x^{2} e^{x}\right )}}{x^{3} e^{x} - e^{x} \log \left (x\right ) + 3 \, e^{x}} \]
integrate((((10*exp(x)*x^2+20*x)*log(x)+(-10*x^5-30*x^2)*exp(x)+10*x^4-70* x)*exp(exp(x))+(-40*x^3+40*x)*log(x)+10*x^6+20*x^4+130*x^3-140*x)/(log(x)^ 2+(-2*x^3-6)*log(x)+x^6+6*x^3+9),x, algorithm=\
Time = 14.34 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.71 \[ \int \frac {-140 x+130 x^3+20 x^4+10 x^6+\left (40 x-40 x^3\right ) \log (x)+e^{e^x} \left (-70 x+10 x^4+e^x \left (-30 x^2-10 x^5\right )+\left (20 x+10 e^x x^2\right ) \log (x)\right )}{9+6 x^3+x^6+\left (-6-2 x^3\right ) \log (x)+\log ^2(x)} \, dx=\frac {40\,x}{3}+\frac {\frac {40\,x}{9}-\frac {40\,x^2}{3}}{x^3-\frac {1}{3}}-\frac {\frac {10\,x^2\,\left (x^5+2\,x^3+13\,x^2-14\right )}{3\,x^3-1}-\frac {40\,x^2\,\ln \left (x\right )\,\left (x^2-1\right )}{3\,x^3-1}}{x^3-\ln \left (x\right )+3}-\frac {10\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}}{x^3-\ln \left (x\right )+3} \]
int((log(x)*(40*x - 40*x^3) - exp(exp(x))*(70*x + exp(x)*(30*x^2 + 10*x^5) - log(x)*(20*x + 10*x^2*exp(x)) - 10*x^4) - 140*x + 130*x^3 + 20*x^4 + 10 *x^6)/(log(x)^2 + 6*x^3 + x^6 - log(x)*(2*x^3 + 6) + 9),x)