Integrand size = 139, antiderivative size = 26 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {e^x}{x}+e^3 x \log \left (10-x+x^2+\log (2+x)\right ) \]
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {e^x}{x}+e^3 x \log \left (10-x+x^2+\log (2+x)\right ) \]
Integrate[(E^x*(-20 + 12*x + 7*x^2 + x^4) + E^3*(-x^3 + 3*x^4 + 2*x^5) + E ^x*(-2 + x + x^2)*Log[2 + x] + (E^3*(20*x^2 + 8*x^3 + x^4 + x^5) + E^3*(2* x^2 + x^3)*Log[2 + x])*Log[10 - x + x^2 + Log[2 + x]])/(20*x^2 + 8*x^3 + x ^4 + x^5 + (2*x^2 + x^3)*Log[2 + x]),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.
\(\displaystyle \int \frac {e^x \left (x^2+x-2\right ) \log (x+2)+e^x \left (x^4+7 x^2+12 x-20\right )+e^3 \left (2 x^5+3 x^4-x^3\right )+\left (e^3 \left (x^3+2 x^2\right ) \log (x+2)+e^3 \left (x^5+x^4+8 x^3+20 x^2\right )\right ) \log \left (x^2-x+\log (x+2)+10\right )}{x^5+x^4+8 x^3+20 x^2+\left (x^3+2 x^2\right ) \log (x+2)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x \left (x^2+x-2\right ) \log (x+2)+e^x \left (x^4+7 x^2+12 x-20\right )+e^3 \left (2 x^5+3 x^4-x^3\right )+\left (e^3 \left (x^3+2 x^2\right ) \log (x+2)+e^3 \left (x^5+x^4+8 x^3+20 x^2\right )\right ) \log \left (x^2-x+\log (x+2)+10\right )}{x^2 (x+2) \left (x^2-x+\log (x+2)+10\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^x (x-1)}{x^2}+\frac {e^3 \left (2 x^3+3 x^2+x^2 \log \left (x^2-x+\log (x+2)+10\right )+x \log (x+2) \log \left (x^2-x+\log (x+2)+10\right )+8 x \log \left (x^2-x+\log (x+2)+10\right )+2 \log (x+2) \log \left (x^2-x+\log (x+2)+10\right )+20 \log \left (x^2-x+\log (x+2)+10\right )+x^3 \log \left (x^2-x+\log (x+2)+10\right )-x\right )}{(x+2) \left (x^2-x+\log (x+2)+10\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e^3 \int \frac {1}{x^2-x+\log (x+2)+10}dx-e^3 \int \frac {x}{x^2-x+\log (x+2)+10}dx+2 e^3 \int \frac {x^2}{x^2-x+\log (x+2)+10}dx-2 e^3 \int \frac {1}{(x+2) \left (x^2-x+\log (x+2)+10\right )}dx+e^3 \int \log \left (x^2-x+\log (x+2)+10\right )dx+\frac {e^x}{x}\) |
Int[(E^x*(-20 + 12*x + 7*x^2 + x^4) + E^3*(-x^3 + 3*x^4 + 2*x^5) + E^x*(-2 + x + x^2)*Log[2 + x] + (E^3*(20*x^2 + 8*x^3 + x^4 + x^5) + E^3*(2*x^2 + x^3)*Log[2 + x])*Log[10 - x + x^2 + Log[2 + x]])/(20*x^2 + 8*x^3 + x^4 + x ^5 + (2*x^2 + x^3)*Log[2 + x]),x]
3.7.37.3.1 Defintions of rubi rules used
Time = 32.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96
method | result | size |
risch | \(\frac {{\mathrm e}^{x}}{x}+{\mathrm e}^{3} \ln \left (\ln \left (2+x \right )+x^{2}-x +10\right ) x\) | \(25\) |
parallelrisch | \(\frac {\ln \left (\ln \left (2+x \right )+x^{2}-x +10\right ) {\mathrm e}^{3} x^{2}+{\mathrm e}^{x}}{x}\) | \(27\) |
int((((x^3+2*x^2)*exp(3)*ln(2+x)+(x^5+x^4+8*x^3+20*x^2)*exp(3))*ln(ln(2+x) +x^2-x+10)+(x^2+x-2)*exp(x)*ln(2+x)+(x^4+7*x^2+12*x-20)*exp(x)+(2*x^5+3*x^ 4-x^3)*exp(3))/((x^3+2*x^2)*ln(2+x)+x^5+x^4+8*x^3+20*x^2),x,method=_RETURN VERBOSE)
Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {x^{2} e^{3} \log \left (x^{2} - x + \log \left (x + 2\right ) + 10\right ) + e^{x}}{x} \]
integrate((((x^3+2*x^2)*exp(3)*log(2+x)+(x^5+x^4+8*x^3+20*x^2)*exp(3))*log (log(2+x)+x^2-x+10)+(x^2+x-2)*exp(x)*log(2+x)+(x^4+7*x^2+12*x-20)*exp(x)+( 2*x^5+3*x^4-x^3)*exp(3))/((x^3+2*x^2)*log(2+x)+x^5+x^4+8*x^3+20*x^2),x, al gorithm=\
Time = 0.81 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\left (x e^{3} + e^{3}\right ) \log {\left (x^{2} - x + \log {\left (x + 2 \right )} + 10 \right )} - e^{3} \log {\left (x^{2} - x + \log {\left (x + 2 \right )} + 10 \right )} + \frac {e^{x}}{x} \]
integrate((((x**3+2*x**2)*exp(3)*ln(2+x)+(x**5+x**4+8*x**3+20*x**2)*exp(3) )*ln(ln(2+x)+x**2-x+10)+(x**2+x-2)*exp(x)*ln(2+x)+(x**4+7*x**2+12*x-20)*ex p(x)+(2*x**5+3*x**4-x**3)*exp(3))/((x**3+2*x**2)*ln(2+x)+x**5+x**4+8*x**3+ 20*x**2),x)
(x*exp(3) + exp(3))*log(x**2 - x + log(x + 2) + 10) - exp(3)*log(x**2 - x + log(x + 2) + 10) + exp(x)/x
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {x^{2} e^{3} \log \left (x^{2} - x + \log \left (x + 2\right ) + 10\right ) + e^{x}}{x} \]
integrate((((x^3+2*x^2)*exp(3)*log(2+x)+(x^5+x^4+8*x^3+20*x^2)*exp(3))*log (log(2+x)+x^2-x+10)+(x^2+x-2)*exp(x)*log(2+x)+(x^4+7*x^2+12*x-20)*exp(x)+( 2*x^5+3*x^4-x^3)*exp(3))/((x^3+2*x^2)*log(2+x)+x^5+x^4+8*x^3+20*x^2),x, al gorithm=\
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {x^{2} e^{3} \log \left (x^{2} - x + \log \left (x + 2\right ) + 10\right ) + e^{x}}{x} \]
integrate((((x^3+2*x^2)*exp(3)*log(2+x)+(x^5+x^4+8*x^3+20*x^2)*exp(3))*log (log(2+x)+x^2-x+10)+(x^2+x-2)*exp(x)*log(2+x)+(x^4+7*x^2+12*x-20)*exp(x)+( 2*x^5+3*x^4-x^3)*exp(3))/((x^3+2*x^2)*log(2+x)+x^5+x^4+8*x^3+20*x^2),x, al gorithm=\
Time = 0.41 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^x \left (-20+12 x+7 x^2+x^4\right )+e^3 \left (-x^3+3 x^4+2 x^5\right )+e^x \left (-2+x+x^2\right ) \log (2+x)+\left (e^3 \left (20 x^2+8 x^3+x^4+x^5\right )+e^3 \left (2 x^2+x^3\right ) \log (2+x)\right ) \log \left (10-x+x^2+\log (2+x)\right )}{20 x^2+8 x^3+x^4+x^5+\left (2 x^2+x^3\right ) \log (2+x)} \, dx=\frac {{\mathrm {e}}^x}{x}+x\,{\mathrm {e}}^3\,\ln \left (\ln \left (x+2\right )-x+x^2+10\right ) \]