Integrand size = 92, antiderivative size = 27 \[ \int \frac {-10-20 x+\left (-5-10 x-10 x^2\right ) \log (3)+e^x \left (-20-10 x+\left (-5-20 x-5 x^2+5 x^3\right ) \log (3)\right )}{x^2+2 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (2 x+4 x^2+2 x^3\right )} \, dx=\frac {5 \left (2+\left (1+x-x^2\right ) \log (3)\right )}{(1+x) \left (e^x+x\right )} \] Output:
5/(1+x)*((-x^2+x+1)*ln(3)+2)/(exp(x)+x)
Time = 0.47 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {-10-20 x+\left (-5-10 x-10 x^2\right ) \log (3)+e^x \left (-20-10 x+\left (-5-20 x-5 x^2+5 x^3\right ) \log (3)\right )}{x^2+2 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (2 x+4 x^2+2 x^3\right )} \, dx=-\frac {5 \left (2+\log (3)+x \log (3)-x^3 \log (3)+x^4 \log (3)-x^2 (2+\log (9))\right )}{(-1+x) (1+x)^2 \left (e^x+x\right )} \] Input:
Integrate[(-10 - 20*x + (-5 - 10*x - 10*x^2)*Log[3] + E^x*(-20 - 10*x + (- 5 - 20*x - 5*x^2 + 5*x^3)*Log[3]))/(x^2 + 2*x^3 + x^4 + E^(2*x)*(1 + 2*x + x^2) + E^x*(2*x + 4*x^2 + 2*x^3)),x]
Output:
(-5*(2 + Log[3] + x*Log[3] - x^3*Log[3] + x^4*Log[3] - x^2*(2 + Log[9])))/ ((-1 + x)*(1 + x)^2*(E^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.
\(\displaystyle \int \frac {\left (-10 x^2-10 x-5\right ) \log (3)+e^x \left (\left (5 x^3-5 x^2-20 x-5\right ) \log (3)-10 x-20\right )-20 x-10}{x^4+2 x^3+x^2+e^{2 x} \left (x^2+2 x+1\right )+e^x \left (2 x^3+4 x^2+2 x\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {\left (-10 x^2-10 x-5\right ) \log (3)+e^x \left (\left (5 x^3-5 x^2-20 x-5\right ) \log (3)-10 x-20\right )-20 x-10}{(x+1)^2 \left (x+e^x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5 \left (x^3 \log (3)-x^2 \log (3)-2 x (1+\log (9))-4-\log (3)\right )}{(x+1)^2 \left (x+e^x\right )}-\frac {5 \left (x^3 \log (3)-x^2 \log (9)-2 x+2+\log (3)\right )}{(x+1) \left (x+e^x\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \log (3) \int \frac {x^2}{\left (x+e^x\right )^2}dx+5 \log (27) \int \frac {1}{\left (x+e^x\right )^2}dx+5 (2-\log (27)) \int \frac {1}{\left (x+e^x\right )^2}dx-5 (4-\log (9)) \int \frac {1}{(x+1) \left (x+e^x\right )^2}dx+5 \log (27) \int \frac {1}{x+e^x}dx-15 \log (3) \int \frac {1}{x+e^x}dx+5 \log (3) \int \frac {x}{x+e^x}dx-5 (2-\log (3)) \int \frac {1}{(x+1)^2 \left (x+e^x\right )}dx-5 (2-\log (3)) \int \frac {1}{(x+1) \left (x+e^x\right )}dx+\frac {5 \log (27)}{x+e^x}\) |
Input:
Int[(-10 - 20*x + (-5 - 10*x - 10*x^2)*Log[3] + E^x*(-20 - 10*x + (-5 - 20 *x - 5*x^2 + 5*x^3)*Log[3]))/(x^2 + 2*x^3 + x^4 + E^(2*x)*(1 + 2*x + x^2) + E^x*(2*x + 4*x^2 + 2*x^3)),x]
Output:
$Aborted
Time = 0.44 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15
method | result | size |
risch | \(-\frac {5 \left (x^{2} \ln \left (3\right )-x \ln \left (3\right )-\ln \left (3\right )-2\right )}{\left (1+x \right ) \left ({\mathrm e}^{x}+x \right )}\) | \(31\) |
parallelrisch | \(-\frac {5 x^{2} \ln \left (3\right )-10-5 x \ln \left (3\right )-5 \ln \left (3\right )}{{\mathrm e}^{x} x +x^{2}+{\mathrm e}^{x}+x}\) | \(34\) |
norman | \(\frac {5 \,{\mathrm e}^{x} \ln \left (3\right )+10 x \ln \left (3\right )+5 x \ln \left (3\right ) {\mathrm e}^{x}+10+5 \ln \left (3\right )}{{\mathrm e}^{x} x +x^{2}+{\mathrm e}^{x}+x}\) | \(39\) |
Input:
int((((5*x^3-5*x^2-20*x-5)*ln(3)-10*x-20)*exp(x)+(-10*x^2-10*x-5)*ln(3)-20 *x-10)/((x^2+2*x+1)*exp(x)^2+(2*x^3+4*x^2+2*x)*exp(x)+x^4+2*x^3+x^2),x,met hod=_RETURNVERBOSE)
Output:
-5*(x^2*ln(3)-x*ln(3)-ln(3)-2)/(1+x)/(exp(x)+x)
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-10-20 x+\left (-5-10 x-10 x^2\right ) \log (3)+e^x \left (-20-10 x+\left (-5-20 x-5 x^2+5 x^3\right ) \log (3)\right )}{x^2+2 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (2 x+4 x^2+2 x^3\right )} \, dx=-\frac {5 \, {\left ({\left (x^{2} - x - 1\right )} \log \left (3\right ) - 2\right )}}{x^{2} + {\left (x + 1\right )} e^{x} + x} \] Input:
integrate((((5*x^3-5*x^2-20*x-5)*log(3)-10*x-20)*exp(x)+(-10*x^2-10*x-5)*l og(3)-20*x-10)/((x^2+2*x+1)*exp(x)^2+(2*x^3+4*x^2+2*x)*exp(x)+x^4+2*x^3+x^ 2),x, algorithm="fricas")
Output:
-5*((x^2 - x - 1)*log(3) - 2)/(x^2 + (x + 1)*e^x + x)
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {-10-20 x+\left (-5-10 x-10 x^2\right ) \log (3)+e^x \left (-20-10 x+\left (-5-20 x-5 x^2+5 x^3\right ) \log (3)\right )}{x^2+2 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (2 x+4 x^2+2 x^3\right )} \, dx=\frac {- 5 x^{2} \log {\left (3 \right )} + 5 x \log {\left (3 \right )} + 5 \log {\left (3 \right )} + 10}{x^{2} + x + \left (x + 1\right ) e^{x}} \] Input:
integrate((((5*x**3-5*x**2-20*x-5)*ln(3)-10*x-20)*exp(x)+(-10*x**2-10*x-5) *ln(3)-20*x-10)/((x**2+2*x+1)*exp(x)**2+(2*x**3+4*x**2+2*x)*exp(x)+x**4+2* x**3+x**2),x)
Output:
(-5*x**2*log(3) + 5*x*log(3) + 5*log(3) + 10)/(x**2 + x + (x + 1)*exp(x))
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {-10-20 x+\left (-5-10 x-10 x^2\right ) \log (3)+e^x \left (-20-10 x+\left (-5-20 x-5 x^2+5 x^3\right ) \log (3)\right )}{x^2+2 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (2 x+4 x^2+2 x^3\right )} \, dx=-\frac {5 \, {\left (x^{2} \log \left (3\right ) - x \log \left (3\right ) - \log \left (3\right ) - 2\right )}}{x^{2} + {\left (x + 1\right )} e^{x} + x} \] Input:
integrate((((5*x^3-5*x^2-20*x-5)*log(3)-10*x-20)*exp(x)+(-10*x^2-10*x-5)*l og(3)-20*x-10)/((x^2+2*x+1)*exp(x)^2+(2*x^3+4*x^2+2*x)*exp(x)+x^4+2*x^3+x^ 2),x, algorithm="maxima")
Output:
-5*(x^2*log(3) - x*log(3) - log(3) - 2)/(x^2 + (x + 1)*e^x + x)
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {-10-20 x+\left (-5-10 x-10 x^2\right ) \log (3)+e^x \left (-20-10 x+\left (-5-20 x-5 x^2+5 x^3\right ) \log (3)\right )}{x^2+2 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (2 x+4 x^2+2 x^3\right )} \, dx=-\frac {5 \, {\left (x^{2} \log \left (3\right ) - x \log \left (3\right ) - \log \left (3\right ) - 2\right )}}{x^{2} + x e^{x} + x + e^{x}} \] Input:
integrate((((5*x^3-5*x^2-20*x-5)*log(3)-10*x-20)*exp(x)+(-10*x^2-10*x-5)*l og(3)-20*x-10)/((x^2+2*x+1)*exp(x)^2+(2*x^3+4*x^2+2*x)*exp(x)+x^4+2*x^3+x^ 2),x, algorithm="giac")
Output:
-5*(x^2*log(3) - x*log(3) - log(3) - 2)/(x^2 + x*e^x + x + e^x)
Time = 3.79 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-10-20 x+\left (-5-10 x-10 x^2\right ) \log (3)+e^x \left (-20-10 x+\left (-5-20 x-5 x^2+5 x^3\right ) \log (3)\right )}{x^2+2 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (2 x+4 x^2+2 x^3\right )} \, dx=\frac {-\ln \left (243\right )\,x^2+\ln \left (243\right )\,x+\ln \left (243\right )+10}{\left (x+{\mathrm {e}}^x\right )\,\left (x+1\right )} \] Input:
int(-(20*x + log(3)*(10*x + 10*x^2 + 5) + exp(x)*(10*x + log(3)*(20*x + 5* x^2 - 5*x^3 + 5) + 20) + 10)/(exp(2*x)*(2*x + x^2 + 1) + x^2 + 2*x^3 + x^4 + exp(x)*(2*x + 4*x^2 + 2*x^3)),x)
Output:
(log(243) + x*log(243) - x^2*log(243) + 10)/((x + exp(x))*(x + 1))
Time = 0.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {-10-20 x+\left (-5-10 x-10 x^2\right ) \log (3)+e^x \left (-20-10 x+\left (-5-20 x-5 x^2+5 x^3\right ) \log (3)\right )}{x^2+2 x^3+x^4+e^{2 x} \left (1+2 x+x^2\right )+e^x \left (2 x+4 x^2+2 x^3\right )} \, dx=\frac {-5 \,\mathrm {log}\left (3\right ) x^{2}+5 \,\mathrm {log}\left (3\right ) x +5 \,\mathrm {log}\left (3\right )+10}{e^{x} x +e^{x}+x^{2}+x} \] Input:
int((((5*x^3-5*x^2-20*x-5)*log(3)-10*x-20)*exp(x)+(-10*x^2-10*x-5)*log(3)- 20*x-10)/((x^2+2*x+1)*exp(x)^2+(2*x^3+4*x^2+2*x)*exp(x)+x^4+2*x^3+x^2),x)
Output:
(5*( - log(3)*x**2 + log(3)*x + log(3) + 2))/(e**x*x + e**x + x**2 + x)