Integrand size = 72, antiderivative size = 24 \[ \int \frac {2 x^2-4 x^3+e \left (-8-2 x^2\right )+e^x \left (e (-4-x)+x^2-2 x^3\right )+\left (2 e+e^{1+x}\right ) \log \left (2+e^x\right )}{2 x^2+e^x x^2} \, dx=(-1+x) \left (-x+\frac {e \left (-4-x+\log \left (2+e^x\right )\right )}{x}\right ) \] Output:
(exp(1)/x*(ln(2+exp(x))-4-x)-x)*(-1+x)
Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {2 x^2-4 x^3+e \left (-8-2 x^2\right )+e^x \left (e (-4-x)+x^2-2 x^3\right )+\left (2 e+e^{1+x}\right ) \log \left (2+e^x\right )}{2 x^2+e^x x^2} \, dx=\frac {4 e}{x}+x-e x-x^2+\frac {e (-1+x) \log \left (2+e^x\right )}{x} \] Input:
Integrate[(2*x^2 - 4*x^3 + E*(-8 - 2*x^2) + E^x*(E*(-4 - x) + x^2 - 2*x^3) + (2*E + E^(1 + x))*Log[2 + E^x])/(2*x^2 + E^x*x^2),x]
Output:
(4*E)/x + x - E*x - x^2 + (E*(-1 + x)*Log[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.
\(\displaystyle \int \frac {-4 x^3+2 x^2+e \left (-2 x^2-8\right )+e^x \left (-2 x^3+x^2+e (-x-4)\right )+\left (e^{x+1}+2 e\right ) \log \left (e^x+2\right )}{e^x x^2+2 x^2} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-2 x^3+x^2-e x+e \log \left (e^x+2\right )-4 e}{x^2}-\frac {2 e (x-1)}{\left (e^x+2\right ) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 e \int \frac {1}{\left (2+e^x\right ) x}dx+e \int \frac {e^x}{\left (2+e^x\right ) x}dx-x^2-e x+x+\frac {4 e}{x}+e \log \left (e^x+2\right )-e \log (x)-\frac {e \log \left (e^x+2\right )}{x}\) |
Input:
Int[(2*x^2 - 4*x^3 + E*(-8 - 2*x^2) + E^x*(E*(-4 - x) + x^2 - 2*x^3) + (2* E + E^(1 + x))*Log[2 + E^x])/(2*x^2 + E^x*x^2),x]
Output:
$Aborted
Time = 0.40 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79
method | result | size |
norman | \(\frac {\left (1-{\mathrm e}\right ) x^{2}+{\mathrm e} x \ln \left ({\mathrm e}^{x}+2\right )-x^{3}-{\mathrm e} \ln \left ({\mathrm e}^{x}+2\right )+4 \,{\mathrm e}}{x}\) | \(43\) |
parallelrisch | \(-\frac {x^{2} {\mathrm e}-{\mathrm e} x \ln \left ({\mathrm e}^{x}+2\right )+x^{3}+{\mathrm e} \ln \left ({\mathrm e}^{x}+2\right )-x^{2}-4 \,{\mathrm e}}{x}\) | \(43\) |
risch | \(-\frac {{\mathrm e} \ln \left ({\mathrm e}^{x}+2\right )}{x}+\frac {{\mathrm e} x \ln \left ({\mathrm e}^{x}+2\right )-x^{2} {\mathrm e}-x^{3}+x^{2}+4 \,{\mathrm e}}{x}\) | \(47\) |
Input:
int(((exp(1)*exp(x)+2*exp(1))*ln(exp(x)+2)+((-4-x)*exp(1)-2*x^3+x^2)*exp(x )+(-2*x^2-8)*exp(1)-4*x^3+2*x^2)/(exp(x)*x^2+2*x^2),x,method=_RETURNVERBOS E)
Output:
((1-exp(1))*x^2+exp(1)*x*ln(exp(x)+2)-x^3-exp(1)*ln(exp(x)+2)+4*exp(1))/x
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {2 x^2-4 x^3+e \left (-8-2 x^2\right )+e^x \left (e (-4-x)+x^2-2 x^3\right )+\left (2 e+e^{1+x}\right ) \log \left (2+e^x\right )}{2 x^2+e^x x^2} \, dx=-\frac {x^{3} - {\left (x - 1\right )} e \log \left ({\left (2 \, e + e^{\left (x + 1\right )}\right )} e^{\left (-1\right )}\right ) - x^{2} + {\left (x^{2} - 4\right )} e}{x} \] Input:
integrate(((exp(1)*exp(x)+2*exp(1))*log(2+exp(x))+((-4-x)*exp(1)-2*x^3+x^2 )*exp(x)+(-2*x^2-8)*exp(1)-4*x^3+2*x^2)/(exp(x)*x^2+2*x^2),x, algorithm="f ricas")
Output:
-(x^3 - (x - 1)*e*log((2*e + e^(x + 1))*e^(-1)) - x^2 + (x^2 - 4)*e)/x
Time = 0.14 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {2 x^2-4 x^3+e \left (-8-2 x^2\right )+e^x \left (e (-4-x)+x^2-2 x^3\right )+\left (2 e+e^{1+x}\right ) \log \left (2+e^x\right )}{2 x^2+e^x x^2} \, dx=- x^{2} - x \left (-1 + e\right ) + e \log {\left (e^{x} + 2 \right )} - \frac {e \log {\left (e^{x} + 2 \right )}}{x} + \frac {4 e}{x} \] Input:
integrate(((exp(1)*exp(x)+2*exp(1))*ln(2+exp(x))+((-4-x)*exp(1)-2*x**3+x** 2)*exp(x)+(-2*x**2-8)*exp(1)-4*x**3+2*x**2)/(exp(x)*x**2+2*x**2),x)
Output:
-x**2 - x*(-1 + E) + E*log(exp(x) + 2) - E*log(exp(x) + 2)/x + 4*E/x
Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {2 x^2-4 x^3+e \left (-8-2 x^2\right )+e^x \left (e (-4-x)+x^2-2 x^3\right )+\left (2 e+e^{1+x}\right ) \log \left (2+e^x\right )}{2 x^2+e^x x^2} \, dx=-\frac {x^{3} + x^{2} {\left (e - 1\right )} - {\left (x e - e\right )} \log \left (e^{x} + 2\right ) - 4 \, e}{x} \] Input:
integrate(((exp(1)*exp(x)+2*exp(1))*log(2+exp(x))+((-4-x)*exp(1)-2*x^3+x^2 )*exp(x)+(-2*x^2-8)*exp(1)-4*x^3+2*x^2)/(exp(x)*x^2+2*x^2),x, algorithm="m axima")
Output:
-(x^3 + x^2*(e - 1) - (x*e - e)*log(e^x + 2) - 4*e)/x
Time = 0.11 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.75 \[ \int \frac {2 x^2-4 x^3+e \left (-8-2 x^2\right )+e^x \left (e (-4-x)+x^2-2 x^3\right )+\left (2 e+e^{1+x}\right ) \log \left (2+e^x\right )}{2 x^2+e^x x^2} \, dx=-\frac {x^{3} + x^{2} e - x e \log \left (e^{x} + 2\right ) - x^{2} + e \log \left (e^{x} + 2\right ) - 4 \, e}{x} \] Input:
integrate(((exp(1)*exp(x)+2*exp(1))*log(2+exp(x))+((-4-x)*exp(1)-2*x^3+x^2 )*exp(x)+(-2*x^2-8)*exp(1)-4*x^3+2*x^2)/(exp(x)*x^2+2*x^2),x, algorithm="g iac")
Output:
-(x^3 + x^2*e - x*e*log(e^x + 2) - x^2 + e*log(e^x + 2) - 4*e)/x
Time = 2.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 6.67 \[ \int \frac {2 x^2-4 x^3+e \left (-8-2 x^2\right )+e^x \left (e (-4-x)+x^2-2 x^3\right )+\left (2 e+e^{1+x}\right ) \log \left (2+e^x\right )}{2 x^2+e^x x^2} \, dx=-\frac {4\,x^3\,{\mathrm {e}}^x-16\,\mathrm {e}-4\,{\mathrm {e}}^{2\,x+1}-16\,{\mathrm {e}}^{x+1}+4\,{\mathrm {e}}^{x+1}\,\ln \left ({\mathrm {e}}^x+2\right )+x^2\,\left (4\,\mathrm {e}-4\right )+4\,\mathrm {e}\,\ln \left ({\mathrm {e}}^x+2\right )+x^3\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{2\,x+1}\,\ln \left ({\mathrm {e}}^x+2\right )+4\,x^3-x\,{\mathrm {e}}^{2\,x+1}\,\ln \left ({\mathrm {e}}^x+2\right )+x^2\,{\mathrm {e}}^{2\,x}\,\left (\mathrm {e}-1\right )+x^2\,{\mathrm {e}}^x\,\left (4\,\mathrm {e}-4\right )-4\,x\,{\mathrm {e}}^{x+1}\,\ln \left ({\mathrm {e}}^x+2\right )-4\,x\,\mathrm {e}\,\ln \left ({\mathrm {e}}^x+2\right )}{4\,x+x\,{\mathrm {e}}^{2\,x}+4\,x\,{\mathrm {e}}^x} \] Input:
int(-(exp(1)*(2*x^2 + 8) + exp(x)*(exp(1)*(x + 4) - x^2 + 2*x^3) - log(exp (x) + 2)*(2*exp(1) + exp(1)*exp(x)) - 2*x^2 + 4*x^3)/(x^2*exp(x) + 2*x^2), x)
Output:
-(4*x^3*exp(x) - 16*exp(1) - 4*exp(2*x + 1) - 16*exp(x + 1) + 4*exp(x + 1) *log(exp(x) + 2) + x^2*(4*exp(1) - 4) + 4*exp(1)*log(exp(x) + 2) + x^3*exp (2*x) + exp(2*x + 1)*log(exp(x) + 2) + 4*x^3 - x*exp(2*x + 1)*log(exp(x) + 2) + x^2*exp(2*x)*(exp(1) - 1) + x^2*exp(x)*(4*exp(1) - 4) - 4*x*exp(x + 1)*log(exp(x) + 2) - 4*x*exp(1)*log(exp(x) + 2))/(4*x + x*exp(2*x) + 4*x*e xp(x))
Time = 0.22 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {2 x^2-4 x^3+e \left (-8-2 x^2\right )+e^x \left (e (-4-x)+x^2-2 x^3\right )+\left (2 e+e^{1+x}\right ) \log \left (2+e^x\right )}{2 x^2+e^x x^2} \, dx=\frac {\mathrm {log}\left (e^{x}+2\right ) e x -\mathrm {log}\left (e^{x}+2\right ) e -e \,x^{2}+4 e -x^{3}+x^{2}}{x} \] Input:
int(((exp(1)*exp(x)+2*exp(1))*log(2+exp(x))+((-4-x)*exp(1)-2*x^3+x^2)*exp( x)+(-2*x^2-8)*exp(1)-4*x^3+2*x^2)/(exp(x)*x^2+2*x^2),x)
Output:
(log(e**x + 2)*e*x - log(e**x + 2)*e - e*x**2 + 4*e - x**3 + x**2)/x