Integrand size = 102, antiderivative size = 21 \[ \int \frac {e^{2 x} (-1-x)+e^x \left (-2 x-3 x^2\right )+\left (e^{2 x} (-1+x)+e^x \left (-x+x^3\right )\right ) \log \left (3 e^x x+3 x^2+3 x^3\right )}{\left (e^x x^2+x^3+x^4\right ) \log ^2\left (3 e^x x+3 x^2+3 x^3\right )} \, dx=\frac {e^x}{x \log \left (3 x \left (e^x+x+x^2\right )\right )} \]
Time = 5.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {e^{2 x} (-1-x)+e^x \left (-2 x-3 x^2\right )+\left (e^{2 x} (-1+x)+e^x \left (-x+x^3\right )\right ) \log \left (3 e^x x+3 x^2+3 x^3\right )}{\left (e^x x^2+x^3+x^4\right ) \log ^2\left (3 e^x x+3 x^2+3 x^3\right )} \, dx=\frac {e^x}{x \log \left (3 x \left (e^x+x+x^2\right )\right )} \]
Integrate[(E^(2*x)*(-1 - x) + E^x*(-2*x - 3*x^2) + (E^(2*x)*(-1 + x) + E^x *(-x + x^3))*Log[3*E^x*x + 3*x^2 + 3*x^3])/((E^x*x^2 + x^3 + x^4)*Log[3*E^ x*x + 3*x^2 + 3*x^3]^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.
\(\displaystyle \int \frac {e^x \left (-3 x^2-2 x\right )+\left (e^x \left (x^3-x\right )+e^{2 x} (x-1)\right ) \log \left (3 x^3+3 x^2+3 e^x x\right )+e^{2 x} (-x-1)}{\left (x^4+x^3+e^x x^2\right ) \log ^2\left (3 x^3+3 x^2+3 e^x x\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^x \left (-3 x^2-2 x\right )+\left (e^x \left (x^3-x\right )+e^{2 x} (x-1)\right ) \log \left (3 x^3+3 x^2+3 e^x x\right )+e^{2 x} (-x-1)}{\left (x^4+x^3+e^x x^2\right ) \log ^2\left (3 x \left (x^2+x+e^x\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2-x-1}{x \log ^2\left (3 x \left (x^2+x+e^x\right )\right )}+\frac {e^x \left (x \log \left (3 x \left (x^2+x+e^x\right )\right )-\log \left (3 x \left (x^2+x+e^x\right )\right )-x-1\right )}{x^2 \log ^2\left (3 x \left (x^2+x+e^x\right )\right )}-\frac {x^3-2 x-1}{\left (x^2+x+e^x\right ) \log ^2\left (3 x \left (x^2+x+e^x\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {1}{\log ^2\left (3 x \left (x^2+x+e^x\right )\right )}dx-\int \frac {e^x}{x^2 \log ^2\left (3 x \left (x^2+x+e^x\right )\right )}dx-\int \frac {1}{x \log ^2\left (3 x \left (x^2+x+e^x\right )\right )}dx-\int \frac {e^x}{x \log ^2\left (3 x \left (x^2+x+e^x\right )\right )}dx+\int \frac {x}{\log ^2\left (3 x \left (x^2+x+e^x\right )\right )}dx+\int \frac {1}{\left (x^2+x+e^x\right ) \log ^2\left (3 x \left (x^2+x+e^x\right )\right )}dx+2 \int \frac {x}{\left (x^2+x+e^x\right ) \log ^2\left (3 x \left (x^2+x+e^x\right )\right )}dx-\int \frac {e^x}{x^2 \log \left (3 x \left (x^2+x+e^x\right )\right )}dx+\int \frac {e^x}{x \log \left (3 x \left (x^2+x+e^x\right )\right )}dx-\int \frac {x^3}{\left (x^2+x+e^x\right ) \log ^2\left (3 x \left (x^2+x+e^x\right )\right )}dx\) |
Int[(E^(2*x)*(-1 - x) + E^x*(-2*x - 3*x^2) + (E^(2*x)*(-1 + x) + E^x*(-x + x^3))*Log[3*E^x*x + 3*x^2 + 3*x^3])/((E^x*x^2 + x^3 + x^4)*Log[3*E^x*x + 3*x^2 + 3*x^3]^2),x]
3.22.58.3.1 Defintions of rubi rules used
Time = 0.31 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{x}}{x \ln \left (3 \left ({\mathrm e}^{x}+x^{2}+x \right ) x \right )}\) | \(20\) |
risch | \(\frac {2 i {\mathrm e}^{x}}{x \left (\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+x^{2}+x \right )\right ) \operatorname {csgn}\left (i x \left ({\mathrm e}^{x}+x^{2}+x \right )\right )-\pi \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left ({\mathrm e}^{x}+x^{2}+x \right )\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+x^{2}+x \right )\right ) {\operatorname {csgn}\left (i x \left ({\mathrm e}^{x}+x^{2}+x \right )\right )}^{2}+\pi {\operatorname {csgn}\left (i x \left ({\mathrm e}^{x}+x^{2}+x \right )\right )}^{3}+2 i \ln \left (3\right )+2 i \ln \left (x \right )+2 i \ln \left ({\mathrm e}^{x}+x^{2}+x \right )\right )}\) | \(129\) |
int((((-1+x)*exp(x)^2+(x^3-x)*exp(x))*ln(3*exp(x)*x+3*x^3+3*x^2)+(-1-x)*ex p(x)^2+(-3*x^2-2*x)*exp(x))/(exp(x)*x^2+x^4+x^3)/ln(3*exp(x)*x+3*x^3+3*x^2 )^2,x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2 x} (-1-x)+e^x \left (-2 x-3 x^2\right )+\left (e^{2 x} (-1+x)+e^x \left (-x+x^3\right )\right ) \log \left (3 e^x x+3 x^2+3 x^3\right )}{\left (e^x x^2+x^3+x^4\right ) \log ^2\left (3 e^x x+3 x^2+3 x^3\right )} \, dx=\frac {e^{x}}{x \log \left (3 \, x^{3} + 3 \, x^{2} + 3 \, x e^{x}\right )} \]
integrate((((-1+x)*exp(x)^2+(x^3-x)*exp(x))*log(3*exp(x)*x+3*x^3+3*x^2)+(- 1-x)*exp(x)^2+(-3*x^2-2*x)*exp(x))/(exp(x)*x^2+x^4+x^3)/log(3*exp(x)*x+3*x ^3+3*x^2)^2,x, algorithm=\
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {e^{2 x} (-1-x)+e^x \left (-2 x-3 x^2\right )+\left (e^{2 x} (-1+x)+e^x \left (-x+x^3\right )\right ) \log \left (3 e^x x+3 x^2+3 x^3\right )}{\left (e^x x^2+x^3+x^4\right ) \log ^2\left (3 e^x x+3 x^2+3 x^3\right )} \, dx=\frac {e^{x}}{x \log {\left (3 x^{3} + 3 x^{2} + 3 x e^{x} \right )}} \]
integrate((((-1+x)*exp(x)**2+(x**3-x)*exp(x))*ln(3*exp(x)*x+3*x**3+3*x**2) +(-1-x)*exp(x)**2+(-3*x**2-2*x)*exp(x))/(exp(x)*x**2+x**4+x**3)/ln(3*exp(x )*x+3*x**3+3*x**2)**2,x)
Time = 0.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {e^{2 x} (-1-x)+e^x \left (-2 x-3 x^2\right )+\left (e^{2 x} (-1+x)+e^x \left (-x+x^3\right )\right ) \log \left (3 e^x x+3 x^2+3 x^3\right )}{\left (e^x x^2+x^3+x^4\right ) \log ^2\left (3 e^x x+3 x^2+3 x^3\right )} \, dx=\frac {e^{x}}{x \log \left (3\right ) + x \log \left (x^{2} + x + e^{x}\right ) + x \log \left (x\right )} \]
integrate((((-1+x)*exp(x)^2+(x^3-x)*exp(x))*log(3*exp(x)*x+3*x^3+3*x^2)+(- 1-x)*exp(x)^2+(-3*x^2-2*x)*exp(x))/(exp(x)*x^2+x^4+x^3)/log(3*exp(x)*x+3*x ^3+3*x^2)^2,x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2 x} (-1-x)+e^x \left (-2 x-3 x^2\right )+\left (e^{2 x} (-1+x)+e^x \left (-x+x^3\right )\right ) \log \left (3 e^x x+3 x^2+3 x^3\right )}{\left (e^x x^2+x^3+x^4\right ) \log ^2\left (3 e^x x+3 x^2+3 x^3\right )} \, dx=\frac {e^{x}}{x \log \left (3 \, x^{3} + 3 \, x^{2} + 3 \, x e^{x}\right )} \]
integrate((((-1+x)*exp(x)^2+(x^3-x)*exp(x))*log(3*exp(x)*x+3*x^3+3*x^2)+(- 1-x)*exp(x)^2+(-3*x^2-2*x)*exp(x))/(exp(x)*x^2+x^4+x^3)/log(3*exp(x)*x+3*x ^3+3*x^2)^2,x, algorithm=\
Timed out. \[ \int \frac {e^{2 x} (-1-x)+e^x \left (-2 x-3 x^2\right )+\left (e^{2 x} (-1+x)+e^x \left (-x+x^3\right )\right ) \log \left (3 e^x x+3 x^2+3 x^3\right )}{\left (e^x x^2+x^3+x^4\right ) \log ^2\left (3 e^x x+3 x^2+3 x^3\right )} \, dx=\int -\frac {\ln \left (3\,x\,{\mathrm {e}}^x+3\,x^2+3\,x^3\right )\,\left ({\mathrm {e}}^x\,\left (x-x^3\right )-{\mathrm {e}}^{2\,x}\,\left (x-1\right )\right )+{\mathrm {e}}^{2\,x}\,\left (x+1\right )+{\mathrm {e}}^x\,\left (3\,x^2+2\,x\right )}{{\ln \left (3\,x\,{\mathrm {e}}^x+3\,x^2+3\,x^3\right )}^2\,\left (x^2\,{\mathrm {e}}^x+x^3+x^4\right )} \,d x \]
int(-(log(3*x*exp(x) + 3*x^2 + 3*x^3)*(exp(x)*(x - x^3) - exp(2*x)*(x - 1) ) + exp(2*x)*(x + 1) + exp(x)*(2*x + 3*x^2))/(log(3*x*exp(x) + 3*x^2 + 3*x ^3)^2*(x^2*exp(x) + x^3 + x^4)),x)