Integrand size = 129, antiderivative size = 32 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-x+\frac {-e^{e^x}+\log \left (x^2\right )}{\log \left (\frac {(2+x) \left (x+x^2\right )}{x}\right )} \]
Time = 0.43 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-x-\frac {e^{e^x}}{\log \left (2+3 x+x^2\right )}+\frac {\log \left (x^2\right )}{\log \left (2+3 x+x^2\right )} \]
Integrate[((-3*x - 2*x^2)*Log[x^2] + (4 + 6*x + 2*x^2)*Log[2 + 3*x + x^2] + (-2*x - 3*x^2 - x^3)*Log[2 + 3*x + x^2]^2 + E^E^x*(3*x + 2*x^2 + E^x*(-2 *x - 3*x^2 - x^3)*Log[2 + 3*x + x^2]))/((2*x + 3*x^2 + x^3)*Log[2 + 3*x + x^2]^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.
\(\displaystyle \int \frac {\left (2 x^2+6 x+4\right ) \log \left (x^2+3 x+2\right )+\left (-2 x^2-3 x\right ) \log \left (x^2\right )+\left (-x^3-3 x^2-2 x\right ) \log ^2\left (x^2+3 x+2\right )+e^{e^x} \left (2 x^2+e^x \left (-x^3-3 x^2-2 x\right ) \log \left (x^2+3 x+2\right )+3 x\right )}{\left (x^3+3 x^2+2 x\right ) \log ^2\left (x^2+3 x+2\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (2 x^2+6 x+4\right ) \log \left (x^2+3 x+2\right )+\left (-2 x^2-3 x\right ) \log \left (x^2\right )+\left (-x^3-3 x^2-2 x\right ) \log ^2\left (x^2+3 x+2\right )+e^{e^x} \left (2 x^2+e^x \left (-x^3-3 x^2-2 x\right ) \log \left (x^2+3 x+2\right )+3 x\right )}{x \left (x^2+3 x+2\right ) \log ^2\left (x^2+3 x+2\right )}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {(2 x+3) \left (e^{e^x}-\log \left (x^2\right )\right )}{\left (x^2+3 x+2\right ) \log ^2\left (x^2+3 x+2\right )}+\frac {2-e^{x+e^x} x}{x \log \left (x^2+3 x+2\right )}-1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {e^{e^x}}{(x+1) \log ^2\left (x^2+3 x+2\right )}dx+\int \frac {e^{e^x}}{(x+2) \log ^2\left (x^2+3 x+2\right )}dx-\int \frac {\log \left (x^2\right )}{(x+1) \log ^2\left (x^2+3 x+2\right )}dx-\int \frac {\log \left (x^2\right )}{(x+2) \log ^2\left (x^2+3 x+2\right )}dx-\int \frac {e^{x+e^x}}{\log \left (x^2+3 x+2\right )}dx+2 \int \frac {1}{x \log \left (x^2+3 x+2\right )}dx-x\) |
Int[((-3*x - 2*x^2)*Log[x^2] + (4 + 6*x + 2*x^2)*Log[2 + 3*x + x^2] + (-2* x - 3*x^2 - x^3)*Log[2 + 3*x + x^2]^2 + E^E^x*(3*x + 2*x^2 + E^x*(-2*x - 3 *x^2 - x^3)*Log[2 + 3*x + x^2]))/((2*x + 3*x^2 + x^3)*Log[2 + 3*x + x^2]^2 ),x]
3.21.27.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 59.27 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53
method | result | size |
parallelrisch | \(\frac {-6 x \ln \left (x^{2}+3 x +2\right )+6 \ln \left (x^{2}\right )-6 \,{\mathrm e}^{{\mathrm e}^{x}}+13 \ln \left (x^{2}+3 x +2\right )}{6 \ln \left (x^{2}+3 x +2\right )}\) | \(49\) |
risch | \(-x +\frac {4 \ln \left (x \right )-i \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+2 i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2 \ln \left (x^{2}+3 x +2\right )}-\frac {{\mathrm e}^{{\mathrm e}^{x}}}{\ln \left (x^{2}+3 x +2\right )}\) | \(88\) |
int((((-x^3-3*x^2-2*x)*exp(x)*ln(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+(-x^3-3 *x^2-2*x)*ln(x^2+3*x+2)^2+(2*x^2+6*x+4)*ln(x^2+3*x+2)+(-2*x^2-3*x)*ln(x^2) )/(x^3+3*x^2+2*x)/ln(x^2+3*x+2)^2,x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x^{2} + 3 \, x + 2\right ) + e^{\left (e^{x}\right )} - \log \left (x^{2}\right )}{\log \left (x^{2} + 3 \, x + 2\right )} \]
integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+ (-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x )*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm=\
Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=- x - \frac {e^{e^{x}}}{\log {\left (x^{2} + 3 x + 2 \right )}} + \frac {\log {\left (x^{2} \right )}}{\log {\left (x^{2} + 3 x + 2 \right )}} \]
integrate((((-x**3-3*x**2-2*x)*exp(x)*ln(x**2+3*x+2)+2*x**2+3*x)*exp(exp(x ))+(-x**3-3*x**2-2*x)*ln(x**2+3*x+2)**2+(2*x**2+6*x+4)*ln(x**2+3*x+2)+(-2* x**2-3*x)*ln(x**2))/(x**3+3*x**2+2*x)/ln(x**2+3*x+2)**2,x)
Time = 0.24 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.03 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x + 2\right ) + x \log \left (x + 1\right ) + e^{\left (e^{x}\right )} - 2 \, \log \left (x\right )}{\log \left (x + 2\right ) + \log \left (x + 1\right )} \]
integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+ (-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x )*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm=\
Time = 0.37 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=-\frac {x \log \left (x^{2} + 3 \, x + 2\right ) + e^{\left (e^{x}\right )} - \log \left (x^{2}\right )}{\log \left (x^{2} + 3 \, x + 2\right )} \]
integrate((((-x^3-3*x^2-2*x)*exp(x)*log(x^2+3*x+2)+2*x^2+3*x)*exp(exp(x))+ (-x^3-3*x^2-2*x)*log(x^2+3*x+2)^2+(2*x^2+6*x+4)*log(x^2+3*x+2)+(-2*x^2-3*x )*log(x^2))/(x^3+3*x^2+2*x)/log(x^2+3*x+2)^2,x, algorithm=\
Timed out. \[ \int \frac {\left (-3 x-2 x^2\right ) \log \left (x^2\right )+\left (4+6 x+2 x^2\right ) \log \left (2+3 x+x^2\right )+\left (-2 x-3 x^2-x^3\right ) \log ^2\left (2+3 x+x^2\right )+e^{e^x} \left (3 x+2 x^2+e^x \left (-2 x-3 x^2-x^3\right ) \log \left (2+3 x+x^2\right )\right )}{\left (2 x+3 x^2+x^3\right ) \log ^2\left (2+3 x+x^2\right )} \, dx=\int \frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (3\,x+2\,x^2-{\mathrm {e}}^x\,\ln \left (x^2+3\,x+2\right )\,\left (x^3+3\,x^2+2\,x\right )\right )-\ln \left (x^2\right )\,\left (2\,x^2+3\,x\right )+\ln \left (x^2+3\,x+2\right )\,\left (2\,x^2+6\,x+4\right )-{\ln \left (x^2+3\,x+2\right )}^2\,\left (x^3+3\,x^2+2\,x\right )}{{\ln \left (x^2+3\,x+2\right )}^2\,\left (x^3+3\,x^2+2\,x\right )} \,d x \]
int((exp(exp(x))*(3*x + 2*x^2 - exp(x)*log(3*x + x^2 + 2)*(2*x + 3*x^2 + x ^3)) - log(x^2)*(3*x + 2*x^2) + log(3*x + x^2 + 2)*(6*x + 2*x^2 + 4) - log (3*x + x^2 + 2)^2*(2*x + 3*x^2 + x^3))/(log(3*x + x^2 + 2)^2*(2*x + 3*x^2 + x^3)),x)