Integrand size = 129, antiderivative size = 28 \[ \int \frac {\log (e x) \left (e^{x+x^2} (-4+2 x)-2 e^{x+x^2} \log \left (x^2\right )\right )+\log ^2(e x) \left (e^{x+x^2} \left (4-4 x-3 x^2+2 x^3\right )+e^{x+x^2} \left (1-x-2 x^2\right ) \log \left (x^2\right )\right )}{4 x^2-4 x^3+x^4+\left (4 x^2-2 x^3\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=\frac {e^{x (1+x)} \log ^2(e x)}{x \left (-2+x-\log \left (x^2\right )\right )} \]
Time = 5.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {\log (e x) \left (e^{x+x^2} (-4+2 x)-2 e^{x+x^2} \log \left (x^2\right )\right )+\log ^2(e x) \left (e^{x+x^2} \left (4-4 x-3 x^2+2 x^3\right )+e^{x+x^2} \left (1-x-2 x^2\right ) \log \left (x^2\right )\right )}{4 x^2-4 x^3+x^4+\left (4 x^2-2 x^3\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=\frac {e^{x+x^2} (1+\log (x))^2}{x \left (-2+x-\log \left (x^2\right )\right )} \]
Integrate[(Log[E*x]*(E^(x + x^2)*(-4 + 2*x) - 2*E^(x + x^2)*Log[x^2]) + Lo g[E*x]^2*(E^(x + x^2)*(4 - 4*x - 3*x^2 + 2*x^3) + E^(x + x^2)*(1 - x - 2*x ^2)*Log[x^2]))/(4*x^2 - 4*x^3 + x^4 + (4*x^2 - 2*x^3)*Log[x^2] + x^2*Log[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.
| \(\displaystyle \int \frac {\left (e^{x^2+x} (2 x-4)-2 e^{x^2+x} \log \left (x^2\right )\right ) \log (e x)+\left (e^{x^2+x} \left (-2 x^2-x+1\right ) \log \left (x^2\right )+e^{x^2+x} \left (2 x^3-3 x^2-4 x+4\right )\right ) \log ^2(e x)}{x^4-4 x^3+4 x^2+x^2 \log ^2\left (x^2\right )+\left (4 x^2-2 x^3\right ) \log \left (x^2\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^{x^2+x} (\log (x)+1) \left (2 \left (-\log \left (x^2\right )+x-2\right )+(\log (x)+1) \left (2 x^3-3 x^2-\left (2 x^2+x-1\right ) \log \left (x^2\right )-4 x+4\right )\right )}{x^2 \left (\log \left (x^2\right )-x+2\right )^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {e^{x^2+x} (\log (x)+1) \left (2 x^2+2 x^2 \log (x)+x+x \log (x)-\log (x)+1\right )}{x^2 \left (-\log \left (x^2\right )+x-2\right )}-\frac {e^{x^2+x} (x-2) (\log (x)+1)^2}{x^2 \left (-\log \left (x^2\right )+x-2\right )^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 2 \int \frac {e^{x^2+x} \log ^2(x)}{x^2 \left (x-\log \left (x^2\right )-2\right )^2}dx-\int \frac {e^{x^2+x} \log ^2(x)}{x \left (x-\log \left (x^2\right )-2\right )^2}dx+2 \int \frac {e^{x^2+x} \log ^2(x)}{x-\log \left (x^2\right )-2}dx-\int \frac {e^{x^2+x} \log ^2(x)}{x^2 \left (x-\log \left (x^2\right )-2\right )}dx+\int \frac {e^{x^2+x} \log ^2(x)}{x \left (x-\log \left (x^2\right )-2\right )}dx+2 \int \frac {e^{x^2+x}}{x^2 \left (x-\log \left (x^2\right )-2\right )^2}dx-\int \frac {e^{x^2+x}}{x \left (x-\log \left (x^2\right )-2\right )^2}dx+4 \int \frac {e^{x^2+x} \log (x)}{x^2 \left (x-\log \left (x^2\right )-2\right )^2}dx-2 \int \frac {e^{x^2+x} \log (x)}{x \left (x-\log \left (x^2\right )-2\right )^2}dx+2 \int \frac {e^{x^2+x}}{x-\log \left (x^2\right )-2}dx+\int \frac {e^{x^2+x}}{x^2 \left (x-\log \left (x^2\right )-2\right )}dx+\int \frac {e^{x^2+x}}{x \left (x-\log \left (x^2\right )-2\right )}dx+4 \int \frac {e^{x^2+x} \log (x)}{x-\log \left (x^2\right )-2}dx+2 \int \frac {e^{x^2+x} \log (x)}{x \left (x-\log \left (x^2\right )-2\right )}dx\) |
Int[(Log[E*x]*(E^(x + x^2)*(-4 + 2*x) - 2*E^(x + x^2)*Log[x^2]) + Log[E*x] ^2*(E^(x + x^2)*(4 - 4*x - 3*x^2 + 2*x^3) + E^(x + x^2)*(1 - x - 2*x^2)*Lo g[x^2]))/(4*x^2 - 4*x^3 + x^4 + (4*x^2 - 2*x^3)*Log[x^2] + x^2*Log[x^2]^2) ,x]
3.29.19.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04
| method | result | size |
| parallelrisch | \(\frac {\ln \left (x \,{\mathrm e}\right )^{2} {\mathrm e}^{x^{2}+x}}{x \left (x -2-\ln \left (x^{2}\right )\right )}\) | \(29\) |
| risch | \(-\frac {{\mathrm e}^{\left (1+x \right ) x} \ln \left (x \right )}{2 x}-\frac {\left (4+i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 x \right ) {\mathrm e}^{\left (1+x \right ) x}}{8 x}+\frac {\left (4 x^{2}+4 i x \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-8 i x \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+4 i x \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-\pi ^{2} \operatorname {csgn}\left (i x \right )^{4} \operatorname {csgn}\left (i x^{2}\right )^{2}+4 \pi ^{2} \operatorname {csgn}\left (i x \right )^{3} \operatorname {csgn}\left (i x^{2}\right )^{3}-6 \pi ^{2} \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )^{4}+4 \pi ^{2} \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{5}-\pi ^{2} \operatorname {csgn}\left (i x^{2}\right )^{6}\right ) {\mathrm e}^{\left (1+x \right ) x}}{8 x \left (i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}-2 i \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )+i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 x -4 \ln \left (x \right )-4\right )}\) | \(305\) |
int((((-2*x^2-x+1)*exp(x^2+x)*ln(x^2)+(2*x^3-3*x^2-4*x+4)*exp(x^2+x))*ln(x *exp(1))^2+(-2*exp(x^2+x)*ln(x^2)+(2*x-4)*exp(x^2+x))*ln(x*exp(1)))/(x^2*l n(x^2)^2+(-2*x^3+4*x^2)*ln(x^2)+x^4-4*x^3+4*x^2),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.86 \[ \int \frac {\log (e x) \left (e^{x+x^2} (-4+2 x)-2 e^{x+x^2} \log \left (x^2\right )\right )+\log ^2(e x) \left (e^{x+x^2} \left (4-4 x-3 x^2+2 x^3\right )+e^{x+x^2} \left (1-x-2 x^2\right ) \log \left (x^2\right )\right )}{4 x^2-4 x^3+x^4+\left (4 x^2-2 x^3\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=\frac {e^{\left (x^{2} + x\right )} \log \left (x^{2}\right )^{2} + 4 \, e^{\left (x^{2} + x\right )} \log \left (x^{2}\right ) + 4 \, e^{\left (x^{2} + x\right )}}{4 \, {\left (x^{2} - x \log \left (x^{2}\right ) - 2 \, x\right )}} \]
integrate((((-2*x^2-x+1)*exp(x^2+x)*log(x^2)+(2*x^3-3*x^2-4*x+4)*exp(x^2+x ))*log(x*exp(1))^2+(-2*exp(x^2+x)*log(x^2)+(2*x-4)*exp(x^2+x))*log(x*exp(1 )))/(x^2*log(x^2)^2+(-2*x^3+4*x^2)*log(x^2)+x^4-4*x^3+4*x^2),x, algorithm= \
1/4*(e^(x^2 + x)*log(x^2)^2 + 4*e^(x^2 + x)*log(x^2) + 4*e^(x^2 + x))/(x^2 - x*log(x^2) - 2*x)
Time = 0.17 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {\log (e x) \left (e^{x+x^2} (-4+2 x)-2 e^{x+x^2} \log \left (x^2\right )\right )+\log ^2(e x) \left (e^{x+x^2} \left (4-4 x-3 x^2+2 x^3\right )+e^{x+x^2} \left (1-x-2 x^2\right ) \log \left (x^2\right )\right )}{4 x^2-4 x^3+x^4+\left (4 x^2-2 x^3\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=\frac {\left (\log {\left (x^{2} \right )}^{2} + 4 \log {\left (x^{2} \right )} + 4\right ) e^{x^{2} + x}}{4 x^{2} - 4 x \log {\left (x^{2} \right )} - 8 x} \]
integrate((((-2*x**2-x+1)*exp(x**2+x)*ln(x**2)+(2*x**3-3*x**2-4*x+4)*exp(x **2+x))*ln(x*exp(1))**2+(-2*exp(x**2+x)*ln(x**2)+(2*x-4)*exp(x**2+x))*ln(x *exp(1)))/(x**2*ln(x**2)**2+(-2*x**3+4*x**2)*ln(x**2)+x**4-4*x**3+4*x**2), x)
Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {\log (e x) \left (e^{x+x^2} (-4+2 x)-2 e^{x+x^2} \log \left (x^2\right )\right )+\log ^2(e x) \left (e^{x+x^2} \left (4-4 x-3 x^2+2 x^3\right )+e^{x+x^2} \left (1-x-2 x^2\right ) \log \left (x^2\right )\right )}{4 x^2-4 x^3+x^4+\left (4 x^2-2 x^3\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=\frac {{\left (\log \left (x\right )^{2} + 2 \, \log \left (x\right ) + 1\right )} e^{\left (x^{2} + x\right )}}{x^{2} - 2 \, x \log \left (x\right ) - 2 \, x} \]
integrate((((-2*x^2-x+1)*exp(x^2+x)*log(x^2)+(2*x^3-3*x^2-4*x+4)*exp(x^2+x ))*log(x*exp(1))^2+(-2*exp(x^2+x)*log(x^2)+(2*x-4)*exp(x^2+x))*log(x*exp(1 )))/(x^2*log(x^2)^2+(-2*x^3+4*x^2)*log(x^2)+x^4-4*x^3+4*x^2),x, algorithm= \
Time = 0.34 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {\log (e x) \left (e^{x+x^2} (-4+2 x)-2 e^{x+x^2} \log \left (x^2\right )\right )+\log ^2(e x) \left (e^{x+x^2} \left (4-4 x-3 x^2+2 x^3\right )+e^{x+x^2} \left (1-x-2 x^2\right ) \log \left (x^2\right )\right )}{4 x^2-4 x^3+x^4+\left (4 x^2-2 x^3\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=\frac {e^{\left (x^{2} + x\right )} \log \left (x\right )^{2} + 2 \, e^{\left (x^{2} + x\right )} \log \left (x\right ) + e^{\left (x^{2} + x\right )}}{x^{2} - 2 \, x \log \left (x\right ) - 2 \, x} \]
integrate((((-2*x^2-x+1)*exp(x^2+x)*log(x^2)+(2*x^3-3*x^2-4*x+4)*exp(x^2+x ))*log(x*exp(1))^2+(-2*exp(x^2+x)*log(x^2)+(2*x-4)*exp(x^2+x))*log(x*exp(1 )))/(x^2*log(x^2)^2+(-2*x^3+4*x^2)*log(x^2)+x^4-4*x^3+4*x^2),x, algorithm= \
Timed out. \[ \int \frac {\log (e x) \left (e^{x+x^2} (-4+2 x)-2 e^{x+x^2} \log \left (x^2\right )\right )+\log ^2(e x) \left (e^{x+x^2} \left (4-4 x-3 x^2+2 x^3\right )+e^{x+x^2} \left (1-x-2 x^2\right ) \log \left (x^2\right )\right )}{4 x^2-4 x^3+x^4+\left (4 x^2-2 x^3\right ) \log \left (x^2\right )+x^2 \log ^2\left (x^2\right )} \, dx=\int -\frac {{\ln \left (x\,\mathrm {e}\right )}^2\,\left ({\mathrm {e}}^{x^2+x}\,\left (-2\,x^3+3\,x^2+4\,x-4\right )+\ln \left (x^2\right )\,{\mathrm {e}}^{x^2+x}\,\left (2\,x^2+x-1\right )\right )-\ln \left (x\,\mathrm {e}\right )\,\left ({\mathrm {e}}^{x^2+x}\,\left (2\,x-4\right )-2\,\ln \left (x^2\right )\,{\mathrm {e}}^{x^2+x}\right )}{\ln \left (x^2\right )\,\left (4\,x^2-2\,x^3\right )+4\,x^2-4\,x^3+x^4+x^2\,{\ln \left (x^2\right )}^2} \,d x \]
int(-(log(x*exp(1))^2*(exp(x + x^2)*(4*x + 3*x^2 - 2*x^3 - 4) + log(x^2)*e xp(x + x^2)*(x + 2*x^2 - 1)) - log(x*exp(1))*(exp(x + x^2)*(2*x - 4) - 2*l og(x^2)*exp(x + x^2)))/(log(x^2)*(4*x^2 - 2*x^3) + 4*x^2 - 4*x^3 + x^4 + x ^2*log(x^2)^2),x)