Integrand size = 113, antiderivative size = 31 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=e^x \left (5-\log \left (e^{-1/e} \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right )\right ) \]
Time = 0.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=e^{-1+x} \left (1+5 e-e \log \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right ) \]
Integrate[(3*E^(x + x^2) + E^(x + x^2)*(-3 + 15*x - 6*x^2)*Log[x] + E^x*(- 1 + 5*x)*Log[x]^2 + (-3*E^(x + x^2)*x*Log[x] - E^x*x*Log[x]^2)*Log[(3*E^x^ 2*x + x*Log[x])/(E^E^(-1)*Log[x])])/(3*E^x^2*x*Log[x] + x*Log[x]^2),x]
Time = 4.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {7292, 7293, 2009}
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.
\(\displaystyle \int \frac {3 e^{x^2+x}+\left (-3 e^{x^2+x} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )+e^{x^2+x} \left (-6 x^2+15 x-3\right ) \log (x)+e^x (5 x-1) \log ^2(x)}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {3 e^{x^2+x}+\left (-3 e^{x^2+x} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )+e^{x^2+x} \left (-6 x^2+15 x-3\right ) \log (x)+e^x (5 x-1) \log ^2(x)}{x \log (x) \left (3 e^{x^2}+\log (x)\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^x \left (2 x^2 \log (x)-1\right )}{x \left (3 e^{x^2}+\log (x)\right )}+\frac {e^{x-1} \left (-2 e x^2 \log (x)-e x \log (x) \log \left (\frac {3 e^{x^2} x}{\log (x)}+x\right )+(1+5 e) x \log (x)-e \log (x)+e\right )}{x \log (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle (1+5 e) e^{x-1}-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )\) |
Int[(3*E^(x + x^2) + E^(x + x^2)*(-3 + 15*x - 6*x^2)*Log[x] + E^x*(-1 + 5* x)*Log[x]^2 + (-3*E^(x + x^2)*x*Log[x] - E^x*x*Log[x]^2)*Log[(3*E^x^2*x + x*Log[x])/(E^E^(-1)*Log[x])])/(3*E^x^2*x*Log[x] + x*Log[x]^2),x]
3.3.93.3.1 Defintions of rubi rules used
Time = 6.43 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06
method | result | size |
parallelrisch | \(-{\mathrm e}^{x} \ln \left (\frac {x \left (3 \,{\mathrm e}^{x^{2}}+\ln \left (x \right )\right ) {\mathrm e}^{-{\mathrm e}^{-1}}}{\ln \left (x \right )}\right )+5 \,{\mathrm e}^{x}\) | \(33\) |
risch | \(-{\mathrm e}^{x} \ln \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )+\frac {\left (i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{3} {\mathrm e}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right ) {\mathrm e}+i \pi {\operatorname {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{3} {\mathrm e}-i \pi \,\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right ) {\operatorname {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{2} {\mathrm e}-i \pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{2} {\mathrm e}+i \pi \,\operatorname {csgn}\left (\frac {i}{\ln \left (x \right )}\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right ) {\mathrm e}-i \pi \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{2} {\mathrm e}-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \left (x \right )}{3}\right )}{\ln \left (x \right )}\right )}^{2} {\mathrm e}+2-2 \ln \left (3\right ) {\mathrm e}-2 \,{\mathrm e} \ln \left (x \right )+2 \ln \left (\ln \left (x \right )\right ) {\mathrm e}+10 \,{\mathrm e}\right ) {\mathrm e}^{-1+x}}{2}\) | \(333\) |
int(((-x*exp(x)*ln(x)^2-3*x*exp(x)*exp(x^2)*ln(x))*ln((x*ln(x)+3*exp(x^2)* x)/exp(1/exp(1))/ln(x))+(5*x-1)*exp(x)*ln(x)^2+(-6*x^2+15*x-3)*exp(x)*exp( x^2)*ln(x)+3*exp(x)*exp(x^2))/(x*ln(x)^2+3*x*exp(x^2)*ln(x)),x,method=_RET URNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.52 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=-{\left (e^{\left (x^{2} + x\right )} \log \left (\frac {{\left (3 \, x e^{\left (x^{2}\right )} + x \log \left (x\right )\right )} e^{\left (-e^{\left (-1\right )}\right )}}{\log \left (x\right )}\right ) - 5 \, e^{\left (x^{2} + x\right )}\right )} e^{\left (-x^{2}\right )} \]
integrate(((-x*exp(x)*log(x)^2-3*x*exp(x)*exp(x^2)*log(x))*log((x*log(x)+3 *exp(x^2)*x)/exp(1/exp(1))/log(x))+(5*x-1)*exp(x)*log(x)^2+(-6*x^2+15*x-3) *exp(x)*exp(x^2)*log(x)+3*exp(x)*exp(x^2))/(x*log(x)^2+3*x*exp(x^2)*log(x) ),x, algorithm=\
Timed out. \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=\text {Timed out} \]
integrate(((-x*exp(x)*ln(x)**2-3*x*exp(x)*exp(x**2)*ln(x))*ln((x*ln(x)+3*e xp(x**2)*x)/exp(1/exp(1))/ln(x))+(5*x-1)*exp(x)*ln(x)**2+(-6*x**2+15*x-3)* exp(x)*exp(x**2)*ln(x)+3*exp(x)*exp(x**2))/(x*ln(x)**2+3*x*exp(x**2)*ln(x) ),x)
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=-{\left ({\left (e \log \left (x\right ) - 5 \, e - 1\right )} e^{x} + e^{\left (x + 1\right )} \log \left (3 \, e^{\left (x^{2}\right )} + \log \left (x\right )\right ) - e^{\left (x + 1\right )} \log \left (\log \left (x\right )\right )\right )} e^{\left (-1\right )} \]
integrate(((-x*exp(x)*log(x)^2-3*x*exp(x)*exp(x^2)*log(x))*log((x*log(x)+3 *exp(x^2)*x)/exp(1/exp(1))/log(x))+(5*x-1)*exp(x)*log(x)^2+(-6*x^2+15*x-3) *exp(x)*exp(x^2)*log(x)+3*exp(x)*exp(x^2))/(x*log(x)^2+3*x*exp(x^2)*log(x) ),x, algorithm=\
-((e*log(x) - 5*e - 1)*e^x + e^(x + 1)*log(3*e^(x^2) + log(x)) - e^(x + 1) *log(log(x)))*e^(-1)
Time = 0.32 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=-{\left (e^{\left (x + 1\right )} \log \left (x\right ) + e^{\left (x + 1\right )} \log \left (3 \, e^{\left (x^{2}\right )} + \log \left (x\right )\right ) - e^{\left (x + 1\right )} \log \left (\log \left (x\right )\right ) - 5 \, e^{\left (x + 1\right )} - e^{x}\right )} e^{\left (-1\right )} \]
integrate(((-x*exp(x)*log(x)^2-3*x*exp(x)*exp(x^2)*log(x))*log((x*log(x)+3 *exp(x^2)*x)/exp(1/exp(1))/log(x))+(5*x-1)*exp(x)*log(x)^2+(-6*x^2+15*x-3) *exp(x)*exp(x^2)*log(x)+3*exp(x)*exp(x^2))/(x*log(x)^2+3*x*exp(x^2)*log(x) ),x, algorithm=\
-(e^(x + 1)*log(x) + e^(x + 1)*log(3*e^(x^2) + log(x)) - e^(x + 1)*log(log (x)) - 5*e^(x + 1) - e^x)*e^(-1)
Time = 11.73 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{3 e^{x^2} x \log (x)+x \log ^2(x)} \, dx=-{\mathrm {e}}^x\,\left (\ln \left (\frac {{\mathrm {e}}^{-{\mathrm {e}}^{-1}}\,\left (3\,x\,{\mathrm {e}}^{x^2}+x\,\ln \left (x\right )\right )}{\ln \left (x\right )}\right )-5\right ) \]