Integrand size = 108, antiderivative size = 25 \[ \int \frac {-2+e^{\frac {1}{2} \left (x+x^3\right )} \left (-6+3 x-x^2+(9-3 x) x^3\right )+2 \log (x)+\left (e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)\right ) \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )}{e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)} \, dx=x \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right ) \]
Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {-2+e^{\frac {1}{2} \left (x+x^3\right )} \left (-6+3 x-x^2+(9-3 x) x^3\right )+2 \log (x)+\left (e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)\right ) \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )}{e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)} \, dx=x \log \left (\frac {x}{e^{\frac {x}{2}+\frac {x^3}{2}} (-3+x)+\log (x)}\right ) \]
Integrate[(-2 + E^((x + x^3)/2)*(-6 + 3*x - x^2 + (9 - 3*x)*x^3) + 2*Log[x ] + (E^((x + x^3)/2)*(-6 + 2*x) + 2*Log[x])*Log[x/(E^((x + x^3)/2)*(-3 + x ) + Log[x])])/(E^((x + x^3)/2)*(-6 + 2*x) + 2*Log[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.
\(\displaystyle \int \frac {\left (e^{\frac {1}{2} \left (x^3+x\right )} (2 x-6)+2 \log (x)\right ) \log \left (\frac {x}{e^{\frac {1}{2} \left (x^3+x\right )} (x-3)+\log (x)}\right )+e^{\frac {1}{2} \left (x^3+x\right )} \left ((9-3 x) x^3-x^2+3 x-6\right )+2 \log (x)-2}{e^{\frac {1}{2} \left (x^3+x\right )} (2 x-6)+2 \log (x)} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {3 x^4 \log (x)-9 x^3 \log (x)+x^2 \log (x)-2 x-x \log (x)+6}{2 (x-3) \left (e^{\frac {x^3}{2}+\frac {x}{2}} x-3 e^{\frac {x^3}{2}+\frac {x}{2}}+\log (x)\right )}+\frac {-3 x^4+9 x^3+2 x \log \left (\frac {x}{e^{\frac {1}{2} \left (x^3+x\right )} (x-3)+\log (x)}\right )-6 \log \left (\frac {x}{e^{\frac {1}{2} \left (x^3+x\right )} (x-3)+\log (x)}\right )-x^2+3 x-6}{2 (x-3)}\right )dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \left (\frac {3 x^4 \log (x)-9 x^3 \log (x)+x^2 \log (x)-2 x-x \log (x)+6}{2 (x-3) \left (e^{\frac {x^3}{2}+\frac {x}{2}} x-3 e^{\frac {x^3}{2}+\frac {x}{2}}+\log (x)\right )}+\frac {-3 x^4+9 x^3+2 x \log \left (\frac {x}{e^{\frac {1}{2} \left (x^3+x\right )} (x-3)+\log (x)}\right )-6 \log \left (\frac {x}{e^{\frac {1}{2} \left (x^3+x\right )} (x-3)+\log (x)}\right )-x^2+3 x-6}{2 (x-3)}\right )dx\) |
Int[(-2 + E^((x + x^3)/2)*(-6 + 3*x - x^2 + (9 - 3*x)*x^3) + 2*Log[x] + (E ^((x + x^3)/2)*(-6 + 2*x) + 2*Log[x])*Log[x/(E^((x + x^3)/2)*(-3 + x) + Lo g[x])])/(E^((x + x^3)/2)*(-6 + 2*x) + 2*Log[x]),x]
3.30.54.3.1 Defintions of rubi rules used
Leaf count of result is larger than twice the leaf count of optimal. \(103\) vs. \(2(24)=48\).
Time = 3.24 (sec) , antiderivative size = 104, normalized size of antiderivative = 4.16
method | result | size |
parallelrisch | \(x \ln \left (\frac {x}{{\mathrm e}^{\frac {1}{2} x^{3}+\frac {1}{2} x} x +\ln \left (x \right )-3 \,{\mathrm e}^{\frac {1}{2} x^{3}+\frac {1}{2} x}}\right )+3 \ln \left ({\mathrm e}^{\frac {1}{2} x^{3}+\frac {1}{2} x} x +\ln \left (x \right )-3 \,{\mathrm e}^{\frac {1}{2} x^{3}+\frac {1}{2} x}\right )-3 \ln \left (x \right )+3 \ln \left (\frac {x}{{\mathrm e}^{\frac {1}{2} x^{3}+\frac {1}{2} x} x +\ln \left (x \right )-3 \,{\mathrm e}^{\frac {1}{2} x^{3}+\frac {1}{2} x}}\right )\) | \(104\) |
risch | \(-x \ln \left ({\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \left (x \right )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}\right )+x \ln \left (x \right )+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \left (x \right )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right ) \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \left (x \right )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right )^{2}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \left (x \right )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right ) \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \left (x \right )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right ) \operatorname {csgn}\left (i x \right )}{2}-\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \left (x \right )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right )^{3}}{2}+\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}} x +\ln \left (x \right )-3 \,{\mathrm e}^{\frac {x \left (x^{2}+1\right )}{2}}}\right )^{2} \operatorname {csgn}\left (i x \right )}{2}\) | \(261\) |
int((((2*x-6)*exp(1/2*x^3+1/2*x)+2*ln(x))*ln(x/((-3+x)*exp(1/2*x^3+1/2*x)+ ln(x)))+((-3*x+9)*x^3-x^2+3*x-6)*exp(1/2*x^3+1/2*x)+2*ln(x)-2)/((2*x-6)*ex p(1/2*x^3+1/2*x)+2*ln(x)),x,method=_RETURNVERBOSE)
x*ln(x/(exp(1/2*x^3+1/2*x)*x+ln(x)-3*exp(1/2*x^3+1/2*x)))+3*ln(exp(1/2*x^3 +1/2*x)*x+ln(x)-3*exp(1/2*x^3+1/2*x))-3*ln(x)+3*ln(x/(exp(1/2*x^3+1/2*x)*x +ln(x)-3*exp(1/2*x^3+1/2*x)))
Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-2+e^{\frac {1}{2} \left (x+x^3\right )} \left (-6+3 x-x^2+(9-3 x) x^3\right )+2 \log (x)+\left (e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)\right ) \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )}{e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)} \, dx=x \log \left (\frac {x}{{\left (x - 3\right )} e^{\left (\frac {1}{2} \, x^{3} + \frac {1}{2} \, x\right )} + \log \left (x\right )}\right ) \]
integrate((((2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x))*log(x/((-3+x)*exp(1/2*x^3 +1/2*x)+log(x)))+((-3*x+9)*x^3-x^2+3*x-6)*exp(1/2*x^3+1/2*x)+2*log(x)-2)/( (2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x)),x, algorithm=\
Time = 0.90 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-2+e^{\frac {1}{2} \left (x+x^3\right )} \left (-6+3 x-x^2+(9-3 x) x^3\right )+2 \log (x)+\left (e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)\right ) \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )}{e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)} \, dx=x \log {\left (\frac {x}{\left (x - 3\right ) e^{\frac {x^{3}}{2} + \frac {x}{2}} + \log {\left (x \right )}} \right )} \]
integrate((((2*x-6)*exp(1/2*x**3+1/2*x)+2*ln(x))*ln(x/((-3+x)*exp(1/2*x**3 +1/2*x)+ln(x)))+((-3*x+9)*x**3-x**2+3*x-6)*exp(1/2*x**3+1/2*x)+2*ln(x)-2)/ ((2*x-6)*exp(1/2*x**3+1/2*x)+2*ln(x)),x)
Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-2+e^{\frac {1}{2} \left (x+x^3\right )} \left (-6+3 x-x^2+(9-3 x) x^3\right )+2 \log (x)+\left (e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)\right ) \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )}{e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)} \, dx=-x \log \left ({\left (x - 3\right )} e^{\left (\frac {1}{2} \, x^{3} + \frac {1}{2} \, x\right )} + \log \left (x\right )\right ) + x \log \left (x\right ) \]
integrate((((2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x))*log(x/((-3+x)*exp(1/2*x^3 +1/2*x)+log(x)))+((-3*x+9)*x^3-x^2+3*x-6)*exp(1/2*x^3+1/2*x)+2*log(x)-2)/( (2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x)),x, algorithm=\
Time = 0.49 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.44 \[ \int \frac {-2+e^{\frac {1}{2} \left (x+x^3\right )} \left (-6+3 x-x^2+(9-3 x) x^3\right )+2 \log (x)+\left (e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)\right ) \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )}{e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)} \, dx=-x \log \left (x e^{\left (\frac {1}{2} \, x^{3} + \frac {1}{2} \, x\right )} - 3 \, e^{\left (\frac {1}{2} \, x^{3} + \frac {1}{2} \, x\right )} + \log \left (x\right )\right ) + x \log \left (x\right ) \]
integrate((((2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x))*log(x/((-3+x)*exp(1/2*x^3 +1/2*x)+log(x)))+((-3*x+9)*x^3-x^2+3*x-6)*exp(1/2*x^3+1/2*x)+2*log(x)-2)/( (2*x-6)*exp(1/2*x^3+1/2*x)+2*log(x)),x, algorithm=\
Time = 13.16 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {-2+e^{\frac {1}{2} \left (x+x^3\right )} \left (-6+3 x-x^2+(9-3 x) x^3\right )+2 \log (x)+\left (e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)\right ) \log \left (\frac {x}{e^{\frac {1}{2} \left (x+x^3\right )} (-3+x)+\log (x)}\right )}{e^{\frac {1}{2} \left (x+x^3\right )} (-6+2 x)+2 \log (x)} \, dx=x\,\ln \left (\frac {x}{\ln \left (x\right )+{\mathrm {e}}^{x/2}\,{\mathrm {e}}^{\frac {x^3}{2}}\,\left (x-3\right )}\right ) \]