Integrand size = 54, antiderivative size = 21 \[ \int \frac {1}{12} e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \left (-3+e^x (10+6 x)+e^x (2+2 x) \log \left (x^2\right )\right ) \, dx=e^{x \left (-\frac {1}{4}+\frac {1}{6} e^x \left (3+\log \left (x^2\right )\right )\right )} \]
Time = 1.42 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {1}{12} e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \left (-3+e^x (10+6 x)+e^x (2+2 x) \log \left (x^2\right )\right ) \, dx=e^{-\frac {x}{4}+\frac {e^x x}{2}} \left (x^2\right )^{\frac {e^x x}{6}} \]
Integrate[(E^((-3*x + 6*E^x*x + 2*E^x*x*Log[x^2])/12)*(-3 + E^x*(10 + 6*x) + E^x*(2 + 2*x)*Log[x^2]))/12,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.
\(\displaystyle \int \frac {1}{12} e^{\frac {1}{12} \left (2 e^x x \log \left (x^2\right )+6 e^x x-3 x\right )} \left (e^x (2 x+2) \log \left (x^2\right )+e^x (6 x+10)-3\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{12} \int -e^{\frac {1}{4} \left (2 e^x x-x\right )} \left (x^2\right )^{\frac {e^x x}{6}} \left (-2 e^x (3 x+5)-2 e^x (x+1) \log \left (x^2\right )+3\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{12} \int e^{\frac {1}{4} \left (2 e^x x-x\right )} \left (x^2\right )^{\frac {e^x x}{6}} \left (-2 e^x (3 x+5)-2 e^x (x+1) \log \left (x^2\right )+3\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{12} \int \left (3 e^{\frac {1}{4} \left (2 e^x x-x\right )} \left (x^2\right )^{\frac {e^x x}{6}}-2 e^{x+\frac {1}{4} \left (2 e^x x-x\right )} (3 x+5) \left (x^2\right )^{\frac {e^x x}{6}}-2 e^{x+\frac {1}{4} \left (2 e^x x-x\right )} (x+1) \log \left (x^2\right ) \left (x^2\right )^{\frac {e^x x}{6}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{12} \left (10 \int e^{\frac {1}{4} \left (3+2 e^x\right ) x} \left (x^2\right )^{\frac {e^x x}{6}}dx-3 \int e^{\frac {1}{4} \left (2 e^x x-x\right )} \left (x^2\right )^{\frac {e^x x}{6}}dx+6 \int e^{\frac {1}{4} \left (3+2 e^x\right ) x} x \left (x^2\right )^{\frac {e^x x}{6}}dx-4 \int \frac {\int e^{\frac {1}{4} \left (3+2 e^x\right ) x} \left (x^2\right )^{\frac {e^x x}{6}}dx}{x}dx-4 \int \frac {\int e^{\frac {1}{4} \left (3+2 e^x\right ) x} x \left (x^2\right )^{\frac {e^x x}{6}}dx}{x}dx+2 \log \left (x^2\right ) \int e^{\frac {1}{4} \left (3+2 e^x\right ) x} \left (x^2\right )^{\frac {e^x x}{6}}dx+2 \log \left (x^2\right ) \int e^{\frac {1}{4} \left (3+2 e^x\right ) x} x \left (x^2\right )^{\frac {e^x x}{6}}dx\right )\) |
Int[(E^((-3*x + 6*E^x*x + 2*E^x*x*Log[x^2])/12)*(-3 + E^x*(10 + 6*x) + E^x *(2 + 2*x)*Log[x^2]))/12,x]
3.18.22.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.38 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {x \left (2 \,{\mathrm e}^{x} \ln \left (x^{2}\right )+6 \,{\mathrm e}^{x}-3\right )}{12}}\) | \(19\) |
default | \({\mathrm e}^{\frac {x \,{\mathrm e}^{x} \ln \left (x^{2}\right )}{6}+\frac {{\mathrm e}^{x} x}{2}-\frac {x}{4}}\) | \(20\) |
norman | \({\mathrm e}^{\frac {x \,{\mathrm e}^{x} \ln \left (x^{2}\right )}{6}+\frac {{\mathrm e}^{x} x}{2}-\frac {x}{4}}\) | \(20\) |
risch | \({\mathrm e}^{\frac {x \left (-i {\mathrm e}^{x} \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+2 i {\mathrm e}^{x} \pi \operatorname {csgn}\left (i x^{2}\right )^{2} \operatorname {csgn}\left (i x \right )-i {\mathrm e}^{x} \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}+4 \,{\mathrm e}^{x} \ln \left (x \right )+6 \,{\mathrm e}^{x}-3\right )}{12}}\) | \(72\) |
int(1/12*((2+2*x)*exp(x)*ln(x^2)+(6*x+10)*exp(x)-3)*exp(1/6*x*exp(x)*ln(x^ 2)+1/2*exp(x)*x-1/4*x),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{12} e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \left (-3+e^x (10+6 x)+e^x (2+2 x) \log \left (x^2\right )\right ) \, dx=e^{\left (\frac {1}{6} \, x e^{x} \log \left (x^{2}\right ) + \frac {1}{2} \, x e^{x} - \frac {1}{4} \, x\right )} \]
integrate(1/12*((2+2*x)*exp(x)*log(x^2)+(6*x+10)*exp(x)-3)*exp(1/6*x*exp(x )*log(x^2)+1/2*exp(x)*x-1/4*x),x, algorithm=\
Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {1}{12} e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \left (-3+e^x (10+6 x)+e^x (2+2 x) \log \left (x^2\right )\right ) \, dx=e^{\frac {x e^{x} \log {\left (x^{2} \right )}}{6} + \frac {x e^{x}}{2} - \frac {x}{4}} \]
integrate(1/12*((2+2*x)*exp(x)*ln(x**2)+(6*x+10)*exp(x)-3)*exp(1/6*x*exp(x )*ln(x**2)+1/2*exp(x)*x-1/4*x),x)
\[ \int \frac {1}{12} e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \left (-3+e^x (10+6 x)+e^x (2+2 x) \log \left (x^2\right )\right ) \, dx=\int { \frac {1}{12} \, {\left (2 \, {\left (x + 1\right )} e^{x} \log \left (x^{2}\right ) + 2 \, {\left (3 \, x + 5\right )} e^{x} - 3\right )} e^{\left (\frac {1}{6} \, x e^{x} \log \left (x^{2}\right ) + \frac {1}{2} \, x e^{x} - \frac {1}{4} \, x\right )} \,d x } \]
integrate(1/12*((2+2*x)*exp(x)*log(x^2)+(6*x+10)*exp(x)-3)*exp(1/6*x*exp(x )*log(x^2)+1/2*exp(x)*x-1/4*x),x, algorithm=\
1/12*integrate((2*(x + 1)*e^x*log(x^2) + 2*(3*x + 5)*e^x - 3)*e^(1/6*x*e^x *log(x^2) + 1/2*x*e^x - 1/4*x), x)
Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {1}{12} e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \left (-3+e^x (10+6 x)+e^x (2+2 x) \log \left (x^2\right )\right ) \, dx=e^{\left (\frac {1}{6} \, x e^{x} \log \left (x^{2}\right ) + \frac {1}{2} \, x e^{x} - \frac {1}{4} \, x\right )} \]
integrate(1/12*((2+2*x)*exp(x)*log(x^2)+(6*x+10)*exp(x)-3)*exp(1/6*x*exp(x )*log(x^2)+1/2*exp(x)*x-1/4*x),x, algorithm=\
Time = 9.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {1}{12} e^{\frac {1}{12} \left (-3 x+6 e^x x+2 e^x x \log \left (x^2\right )\right )} \left (-3+e^x (10+6 x)+e^x (2+2 x) \log \left (x^2\right )\right ) \, dx={\mathrm {e}}^{\frac {x\,{\mathrm {e}}^x}{2}-\frac {x}{4}}\,{\left (x^2\right )}^{\frac {x\,{\mathrm {e}}^x}{6}} \]