Integrand size = 74, antiderivative size = 29 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{7+x+\frac {1}{3} x \left (x-\left (\frac {3}{5}+x\right ) \log \left (e^{-e^x} x\right )\right )} \]
Time = 3.12 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{7+x+\frac {x^2}{3}} \left (e^{-e^x} x\right )^{-\frac {1}{15} x (3+5 x)} \]
Integrate[(E^((105 + 15*x + 5*x^2 + (-3*x - 5*x^2)*Log[x/E^E^x])/15)*(12 + 5*x + E^x*(3*x + 5*x^2) + (-3 - 10*x)*Log[x/E^E^x]))/15,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.
\(\displaystyle \int \frac {1}{15} \left (e^x \left (5 x^2+3 x\right )+5 x+(-10 x-3) \log \left (e^{-e^x} x\right )+12\right ) \exp \left (\frac {1}{15} \left (5 x^2+\left (-5 x^2-3 x\right ) \log \left (e^{-e^x} x\right )+15 x+105\right )\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{15} \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} \left (e^{-e^x} x\right )^{\frac {1}{15} \left (-5 x^2-3 x\right )} \left (5 x+e^x \left (5 x^2+3 x\right )-(10 x+3) \log \left (e^{-e^x} x\right )+12\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {1}{15} \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x} \left (5 x+e^x \left (5 x^2+3 x\right )-(10 x+3) \log \left (e^{-e^x} x\right )+12\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{15} \int \left (12 e^{\frac {1}{3} \left (x^2+3 x+21\right )} \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}+5 e^{\frac {1}{3} \left (x^2+3 x+21\right )} x \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}+e^{x+\frac {1}{3} \left (x^2+3 x+21\right )} x (5 x+3) \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}-e^{\frac {1}{3} \left (x^2+3 x+21\right )} (10 x+3) \log \left (e^{-e^x} x\right ) \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{15} \left (12 \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx+3 \int e^{\frac {x^2}{3}+2 x+7} x \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx+5 \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} x \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx+5 \int e^{\frac {x^2}{3}+2 x+7} x^2 \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx-3 \int e^x \int e^{\frac {x^2}{3}+x+7} \left (e^{-e^x} x\right )^{-\frac {1}{15} x (5 x+3)}dxdx+3 \int \frac {\int e^{\frac {x^2}{3}+x+7} \left (e^{-e^x} x\right )^{-\frac {1}{15} x (5 x+3)}dx}{x}dx-10 \int e^x \int e^{\frac {x^2}{3}+x+7} x \left (e^{-e^x} x\right )^{-\frac {1}{15} x (5 x+3)}dxdx+10 \int \frac {\int e^{\frac {x^2}{3}+x+7} x \left (e^{-e^x} x\right )^{-\frac {1}{15} x (5 x+3)}dx}{x}dx-3 \log \left (e^{-e^x} x\right ) \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx-10 \log \left (e^{-e^x} x\right ) \int e^{\frac {1}{3} \left (x^2+3 x+21\right )} x \left (e^{-e^x} x\right )^{\frac {1}{15} (-5 x-3) x}dx\right )\) |
Int[(E^((105 + 15*x + 5*x^2 + (-3*x - 5*x^2)*Log[x/E^E^x])/15)*(12 + 5*x + E^x*(3*x + 5*x^2) + (-3 - 10*x)*Log[x/E^E^x]))/15,x]
3.13.26.3.1 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 = 1.59 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {\left (-5 x^{2}-3 x \right ) \ln \left (x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )}{15}+\frac {x^{2}}{3}+x +7}\) | \(29\) |
risch | \(x^{-\frac {x^{2}}{3}} x^{-\frac {x}{5}} \left ({\mathrm e}^{{\mathrm e}^{x}}\right )^{\frac {x^{2}}{3}} \left ({\mathrm e}^{{\mathrm e}^{x}}\right )^{\frac {x}{5}} {\mathrm e}^{7+\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{3} x^{2}}{6}+\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{3} x}{10}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i x \right ) x^{2}}{6}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i x \right ) x}{10}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x^{2}}{6}-\frac {i \pi \operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x}{10}+\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x^{2}}{6}+\frac {i \pi \,\operatorname {csgn}\left (i x \,{\mathrm e}^{-{\mathrm e}^{x}}\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i {\mathrm e}^{-{\mathrm e}^{x}}\right ) x}{10}+\frac {x^{2}}{3}+x}\) | \(235\) |
int(1/15*((-10*x-3)*ln(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*exp(1/15* (-5*x^2-3*x)*ln(x/exp(exp(x)))+1/3*x^2+x+7),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (\frac {1}{3} \, x^{2} - \frac {1}{15} \, {\left (5 \, x^{2} + 3 \, x\right )} \log \left (x e^{\left (-e^{x}\right )}\right ) + x + 7\right )} \]
integrate(1/15*((-10*x-3)*log(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*ex p(1/15*(-5*x^2-3*x)*log(x/exp(exp(x)))+1/3*x^2+x+7),x, algorithm=\
Timed out. \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=\text {Timed out} \]
integrate(1/15*((-10*x-3)*ln(x/exp(exp(x)))+(5*x**2+3*x)*exp(x)+5*x+12)*ex p(1/15*(-5*x**2-3*x)*ln(x/exp(exp(x)))+1/3*x**2+x+7),x)
Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (\frac {1}{3} \, x^{2} e^{x} - \frac {1}{3} \, x^{2} \log \left (x\right ) + \frac {1}{3} \, x^{2} + \frac {1}{5} \, x e^{x} - \frac {1}{5} \, x \log \left (x\right ) + x + 7\right )} \]
integrate(1/15*((-10*x-3)*log(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*ex p(1/15*(-5*x^2-3*x)*log(x/exp(exp(x)))+1/3*x^2+x+7),x, algorithm=\
Time = 0.32 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=e^{\left (-\frac {1}{3} \, x^{2} \log \left (x e^{\left (-e^{x}\right )}\right ) + \frac {1}{3} \, x^{2} - \frac {1}{5} \, x \log \left (x e^{\left (-e^{x}\right )}\right ) + x + 7\right )} \]
integrate(1/15*((-10*x-3)*log(x/exp(exp(x)))+(5*x^2+3*x)*exp(x)+5*x+12)*ex p(1/15*(-5*x^2-3*x)*log(x/exp(exp(x)))+1/3*x^2+x+7),x, algorithm=\
Time = 11.97 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {1}{15} e^{\frac {1}{15} \left (105+15 x+5 x^2+\left (-3 x-5 x^2\right ) \log \left (e^{-e^x} x\right )\right )} \left (12+5 x+e^x \left (3 x+5 x^2\right )+(-3-10 x) \log \left (e^{-e^x} x\right )\right ) \, dx=\frac {{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^x}{5}}\,{\mathrm {e}}^7\,{\mathrm {e}}^{\frac {x^2\,{\mathrm {e}}^x}{3}}\,{\mathrm {e}}^{\frac {x^2}{3}}\,{\mathrm {e}}^x}{x^{\frac {x^2}{3}+\frac {x}{5}}} \]