Integrand size = 69, antiderivative size = 27 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=e^{\frac {\left (-1+\frac {5 e^x (2-x)}{x}\right ) (1+x)}{\log (3)}}+x \]
Time = 2.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=e^{-\frac {1}{\log (3)}-\frac {x}{\log (3)}-\frac {5 e^x (-2+x) (1+x)}{x \log (3)}}+x \]
Integrate[(E^((-x - x^2 + E^x*(10 + 5*x - 5*x^2))/(x*Log[3]))*(-x^2 + E^x* (-10 + 10*x - 5*x^3)) + x^2*Log[3])/(x^2*Log[3]),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.
\(\displaystyle \int \frac {\left (e^x \left (-5 x^3+10 x-10\right )-x^2\right ) \exp \left (\frac {-x^2+e^x \left (-5 x^2+5 x+10\right )-x}{x \log (3)}\right )+x^2 \log (3)}{x^2 \log (3)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {e^{-\frac {x^2+x-5 e^x \left (-x^2+x+2\right )}{x \log (3)}} \left (x^2+5 e^x \left (x^3-2 x+2\right )\right )-x^2 \log (3)}{x^2}dx}{\log (3)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {e^{-\frac {x^2+x-5 e^x \left (-x^2+x+2\right )}{x \log (3)}} \left (x^2+5 e^x \left (x^3-2 x+2\right )\right )-x^2 \log (3)}{x^2}dx}{\log (3)}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {\int \left (\frac {\exp \left (-\frac {(x+1) \left (5 e^x (x-2)+x\right )}{x \log (3)}\right ) \left (5 e^x x^3+x^2-10 e^x x+10 e^x\right )}{x^2}-\log (3)\right )dx}{\log (3)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {10 \int \frac {\exp \left (x-\frac {(x+1) \left (5 e^x (x-2)+x\right )}{x \log (3)}\right )}{x^2}dx+\int \exp \left (-\frac {(x+1) \left (5 e^x (x-2)+x\right )}{x \log (3)}\right )dx-10 \int \frac {\exp \left (x-\frac {(x+1) \left (5 e^x (x-2)+x\right )}{x \log (3)}\right )}{x}dx+5 \int \exp \left (x-\frac {(x+1) \left (5 e^x (x-2)+x\right )}{x \log (3)}\right ) xdx+x (-\log (3))}{\log (3)}\) |
Int[(E^((-x - x^2 + E^x*(10 + 5*x - 5*x^2))/(x*Log[3]))*(-x^2 + E^x*(-10 + 10*x - 5*x^3)) + x^2*Log[3])/(x^2*Log[3]),x]
3.20.97.3.1 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]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.62 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00
method | result | size |
risch | \(x +{\mathrm e}^{-\frac {\left (1+x \right ) \left (5 \,{\mathrm e}^{x} x -10 \,{\mathrm e}^{x}+x \right )}{\ln \left (3\right ) x}}\) | \(27\) |
norman | \(\frac {x^{2}+x \,{\mathrm e}^{\frac {\left (-5 x^{2}+5 x +10\right ) {\mathrm e}^{x}-x^{2}-x}{x \ln \left (3\right )}}}{x}\) | \(42\) |
parallelrisch | \(\frac {x \ln \left (3\right )+\ln \left (3\right ) {\mathrm e}^{\frac {\left (-5 x^{2}+5 x +10\right ) {\mathrm e}^{x}-x^{2}-x}{x \ln \left (3\right )}}}{\ln \left (3\right )}\) | \(45\) |
int((((-5*x^3+10*x-10)*exp(x)-x^2)*exp(((-5*x^2+5*x+10)*exp(x)-x^2-x)/x/ln (3))+x^2*ln(3))/x^2/ln(3),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=x + e^{\left (-\frac {x^{2} + 5 \, {\left (x^{2} - x - 2\right )} e^{x} + x}{x \log \left (3\right )}\right )} \]
integrate((((-5*x^3+10*x-10)*exp(x)-x^2)*exp(((-5*x^2+5*x+10)*exp(x)-x^2-x )/x/log(3))+x^2*log(3))/x^2/log(3),x, algorithm=\
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=x + e^{\frac {- x^{2} - x + \left (- 5 x^{2} + 5 x + 10\right ) e^{x}}{x \log {\left (3 \right )}}} \]
integrate((((-5*x**3+10*x-10)*exp(x)-x**2)*exp(((-5*x**2+5*x+10)*exp(x)-x* *2-x)/x/ln(3))+x**2*ln(3))/x**2/ln(3),x)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (24) = 48\).
Time = 0.38 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.07 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=\frac {x \log \left (3\right ) + e^{\left (-\frac {5 \, x e^{x}}{\log \left (3\right )} - \frac {x}{\log \left (3\right )} + \frac {5 \, e^{x}}{\log \left (3\right )} + \frac {10 \, e^{x}}{x \log \left (3\right )} - \frac {1}{\log \left (3\right )}\right )} \log \left (3\right )}{\log \left (3\right )} \]
integrate((((-5*x^3+10*x-10)*exp(x)-x^2)*exp(((-5*x^2+5*x+10)*exp(x)-x^2-x )/x/log(3))+x^2*log(3))/x^2/log(3),x, algorithm=\
(x*log(3) + e^(-5*x*e^x/log(3) - x/log(3) + 5*e^x/log(3) + 10*e^x/(x*log(3 )) - 1/log(3))*log(3))/log(3)
\[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=\int { \frac {x^{2} \log \left (3\right ) - {\left (x^{2} + 5 \, {\left (x^{3} - 2 \, x + 2\right )} e^{x}\right )} e^{\left (-\frac {x^{2} + 5 \, {\left (x^{2} - x - 2\right )} e^{x} + x}{x \log \left (3\right )}\right )}}{x^{2} \log \left (3\right )} \,d x } \]
integrate((((-5*x^3+10*x-10)*exp(x)-x^2)*exp(((-5*x^2+5*x+10)*exp(x)-x^2-x )/x/log(3))+x^2*log(3))/x^2/log(3),x, algorithm=\
integrate((x^2*log(3) - (x^2 + 5*(x^3 - 2*x + 2)*e^x)*e^(-(x^2 + 5*(x^2 - x - 2)*e^x + x)/(x*log(3))))/(x^2*log(3)), x)
Time = 11.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.81 \[ \int \frac {e^{\frac {-x-x^2+e^x \left (10+5 x-5 x^2\right )}{x \log (3)}} \left (-x^2+e^x \left (-10+10 x-5 x^3\right )\right )+x^2 \log (3)}{x^2 \log (3)} \, dx=x+{\mathrm {e}}^{-\frac {x}{\ln \left (3\right )}}\,{\mathrm {e}}^{\frac {10\,{\mathrm {e}}^x}{x\,\ln \left (3\right )}}\,{\mathrm {e}}^{\frac {5\,{\mathrm {e}}^x}{\ln \left (3\right )}}\,{\mathrm {e}}^{-\frac {1}{\ln \left (3\right )}}\,{\mathrm {e}}^{-\frac {5\,x\,{\mathrm {e}}^x}{\ln \left (3\right )}} \]
int(-(exp(-(x - exp(x)*(5*x - 5*x^2 + 10) + x^2)/(x*log(3)))*(exp(x)*(5*x^ 3 - 10*x + 10) + x^2) - x^2*log(3))/(x^2*log(3)),x)