Integrand size = 82, antiderivative size = 24 \[ \int \frac {e^{-\frac {x^2}{1+x}} \left (e^{\frac {2 e^{-\frac {x^2}{1+x}}}{\log (4)}} \left (-4 x-2 x^2\right )+e^{\frac {x^2}{1+x}} \left (-1-2 x-x^2\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=e^{\frac {2 e^{-\frac {x^2}{1+x}}}{\log (4)}}-x \]
Time = 0.59 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {e^{-\frac {x^2}{1+x}} \left (e^{\frac {2 e^{-\frac {x^2}{1+x}}}{\log (4)}} \left (-4 x-2 x^2\right )+e^{\frac {x^2}{1+x}} \left (-1-2 x-x^2\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=e^{\frac {2 e^{1-x-\frac {1}{1+x}}}{\log (4)}}-x \]
Integrate[(E^(2/(E^(x^2/(1 + x))*Log[4]))*(-4*x - 2*x^2) + E^(x^2/(1 + x)) *(-1 - 2*x - x^2)*Log[4])/(E^(x^2/(1 + x))*(1 + 2*x + x^2)*Log[4]),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.
\(\displaystyle \int \frac {e^{-\frac {x^2}{x+1}} \left (\left (-2 x^2-4 x\right ) e^{\frac {2 e^{-\frac {x^2}{x+1}}}{\log (4)}}+e^{\frac {x^2}{x+1}} \left (-x^2-2 x-1\right ) \log (4)\right )}{\left (x^2+2 x+1\right ) \log (4)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {e^{-\frac {x^2}{x+1}} \left (2 e^{\frac {2 e^{-\frac {x^2}{x+1}}}{\log (4)}} \left (x^2+2 x\right )+e^{\frac {x^2}{x+1}} \left (x^2+2 x+1\right ) \log (4)\right )}{x^2+2 x+1}dx}{\log (4)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {e^{-\frac {x^2}{x+1}} \left (2 e^{\frac {2 e^{-\frac {x^2}{x+1}}}{\log (4)}} \left (x^2+2 x\right )+e^{\frac {x^2}{x+1}} \left (x^2+2 x+1\right ) \log (4)\right )}{x^2+2 x+1}dx}{\log (4)}\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle -\frac {\int \frac {e^{-\frac {x^2}{x+1}} \left (2 e^{\frac {2 e^{-\frac {x^2}{x+1}}}{\log (4)}} \left (x^2+2 x\right )+e^{\frac {x^2}{x+1}} \left (x^2+2 x+1\right ) \log (4)\right )}{(x+1)^2}dx}{\log (4)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (\frac {2 e^{\frac {2 e^{-\frac {x^2}{x+1}}}{\log (4)}-\frac {x^2}{x+1}} x (x+2)}{(x+1)^2}+\log (4)\right )dx}{\log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \int e^{\frac {2 e^{-\frac {x^2}{x+1}}}{\log (4)}-\frac {x^2}{x+1}}dx-2 \int \frac {e^{\frac {2 e^{-\frac {x^2}{x+1}}}{\log (4)}-\frac {x^2}{x+1}}}{(x+1)^2}dx+x \log (4)}{\log (4)}\) |
Int[(E^(2/(E^(x^2/(1 + x))*Log[4]))*(-4*x - 2*x^2) + E^(x^2/(1 + x))*(-1 - 2*x - x^2)*Log[4])/(E^(x^2/(1 + x))*(1 + 2*x + x^2)*Log[4]),x]
3.27.96.3.1 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_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
Time = 1.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92
method | result | size |
risch | \({\mathrm e}^{\frac {{\mathrm e}^{-\frac {x^{2}}{1+x}}}{\ln \left (2\right )}}-x\) | \(22\) |
parallelrisch | \(-\frac {2 x^{2} \ln \left (2\right )-2 \ln \left (2\right ) x \,{\mathrm e}^{\frac {{\mathrm e}^{-\frac {x^{2}}{1+x}}}{\ln \left (2\right )}}-6 x \ln \left (2\right )-2 \ln \left (2\right ) {\mathrm e}^{\frac {{\mathrm e}^{-\frac {x^{2}}{1+x}}}{\ln \left (2\right )}}-8 \ln \left (2\right )}{2 \ln \left (2\right ) \left (1+x \right )}\) | \(74\) |
parts | \(-x +\frac {\left ({\mathrm e}^{\frac {x^{2}}{1+x}} {\mathrm e}^{\frac {{\mathrm e}^{-\frac {x^{2}}{1+x}}}{\ln \left (2\right )}}+x \,{\mathrm e}^{\frac {x^{2}}{1+x}} {\mathrm e}^{\frac {{\mathrm e}^{-\frac {x^{2}}{1+x}}}{\ln \left (2\right )}}\right ) {\mathrm e}^{-\frac {x^{2}}{1+x}}}{1+x}\) | \(83\) |
norman | \(\frac {\left ({\mathrm e}^{\frac {x^{2}}{1+x}}+{\mathrm e}^{\frac {x^{2}}{1+x}} {\mathrm e}^{\frac {{\mathrm e}^{-\frac {x^{2}}{1+x}}}{\ln \left (2\right )}}+x \,{\mathrm e}^{\frac {x^{2}}{1+x}} {\mathrm e}^{\frac {{\mathrm e}^{-\frac {x^{2}}{1+x}}}{\ln \left (2\right )}}-x^{2} {\mathrm e}^{\frac {x^{2}}{1+x}}\right ) {\mathrm e}^{-\frac {x^{2}}{1+x}}}{1+x}\) | \(104\) |
int(1/2*((-2*x^2-4*x)*exp(1/ln(2)/exp(x^2/(1+x)))+2*(-x^2-2*x-1)*ln(2)*exp (x^2/(1+x)))/(x^2+2*x+1)/ln(2)/exp(x^2/(1+x)),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-\frac {x^2}{1+x}} \left (e^{\frac {2 e^{-\frac {x^2}{1+x}}}{\log (4)}} \left (-4 x-2 x^2\right )+e^{\frac {x^2}{1+x}} \left (-1-2 x-x^2\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=-x + e^{\left (\frac {e^{\left (-\frac {x^{2}}{x + 1}\right )}}{\log \left (2\right )}\right )} \]
integrate(1/2*((-2*x^2-4*x)*exp(1/log(2)/exp(x^2/(1+x)))+2*(-x^2-2*x-1)*lo g(2)*exp(x^2/(1+x)))/(x^2+2*x+1)/log(2)/exp(x^2/(1+x)),x, algorithm=\
Time = 0.17 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {e^{-\frac {x^2}{1+x}} \left (e^{\frac {2 e^{-\frac {x^2}{1+x}}}{\log (4)}} \left (-4 x-2 x^2\right )+e^{\frac {x^2}{1+x}} \left (-1-2 x-x^2\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=- x + e^{\frac {e^{- \frac {x^{2}}{x + 1}}}{\log {\left (2 \right )}}} \]
integrate(1/2*((-2*x**2-4*x)*exp(1/ln(2)/exp(x**2/(1+x)))+2*(-x**2-2*x-1)* ln(2)*exp(x**2/(1+x)))/(x**2+2*x+1)/ln(2)/exp(x**2/(1+x)),x)
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (21) = 42\).
Time = 0.46 (sec) , antiderivative size = 75, normalized size of antiderivative = 3.12 \[ \int \frac {e^{-\frac {x^2}{1+x}} \left (e^{\frac {2 e^{-\frac {x^2}{1+x}}}{\log (4)}} \left (-4 x-2 x^2\right )+e^{\frac {x^2}{1+x}} \left (-1-2 x-x^2\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=-\frac {{\left (\frac {x^{2} + x - 1}{x + 1} - 2 \, \log \left (x + 1\right )\right )} \log \left (2\right ) + 2 \, {\left (\frac {1}{x + 1} + \log \left (x + 1\right )\right )} \log \left (2\right ) - e^{\left (\frac {e^{\left (-x - \frac {1}{x + 1} + 1\right )}}{\log \left (2\right )}\right )} \log \left (2\right ) - \frac {\log \left (2\right )}{x + 1}}{\log \left (2\right )} \]
integrate(1/2*((-2*x^2-4*x)*exp(1/log(2)/exp(x^2/(1+x)))+2*(-x^2-2*x-1)*lo g(2)*exp(x^2/(1+x)))/(x^2+2*x+1)/log(2)/exp(x^2/(1+x)),x, algorithm=\
-(((x^2 + x - 1)/(x + 1) - 2*log(x + 1))*log(2) + 2*(1/(x + 1) + log(x + 1 ))*log(2) - e^(e^(-x - 1/(x + 1) + 1)/log(2))*log(2) - log(2)/(x + 1))/log (2)
\[ \int \frac {e^{-\frac {x^2}{1+x}} \left (e^{\frac {2 e^{-\frac {x^2}{1+x}}}{\log (4)}} \left (-4 x-2 x^2\right )+e^{\frac {x^2}{1+x}} \left (-1-2 x-x^2\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx=\int { -\frac {{\left ({\left (x^{2} + 2 \, x + 1\right )} e^{\left (\frac {x^{2}}{x + 1}\right )} \log \left (2\right ) + {\left (x^{2} + 2 \, x\right )} e^{\left (\frac {e^{\left (-\frac {x^{2}}{x + 1}\right )}}{\log \left (2\right )}\right )}\right )} e^{\left (-\frac {x^{2}}{x + 1}\right )}}{{\left (x^{2} + 2 \, x + 1\right )} \log \left (2\right )} \,d x } \]
integrate(1/2*((-2*x^2-4*x)*exp(1/log(2)/exp(x^2/(1+x)))+2*(-x^2-2*x-1)*lo g(2)*exp(x^2/(1+x)))/(x^2+2*x+1)/log(2)/exp(x^2/(1+x)),x, algorithm=\
integrate(-((x^2 + 2*x + 1)*e^(x^2/(x + 1))*log(2) + (x^2 + 2*x)*e^(e^(-x^ 2/(x + 1))/log(2)))*e^(-x^2/(x + 1))/((x^2 + 2*x + 1)*log(2)), x)
Time = 9.47 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-\frac {x^2}{1+x}} \left (e^{\frac {2 e^{-\frac {x^2}{1+x}}}{\log (4)}} \left (-4 x-2 x^2\right )+e^{\frac {x^2}{1+x}} \left (-1-2 x-x^2\right ) \log (4)\right )}{\left (1+2 x+x^2\right ) \log (4)} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{-\frac {x^2}{x+1}}}{\ln \left (2\right )}}-x \]