Integrand size = 99, antiderivative size = 27 \[ \int \frac {e^{-2 x^2} \left (6 x-24 x^3-4 x^4+e^{2 x^2} \left (36 x^2+12 x^3+x^4\right )+e^{2 x^2} \left (-36-12 x-x^2\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )\right )}{36 x^2+12 x^3+x^4} \, dx=x+\frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x} \]
Time = 5.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2 x^2} \left (6 x-24 x^3-4 x^4+e^{2 x^2} \left (36 x^2+12 x^3+x^4\right )+e^{2 x^2} \left (-36-12 x-x^2\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )\right )}{36 x^2+12 x^3+x^4} \, dx=x+\frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )}{x} \]
Integrate[(6*x - 24*x^3 - 4*x^4 + E^(2*x^2)*(36*x^2 + 12*x^3 + x^4) + E^(2 *x^2)*(-36 - 12*x - x^2)*Log[(55*E^(x/(E^(2*x^2)*(6 + x))))/4])/(E^(2*x^2) *(36*x^2 + 12*x^3 + x^4)),x]
Time = 1.91 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2026, 2007, 7293, 2009}
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.
\(\displaystyle \int \frac {e^{-2 x^2} \left (-4 x^4-24 x^3+e^{2 x^2} \left (-x^2-12 x-36\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{x+6}}\right )+e^{2 x^2} \left (x^4+12 x^3+36 x^2\right )+6 x\right )}{x^4+12 x^3+36 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-2 x^2} \left (-4 x^4-24 x^3+e^{2 x^2} \left (-x^2-12 x-36\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{x+6}}\right )+e^{2 x^2} \left (x^4+12 x^3+36 x^2\right )+6 x\right )}{x^2 \left (x^2+12 x+36\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{-2 x^2} \left (-4 x^4-24 x^3+e^{2 x^2} \left (-x^2-12 x-36\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{x+6}}\right )+e^{2 x^2} \left (x^4+12 x^3+36 x^2\right )+6 x\right )}{x^2 (x+6)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2-\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{x+6}}\right )}{x^2}-\frac {2 e^{-2 x^2} \left (2 x^3+12 x^2-3\right )}{x (x+6)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{x+6}}\right )}{x}+x\) |
Int[(6*x - 24*x^3 - 4*x^4 + E^(2*x^2)*(36*x^2 + 12*x^3 + x^4) + E^(2*x^2)* (-36 - 12*x - x^2)*Log[(55*E^(x/(E^(2*x^2)*(6 + x))))/4])/(E^(2*x^2)*(36*x ^2 + 12*x^3 + x^4)),x]
3.5.72.3.1 Defintions of rubi rules used
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]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 9.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63
method | result | size |
risch | \(\frac {\ln \left ({\mathrm e}^{\frac {x \,{\mathrm e}^{-2 x^{2}}}{6+x}}\right )}{x}-\frac {-2 x^{2}+4 \ln \left (2\right )-2 \ln \left (5\right )-2 \ln \left (11\right )}{2 x}\) | \(44\) |
default | \(\frac {\left (x^{4} {\mathrm e}^{4 x^{2}}+{\mathrm e}^{2 x^{2}} x^{2}+\left (-432+12 \ln \left (\frac {55 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-2 x^{2}}}{6+x}}}{4}\right )-\frac {12 x \,{\mathrm e}^{-2 x^{2}}}{6+x}\right ) x \,{\mathrm e}^{4 x^{2}}+\left (-108+\ln \left (\frac {55 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-2 x^{2}}}{6+x}}}{4}\right )-\frac {x \,{\mathrm e}^{-2 x^{2}}}{6+x}\right ) x^{2} {\mathrm e}^{4 x^{2}}+36 \left (\ln \left (\frac {55 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-2 x^{2}}}{6+x}}}{4}\right )-\frac {x \,{\mathrm e}^{-2 x^{2}}}{6+x}\right ) {\mathrm e}^{4 x^{2}}+6 x \,{\mathrm e}^{2 x^{2}}\right ) {\mathrm e}^{-4 x^{2}}}{x \left (6+x \right )^{2}}\) | \(172\) |
parallelrisch | \(-\frac {\left (-648 x^{3} {\mathrm e}^{4 x^{2}}+9 x^{5} {\mathrm e}^{4 x^{2}}+11664 x^{2} {\mathrm e}^{4 x^{2}}+58320 x \,{\mathrm e}^{4 x^{2}}-108 x^{4} {\mathrm e}^{4 x^{2}}-7776 \ln \left (\frac {55 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-2 x^{2}}}{6+x}}}{4}\right ) {\mathrm e}^{4 x^{2}}-216 \ln \left (\frac {55 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-2 x^{2}}}{6+x}}}{4}\right ) {\mathrm e}^{4 x^{2}} x^{2}-2592 \,{\mathrm e}^{4 x^{2}} x \ln \left (\frac {55 \,{\mathrm e}^{\frac {x \,{\mathrm e}^{-2 x^{2}}}{6+x}}}{4}\right )-9 \ln \left ({\mathrm e}^{x^{2}}\right ) x^{3} {\mathrm e}^{4 x^{2}}-108 \ln \left ({\mathrm e}^{x^{2}}\right ) x^{2} {\mathrm e}^{4 x^{2}}-324 \ln \left ({\mathrm e}^{x^{2}}\right ) {\mathrm e}^{4 x^{2}} x \right ) {\mathrm e}^{-4 x^{2}}}{216 x \left (6+x \right )^{2}}\) | \(196\) |
int(((-x^2-12*x-36)*exp(x^2)^2*ln(55/4*exp(x/(6+x)/exp(x^2)^2))+(x^4+12*x^ 3+36*x^2)*exp(x^2)^2-4*x^4-24*x^3+6*x)/(x^4+12*x^3+36*x^2)/exp(x^2)^2,x,me thod=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-2 x^2} \left (6 x-24 x^3-4 x^4+e^{2 x^2} \left (36 x^2+12 x^3+x^4\right )+e^{2 x^2} \left (-36-12 x-x^2\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )\right )}{36 x^2+12 x^3+x^4} \, dx=\frac {{\left ({\left (x^{3} + 6 \, x^{2} + {\left (x + 6\right )} \log \left (\frac {55}{4}\right )\right )} e^{\left (2 \, x^{2}\right )} + x\right )} e^{\left (-2 \, x^{2}\right )}}{x^{2} + 6 \, x} \]
integrate(((-x^2-12*x-36)*exp(x^2)^2*log(55/4*exp(x/(6+x)/exp(x^2)^2))+(x^ 4+12*x^3+36*x^2)*exp(x^2)^2-4*x^4-24*x^3+6*x)/(x^4+12*x^3+36*x^2)/exp(x^2) ^2,x, algorithm=\
Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {e^{-2 x^2} \left (6 x-24 x^3-4 x^4+e^{2 x^2} \left (36 x^2+12 x^3+x^4\right )+e^{2 x^2} \left (-36-12 x-x^2\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )\right )}{36 x^2+12 x^3+x^4} \, dx=x + \frac {e^{- 2 x^{2}}}{x + 6} + \frac {- 2 \log {\left (2 \right )} + \log {\left (55 \right )}}{x} \]
integrate(((-x**2-12*x-36)*exp(x**2)**2*ln(55/4*exp(x/(6+x)/exp(x**2)**2)) +(x**4+12*x**3+36*x**2)*exp(x**2)**2-4*x**4-24*x**3+6*x)/(x**4+12*x**3+36* x**2)/exp(x**2)**2,x)
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (23) = 46\).
Time = 0.31 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {e^{-2 x^2} \left (6 x-24 x^3-4 x^4+e^{2 x^2} \left (36 x^2+12 x^3+x^4\right )+e^{2 x^2} \left (-36-12 x-x^2\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )\right )}{36 x^2+12 x^3+x^4} \, dx=\frac {x^{3} + 6 \, x^{2} + x {\left (\log \left (11\right ) + \log \left (5\right ) - 2 \, \log \left (2\right )\right )} + x e^{\left (-2 \, x^{2}\right )} + 6 \, \log \left (11\right ) + 6 \, \log \left (5\right ) - 12 \, \log \left (2\right )}{x^{2} + 6 \, x} \]
integrate(((-x^2-12*x-36)*exp(x^2)^2*log(55/4*exp(x/(6+x)/exp(x^2)^2))+(x^ 4+12*x^3+36*x^2)*exp(x^2)^2-4*x^4-24*x^3+6*x)/(x^4+12*x^3+36*x^2)/exp(x^2) ^2,x, algorithm=\
(x^3 + 6*x^2 + x*(log(11) + log(5) - 2*log(2)) + x*e^(-2*x^2) + 6*log(11) + 6*log(5) - 12*log(2))/(x^2 + 6*x)
Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int \frac {e^{-2 x^2} \left (6 x-24 x^3-4 x^4+e^{2 x^2} \left (36 x^2+12 x^3+x^4\right )+e^{2 x^2} \left (-36-12 x-x^2\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )\right )}{36 x^2+12 x^3+x^4} \, dx=\frac {x^{3} + 6 \, x^{2} + x e^{\left (-2 \, x^{2}\right )} + x \log \left (55\right ) - 2 \, x \log \left (2\right ) + 6 \, \log \left (55\right ) - 12 \, \log \left (2\right )}{x^{2} + 6 \, x} \]
integrate(((-x^2-12*x-36)*exp(x^2)^2*log(55/4*exp(x/(6+x)/exp(x^2)^2))+(x^ 4+12*x^3+36*x^2)*exp(x^2)^2-4*x^4-24*x^3+6*x)/(x^4+12*x^3+36*x^2)/exp(x^2) ^2,x, algorithm=\
Time = 9.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-2 x^2} \left (6 x-24 x^3-4 x^4+e^{2 x^2} \left (36 x^2+12 x^3+x^4\right )+e^{2 x^2} \left (-36-12 x-x^2\right ) \log \left (\frac {55}{4} e^{\frac {e^{-2 x^2} x}{6+x}}\right )\right )}{36 x^2+12 x^3+x^4} \, dx=x+\frac {1}{6\,{\mathrm {e}}^{2\,x^2}+x\,{\mathrm {e}}^{2\,x^2}}-\frac {2\,\ln \left (2\right )}{x}+\frac {\ln \left (55\right )}{x} \]