Integrand size = 74, antiderivative size = 21 \[ \int \frac {3 e^{x-\frac {e^{e^x}}{4+x}} \left (\left (-80-120 x-45 x^2-5 x^3\right ) \log (2)+e^{e^x} \left (-5 x \log (2)+e^x \left (20 x+5 x^2\right ) \log (2)\right )\right )}{16+8 x+x^2} \, dx=-15 e^{x-\frac {e^{e^x}}{4+x}} x \log (2) \]
Time = 5.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {3 e^{x-\frac {e^{e^x}}{4+x}} \left (\left (-80-120 x-45 x^2-5 x^3\right ) \log (2)+e^{e^x} \left (-5 x \log (2)+e^x \left (20 x+5 x^2\right ) \log (2)\right )\right )}{16+8 x+x^2} \, dx=-15 e^{x-\frac {e^{e^x}}{4+x}} x \log (2) \]
Integrate[(3*E^(x - E^E^x/(4 + x))*((-80 - 120*x - 45*x^2 - 5*x^3)*Log[2] + E^E^x*(-5*x*Log[2] + E^x*(20*x + 5*x^2)*Log[2])))/(16 + 8*x + x^2),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.
\(\displaystyle \int \frac {3 e^{x-\frac {e^{e^x}}{x+4}} \left (e^{e^x} \left (e^x \left (5 x^2+20 x\right ) \log (2)-5 x \log (2)\right )+\left (-5 x^3-45 x^2-120 x-80\right ) \log (2)\right )}{x^2+8 x+16} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 3 \int -\frac {5 e^{x-\frac {e^{e^x}}{x+4}} \left (\log (2) \left (x^3+9 x^2+24 x+16\right )+e^{e^x} \left (x \log (2)-e^x \left (x^2+4 x\right ) \log (2)\right )\right )}{x^2+8 x+16}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -15 \int \frac {e^{x-\frac {e^{e^x}}{x+4}} \left (\log (2) \left (x^3+9 x^2+24 x+16\right )+e^{e^x} \left (x \log (2)-e^x \left (x^2+4 x\right ) \log (2)\right )\right )}{x^2+8 x+16}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle -15 \int \frac {e^{x-\frac {e^{e^x}}{x+4}} \left (\log (2) \left (x^3+9 x^2+24 x+16\right )+e^{e^x} \left (x \log (2)-e^x \left (x^2+4 x\right ) \log (2)\right )\right )}{(x+4)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -15 \int \left (\frac {e^{x-\frac {e^{e^x}}{x+4}} \left (x^3+9 x^2+e^{e^x} x+24 x+16\right ) \log (2)}{(x+4)^2}-\frac {e^{2 x+e^x-\frac {e^{e^x}}{x+4}} x \log (2)}{x+4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -15 \left (\log (2) \int e^{x-\frac {e^{e^x}}{x+4}}dx-\log (2) \int e^{2 x+e^x-\frac {e^{e^x}}{x+4}}dx+\log (2) \int e^{x-\frac {e^{e^x}}{x+4}} xdx-4 \log (2) \int \frac {e^{x+e^x-\frac {e^{e^x}}{x+4}}}{(x+4)^2}dx+\log (2) \int \frac {e^{x+e^x-\frac {e^{e^x}}{x+4}}}{x+4}dx+4 \log (2) \int \frac {e^{2 x+e^x-\frac {e^{e^x}}{x+4}}}{x+4}dx\right )\) |
Int[(3*E^(x - E^E^x/(4 + x))*((-80 - 120*x - 45*x^2 - 5*x^3)*Log[2] + E^E^ x*(-5*x*Log[2] + E^x*(20*x + 5*x^2)*Log[2])))/(16 + 8*x + x^2),x]
3.12.54.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(131\) vs. \(2(23)=46\).
Time = 183.65 (sec) , antiderivative size = 132, normalized size of antiderivative = 6.29
method | result | size |
parallelrisch | \(-\frac {1280 \,{\mathrm e}^{\ln \left (3 \,{\mathrm e}^{-\frac {{\mathrm e}^{{\mathrm e}^{x}}}{4+x}}\right )+x} \ln \left (2\right ) x +1280 \ln \left (2\right ) {\mathrm e}^{\ln \left (3 \,{\mathrm e}^{-\frac {{\mathrm e}^{{\mathrm e}^{x}}}{4+x}}\right )+x} x^{2}+480 \ln \left (2\right ) {\mathrm e}^{\ln \left (3 \,{\mathrm e}^{-\frac {{\mathrm e}^{{\mathrm e}^{x}}}{4+x}}\right )+x} x^{3}+80 \ln \left (2\right ) {\mathrm e}^{\ln \left (3 \,{\mathrm e}^{-\frac {{\mathrm e}^{{\mathrm e}^{x}}}{4+x}}\right )+x} x^{4}+5 \ln \left (2\right ) {\mathrm e}^{\ln \left (3 \,{\mathrm e}^{-\frac {{\mathrm e}^{{\mathrm e}^{x}}}{4+x}}\right )+x} x^{5}}{\left (4+x \right )^{4}}\) | \(132\) |
int((((5*x^2+20*x)*ln(2)*exp(x)-5*x*ln(2))*exp(exp(x))+(-5*x^3-45*x^2-120* x-80)*ln(2))*exp(ln(3/exp(exp(exp(x))/(4+x)))+x)/(x^2+8*x+16),x,method=_RE TURNVERBOSE)
-(1280*exp(ln(3/exp(exp(exp(x))/(4+x)))+x)*ln(2)*x+1280*ln(2)*exp(ln(3/exp (exp(exp(x))/(4+x)))+x)*x^2+480*ln(2)*exp(ln(3/exp(exp(exp(x))/(4+x)))+x)* x^3+80*ln(2)*exp(ln(3/exp(exp(exp(x))/(4+x)))+x)*x^4+5*ln(2)*exp(ln(3/exp( exp(exp(x))/(4+x)))+x)*x^5)/(4+x)^4
Time = 0.27 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.43 \[ \int \frac {3 e^{x-\frac {e^{e^x}}{4+x}} \left (\left (-80-120 x-45 x^2-5 x^3\right ) \log (2)+e^{e^x} \left (-5 x \log (2)+e^x \left (20 x+5 x^2\right ) \log (2)\right )\right )}{16+8 x+x^2} \, dx=-5 \, x e^{\left (\frac {x^{2} + {\left (x + 4\right )} \log \left (3\right ) + 4 \, x - e^{\left (e^{x}\right )}}{x + 4}\right )} \log \left (2\right ) \]
integrate((((5*x^2+20*x)*log(2)*exp(x)-5*x*log(2))*exp(exp(x))+(-5*x^3-45* x^2-120*x-80)*log(2))*exp(log(3/exp(exp(exp(x))/(4+x)))+x)/(x^2+8*x+16),x, algorithm=\
Time = 6.56 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {3 e^{x-\frac {e^{e^x}}{4+x}} \left (\left (-80-120 x-45 x^2-5 x^3\right ) \log (2)+e^{e^x} \left (-5 x \log (2)+e^x \left (20 x+5 x^2\right ) \log (2)\right )\right )}{16+8 x+x^2} \, dx=- 15 x e^{x} e^{- \frac {e^{e^{x}}}{x + 4}} \log {\left (2 \right )} \]
integrate((((5*x**2+20*x)*ln(2)*exp(x)-5*x*ln(2))*exp(exp(x))+(-5*x**3-45* x**2-120*x-80)*ln(2))*exp(ln(3/exp(exp(exp(x))/(4+x)))+x)/(x**2+8*x+16),x)
Time = 0.34 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {3 e^{x-\frac {e^{e^x}}{4+x}} \left (\left (-80-120 x-45 x^2-5 x^3\right ) \log (2)+e^{e^x} \left (-5 x \log (2)+e^x \left (20 x+5 x^2\right ) \log (2)\right )\right )}{16+8 x+x^2} \, dx=-15 \, x e^{\left (x - \frac {e^{\left (e^{x}\right )}}{x + 4}\right )} \log \left (2\right ) \]
integrate((((5*x^2+20*x)*log(2)*exp(x)-5*x*log(2))*exp(exp(x))+(-5*x^3-45* x^2-120*x-80)*log(2))*exp(log(3/exp(exp(exp(x))/(4+x)))+x)/(x^2+8*x+16),x, algorithm=\
\[ \int \frac {3 e^{x-\frac {e^{e^x}}{4+x}} \left (\left (-80-120 x-45 x^2-5 x^3\right ) \log (2)+e^{e^x} \left (-5 x \log (2)+e^x \left (20 x+5 x^2\right ) \log (2)\right )\right )}{16+8 x+x^2} \, dx=\int { \frac {5 \, {\left ({\left ({\left (x^{2} + 4 \, x\right )} e^{x} \log \left (2\right ) - x \log \left (2\right )\right )} e^{\left (e^{x}\right )} - {\left (x^{3} + 9 \, x^{2} + 24 \, x + 16\right )} \log \left (2\right )\right )} e^{\left (x + \log \left (3 \, e^{\left (-\frac {e^{\left (e^{x}\right )}}{x + 4}\right )}\right )\right )}}{x^{2} + 8 \, x + 16} \,d x } \]
integrate((((5*x^2+20*x)*log(2)*exp(x)-5*x*log(2))*exp(exp(x))+(-5*x^3-45* x^2-120*x-80)*log(2))*exp(log(3/exp(exp(exp(x))/(4+x)))+x)/(x^2+8*x+16),x, algorithm=\
integrate(5*(((x^2 + 4*x)*e^x*log(2) - x*log(2))*e^(e^x) - (x^3 + 9*x^2 + 24*x + 16)*log(2))*e^(x + log(3*e^(-e^(e^x)/(x + 4))))/(x^2 + 8*x + 16), x )
Time = 0.55 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {3 e^{x-\frac {e^{e^x}}{4+x}} \left (\left (-80-120 x-45 x^2-5 x^3\right ) \log (2)+e^{e^x} \left (-5 x \log (2)+e^x \left (20 x+5 x^2\right ) \log (2)\right )\right )}{16+8 x+x^2} \, dx=-15\,x\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{{\mathrm {e}}^x}}{x+4}}\,{\mathrm {e}}^x\,\ln \left (2\right ) \]