Integrand size = 104, antiderivative size = 29 \begin {dmath*} \int \frac {e^{-\frac {2}{\log (4-x)}} \left (40 x^2+40 x^3+10 x^4+e^{\frac {8}{2+x}+\frac {2}{\log (4-x)}} (-32+8 x) \log ^2(4-x)+\left (-160 x-120 x^2+10 x^4\right ) \log ^2(4-x)\right )}{\left (-16-12 x+x^3\right ) \log ^2(4-x)} \, dx=-e^{\frac {8}{2+x}}+5 e^{-\frac {2}{\log (4-x)}} x^2 \end {dmath*}
Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \begin {dmath*} \int \frac {e^{-\frac {2}{\log (4-x)}} \left (40 x^2+40 x^3+10 x^4+e^{\frac {8}{2+x}+\frac {2}{\log (4-x)}} (-32+8 x) \log ^2(4-x)+\left (-160 x-120 x^2+10 x^4\right ) \log ^2(4-x)\right )}{\left (-16-12 x+x^3\right ) \log ^2(4-x)} \, dx=2 \left (-\frac {1}{2} e^{\frac {8}{2+x}}+\frac {5}{2} e^{-\frac {2}{\log (4-x)}} x^2\right ) \end {dmath*}
Integrate[(40*x^2 + 40*x^3 + 10*x^4 + E^(8/(2 + x) + 2/Log[4 - x])*(-32 + 8*x)*Log[4 - x]^2 + (-160*x - 120*x^2 + 10*x^4)*Log[4 - x]^2)/(E^(2/Log[4 - x])*(-16 - 12*x + x^3)*Log[4 - 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.
\(\displaystyle \int \frac {e^{-\frac {2}{\log (4-x)}} \left (10 x^4+40 x^3+40 x^2+\left (10 x^4-120 x^2-160 x\right ) \log ^2(4-x)+(8 x-32) e^{\frac {8}{x+2}+\frac {2}{\log (4-x)}} \log ^2(4-x)\right )}{\left (x^3-12 x-16\right ) \log ^2(4-x)} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {e^{-\frac {2}{\log (4-x)}} \left (10 x^4+40 x^3+40 x^2+\left (10 x^4-120 x^2-160 x\right ) \log ^2(4-x)+(8 x-32) e^{\frac {8}{x+2}+\frac {2}{\log (4-x)}} \log ^2(4-x)\right )}{36 (x-4) \log ^2(4-x)}-\frac {e^{-\frac {2}{\log (4-x)}} \left (10 x^4+40 x^3+40 x^2+\left (10 x^4-120 x^2-160 x\right ) \log ^2(4-x)+(8 x-32) e^{\frac {8}{x+2}+\frac {2}{\log (4-x)}} \log ^2(4-x)\right )}{36 (x+2) \log ^2(4-x)}-\frac {e^{-\frac {2}{\log (4-x)}} \left (10 x^4+40 x^3+40 x^2+\left (10 x^4-120 x^2-160 x\right ) \log ^2(4-x)+(8 x-32) e^{\frac {8}{x+2}+\frac {2}{\log (4-x)}} \log ^2(4-x)\right )}{6 (x+2)^2 \log ^2(4-x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -80 \text {Subst}\left (\int \frac {e^{-\frac {2}{\log (x)}}}{\log ^2(x)}dx,x,4-x\right )+10 \text {Subst}\left (\int \frac {e^{-\frac {2}{\log (x)}} x}{\log ^2(x)}dx,x,4-x\right )+10 \int e^{-\frac {2}{\log (4-x)}} xdx-e^{\frac {8}{x+2}}+80 e^{-\frac {2}{\log (4-x)}}\) |
Int[(40*x^2 + 40*x^3 + 10*x^4 + E^(8/(2 + x) + 2/Log[4 - x])*(-32 + 8*x)*L og[4 - x]^2 + (-160*x - 120*x^2 + 10*x^4)*Log[4 - x]^2)/(E^(2/Log[4 - x])* (-16 - 12*x + x^3)*Log[4 - x]^2),x]
3.7.85.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 13.46 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97
method | result | size |
risch | \(5 x^{2} {\mathrm e}^{-\frac {2}{\ln \left (-x +4\right )}}-{\mathrm e}^{\frac {8}{2+x}}\) | \(28\) |
default | \(5 x^{2} {\mathrm e}^{-\frac {2}{\ln \left (-x +4\right )}}-{\mathrm e}^{\frac {8}{2+x}}\) | \(32\) |
parts | \(5 x^{2} {\mathrm e}^{-\frac {2}{\ln \left (-x +4\right )}}-{\mathrm e}^{\frac {8}{2+x}}\) | \(32\) |
int(((8*x-32)*exp(4/(2+x))^2*ln(-x+4)^2*exp(2/ln(-x+4))+(10*x^4-120*x^2-16 0*x)*ln(-x+4)^2+10*x^4+40*x^3+40*x^2)/(x^3-12*x-16)/ln(-x+4)^2/exp(2/ln(-x +4)),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 57 vs. \(2 (27) = 54\).
Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.97 \begin {dmath*} \int \frac {e^{-\frac {2}{\log (4-x)}} \left (40 x^2+40 x^3+10 x^4+e^{\frac {8}{2+x}+\frac {2}{\log (4-x)}} (-32+8 x) \log ^2(4-x)+\left (-160 x-120 x^2+10 x^4\right ) \log ^2(4-x)\right )}{\left (-16-12 x+x^3\right ) \log ^2(4-x)} \, dx=5 \, x^{2} e^{\left (-\frac {2}{\log \left (-x + 4\right )}\right )} - e^{\left (\frac {2 \, {\left (x + 4 \, \log \left (-x + 4\right ) + 2\right )}}{{\left (x + 2\right )} \log \left (-x + 4\right )} - \frac {2}{\log \left (-x + 4\right )}\right )} \end {dmath*}
integrate(((8*x-32)*exp(4/(2+x))^2*log(-x+4)^2*exp(2/log(-x+4))+(10*x^4-12 0*x^2-160*x)*log(-x+4)^2+10*x^4+40*x^3+40*x^2)/(x^3-12*x-16)/log(-x+4)^2/e xp(2/log(-x+4)),x, algorithm=\
5*x^2*e^(-2/log(-x + 4)) - e^(2*(x + 4*log(-x + 4) + 2)/((x + 2)*log(-x + 4)) - 2/log(-x + 4))
Time = 0.94 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66 \begin {dmath*} \int \frac {e^{-\frac {2}{\log (4-x)}} \left (40 x^2+40 x^3+10 x^4+e^{\frac {8}{2+x}+\frac {2}{\log (4-x)}} (-32+8 x) \log ^2(4-x)+\left (-160 x-120 x^2+10 x^4\right ) \log ^2(4-x)\right )}{\left (-16-12 x+x^3\right ) \log ^2(4-x)} \, dx=5 x^{2} e^{- \frac {2}{\log {\left (4 - x \right )}}} - e^{\frac {8}{x + 2}} \end {dmath*}
integrate(((8*x-32)*exp(4/(2+x))**2*ln(-x+4)**2*exp(2/ln(-x+4))+(10*x**4-1 20*x**2-160*x)*ln(-x+4)**2+10*x**4+40*x**3+40*x**2)/(x**3-12*x-16)/ln(-x+4 )**2/exp(2/ln(-x+4)),x)
\begin {dmath*} \int \frac {e^{-\frac {2}{\log (4-x)}} \left (40 x^2+40 x^3+10 x^4+e^{\frac {8}{2+x}+\frac {2}{\log (4-x)}} (-32+8 x) \log ^2(4-x)+\left (-160 x-120 x^2+10 x^4\right ) \log ^2(4-x)\right )}{\left (-16-12 x+x^3\right ) \log ^2(4-x)} \, dx=\int { \frac {2 \, {\left (5 \, x^{4} + 4 \, {\left (x - 4\right )} e^{\left (\frac {8}{x + 2} + \frac {2}{\log \left (-x + 4\right )}\right )} \log \left (-x + 4\right )^{2} + 20 \, x^{3} + 5 \, {\left (x^{4} - 12 \, x^{2} - 16 \, x\right )} \log \left (-x + 4\right )^{2} + 20 \, x^{2}\right )} e^{\left (-\frac {2}{\log \left (-x + 4\right )}\right )}}{{\left (x^{3} - 12 \, x - 16\right )} \log \left (-x + 4\right )^{2}} \,d x } \end {dmath*}
integrate(((8*x-32)*exp(4/(2+x))^2*log(-x+4)^2*exp(2/log(-x+4))+(10*x^4-12 0*x^2-160*x)*log(-x+4)^2+10*x^4+40*x^3+40*x^2)/(x^3-12*x-16)/log(-x+4)^2/e xp(2/log(-x+4)),x, algorithm=\
-e^(8/(x + 2)) + 2*integrate(5*((x^2 - 4*x)*log(-x + 4)^2 + x^2)*e^(-2/log (-x + 4))/((x - 4)*log(-x + 4)^2), x)
Exception generated. \begin {dmath*} \int \frac {e^{-\frac {2}{\log (4-x)}} \left (40 x^2+40 x^3+10 x^4+e^{\frac {8}{2+x}+\frac {2}{\log (4-x)}} (-32+8 x) \log ^2(4-x)+\left (-160 x-120 x^2+10 x^4\right ) \log ^2(4-x)\right )}{\left (-16-12 x+x^3\right ) \log ^2(4-x)} \, dx=\text {Exception raised: TypeError} \end {dmath*}
integrate(((8*x-32)*exp(4/(2+x))^2*log(-x+4)^2*exp(2/log(-x+4))+(10*x^4-12 0*x^2-160*x)*log(-x+4)^2+10*x^4+40*x^3+40*x^2)/(x^3-12*x-16)/log(-x+4)^2/e xp(2/log(-x+4)),x, algorithm=\
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{8,[0,19]%%%}+%%%{64,[0,18]%%%}+%%%{-288,[0,17]%%%}+%%%{-42 24,[0,16]
Timed out. \begin {dmath*} \int \frac {e^{-\frac {2}{\log (4-x)}} \left (40 x^2+40 x^3+10 x^4+e^{\frac {8}{2+x}+\frac {2}{\log (4-x)}} (-32+8 x) \log ^2(4-x)+\left (-160 x-120 x^2+10 x^4\right ) \log ^2(4-x)\right )}{\left (-16-12 x+x^3\right ) \log ^2(4-x)} \, dx=\int -\frac {{\mathrm {e}}^{-\frac {2}{\ln \left (4-x\right )}}\,\left (40\,x^2-{\ln \left (4-x\right )}^2\,\left (-10\,x^4+120\,x^2+160\,x\right )+40\,x^3+10\,x^4+{\mathrm {e}}^{\frac {2}{\ln \left (4-x\right )}}\,{\mathrm {e}}^{\frac {8}{x+2}}\,{\ln \left (4-x\right )}^2\,\left (8\,x-32\right )\right )}{{\ln \left (4-x\right )}^2\,\left (-x^3+12\,x+16\right )} \,d x \end {dmath*}
int(-(exp(-2/log(4 - x))*(40*x^2 - log(4 - x)^2*(160*x + 120*x^2 - 10*x^4) + 40*x^3 + 10*x^4 + exp(2/log(4 - x))*exp(8/(x + 2))*log(4 - x)^2*(8*x - 32)))/(log(4 - x)^2*(12*x - x^3 + 16)),x)