Integrand size = 107, antiderivative size = 24 \[ \int \frac {e^{-8/x} \left (e^{4/x} \left (-2560 x-1280 x^2-160 x^3\right ) \log (x)+e^{4/x} \left (-5120-2560 x+80 x^3\right ) \log ^2(x)+\left (64 x+16 x^2\right ) \log ^3(x)+\left (128+32 x-8 x^2\right ) \log ^4(x)\right )}{1600 x^2+1200 x^3+300 x^4+25 x^5} \, dx=\left (4-\frac {2 e^{-4/x} \log ^2(x)}{5 (4+x)}\right )^2 \]
Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-8/x} \left (e^{4/x} \left (-2560 x-1280 x^2-160 x^3\right ) \log (x)+e^{4/x} \left (-5120-2560 x+80 x^3\right ) \log ^2(x)+\left (64 x+16 x^2\right ) \log ^3(x)+\left (128+32 x-8 x^2\right ) \log ^4(x)\right )}{1600 x^2+1200 x^3+300 x^4+25 x^5} \, dx=-\frac {4 e^{-8/x} \left (20 e^{4/x} (4+x) \log ^2(x)-\log ^4(x)\right )}{25 (4+x)^2} \]
Integrate[(E^(4/x)*(-2560*x - 1280*x^2 - 160*x^3)*Log[x] + E^(4/x)*(-5120 - 2560*x + 80*x^3)*Log[x]^2 + (64*x + 16*x^2)*Log[x]^3 + (128 + 32*x - 8*x ^2)*Log[x]^4)/(E^(8/x)*(1600*x^2 + 1200*x^3 + 300*x^4 + 25*x^5)),x]
Leaf count is larger than twice the leaf count of optimal. \(59\) vs. \(2(24)=48\).
Time = 5.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.46, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2026, 2007, 7239, 27, 25, 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.
\(\displaystyle \int \frac {e^{-8/x} \left (e^{4/x} \left (80 x^3-2560 x-5120\right ) \log ^2(x)+\left (-8 x^2+32 x+128\right ) \log ^4(x)+\left (16 x^2+64 x\right ) \log ^3(x)+e^{4/x} \left (-160 x^3-1280 x^2-2560 x\right ) \log (x)\right )}{25 x^5+300 x^4+1200 x^3+1600 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-8/x} \left (e^{4/x} \left (80 x^3-2560 x-5120\right ) \log ^2(x)+\left (-8 x^2+32 x+128\right ) \log ^4(x)+\left (16 x^2+64 x\right ) \log ^3(x)+e^{4/x} \left (-160 x^3-1280 x^2-2560 x\right ) \log (x)\right )}{x^2 \left (25 x^3+300 x^2+1200 x+1600\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{-8/x} \left (e^{4/x} \left (80 x^3-2560 x-5120\right ) \log ^2(x)+\left (-8 x^2+32 x+128\right ) \log ^4(x)+\left (16 x^2+64 x\right ) \log ^3(x)+e^{4/x} \left (-160 x^3-1280 x^2-2560 x\right ) \log (x)\right )}{x^2 \left (5^{2/3} x+4\ 5^{2/3}\right )^3}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {8 e^{-8/x} \log (x) \left (2 x (x+4)-\left (x^2-4 x-16\right ) \log (x)\right ) \left (\log ^2(x)-10 e^{4/x} (x+4)\right )}{25 x^2 (x+4)^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {8}{25} \int -\frac {e^{-8/x} \log (x) \left (2 x (x+4)+\left (-x^2+4 x+16\right ) \log (x)\right ) \left (10 e^{4/x} (x+4)-\log ^2(x)\right )}{x^2 (x+4)^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {8}{25} \int \frac {e^{-8/x} \log (x) \left (2 x (x+4)+\left (-x^2+4 x+16\right ) \log (x)\right ) \left (10 e^{4/x} (x+4)-\log ^2(x)\right )}{x^2 (x+4)^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {8}{25} \int \left (\frac {e^{-8/x} \log ^3(x) \left (\log (x) x^2-2 x^2-4 \log (x) x-8 x-16 \log (x)\right )}{x^2 (x+4)^3}-\frac {10 e^{-4/x} \log (x) \left (\log (x) x^2-2 x^2-4 \log (x) x-8 x-16 \log (x)\right )}{x^2 (x+4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {8}{25} \left (\frac {10 e^{-4/x} \log (x) (x \log (x)+4 \log (x))}{(x+4)^2}-\frac {e^{-8/x} \log ^3(x) (x \log (x)+4 \log (x))}{2 (x+4)^3}\right )\) |
Int[(E^(4/x)*(-2560*x - 1280*x^2 - 160*x^3)*Log[x] + E^(4/x)*(-5120 - 2560 *x + 80*x^3)*Log[x]^2 + (64*x + 16*x^2)*Log[x]^3 + (128 + 32*x - 8*x^2)*Lo g[x]^4)/(E^(8/x)*(1600*x^2 + 1200*x^3 + 300*x^4 + 25*x^5)),x]
(-8*((10*Log[x]*(4*Log[x] + x*Log[x]))/(E^(4/x)*(4 + x)^2) - (Log[x]^3*(4* Log[x] + x*Log[x]))/(2*E^(8/x)*(4 + x)^3)))/25
3.22.83.3.1 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]
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])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.40 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50
method | result | size |
risch | \(\frac {4 \,{\mathrm e}^{-\frac {8}{x}} \ln \left (x \right )^{4}}{25 \left (4+x \right )^{2}}-\frac {16 \,{\mathrm e}^{-\frac {4}{x}} \ln \left (x \right )^{2}}{5 \left (4+x \right )}\) | \(36\) |
parallelrisch | \(-\frac {\left (10240 \,{\mathrm e}^{\frac {4}{x}} \ln \left (x \right )^{2} x^{2}-512 x \ln \left (x \right )^{4}+40960 \,{\mathrm e}^{\frac {4}{x}} \ln \left (x \right )^{2} x \right ) {\mathrm e}^{-\frac {8}{x}}}{3200 x \left (x^{2}+8 x +16\right )}\) | \(60\) |
int(((-8*x^2+32*x+128)*ln(x)^4+(16*x^2+64*x)*ln(x)^3+(80*x^3-2560*x-5120)* exp(4/x)*ln(x)^2+(-160*x^3-1280*x^2-2560*x)*exp(4/x)*ln(x))/(25*x^5+300*x^ 4+1200*x^3+1600*x^2)/exp(4/x)^2,x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.67 \[ \int \frac {e^{-8/x} \left (e^{4/x} \left (-2560 x-1280 x^2-160 x^3\right ) \log (x)+e^{4/x} \left (-5120-2560 x+80 x^3\right ) \log ^2(x)+\left (64 x+16 x^2\right ) \log ^3(x)+\left (128+32 x-8 x^2\right ) \log ^4(x)\right )}{1600 x^2+1200 x^3+300 x^4+25 x^5} \, dx=-\frac {4 \, {\left (20 \, {\left (x + 4\right )} e^{\frac {4}{x}} \log \left (x\right )^{2} - \log \left (x\right )^{4}\right )} e^{\left (-\frac {8}{x}\right )}}{25 \, {\left (x^{2} + 8 \, x + 16\right )}} \]
integrate(((-8*x^2+32*x+128)*log(x)^4+(16*x^2+64*x)*log(x)^3+(80*x^3-2560* x-5120)*exp(4/x)*log(x)^2+(-160*x^3-1280*x^2-2560*x)*exp(4/x)*log(x))/(25* x^5+300*x^4+1200*x^3+1600*x^2)/exp(4/x)^2,x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (19) = 38\).
Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.75 \[ \int \frac {e^{-8/x} \left (e^{4/x} \left (-2560 x-1280 x^2-160 x^3\right ) \log (x)+e^{4/x} \left (-5120-2560 x+80 x^3\right ) \log ^2(x)+\left (64 x+16 x^2\right ) \log ^3(x)+\left (128+32 x-8 x^2\right ) \log ^4(x)\right )}{1600 x^2+1200 x^3+300 x^4+25 x^5} \, dx=\frac {\left (20 x \log {\left (x \right )}^{4} + 80 \log {\left (x \right )}^{4}\right ) e^{- \frac {8}{x}} + \left (- 400 x^{2} \log {\left (x \right )}^{2} - 3200 x \log {\left (x \right )}^{2} - 6400 \log {\left (x \right )}^{2}\right ) e^{- \frac {4}{x}}}{125 x^{3} + 1500 x^{2} + 6000 x + 8000} \]
integrate(((-8*x**2+32*x+128)*ln(x)**4+(16*x**2+64*x)*ln(x)**3+(80*x**3-25 60*x-5120)*exp(4/x)*ln(x)**2+(-160*x**3-1280*x**2-2560*x)*exp(4/x)*ln(x))/ (25*x**5+300*x**4+1200*x**3+1600*x**2)/exp(4/x)**2,x)
((20*x*log(x)**4 + 80*log(x)**4)*exp(-8/x) + (-400*x**2*log(x)**2 - 3200*x *log(x)**2 - 6400*log(x)**2)*exp(-4/x))/(125*x**3 + 1500*x**2 + 6000*x + 8 000)
Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {e^{-8/x} \left (e^{4/x} \left (-2560 x-1280 x^2-160 x^3\right ) \log (x)+e^{4/x} \left (-5120-2560 x+80 x^3\right ) \log ^2(x)+\left (64 x+16 x^2\right ) \log ^3(x)+\left (128+32 x-8 x^2\right ) \log ^4(x)\right )}{1600 x^2+1200 x^3+300 x^4+25 x^5} \, dx=\frac {4 \, {\left (e^{\left (-\frac {8}{x}\right )} \log \left (x\right )^{4} - 20 \, {\left (x + 4\right )} e^{\left (-\frac {4}{x}\right )} \log \left (x\right )^{2}\right )}}{25 \, {\left (x^{2} + 8 \, x + 16\right )}} \]
integrate(((-8*x^2+32*x+128)*log(x)^4+(16*x^2+64*x)*log(x)^3+(80*x^3-2560* x-5120)*exp(4/x)*log(x)^2+(-160*x^3-1280*x^2-2560*x)*exp(4/x)*log(x))/(25* x^5+300*x^4+1200*x^3+1600*x^2)/exp(4/x)^2,x, algorithm=\
Timed out. \[ \int \frac {e^{-8/x} \left (e^{4/x} \left (-2560 x-1280 x^2-160 x^3\right ) \log (x)+e^{4/x} \left (-5120-2560 x+80 x^3\right ) \log ^2(x)+\left (64 x+16 x^2\right ) \log ^3(x)+\left (128+32 x-8 x^2\right ) \log ^4(x)\right )}{1600 x^2+1200 x^3+300 x^4+25 x^5} \, dx=\text {Timed out} \]
integrate(((-8*x^2+32*x+128)*log(x)^4+(16*x^2+64*x)*log(x)^3+(80*x^3-2560* x-5120)*exp(4/x)*log(x)^2+(-160*x^3-1280*x^2-2560*x)*exp(4/x)*log(x))/(25* x^5+300*x^4+1200*x^3+1600*x^2)/exp(4/x)^2,x, algorithm=\
Timed out. \[ \int \frac {e^{-8/x} \left (e^{4/x} \left (-2560 x-1280 x^2-160 x^3\right ) \log (x)+e^{4/x} \left (-5120-2560 x+80 x^3\right ) \log ^2(x)+\left (64 x+16 x^2\right ) \log ^3(x)+\left (128+32 x-8 x^2\right ) \log ^4(x)\right )}{1600 x^2+1200 x^3+300 x^4+25 x^5} \, dx=\int \frac {{\mathrm {e}}^{-\frac {8}{x}}\,\left (\left (-8\,x^2+32\,x+128\right )\,{\ln \left (x\right )}^4+\left (16\,x^2+64\,x\right )\,{\ln \left (x\right )}^3-{\mathrm {e}}^{4/x}\,\left (-80\,x^3+2560\,x+5120\right )\,{\ln \left (x\right )}^2-{\mathrm {e}}^{4/x}\,\left (160\,x^3+1280\,x^2+2560\,x\right )\,\ln \left (x\right )\right )}{25\,x^5+300\,x^4+1200\,x^3+1600\,x^2} \,d x \]
int((exp(-8/x)*(log(x)^3*(64*x + 16*x^2) + log(x)^4*(32*x - 8*x^2 + 128) - exp(4/x)*log(x)*(2560*x + 1280*x^2 + 160*x^3) - exp(4/x)*log(x)^2*(2560*x - 80*x^3 + 5120)))/(1600*x^2 + 1200*x^3 + 300*x^4 + 25*x^5),x)