Integrand size = 116, antiderivative size = 31 \[ \int \frac {e^{-x} \left (20+40 x+20 x^2+e^x \left (-20 x-32 x^2-16 x^3\right )+\left (-20 x-40 x^2-20 x^3\right ) \log (x)+\left (-5-10 x-5 x^2+e^x \left (5 x+8 x^2+4 x^3\right )+\left (5 x+10 x^2+5 x^3\right ) \log (x)\right ) \log (\log (2))\right )}{5 x+10 x^2+5 x^3} \, dx=\left (-\frac {4 x}{5}+\frac {1}{5 (1+x)}+e^{-x} \log (x)\right ) (4-\log (\log (2))) \] Output:
(4-ln(ln(2)))*(-4/5*x+ln(x)/exp(x)+1/5/(1+x))
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-x} \left (20+40 x+20 x^2+e^x \left (-20 x-32 x^2-16 x^3\right )+\left (-20 x-40 x^2-20 x^3\right ) \log (x)+\left (-5-10 x-5 x^2+e^x \left (5 x+8 x^2+4 x^3\right )+\left (5 x+10 x^2+5 x^3\right ) \log (x)\right ) \log (\log (2))\right )}{5 x+10 x^2+5 x^3} \, dx=\frac {1}{5} \left (4 x-\frac {1}{1+x}-5 e^{-x} \log (x)\right ) (-4+\log (\log (2))) \] Input:
Integrate[(20 + 40*x + 20*x^2 + E^x*(-20*x - 32*x^2 - 16*x^3) + (-20*x - 4 0*x^2 - 20*x^3)*Log[x] + (-5 - 10*x - 5*x^2 + E^x*(5*x + 8*x^2 + 4*x^3) + (5*x + 10*x^2 + 5*x^3)*Log[x])*Log[Log[2]])/(E^x*(5*x + 10*x^2 + 5*x^3)),x ]
Output:
((4*x - (1 + x)^(-1) - (5*Log[x])/E^x)*(-4 + Log[Log[2]]))/5
Time = 3.83 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {2026, 2007, 7239, 27, 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.
\(\displaystyle \int \frac {e^{-x} \left (20 x^2+e^x \left (-16 x^3-32 x^2-20 x\right )+\left (-20 x^3-40 x^2-20 x\right ) \log (x)+\log (\log (2)) \left (-5 x^2+e^x \left (4 x^3+8 x^2+5 x\right )+\left (5 x^3+10 x^2+5 x\right ) \log (x)-10 x-5\right )+40 x+20\right )}{5 x^3+10 x^2+5 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-x} \left (20 x^2+e^x \left (-16 x^3-32 x^2-20 x\right )+\left (-20 x^3-40 x^2-20 x\right ) \log (x)+\log (\log (2)) \left (-5 x^2+e^x \left (4 x^3+8 x^2+5 x\right )+\left (5 x^3+10 x^2+5 x\right ) \log (x)-10 x-5\right )+40 x+20\right )}{x \left (5 x^2+10 x+5\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {e^{-x} \left (20 x^2+e^x \left (-16 x^3-32 x^2-20 x\right )+\left (-20 x^3-40 x^2-20 x\right ) \log (x)+\log (\log (2)) \left (-5 x^2+e^x \left (4 x^3+8 x^2+5 x\right )+\left (5 x^3+10 x^2+5 x\right ) \log (x)-10 x-5\right )+40 x+20\right )}{x \left (\sqrt {5} x+\sqrt {5}\right )^2}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{-x} (4-\log (\log (2))) \left (-4 e^x x^3-\left (8 e^x-5\right ) x^2-5 \left (e^x-2\right ) x-5 (x+1)^2 x \log (x)+5\right )}{5 x (x+1)^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} (4-\log (\log (2))) \int \frac {e^{-x} \left (-4 e^x x^3+\left (5-8 e^x\right ) x^2+5 \left (2-e^x\right ) x-5 (x+1)^2 \log (x) x+5\right )}{x (x+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} (4-\log (\log (2))) \int \left (-\frac {4 x^2+8 x+5}{(x+1)^2}-\frac {5 e^{-x} (x \log (x)-1)}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} (4-\log (\log (2))) \left (-4 x+\frac {1}{x+1}+5 e^{-x} \log (x)\right )\) |
Input:
Int[(20 + 40*x + 20*x^2 + E^x*(-20*x - 32*x^2 - 16*x^3) + (-20*x - 40*x^2 - 20*x^3)*Log[x] + (-5 - 10*x - 5*x^2 + E^x*(5*x + 8*x^2 + 4*x^3) + (5*x + 10*x^2 + 5*x^3)*Log[x])*Log[Log[2]])/(E^x*(5*x + 10*x^2 + 5*x^3)),x]
Output:
((-4*x + (1 + x)^(-1) + (5*Log[x])/E^x)*(4 - Log[Log[2]]))/5
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 = 0.95 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10
method | result | size |
parts | \(\left (4-\ln \left (\ln \left (2\right )\right )\right ) \ln \left (x \right ) {\mathrm e}^{-x}+\frac {\left (\ln \left (\ln \left (2\right )\right )-4\right ) \left (4 x -\frac {1}{1+x}\right )}{5}\) | \(34\) |
default | \(\frac {\left (-5 \ln \left (\ln \left (2\right )\right )+20\right ) \ln \left (x \right ) {\mathrm e}^{-x}}{5}+\frac {\left (\ln \left (\ln \left (2\right )\right )-4\right ) \left (4 x -\frac {1}{1+x}\right )}{5}\) | \(35\) |
risch | \(-\left (\ln \left (\ln \left (2\right )\right )-4\right ) {\mathrm e}^{-x} \ln \left (x \right )+\frac {4 x^{2} \ln \left (\ln \left (2\right )\right )+4 x \ln \left (\ln \left (2\right )\right )-16 x^{2}-\ln \left (\ln \left (2\right )\right )-16 x +4}{5 x +5}\) | \(51\) |
norman | \(\frac {\left (\left (4-\ln \left (\ln \left (2\right )\right )\right ) \ln \left (x \right )+\left (4-\ln \left (\ln \left (2\right )\right )\right ) {\mathrm e}^{x}+\left (4-\ln \left (\ln \left (2\right )\right )\right ) x \ln \left (x \right )+\left (-\frac {16}{5}+\frac {4 \ln \left (\ln \left (2\right )\right )}{5}\right ) x^{2} {\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{1+x}\) | \(56\) |
parallelrisch | \(\frac {\left (4 \,{\mathrm e}^{x} \ln \left (\ln \left (2\right )\right ) x^{2}-5 \ln \left (x \right ) \ln \left (\ln \left (2\right )\right ) x -16 \,{\mathrm e}^{x} x^{2}-5 \ln \left (x \right ) \ln \left (\ln \left (2\right )\right )-5 \,{\mathrm e}^{x} \ln \left (\ln \left (2\right )\right )+20 x \ln \left (x \right )+20 \ln \left (x \right )+20 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{5 x +5}\) | \(65\) |
orering | \(\text {Expression too large to display}\) | \(1691\) |
Input:
int((((5*x^3+10*x^2+5*x)*ln(x)+(4*x^3+8*x^2+5*x)*exp(x)-5*x^2-10*x-5)*ln(l n(2))+(-20*x^3-40*x^2-20*x)*ln(x)+(-16*x^3-32*x^2-20*x)*exp(x)+20*x^2+40*x +20)/(5*x^3+10*x^2+5*x)/exp(x),x,method=_RETURNVERBOSE)
Output:
(4-ln(ln(2)))*ln(x)/exp(x)+1/5*(ln(ln(2))-4)*(4*x-1/(1+x))
Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (24) = 48\).
Time = 0.09 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.90 \[ \int \frac {e^{-x} \left (20+40 x+20 x^2+e^x \left (-20 x-32 x^2-16 x^3\right )+\left (-20 x-40 x^2-20 x^3\right ) \log (x)+\left (-5-10 x-5 x^2+e^x \left (5 x+8 x^2+4 x^3\right )+\left (5 x+10 x^2+5 x^3\right ) \log (x)\right ) \log (\log (2))\right )}{5 x+10 x^2+5 x^3} \, dx=-\frac {{\left (4 \, {\left (4 \, x^{2} + 4 \, x - 1\right )} e^{x} - 20 \, {\left (x + 1\right )} \log \left (x\right ) - {\left ({\left (4 \, x^{2} + 4 \, x - 1\right )} e^{x} - 5 \, {\left (x + 1\right )} \log \left (x\right )\right )} \log \left (\log \left (2\right )\right )\right )} e^{\left (-x\right )}}{5 \, {\left (x + 1\right )}} \] Input:
integrate((((5*x^3+10*x^2+5*x)*log(x)+(4*x^3+8*x^2+5*x)*exp(x)-5*x^2-10*x- 5)*log(log(2))+(-20*x^3-40*x^2-20*x)*log(x)+(-16*x^3-32*x^2-20*x)*exp(x)+2 0*x^2+40*x+20)/(5*x^3+10*x^2+5*x)/exp(x),x, algorithm="fricas")
Output:
-1/5*(4*(4*x^2 + 4*x - 1)*e^x - 20*(x + 1)*log(x) - ((4*x^2 + 4*x - 1)*e^x - 5*(x + 1)*log(x))*log(log(2)))*e^(-x)/(x + 1)
Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {e^{-x} \left (20+40 x+20 x^2+e^x \left (-20 x-32 x^2-16 x^3\right )+\left (-20 x-40 x^2-20 x^3\right ) \log (x)+\left (-5-10 x-5 x^2+e^x \left (5 x+8 x^2+4 x^3\right )+\left (5 x+10 x^2+5 x^3\right ) \log (x)\right ) \log (\log (2))\right )}{5 x+10 x^2+5 x^3} \, dx=- x \left (\frac {16}{5} - \frac {4 \log {\left (\log {\left (2 \right )} \right )}}{5}\right ) + \left (- \log {\left (x \right )} \log {\left (\log {\left (2 \right )} \right )} + 4 \log {\left (x \right )}\right ) e^{- x} - \frac {-4 + \log {\left (\log {\left (2 \right )} \right )}}{5 x + 5} \] Input:
integrate((((5*x**3+10*x**2+5*x)*ln(x)+(4*x**3+8*x**2+5*x)*exp(x)-5*x**2-1 0*x-5)*ln(ln(2))+(-20*x**3-40*x**2-20*x)*ln(x)+(-16*x**3-32*x**2-20*x)*exp (x)+20*x**2+40*x+20)/(5*x**3+10*x**2+5*x)/exp(x),x)
Output:
-x*(16/5 - 4*log(log(2))/5) + (-log(x)*log(log(2)) + 4*log(x))*exp(-x) - ( -4 + log(log(2)))/(5*x + 5)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (24) = 48\).
Time = 0.19 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68 \[ \int \frac {e^{-x} \left (20+40 x+20 x^2+e^x \left (-20 x-32 x^2-16 x^3\right )+\left (-20 x-40 x^2-20 x^3\right ) \log (x)+\left (-5-10 x-5 x^2+e^x \left (5 x+8 x^2+4 x^3\right )+\left (5 x+10 x^2+5 x^3\right ) \log (x)\right ) \log (\log (2))\right )}{5 x+10 x^2+5 x^3} \, dx=\frac {4 \, x^{2} {\left (\log \left (\log \left (2\right )\right ) - 4\right )} - 5 \, {\left (x {\left (\log \left (\log \left (2\right )\right ) - 4\right )} + \log \left (\log \left (2\right )\right ) - 4\right )} e^{\left (-x\right )} \log \left (x\right ) + 4 \, x {\left (\log \left (\log \left (2\right )\right ) - 4\right )} - \log \left (\log \left (2\right )\right ) + 4}{5 \, {\left (x + 1\right )}} \] Input:
integrate((((5*x^3+10*x^2+5*x)*log(x)+(4*x^3+8*x^2+5*x)*exp(x)-5*x^2-10*x- 5)*log(log(2))+(-20*x^3-40*x^2-20*x)*log(x)+(-16*x^3-32*x^2-20*x)*exp(x)+2 0*x^2+40*x+20)/(5*x^3+10*x^2+5*x)/exp(x),x, algorithm="maxima")
Output:
1/5*(4*x^2*(log(log(2)) - 4) - 5*(x*(log(log(2)) - 4) + log(log(2)) - 4)*e ^(-x)*log(x) + 4*x*(log(log(2)) - 4) - log(log(2)) + 4)/(x + 1)
Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (24) = 48\).
Time = 0.12 (sec) , antiderivative size = 74, normalized size of antiderivative = 2.39 \[ \int \frac {e^{-x} \left (20+40 x+20 x^2+e^x \left (-20 x-32 x^2-16 x^3\right )+\left (-20 x-40 x^2-20 x^3\right ) \log (x)+\left (-5-10 x-5 x^2+e^x \left (5 x+8 x^2+4 x^3\right )+\left (5 x+10 x^2+5 x^3\right ) \log (x)\right ) \log (\log (2))\right )}{5 x+10 x^2+5 x^3} \, dx=-\frac {5 \, x e^{\left (-x\right )} \log \left (x\right ) \log \left (\log \left (2\right )\right ) - 20 \, x e^{\left (-x\right )} \log \left (x\right ) - 4 \, x^{2} \log \left (\log \left (2\right )\right ) + 5 \, e^{\left (-x\right )} \log \left (x\right ) \log \left (\log \left (2\right )\right ) + 16 \, x^{2} - 20 \, e^{\left (-x\right )} \log \left (x\right ) - 4 \, x \log \left (\log \left (2\right )\right ) + 16 \, x + \log \left (\log \left (2\right )\right ) - 4}{5 \, {\left (x + 1\right )}} \] Input:
integrate((((5*x^3+10*x^2+5*x)*log(x)+(4*x^3+8*x^2+5*x)*exp(x)-5*x^2-10*x- 5)*log(log(2))+(-20*x^3-40*x^2-20*x)*log(x)+(-16*x^3-32*x^2-20*x)*exp(x)+2 0*x^2+40*x+20)/(5*x^3+10*x^2+5*x)/exp(x),x, algorithm="giac")
Output:
-1/5*(5*x*e^(-x)*log(x)*log(log(2)) - 20*x*e^(-x)*log(x) - 4*x^2*log(log(2 )) + 5*e^(-x)*log(x)*log(log(2)) + 16*x^2 - 20*e^(-x)*log(x) - 4*x*log(log (2)) + 16*x + log(log(2)) - 4)/(x + 1)
Time = 2.72 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.23 \[ \int \frac {e^{-x} \left (20+40 x+20 x^2+e^x \left (-20 x-32 x^2-16 x^3\right )+\left (-20 x-40 x^2-20 x^3\right ) \log (x)+\left (-5-10 x-5 x^2+e^x \left (5 x+8 x^2+4 x^3\right )+\left (5 x+10 x^2+5 x^3\right ) \log (x)\right ) \log (\log (2))\right )}{5 x+10 x^2+5 x^3} \, dx=\frac {{\mathrm {e}}^{-x}\,\left (\ln \left (\ln \left (2\right )\right )-4\right )\,\left ({\mathrm {e}}^x-5\,\ln \left (x\right )+4\,x\,{\mathrm {e}}^x\right )}{5}-\frac {\frac {\ln \left (\ln \left (2\right )\right )}{5}-\frac {4}{5}}{x+1} \] Input:
int((exp(-x)*(40*x - log(log(2))*(10*x + 5*x^2 - exp(x)*(5*x + 8*x^2 + 4*x ^3) - log(x)*(5*x + 10*x^2 + 5*x^3) + 5) + 20*x^2 - exp(x)*(20*x + 32*x^2 + 16*x^3) - log(x)*(20*x + 40*x^2 + 20*x^3) + 20))/(5*x + 10*x^2 + 5*x^3), x)
Output:
(exp(-x)*(log(log(2)) - 4)*(exp(x) - 5*log(x) + 4*x*exp(x)))/5 - (log(log( 2))/5 - 4/5)/(x + 1)
Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.29 \[ \int \frac {e^{-x} \left (20+40 x+20 x^2+e^x \left (-20 x-32 x^2-16 x^3\right )+\left (-20 x-40 x^2-20 x^3\right ) \log (x)+\left (-5-10 x-5 x^2+e^x \left (5 x+8 x^2+4 x^3\right )+\left (5 x+10 x^2+5 x^3\right ) \log (x)\right ) \log (\log (2))\right )}{5 x+10 x^2+5 x^3} \, dx=\frac {4 e^{x} \mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x^{2}+5 e^{x} \mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x -16 e^{x} x^{2}-20 e^{x} x -5 \,\mathrm {log}\left (\mathrm {log}\left (2\right )\right ) \mathrm {log}\left (x \right ) x -5 \,\mathrm {log}\left (\mathrm {log}\left (2\right )\right ) \mathrm {log}\left (x \right )+20 \,\mathrm {log}\left (x \right ) x +20 \,\mathrm {log}\left (x \right )}{5 e^{x} \left (x +1\right )} \] Input:
int((((5*x^3+10*x^2+5*x)*log(x)+(4*x^3+8*x^2+5*x)*exp(x)-5*x^2-10*x-5)*log (log(2))+(-20*x^3-40*x^2-20*x)*log(x)+(-16*x^3-32*x^2-20*x)*exp(x)+20*x^2+ 40*x+20)/(5*x^3+10*x^2+5*x)/exp(x),x)
Output:
(4*e**x*log(log(2))*x**2 + 5*e**x*log(log(2))*x - 16*e**x*x**2 - 20*e**x*x - 5*log(log(2))*log(x)*x - 5*log(log(2))*log(x) + 20*log(x)*x + 20*log(x) )/(5*e**x*(x + 1))