Integrand size = 59, antiderivative size = 29 \[ \int \frac {e^{-x} \left (-4-4 e^5 x+4 x \log (x)+\left (-9+x^2+e^5 \left (-10 x+x^2\right )+\left (10 x-x^2\right ) \log (x)\right ) \log (\log (2))\right )}{x \log (\log (2))} \, dx=e^{-x} \left (-x+\left (-e^5+\log (x)\right ) \left (-9+x-\frac {4}{\log (\log (2))}\right )\right ) \]
Time = 1.14 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72 \[ \int \frac {e^{-x} \left (-4-4 e^5 x+4 x \log (x)+\left (-9+x^2+e^5 \left (-10 x+x^2\right )+\left (10 x-x^2\right ) \log (x)\right ) \log (\log (2))\right )}{x \log (\log (2))} \, dx=\frac {e^{-x} \left (-x \log (\log (2))+e^5 (4+9 \log (\log (2))-x \log (\log (2)))+\log (x) (-4-9 \log (\log (2))+x \log (\log (2)))\right )}{\log (\log (2))} \]
Integrate[(-4 - 4*E^5*x + 4*x*Log[x] + (-9 + x^2 + E^5*(-10*x + x^2) + (10 *x - x^2)*Log[x])*Log[Log[2]])/(E^x*x*Log[Log[2]]),x]
(-(x*Log[Log[2]]) + E^5*(4 + 9*Log[Log[2]] - x*Log[Log[2]]) + Log[x]*(-4 - 9*Log[Log[2]] + x*Log[Log[2]]))/(E^x*Log[Log[2]])
Leaf count is larger than twice the leaf count of optimal. \(99\) vs. \(2(29)=58\).
Time = 0.64 (sec) , antiderivative size = 99, normalized size of antiderivative = 3.41, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {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.
\(\displaystyle \int \frac {e^{-x} \left (\log (\log (2)) \left (x^2+e^5 \left (x^2-10 x\right )+\left (10 x-x^2\right ) \log (x)-9\right )-4 e^5 x+4 x \log (x)-4\right )}{x \log (\log (2))} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {e^{-x} \left (-4 \log (x) x+4 e^5 x+\left (-x^2+e^5 \left (10 x-x^2\right )-\left (10 x-x^2\right ) \log (x)+9\right ) \log (\log (2))+4\right )}{x}dx}{\log (\log (2))}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {e^{-x} \left (-4 \log (x) x+4 e^5 x+\left (-x^2+e^5 \left (10 x-x^2\right )-\left (10 x-x^2\right ) \log (x)+9\right ) \log (\log (2))+4\right )}{x}dx}{\log (\log (2))}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (e^{-x} \log (x) (\log (\log (2)) x-10 \log (\log (2))-4)+\frac {e^{-x} \left (-\left (\left (1+e^5\right ) \log (\log (2)) x^2\right )+2 e^5 (2+5 \log (\log (2))) x+9 \log (\log (2))+4\right )}{x}\right )dx}{\log (\log (2))}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (1+e^5\right ) e^{-x} x \log (\log (2))-e^{-x} \log (x) (x \log (\log (2))-2 (2+5 \log (\log (2))))-2 e^{5-x} (2+5 \log (\log (2)))-e^{-x} \log (\log (2))-e^{-x} \log (\log (2)) \log (x)+\left (1+e^5\right ) e^{-x} \log (\log (2))}{\log (\log (2))}\) |
Int[(-4 - 4*E^5*x + 4*x*Log[x] + (-9 + x^2 + E^5*(-10*x + x^2) + (10*x - x ^2)*Log[x])*Log[Log[2]])/(E^x*x*Log[Log[2]]),x]
-((-(Log[Log[2]]/E^x) + ((1 + E^5)*Log[Log[2]])/E^x + ((1 + E^5)*x*Log[Log [2]])/E^x - (Log[x]*Log[Log[2]])/E^x - 2*E^(5 - x)*(2 + 5*Log[Log[2]]) - ( Log[x]*(x*Log[Log[2]] - 2*(2 + 5*Log[Log[2]])))/E^x)/Log[Log[2]])
3.4.31.3.1 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]]
Time = 0.10 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.72
method | result | size |
norman | \(\left (x \ln \left (x \right )+\left (-{\mathrm e}^{5}-1\right ) x +\frac {{\mathrm e}^{5} \left (9 \ln \left (\ln \left (2\right )\right )+4\right )}{\ln \left (\ln \left (2\right )\right )}-\frac {\left (9 \ln \left (\ln \left (2\right )\right )+4\right ) \ln \left (x \right )}{\ln \left (\ln \left (2\right )\right )}\right ) {\mathrm e}^{-x}\) | \(50\) |
parallelrisch | \(-\frac {{\mathrm e}^{-x} \left ({\mathrm e}^{5} \ln \left (\ln \left (2\right )\right ) x -\ln \left (x \right ) \ln \left (\ln \left (2\right )\right ) x -9 \,{\mathrm e}^{5} \ln \left (\ln \left (2\right )\right )+9 \ln \left (x \right ) \ln \left (\ln \left (2\right )\right )+x \ln \left (\ln \left (2\right )\right )-4 \,{\mathrm e}^{5}+4 \ln \left (x \right )\right )}{\ln \left (\ln \left (2\right )\right )}\) | \(55\) |
risch | \(\frac {\left (x \ln \left (\ln \left (2\right )\right )-9 \ln \left (\ln \left (2\right )\right )-4\right ) {\mathrm e}^{-x} \ln \left (x \right )}{\ln \left (\ln \left (2\right )\right )}-\frac {\left ({\mathrm e}^{5} \ln \left (\ln \left (2\right )\right ) x -9 \,{\mathrm e}^{5} \ln \left (\ln \left (2\right )\right )+x \ln \left (\ln \left (2\right )\right )-4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-x}}{\ln \left (\ln \left (2\right )\right )}\) | \(61\) |
int((((-x^2+10*x)*ln(x)+(x^2-10*x)*exp(5)+x^2-9)*ln(ln(2))+4*x*ln(x)-4*x*e xp(5)-4)/x/exp(x)/ln(ln(2)),x,method=_RETURNVERBOSE)
(x*ln(x)+(-exp(5)-1)*x+exp(5)*(9*ln(ln(2))+4)/ln(ln(2))-(9*ln(ln(2))+4)/ln (ln(2))*ln(x))/exp(x)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (26) = 52\).
Time = 0.25 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.86 \[ \int \frac {e^{-x} \left (-4-4 e^5 x+4 x \log (x)+\left (-9+x^2+e^5 \left (-10 x+x^2\right )+\left (10 x-x^2\right ) \log (x)\right ) \log (\log (2))\right )}{x \log (\log (2))} \, dx=-\frac {4 \, e^{\left (-x\right )} \log \left (x\right ) - {\left ({\left (x - 9\right )} e^{\left (-x\right )} \log \left (x\right ) - {\left ({\left (x - 9\right )} e^{5} + x\right )} e^{\left (-x\right )}\right )} \log \left (\log \left (2\right )\right ) - 4 \, e^{\left (-x + 5\right )}}{\log \left (\log \left (2\right )\right )} \]
integrate((((-x^2+10*x)*log(x)+(x^2-10*x)*exp(5)+x^2-9)*log(log(2))+4*x*lo g(x)-4*x*exp(5)-4)/x/exp(x)/log(log(2)),x, algorithm=\
-(4*e^(-x)*log(x) - ((x - 9)*e^(-x)*log(x) - ((x - 9)*e^5 + x)*e^(-x))*log (log(2)) - 4*e^(-x + 5))/log(log(2))
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (20) = 40\).
Time = 0.22 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.24 \[ \int \frac {e^{-x} \left (-4-4 e^5 x+4 x \log (x)+\left (-9+x^2+e^5 \left (-10 x+x^2\right )+\left (10 x-x^2\right ) \log (x)\right ) \log (\log (2))\right )}{x \log (\log (2))} \, dx=\frac {\left (x \log {\left (x \right )} \log {\left (\log {\left (2 \right )} \right )} - x \log {\left (\log {\left (2 \right )} \right )} - x e^{5} \log {\left (\log {\left (2 \right )} \right )} - 4 \log {\left (x \right )} - 9 \log {\left (x \right )} \log {\left (\log {\left (2 \right )} \right )} + 9 e^{5} \log {\left (\log {\left (2 \right )} \right )} + 4 e^{5}\right ) e^{- x}}{\log {\left (\log {\left (2 \right )} \right )}} \]
integrate((((-x**2+10*x)*ln(x)+(x**2-10*x)*exp(5)+x**2-9)*ln(ln(2))+4*x*ln (x)-4*x*exp(5)-4)/x/exp(x)/ln(ln(2)),x)
(x*log(x)*log(log(2)) - x*log(log(2)) - x*exp(5)*log(log(2)) - 4*log(x) - 9*log(x)*log(log(2)) + 9*exp(5)*log(log(2)) + 4*exp(5))*exp(-x)/log(log(2) )
\[ \int \frac {e^{-x} \left (-4-4 e^5 x+4 x \log (x)+\left (-9+x^2+e^5 \left (-10 x+x^2\right )+\left (10 x-x^2\right ) \log (x)\right ) \log (\log (2))\right )}{x \log (\log (2))} \, dx=\int { -\frac {{\left (4 \, x e^{5} - 4 \, x \log \left (x\right ) - {\left (x^{2} + {\left (x^{2} - 10 \, x\right )} e^{5} - {\left (x^{2} - 10 \, x\right )} \log \left (x\right ) - 9\right )} \log \left (\log \left (2\right )\right ) + 4\right )} e^{\left (-x\right )}}{x \log \left (\log \left (2\right )\right )} \,d x } \]
integrate((((-x^2+10*x)*log(x)+(x^2-10*x)*exp(5)+x^2-9)*log(log(2))+4*x*lo g(x)-4*x*exp(5)-4)/x/exp(x)/log(log(2)),x, algorithm=\
-((x*e^5 + e^5)*e^(-x)*log(log(2)) + (x + 1)*e^(-x)*log(log(2)) + 10*e^(-x )*log(x)*log(log(2)) + 4*e^(-x)*log(x) - ((x + 1)*e^(-x)*log(x) - integrat e((x + 1)*e^(-x)/x, x))*log(log(2)) - Ei(-x)*log(log(2)) - 10*e^(-x + 5)*l og(log(2)) - 4*e^(-x + 5))/log(log(2))
Leaf count of result is larger than twice the leaf count of optimal. 78 vs. \(2 (26) = 52\).
Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 2.69 \[ \int \frac {e^{-x} \left (-4-4 e^5 x+4 x \log (x)+\left (-9+x^2+e^5 \left (-10 x+x^2\right )+\left (10 x-x^2\right ) \log (x)\right ) \log (\log (2))\right )}{x \log (\log (2))} \, dx=\frac {x e^{\left (-x\right )} \log \left (x\right ) \log \left (\log \left (2\right )\right ) - x e^{\left (-x\right )} \log \left (\log \left (2\right )\right ) - x e^{\left (-x + 5\right )} \log \left (\log \left (2\right )\right ) - 9 \, e^{\left (-x\right )} \log \left (x\right ) \log \left (\log \left (2\right )\right ) - 4 \, e^{\left (-x\right )} \log \left (x\right ) + 9 \, e^{\left (-x + 5\right )} \log \left (\log \left (2\right )\right ) + 4 \, e^{\left (-x + 5\right )}}{\log \left (\log \left (2\right )\right )} \]
integrate((((-x^2+10*x)*log(x)+(x^2-10*x)*exp(5)+x^2-9)*log(log(2))+4*x*lo g(x)-4*x*exp(5)-4)/x/exp(x)/log(log(2)),x, algorithm=\
(x*e^(-x)*log(x)*log(log(2)) - x*e^(-x)*log(log(2)) - x*e^(-x + 5)*log(log (2)) - 9*e^(-x)*log(x)*log(log(2)) - 4*e^(-x)*log(x) + 9*e^(-x + 5)*log(lo g(2)) + 4*e^(-x + 5))/log(log(2))
Time = 7.64 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.38 \[ \int \frac {e^{-x} \left (-4-4 e^5 x+4 x \log (x)+\left (-9+x^2+e^5 \left (-10 x+x^2\right )+\left (10 x-x^2\right ) \log (x)\right ) \log (\log (2))\right )}{x \log (\log (2))} \, dx=\frac {{\mathrm {e}}^{5-x}\,\left (9\,\ln \left (\ln \left (2\right )\right )+4\right )-{\mathrm {e}}^{-x}\,\ln \left (x\right )\,\left (9\,\ln \left (\ln \left (2\right )\right )+4\right )}{\ln \left (\ln \left (2\right )\right )}-\frac {x\,\left ({\mathrm {e}}^{-x}\,\ln \left (\ln \left (2\right )\right )\,\left ({\mathrm {e}}^5+1\right )-{\mathrm {e}}^{-x}\,\ln \left (\ln \left (2\right )\right )\,\ln \left (x\right )\right )}{\ln \left (\ln \left (2\right )\right )} \]
int(-(exp(-x)*(4*x*exp(5) + log(log(2))*(exp(5)*(10*x - x^2) - log(x)*(10* x - x^2) - x^2 + 9) - 4*x*log(x) + 4))/(x*log(log(2))),x)