Integrand size = 135, antiderivative size = 26 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {e^{-x} \log (5)}{(x+\log (4)) \left (\log (4)+\frac {\log (\log (4))}{x}\right )} \] Output:
ln(5)/(ln(2*ln(2))/x+2*ln(2))/(x+2*ln(2))/exp(x)
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {e^{-x} x \log (5)}{(x+\log (4)) (x \log (4)+\log (\log (4)))} \] Input:
Integrate[(((-x^2 - x^3)*Log[4] - x^2*Log[4]^2)*Log[5] + (-x^2 + (1 - x)*L og[4])*Log[5]*Log[Log[4]])/(E^x*(x^4*Log[4]^2 + 2*x^3*Log[4]^3 + x^2*Log[4 ]^4) + E^x*(2*x^3*Log[4] + 4*x^2*Log[4]^2 + 2*x*Log[4]^3)*Log[Log[4]] + E^ x*(x^2 + 2*x*Log[4] + Log[4]^2)*Log[Log[4]]^2),x]
Output:
(x*Log[5])/(E^x*(x + Log[4])*(x*Log[4] + Log[Log[4]]))
Leaf count is larger than twice the leaf count of optimal. \(63\) vs. \(2(26)=52\).
Time = 1.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.42, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {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.
| \(\displaystyle \int \frac {\log (5) \log (\log (4)) \left ((1-x) \log (4)-x^2\right )+\log (5) \left (\left (-x^3-x^2\right ) \log (4)-x^2 \log ^2(4)\right )}{e^x \log ^2(\log (4)) \left (x^2+2 x \log (4)+\log ^2(4)\right )+e^x \log (\log (4)) \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right )+e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {e^{-x} \log (5) \left (x^3 (-\log (4))-x^2 \left (\log ^2(4)+\log (4)+\log (\log (4))\right )-x \log (4) \log (\log (4))+\log (4) \log (\log (4))\right )}{(x+\log (4))^2 (x \log (4)+\log (\log (4)))^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \log (5) \int -\frac {e^{-x} \left (\log (4) x^3+\left (\log (4)+\log ^2(4)+\log (\log (4))\right ) x^2+\log (4) \log (\log (4)) x-\log (4) \log (\log (4))\right )}{(x+\log (4))^2 (\log (4) x+\log (\log (4)))^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\log (5) \int \frac {e^{-x} \left (\log (4) x^3+\left (\log (4)+\log ^2(4)+\log (\log (4))\right ) x^2+\log (4) \log (\log (4)) x-\log (4) \log (\log (4))\right )}{(x+\log (4))^2 (\log (4) x+\log (\log (4)))^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\log (5) \int \left (-\frac {e^{-x} \log (\log (4))}{\left (\log ^2(4)-\log (\log (4))\right ) (\log (4) x+\log (\log (4)))}-\frac {e^{-x} \log (4) \log (\log (4))}{\left (\log ^2(4)-\log (\log (4))\right ) (\log (4) x+\log (\log (4)))^2}+\frac {e^{-x} \log (4)}{\left (\log ^2(4)-\log (\log (4))\right ) (x+\log (4))}+\frac {e^{-x} \log (4)}{\left (\log ^2(4)-\log (\log (4))\right ) (x+\log (4))^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\log (5) \left (\frac {e^{-x} \log (\log (4))}{\left (\log ^2(4)-\log (\log (4))\right ) (x \log (4)+\log (\log (4)))}-\frac {e^{-x} \log (4)}{\left (\log ^2(4)-\log (\log (4))\right ) (x+\log (4))}\right )\) |
Input:
Int[(((-x^2 - x^3)*Log[4] - x^2*Log[4]^2)*Log[5] + (-x^2 + (1 - x)*Log[4]) *Log[5]*Log[Log[4]])/(E^x*(x^4*Log[4]^2 + 2*x^3*Log[4]^3 + x^2*Log[4]^4) + E^x*(2*x^3*Log[4] + 4*x^2*Log[4]^2 + 2*x*Log[4]^3)*Log[Log[4]] + E^x*(x^2 + 2*x*Log[4] + Log[4]^2)*Log[Log[4]]^2),x]
Output:
-(Log[5]*(-(Log[4]/(E^x*(x + Log[4])*(Log[4]^2 - Log[Log[4]]))) + Log[Log[ 4]]/(E^x*(Log[4]^2 - Log[Log[4]])*(x*Log[4] + Log[Log[4]]))))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.50 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62
| method | result | size |
| gosper | \(\frac {x \ln \left (5\right ) {\mathrm e}^{-x}}{4 x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+2 \ln \left (2 \ln \left (2\right )\right ) \ln \left (2\right )+\ln \left (2 \ln \left (2\right )\right ) x}\) | \(42\) |
| norman | \(\frac {x \ln \left (5\right ) {\mathrm e}^{-x}}{4 x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+2 \ln \left (2 \ln \left (2\right )\right ) \ln \left (2\right )+\ln \left (2 \ln \left (2\right )\right ) x}\) | \(42\) |
| parallelrisch | \(\frac {x \ln \left (5\right ) {\mathrm e}^{-x}}{4 x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+2 \ln \left (2 \ln \left (2\right )\right ) \ln \left (2\right )+\ln \left (2 \ln \left (2\right )\right ) x}\) | \(42\) |
| risch | \(\frac {x \ln \left (5\right ) {\mathrm e}^{-x}}{4 x \ln \left (2\right )^{2}+2 x^{2} \ln \left (2\right )+2 \ln \left (2\right )^{2}+2 \ln \left (2\right ) \ln \left (\ln \left (2\right )\right )+x \ln \left (2\right )+x \ln \left (\ln \left (2\right )\right )}\) | \(48\) |
| default | \(\text {Expression too large to display}\) | \(2348\) |
Input:
int(((2*(1-x)*ln(2)-x^2)*ln(5)*ln(2*ln(2))+(-4*x^2*ln(2)^2+2*(-x^3-x^2)*ln (2))*ln(5))/((4*ln(2)^2+4*x*ln(2)+x^2)*exp(x)*ln(2*ln(2))^2+(16*x*ln(2)^3+ 16*x^2*ln(2)^2+4*x^3*ln(2))*exp(x)*ln(2*ln(2))+(16*x^2*ln(2)^4+16*x^3*ln(2 )^3+4*x^4*ln(2)^2)*exp(x)),x,method=_RETURNVERBOSE)
Output:
x*ln(5)/(4*x*ln(2)^2+2*x^2*ln(2)+2*ln(2*ln(2))*ln(2)+ln(2*ln(2))*x)/exp(x)
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {x \log \left (5\right )}{{\left (x + 2 \, \log \left (2\right )\right )} e^{x} \log \left (2 \, \log \left (2\right )\right ) + 2 \, {\left (x^{2} \log \left (2\right ) + 2 \, x \log \left (2\right )^{2}\right )} e^{x}} \] Input:
integrate(((2*(1-x)*log(2)-x^2)*log(5)*log(2*log(2))+(-4*x^2*log(2)^2+2*(- x^3-x^2)*log(2))*log(5))/((4*log(2)^2+4*x*log(2)+x^2)*exp(x)*log(2*log(2)) ^2+(16*x*log(2)^3+16*x^2*log(2)^2+4*x^3*log(2))*exp(x)*log(2*log(2))+(16*x ^2*log(2)^4+16*x^3*log(2)^3+4*x^4*log(2)^2)*exp(x)),x, algorithm="fricas")
Output:
x*log(5)/((x + 2*log(2))*e^x*log(2*log(2)) + 2*(x^2*log(2) + 2*x*log(2)^2) *e^x)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (26) = 52\).
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.04 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {x e^{- x} \log {\left (5 \right )}}{2 x^{2} \log {\left (2 \right )} + x \log {\left (\log {\left (2 \right )} \right )} + x \log {\left (2 \right )} + 4 x \log {\left (2 \right )}^{2} + 2 \log {\left (2 \right )} \log {\left (\log {\left (2 \right )} \right )} + 2 \log {\left (2 \right )}^{2}} \] Input:
integrate(((2*(1-x)*ln(2)-x**2)*ln(5)*ln(2*ln(2))+(-4*x**2*ln(2)**2+2*(-x* *3-x**2)*ln(2))*ln(5))/((4*ln(2)**2+4*x*ln(2)+x**2)*exp(x)*ln(2*ln(2))**2+ (16*x*ln(2)**3+16*x**2*ln(2)**2+4*x**3*ln(2))*exp(x)*ln(2*ln(2))+(16*x**2* ln(2)**4+16*x**3*ln(2)**3+4*x**4*ln(2)**2)*exp(x)),x)
Output:
x*exp(-x)*log(5)/(2*x**2*log(2) + x*log(log(2)) + x*log(2) + 4*x*log(2)**2 + 2*log(2)*log(log(2)) + 2*log(2)**2)
Time = 0.20 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {x e^{\left (-x\right )} \log \left (5\right )}{2 \, x^{2} \log \left (2\right ) + {\left (4 \, \log \left (2\right )^{2} + \log \left (2\right ) + \log \left (\log \left (2\right )\right )\right )} x + 2 \, \log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right )} \] Input:
integrate(((2*(1-x)*log(2)-x^2)*log(5)*log(2*log(2))+(-4*x^2*log(2)^2+2*(- x^3-x^2)*log(2))*log(5))/((4*log(2)^2+4*x*log(2)+x^2)*exp(x)*log(2*log(2)) ^2+(16*x*log(2)^3+16*x^2*log(2)^2+4*x^3*log(2))*exp(x)*log(2*log(2))+(16*x ^2*log(2)^4+16*x^3*log(2)^3+4*x^4*log(2)^2)*exp(x)),x, algorithm="maxima")
Output:
x*e^(-x)*log(5)/(2*x^2*log(2) + (4*log(2)^2 + log(2) + log(log(2)))*x + 2* log(2)^2 + 2*log(2)*log(log(2)))
Time = 0.14 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {x e^{\left (-x\right )} \log \left (5\right )}{2 \, x^{2} \log \left (2\right ) + 4 \, x \log \left (2\right )^{2} + x \log \left (2\right ) + 2 \, \log \left (2\right )^{2} + x \log \left (\log \left (2\right )\right ) + 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right )} \] Input:
integrate(((2*(1-x)*log(2)-x^2)*log(5)*log(2*log(2))+(-4*x^2*log(2)^2+2*(- x^3-x^2)*log(2))*log(5))/((4*log(2)^2+4*x*log(2)+x^2)*exp(x)*log(2*log(2)) ^2+(16*x*log(2)^3+16*x^2*log(2)^2+4*x^3*log(2))*exp(x)*log(2*log(2))+(16*x ^2*log(2)^4+16*x^3*log(2)^3+4*x^4*log(2)^2)*exp(x)),x, algorithm="giac")
Output:
x*e^(-x)*log(5)/(2*x^2*log(2) + 4*x*log(2)^2 + x*log(2) + 2*log(2)^2 + x*l og(log(2)) + 2*log(2)*log(log(2)))
Timed out. \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\int -\frac {\ln \left (5\right )\,\left (4\,x^2\,{\ln \left (2\right )}^2+2\,\ln \left (2\right )\,\left (x^3+x^2\right )\right )+\ln \left (2\,\ln \left (2\right )\right )\,\ln \left (5\right )\,\left (2\,\ln \left (2\right )\,\left (x-1\right )+x^2\right )}{{\mathrm {e}}^x\,\left (4\,{\ln \left (2\right )}^2\,x^4+16\,{\ln \left (2\right )}^3\,x^3+16\,{\ln \left (2\right )}^4\,x^2\right )+\ln \left (2\,\ln \left (2\right )\right )\,{\mathrm {e}}^x\,\left (4\,\ln \left (2\right )\,x^3+16\,{\ln \left (2\right )}^2\,x^2+16\,{\ln \left (2\right )}^3\,x\right )+{\ln \left (2\,\ln \left (2\right )\right )}^2\,{\mathrm {e}}^x\,\left (x^2+4\,\ln \left (2\right )\,x+4\,{\ln \left (2\right )}^2\right )} \,d x \] Input:
int(-(log(5)*(4*x^2*log(2)^2 + 2*log(2)*(x^2 + x^3)) + log(2*log(2))*log(5 )*(2*log(2)*(x - 1) + x^2))/(exp(x)*(16*x^2*log(2)^4 + 16*x^3*log(2)^3 + 4 *x^4*log(2)^2) + log(2*log(2))*exp(x)*(16*x^2*log(2)^2 + 16*x*log(2)^3 + 4 *x^3*log(2)) + log(2*log(2))^2*exp(x)*(4*x*log(2) + 4*log(2)^2 + x^2)),x)
Output:
int(-(log(5)*(4*x^2*log(2)^2 + 2*log(2)*(x^2 + x^3)) + log(2*log(2))*log(5 )*(2*log(2)*(x - 1) + x^2))/(exp(x)*(16*x^2*log(2)^4 + 16*x^3*log(2)^3 + 4 *x^4*log(2)^2) + log(2*log(2))*exp(x)*(16*x^2*log(2)^2 + 16*x*log(2)^3 + 4 *x^3*log(2)) + log(2*log(2))^2*exp(x)*(4*x*log(2) + 4*log(2)^2 + x^2)), x)
Time = 0.17 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int \frac {\left (\left (-x^2-x^3\right ) \log (4)-x^2 \log ^2(4)\right ) \log (5)+\left (-x^2+(1-x) \log (4)\right ) \log (5) \log (\log (4))}{e^x \left (x^4 \log ^2(4)+2 x^3 \log ^3(4)+x^2 \log ^4(4)\right )+e^x \left (2 x^3 \log (4)+4 x^2 \log ^2(4)+2 x \log ^3(4)\right ) \log (\log (4))+e^x \left (x^2+2 x \log (4)+\log ^2(4)\right ) \log ^2(\log (4))} \, dx=\frac {\mathrm {log}\left (5\right ) x}{e^{x} \left (2 \,\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right ) \mathrm {log}\left (2\right )+\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right ) x +4 \mathrm {log}\left (2\right )^{2} x +2 \,\mathrm {log}\left (2\right ) x^{2}\right )} \] Input:
int(((2*(1-x)*log(2)-x^2)*log(5)*log(2*log(2))+(-4*x^2*log(2)^2+2*(-x^3-x^ 2)*log(2))*log(5))/((4*log(2)^2+4*x*log(2)+x^2)*exp(x)*log(2*log(2))^2+(16 *x*log(2)^3+16*x^2*log(2)^2+4*x^3*log(2))*exp(x)*log(2*log(2))+(16*x^2*log (2)^4+16*x^3*log(2)^3+4*x^4*log(2)^2)*exp(x)),x)
Output:
(log(5)*x)/(e**x*(2*log(2*log(2))*log(2) + log(2*log(2))*x + 4*log(2)**2*x + 2*log(2)*x**2))