Integrand size = 56, antiderivative size = 17 \[ \int \frac {(25+x) \log (x)+\log (x) \log (\log (x))+\frac {(25+x+\log (\log (x))) (12+(300+24 x) \log (x)+12 \log (x) \log (\log (x)))}{e^5}}{(25+x) \log (x)+\log (x) \log (\log (x))} \, dx=\frac {17}{5}+x+\frac {12 x (25+x+\log (\log (x)))}{e^5} \] Output:
17/5+12*exp(ln(ln(ln(x))+x+25)-5)*x+x
Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {(25+x) \log (x)+\log (x) \log (\log (x))+\frac {(25+x+\log (\log (x))) (12+(300+24 x) \log (x)+12 \log (x) \log (\log (x)))}{e^5}}{(25+x) \log (x)+\log (x) \log (\log (x))} \, dx=\frac {x \left (e^5+12 (25+x)+12 \log (\log (x))\right )}{e^5} \] Input:
Integrate[((25 + x)*Log[x] + Log[x]*Log[Log[x]] + ((25 + x + Log[Log[x]])* (12 + (300 + 24*x)*Log[x] + 12*Log[x]*Log[Log[x]]))/E^5)/((25 + x)*Log[x] + Log[x]*Log[Log[x]]),x]
Output:
(x*(E^5 + 12*(25 + x) + 12*Log[Log[x]]))/E^5
Time = 0.53 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.35, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {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.
\(\displaystyle \int \frac {(x+25) \log (x)+\log (\log (x)) \log (x)+\frac {(x+\log (\log (x))+25) ((24 x+300) \log (x)+12 \log (\log (x)) \log (x)+12)}{e^5}}{(x+25) \log (x)+\log (\log (x)) \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {\log (x) \left (24 x+12 \log (\log (x))+e^5+300\right )+12}{e^5 \log (x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\log (x) \left (24 x+12 \log (\log (x))+e^5+300\right )+12}{\log (x)}dx}{e^5}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (\frac {24 x \log (x)+300 \left (1+\frac {e^5}{300}\right ) \log (x)+12}{\log (x)}+12 \log (\log (x))\right )dx}{e^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {12 x^2+\left (300+e^5\right ) x+12 x \log (\log (x))}{e^5}\) |
Input:
Int[((25 + x)*Log[x] + Log[x]*Log[Log[x]] + ((25 + x + Log[Log[x]])*(12 + (300 + 24*x)*Log[x] + 12*Log[x]*Log[Log[x]]))/E^5)/((25 + x)*Log[x] + Log[ x]*Log[Log[x]]),x]
Output:
((300 + E^5)*x + 12*x^2 + 12*x*Log[Log[x]])/E^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_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 1.06 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24
method | result | size |
default | \(x +{\mathrm e}^{-5} \left (12 x^{2}+300 x +12 x \ln \left (\ln \left (x \right )\right )\right )\) | \(21\) |
parts | \(x +{\mathrm e}^{-5} \left (12 x^{2}+300 x +12 x \ln \left (\ln \left (x \right )\right )\right )\) | \(21\) |
risch | \(12 \,{\mathrm e}^{-5} x^{2}+12 \,{\mathrm e}^{-5} x \ln \left (\ln \left (x \right )\right )+300 \,{\mathrm e}^{-5} x +x\) | \(23\) |
parallelrisch | \(\frac {-390625+50 \ln \left (\ln \left (x \right )\right )^{2} x -625 \ln \left (\ln \left (x \right )\right )^{2}-9375000 \,{\mathrm e}^{\ln \left (\ln \left (\ln \left (x \right )\right )+x +25\right )-5}+100 x^{2} \ln \left (\ln \left (x \right )\right )+1250 x \ln \left (\ln \left (x \right )\right )-31250 \ln \left (\ln \left (x \right )\right )+50 x^{3}+1875 x^{2}+30000 \,{\mathrm e}^{\ln \left (\ln \left (\ln \left (x \right )\right )+x +25\right )-5} x^{2}+600 \,{\mathrm e}^{\ln \left (\ln \left (\ln \left (x \right )\right )+x +25\right )-5} x^{3}-375000 \,{\mathrm e}^{\ln \left (\ln \left (\ln \left (x \right )\right )+x +25\right )-5} \ln \left (\ln \left (x \right )\right )+30000 \,{\mathrm e}^{\ln \left (\ln \left (\ln \left (x \right )\right )+x +25\right )-5} \ln \left (\ln \left (x \right )\right ) x +1200 \,{\mathrm e}^{\ln \left (\ln \left (\ln \left (x \right )\right )+x +25\right )-5} x^{2} \ln \left (\ln \left (x \right )\right )+600 \,{\mathrm e}^{\ln \left (\ln \left (\ln \left (x \right )\right )+x +25\right )-5} x \ln \left (\ln \left (x \right )\right )^{2}}{50 \left (\ln \left (\ln \left (x \right )\right )+x +25\right )^{2}}\) | \(166\) |
Input:
int(((12*ln(x)*ln(ln(x))+(24*x+300)*ln(x)+12)*exp(ln(ln(ln(x))+x+25)-5)+ln (x)*ln(ln(x))+(x+25)*ln(x))/(ln(x)*ln(ln(x))+(x+25)*ln(x)),x,method=_RETUR NVERBOSE)
Output:
x+exp(-5)*(12*x^2+300*x+12*x*ln(ln(x)))
Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {(25+x) \log (x)+\log (x) \log (\log (x))+\frac {(25+x+\log (\log (x))) (12+(300+24 x) \log (x)+12 \log (x) \log (\log (x)))}{e^5}}{(25+x) \log (x)+\log (x) \log (\log (x))} \, dx={\left (12 \, x^{2} + x e^{5} + 12 \, x \log \left (\log \left (x\right )\right ) + 300 \, x\right )} e^{\left (-5\right )} \] Input:
integrate(((12*log(x)*log(log(x))+(24*x+300)*log(x)+12)*exp(log(log(log(x) )+x+25)-5)+log(x)*log(log(x))+(x+25)*log(x))/(log(x)*log(log(x))+(x+25)*lo g(x)),x, algorithm="fricas")
Output:
(12*x^2 + x*e^5 + 12*x*log(log(x)) + 300*x)*e^(-5)
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.71 \[ \int \frac {(25+x) \log (x)+\log (x) \log (\log (x))+\frac {(25+x+\log (\log (x))) (12+(300+24 x) \log (x)+12 \log (x) \log (\log (x)))}{e^5}}{(25+x) \log (x)+\log (x) \log (\log (x))} \, dx=\frac {12 x^{2}}{e^{5}} + \frac {12 x \log {\left (\log {\left (x \right )} \right )}}{e^{5}} + \frac {x \left (e^{5} + 300\right )}{e^{5}} \] Input:
integrate(((12*ln(x)*ln(ln(x))+(24*x+300)*ln(x)+12)*exp(ln(ln(ln(x))+x+25) -5)+ln(x)*ln(ln(x))+(x+25)*ln(x))/(ln(x)*ln(ln(x))+(x+25)*ln(x)),x)
Output:
12*x**2*exp(-5) + 12*x*exp(-5)*log(log(x)) + x*(exp(5) + 300)*exp(-5)
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.24 \[ \int \frac {(25+x) \log (x)+\log (x) \log (\log (x))+\frac {(25+x+\log (\log (x))) (12+(300+24 x) \log (x)+12 \log (x) \log (\log (x)))}{e^5}}{(25+x) \log (x)+\log (x) \log (\log (x))} \, dx={\left (12 \, x^{2} + x {\left (e^{5} + 300\right )} + 12 \, x \log \left (\log \left (x\right )\right )\right )} e^{\left (-5\right )} \] Input:
integrate(((12*log(x)*log(log(x))+(24*x+300)*log(x)+12)*exp(log(log(log(x) )+x+25)-5)+log(x)*log(log(x))+(x+25)*log(x))/(log(x)*log(log(x))+(x+25)*lo g(x)),x, algorithm="maxima")
Output:
(12*x^2 + x*(e^5 + 300) + 12*x*log(log(x)))*e^(-5)
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.29 \[ \int \frac {(25+x) \log (x)+\log (x) \log (\log (x))+\frac {(25+x+\log (\log (x))) (12+(300+24 x) \log (x)+12 \log (x) \log (\log (x)))}{e^5}}{(25+x) \log (x)+\log (x) \log (\log (x))} \, dx=12 \, x^{2} e^{\left (-5\right )} + 12 \, x e^{\left (-5\right )} \log \left (\log \left (x\right )\right ) + 300 \, x e^{\left (-5\right )} + x \] Input:
integrate(((12*log(x)*log(log(x))+(24*x+300)*log(x)+12)*exp(log(log(log(x) )+x+25)-5)+log(x)*log(log(x))+(x+25)*log(x))/(log(x)*log(log(x))+(x+25)*lo g(x)),x, algorithm="giac")
Output:
12*x^2*e^(-5) + 12*x*e^(-5)*log(log(x)) + 300*x*e^(-5) + x
Time = 0.69 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {(25+x) \log (x)+\log (x) \log (\log (x))+\frac {(25+x+\log (\log (x))) (12+(300+24 x) \log (x)+12 \log (x) \log (\log (x)))}{e^5}}{(25+x) \log (x)+\log (x) \log (\log (x))} \, dx=x\,{\mathrm {e}}^{-5}\,\left (12\,x+12\,\ln \left (\ln \left (x\right )\right )+{\mathrm {e}}^5+300\right ) \] Input:
int((log(x)*(x + 25) + exp(log(x + log(log(x)) + 25) - 5)*(log(x)*(24*x + 300) + 12*log(log(x))*log(x) + 12) + log(log(x))*log(x))/(log(x)*(x + 25) + log(log(x))*log(x)),x)
Output:
x*exp(-5)*(12*x + 12*log(log(x)) + exp(5) + 300)
Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {(25+x) \log (x)+\log (x) \log (\log (x))+\frac {(25+x+\log (\log (x))) (12+(300+24 x) \log (x)+12 \log (x) \log (\log (x)))}{e^5}}{(25+x) \log (x)+\log (x) \log (\log (x))} \, dx=\frac {x \left (12 \,\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+e^{5}+12 x +300\right )}{e^{5}} \] Input:
int(((12*log(x)*log(log(x))+(24*x+300)*log(x)+12)*exp(log(log(log(x))+x+25 )-5)+log(x)*log(log(x))+(x+25)*log(x))/(log(x)*log(log(x))+(x+25)*log(x)), x)
Output:
(x*(12*log(log(x)) + e**5 + 12*x + 300))/e**5