Integrand size = 121, antiderivative size = 24 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log (4+x)+\log \left (\log \left (\log \left (25-2 \left (-5+\frac {2 e^3 x}{\log (3)}\right )\right )\right )\right ) \] Output:
ln(4+x)+ln(ln(ln(35-4*exp(3)*x/ln(3))))
Time = 0.16 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log (4+x)+\log \left (\log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )\right ) \] Input:
Integrate[(E^3*(-16 - 4*x) + (-4*E^3*x + 35*Log[3])*Log[(-4*E^3*x + 35*Log [3])/Log[3]]*Log[Log[(-4*E^3*x + 35*Log[3])/Log[3]]])/((E^3*(-16*x - 4*x^2 ) + (140 + 35*x)*Log[3])*Log[(-4*E^3*x + 35*Log[3])/Log[3]]*Log[Log[(-4*E^ 3*x + 35*Log[3])/Log[3]]]),x]
Output:
Log[4 + x] + Log[Log[Log[35 - (4*E^3*x)/Log[3]]]]
Time = 0.95 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {7239, 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.
\(\displaystyle \int \frac {e^3 (-4 x-16)+\left (35 \log (3)-4 e^3 x\right ) \log \left (\frac {35 \log (3)-4 e^3 x}{\log (3)}\right ) \log \left (\log \left (\frac {35 \log (3)-4 e^3 x}{\log (3)}\right )\right )}{\left (e^3 \left (-4 x^2-16 x\right )+(35 x+140) \log (3)\right ) \log \left (\frac {35 \log (3)-4 e^3 x}{\log (3)}\right ) \log \left (\log \left (\frac {35 \log (3)-4 e^3 x}{\log (3)}\right )\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {1}{x+4}+\frac {4 e^3}{\left (4 e^3 x-35 \log (3)\right ) \log \left (35-\frac {4 e^3 x}{\log (3)}\right ) \log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \log (x+4)+\log \left (\log \left (\log \left (35-\frac {4 e^3 x}{\log (3)}\right )\right )\right )\) |
Input:
Int[(E^3*(-16 - 4*x) + (-4*E^3*x + 35*Log[3])*Log[(-4*E^3*x + 35*Log[3])/L og[3]]*Log[Log[(-4*E^3*x + 35*Log[3])/Log[3]]])/((E^3*(-16*x - 4*x^2) + (1 40 + 35*x)*Log[3])*Log[(-4*E^3*x + 35*Log[3])/Log[3]]*Log[Log[(-4*E^3*x + 35*Log[3])/Log[3]]]),x]
Output:
Log[4 + x] + Log[Log[Log[35 - (4*E^3*x)/Log[3]]]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.72 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\ln \left (\ln \left (\ln \left (\frac {35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}}{\ln \left (3\right )}\right )\right )\right )+\ln \left (4+x \right )\) | \(24\) |
risch | \(\ln \left (\ln \left (\ln \left (\frac {35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}}{\ln \left (3\right )}\right )\right )\right )+\ln \left (4+x \right )\) | \(24\) |
parts | \(\ln \left (\ln \left (\ln \left (\frac {35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}}{\ln \left (3\right )}\right )\right )\right )+\ln \left (4+x \right )\) | \(24\) |
default | \(\ln \left (4+x \right )+\ln \left (\ln \left (-\ln \left (\ln \left (3\right )\right )+\ln \left (35 \ln \left (3\right )-4 x \,{\mathrm e}^{3}\right )\right )\right )\) | \(25\) |
parallelrisch | \(\frac {\left (16 \ln \left (\ln \left (\ln \left (-\frac {4 x \,{\mathrm e}^{3}-35 \ln \left (3\right )}{\ln \left (3\right )}\right )\right )\right ) {\mathrm e}^{6}+16 \ln \left (4+x \right ) {\mathrm e}^{6}\right ) {\mathrm e}^{-6}}{16}\) | \(43\) |
Input:
int(((35*ln(3)-4*x*exp(3))*ln((35*ln(3)-4*x*exp(3))/ln(3))*ln(ln((35*ln(3) -4*x*exp(3))/ln(3)))+(-16-4*x)*exp(3))/((35*x+140)*ln(3)+(-4*x^2-16*x)*exp (3))/ln((35*ln(3)-4*x*exp(3))/ln(3))/ln(ln((35*ln(3)-4*x*exp(3))/ln(3))),x ,method=_RETURNVERBOSE)
Output:
ln(ln(ln((35*ln(3)-4*x*exp(3))/ln(3))))+ln(4+x)
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log \left (x + 4\right ) + \log \left (\log \left (\log \left (-\frac {4 \, x e^{3} - 35 \, \log \left (3\right )}{\log \left (3\right )}\right )\right )\right ) \] Input:
integrate(((35*log(3)-4*x*exp(3))*log((35*log(3)-4*x*exp(3))/log(3))*log(l og((35*log(3)-4*x*exp(3))/log(3)))+(-16-4*x)*exp(3))/((35*x+140)*log(3)+(- 4*x^2-16*x)*exp(3))/log((35*log(3)-4*x*exp(3))/log(3))/log(log((35*log(3)- 4*x*exp(3))/log(3))),x, algorithm="fricas")
Output:
log(x + 4) + log(log(log(-(4*x*e^3 - 35*log(3))/log(3))))
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log {\left (x + 4 \right )} + \log {\left (\log {\left (\log {\left (\frac {- 4 x e^{3} + 35 \log {\left (3 \right )}}{\log {\left (3 \right )}} \right )} \right )} \right )} \] Input:
integrate(((35*ln(3)-4*x*exp(3))*ln((35*ln(3)-4*x*exp(3))/ln(3))*ln(ln((35 *ln(3)-4*x*exp(3))/ln(3)))+(-16-4*x)*exp(3))/((35*x+140)*ln(3)+(-4*x**2-16 *x)*exp(3))/ln((35*ln(3)-4*x*exp(3))/ln(3))/ln(ln((35*ln(3)-4*x*exp(3))/ln (3))),x)
Output:
log(x + 4) + log(log(log((-4*x*exp(3) + 35*log(3))/log(3))))
Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\log \left (x + 4\right ) + \log \left (\log \left (\log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) - \log \left (\log \left (3\right )\right )\right )\right ) \] Input:
integrate(((35*log(3)-4*x*exp(3))*log((35*log(3)-4*x*exp(3))/log(3))*log(l og((35*log(3)-4*x*exp(3))/log(3)))+(-16-4*x)*exp(3))/((35*x+140)*log(3)+(- 4*x^2-16*x)*exp(3))/log((35*log(3)-4*x*exp(3))/log(3))/log(log((35*log(3)- 4*x*exp(3))/log(3))),x, algorithm="maxima")
Output:
log(x + 4) + log(log(log(-4*x*e^3 + 35*log(3)) - log(log(3))))
Leaf count of result is larger than twice the leaf count of optimal. 105 vs. \(2 (19) = 38\).
Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 4.38 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\frac {35 \, \log \left (3\right ) \log \left (4 \, x e^{3} - 35 \, \log \left (3\right )\right ) - 35 \, \log \left (3\right ) \log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) + 16 \, e^{3} \log \left (x + 4\right ) + 35 \, \log \left (3\right ) \log \left (x + 4\right ) + 16 \, e^{3} \log \left (\log \left (\log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) - \log \left (\log \left (3\right )\right )\right )\right ) + 35 \, \log \left (3\right ) \log \left (\log \left (\log \left (-4 \, x e^{3} + 35 \, \log \left (3\right )\right ) - \log \left (\log \left (3\right )\right )\right )\right )}{16 \, e^{3} + 35 \, \log \left (3\right )} \] Input:
integrate(((35*log(3)-4*x*exp(3))*log((35*log(3)-4*x*exp(3))/log(3))*log(l og((35*log(3)-4*x*exp(3))/log(3)))+(-16-4*x)*exp(3))/((35*x+140)*log(3)+(- 4*x^2-16*x)*exp(3))/log((35*log(3)-4*x*exp(3))/log(3))/log(log((35*log(3)- 4*x*exp(3))/log(3))),x, algorithm="giac")
Output:
(35*log(3)*log(4*x*e^3 - 35*log(3)) - 35*log(3)*log(-4*x*e^3 + 35*log(3)) + 16*e^3*log(x + 4) + 35*log(3)*log(x + 4) + 16*e^3*log(log(log(-4*x*e^3 + 35*log(3)) - log(log(3)))) + 35*log(3)*log(log(log(-4*x*e^3 + 35*log(3)) - log(log(3)))))/(16*e^3 + 35*log(3))
Time = 1.41 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\ln \left (x+4\right )+\ln \left (\ln \left (\ln \left (35\,\ln \left (3\right )-4\,x\,{\mathrm {e}}^3\right )-\ln \left (\ln \left (3\right )\right )\right )\right ) \] Input:
int(-(exp(3)*(4*x + 16) - log(log((35*log(3) - 4*x*exp(3))/log(3)))*log((3 5*log(3) - 4*x*exp(3))/log(3))*(35*log(3) - 4*x*exp(3)))/(log(log((35*log( 3) - 4*x*exp(3))/log(3)))*log((35*log(3) - 4*x*exp(3))/log(3))*(log(3)*(35 *x + 140) - exp(3)*(16*x + 4*x^2))),x)
Output:
log(x + 4) + log(log(log(35*log(3) - 4*x*exp(3)) - log(log(3))))
Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {e^3 (-16-4 x)+\left (-4 e^3 x+35 \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )}{\left (e^3 \left (-16 x-4 x^2\right )+(140+35 x) \log (3)\right ) \log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right ) \log \left (\log \left (\frac {-4 e^3 x+35 \log (3)}{\log (3)}\right )\right )} \, dx=\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (\frac {35 \,\mathrm {log}\left (3\right )-4 e^{3} x}{\mathrm {log}\left (3\right )}\right )\right )\right )+\mathrm {log}\left (x +4\right ) \] Input:
int(((35*log(3)-4*x*exp(3))*log((35*log(3)-4*x*exp(3))/log(3))*log(log((35 *log(3)-4*x*exp(3))/log(3)))+(-16-4*x)*exp(3))/((35*x+140)*log(3)+(-4*x^2- 16*x)*exp(3))/log((35*log(3)-4*x*exp(3))/log(3))/log(log((35*log(3)-4*x*ex p(3))/log(3))),x)
Output:
log(log(log((35*log(3) - 4*e**3*x)/log(3)))) + log(x + 4)