Integrand size = 106, antiderivative size = 19 \[ \int \frac {4 e^{5 x}+e^{24 x}+\left (4 e^{5 x} x+e^{24 x} x\right ) \log (x)+\left (e^{24 x} \left (-95 x+19 x^2\right ) \log (x)+19 e^{24 x} x \log (x) \log (\log (x))\right ) \log (5-x-\log (\log (x)))}{e^{5 x} \left (-5 x+x^2\right ) \log (x)+e^{5 x} x \log (x) \log (\log (x))} \, dx=\left (4+e^{19 x}\right ) \log (5-x-\log (\log (x))) \] Output:
ln(-ln(ln(x))+5-x)*(4+exp(24*x)/exp(5*x))
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {4 e^{5 x}+e^{24 x}+\left (4 e^{5 x} x+e^{24 x} x\right ) \log (x)+\left (e^{24 x} \left (-95 x+19 x^2\right ) \log (x)+19 e^{24 x} x \log (x) \log (\log (x))\right ) \log (5-x-\log (\log (x)))}{e^{5 x} \left (-5 x+x^2\right ) \log (x)+e^{5 x} x \log (x) \log (\log (x))} \, dx=\left (4+e^{19 x}\right ) \log (5-x-\log (\log (x))) \] Input:
Integrate[(4*E^(5*x) + E^(24*x) + (4*E^(5*x)*x + E^(24*x)*x)*Log[x] + (E^( 24*x)*(-95*x + 19*x^2)*Log[x] + 19*E^(24*x)*x*Log[x]*Log[Log[x]])*Log[5 - x - Log[Log[x]]])/(E^(5*x)*(-5*x + x^2)*Log[x] + E^(5*x)*x*Log[x]*Log[Log[ x]]),x]
Output:
(4 + E^(19*x))*Log[5 - x - Log[Log[x]]]
Leaf count is larger than twice the leaf count of optimal. \(93\) vs. \(2(19)=38\).
Time = 3.65 (sec) , antiderivative size = 93, normalized size of antiderivative = 4.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {7292, 7239, 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.
\(\displaystyle \int \frac {\left (e^{24 x} \left (19 x^2-95 x\right ) \log (x)+19 e^{24 x} x \log (\log (x)) \log (x)\right ) \log (-x-\log (\log (x))+5)+4 e^{5 x}+e^{24 x}+\left (4 e^{5 x} x+e^{24 x} x\right ) \log (x)}{e^{5 x} \left (x^2-5 x\right ) \log (x)+e^{5 x} x \log (\log (x)) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-5 x} \left (-\left (e^{24 x} \left (19 x^2-95 x\right ) \log (x)+19 e^{24 x} x \log (\log (x)) \log (x)\right ) \log (-x-\log (\log (x))+5)-4 e^{5 x}-e^{24 x}-\left (\left (4 e^{5 x} x+e^{24 x} x\right ) \log (x)\right )\right )}{x \log (x) (-x-\log (\log (x))+5)}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-e^{19 x}-x \log (x) \left (e^{19 x}+19 e^{19 x} (x+\log (\log (x))-5) \log (-x-\log (\log (x))+5)+4\right )-4}{x \log (x) (-x-\log (\log (x))+5)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{19 x} \left (19 x^2 \log (x) \log (-x-\log (\log (x))+5)+x \log (x)-95 x \log (x) \log (-x-\log (\log (x))+5)+19 x \log (x) \log (\log (x)) \log (-x-\log (\log (x))+5)+1\right )}{x \log (x) (x+\log (\log (x))-5)}+\frac {4 (x \log (x)+1)}{x \log (x) (x+\log (\log (x))-5)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{19 x} \left (x^2 (-\log (x)) \log (-x-\log (\log (x))+5)+5 x \log (x) \log (-x-\log (\log (x))+5)-x \log (x) \log (\log (x)) \log (-x-\log (\log (x))+5)\right )}{x \log (x) (-x-\log (\log (x))+5)}+4 \log (-x-\log (\log (x))+5)\) |
Input:
Int[(4*E^(5*x) + E^(24*x) + (4*E^(5*x)*x + E^(24*x)*x)*Log[x] + (E^(24*x)* (-95*x + 19*x^2)*Log[x] + 19*E^(24*x)*x*Log[x]*Log[Log[x]])*Log[5 - x - Lo g[Log[x]]])/(E^(5*x)*(-5*x + x^2)*Log[x] + E^(5*x)*x*Log[x]*Log[Log[x]]),x ]
Output:
4*Log[5 - x - Log[Log[x]]] + (E^(19*x)*(5*x*Log[x]*Log[5 - x - Log[Log[x]] ] - x^2*Log[x]*Log[5 - x - Log[Log[x]]] - x*Log[x]*Log[Log[x]]*Log[5 - x - Log[Log[x]]]))/(x*Log[x]*(5 - x - Log[Log[x]]))
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 45.75 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42
method | result | size |
risch | \({\mathrm e}^{19 x} \ln \left (-\ln \left (\ln \left (x \right )\right )+5-x \right )+4 \ln \left (\ln \left (\ln \left (x \right )\right )+x -5\right )\) | \(27\) |
parallelrisch | \(\frac {\left (40 \ln \left (\ln \left (\ln \left (x \right )\right )+x -5\right ) {\mathrm e}^{5 x}+10 \ln \left (-\ln \left (\ln \left (x \right )\right )+5-x \right ) {\mathrm e}^{24 x}\right ) {\mathrm e}^{-5 x}}{10}\) | \(40\) |
Input:
int(((19*x*exp(24*x)*ln(x)*ln(ln(x))+(19*x^2-95*x)*exp(24*x)*ln(x))*ln(-ln (ln(x))+5-x)+(x*exp(24*x)+4*x*exp(5*x))*ln(x)+exp(24*x)+4*exp(5*x))/(x*exp (5*x)*ln(x)*ln(ln(x))+(x^2-5*x)*exp(5*x)*ln(x)),x,method=_RETURNVERBOSE)
Output:
exp(19*x)*ln(-ln(ln(x))+5-x)+4*ln(ln(ln(x))+x-5)
Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {4 e^{5 x}+e^{24 x}+\left (4 e^{5 x} x+e^{24 x} x\right ) \log (x)+\left (e^{24 x} \left (-95 x+19 x^2\right ) \log (x)+19 e^{24 x} x \log (x) \log (\log (x))\right ) \log (5-x-\log (\log (x)))}{e^{5 x} \left (-5 x+x^2\right ) \log (x)+e^{5 x} x \log (x) \log (\log (x))} \, dx={\left (e^{\left (19 \, x\right )} + 4\right )} \log \left (-x - \log \left (\log \left (x\right )\right ) + 5\right ) \] Input:
integrate(((19*x*exp(24*x)*log(x)*log(log(x))+(19*x^2-95*x)*exp(24*x)*log( x))*log(-log(log(x))+5-x)+(x*exp(24*x)+4*x*exp(5*x))*log(x)+exp(24*x)+4*ex p(5*x))/(x*exp(5*x)*log(x)*log(log(x))+(x^2-5*x)*exp(5*x)*log(x)),x, algor ithm="fricas")
Output:
(e^(19*x) + 4)*log(-x - log(log(x)) + 5)
Time = 6.97 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {4 e^{5 x}+e^{24 x}+\left (4 e^{5 x} x+e^{24 x} x\right ) \log (x)+\left (e^{24 x} \left (-95 x+19 x^2\right ) \log (x)+19 e^{24 x} x \log (x) \log (\log (x))\right ) \log (5-x-\log (\log (x)))}{e^{5 x} \left (-5 x+x^2\right ) \log (x)+e^{5 x} x \log (x) \log (\log (x))} \, dx=e^{19 x} \log {\left (- x - \log {\left (\log {\left (x \right )} \right )} + 5 \right )} + 4 \log {\left (x + \log {\left (\log {\left (x \right )} \right )} - 5 \right )} \] Input:
integrate(((19*x*exp(24*x)*ln(x)*ln(ln(x))+(19*x**2-95*x)*exp(24*x)*ln(x)) *ln(-ln(ln(x))+5-x)+(x*exp(24*x)+4*x*exp(5*x))*ln(x)+exp(24*x)+4*exp(5*x)) /(x*exp(5*x)*ln(x)*ln(ln(x))+(x**2-5*x)*exp(5*x)*ln(x)),x)
Output:
exp(19*x)*log(-x - log(log(x)) + 5) + 4*log(x + log(log(x)) - 5)
Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {4 e^{5 x}+e^{24 x}+\left (4 e^{5 x} x+e^{24 x} x\right ) \log (x)+\left (e^{24 x} \left (-95 x+19 x^2\right ) \log (x)+19 e^{24 x} x \log (x) \log (\log (x))\right ) \log (5-x-\log (\log (x)))}{e^{5 x} \left (-5 x+x^2\right ) \log (x)+e^{5 x} x \log (x) \log (\log (x))} \, dx={\left (e^{\left (19 \, x\right )} + 4\right )} \log \left (-x - \log \left (\log \left (x\right )\right ) + 5\right ) \] Input:
integrate(((19*x*exp(24*x)*log(x)*log(log(x))+(19*x^2-95*x)*exp(24*x)*log( x))*log(-log(log(x))+5-x)+(x*exp(24*x)+4*x*exp(5*x))*log(x)+exp(24*x)+4*ex p(5*x))/(x*exp(5*x)*log(x)*log(log(x))+(x^2-5*x)*exp(5*x)*log(x)),x, algor ithm="maxima")
Output:
(e^(19*x) + 4)*log(-x - log(log(x)) + 5)
Time = 0.16 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {4 e^{5 x}+e^{24 x}+\left (4 e^{5 x} x+e^{24 x} x\right ) \log (x)+\left (e^{24 x} \left (-95 x+19 x^2\right ) \log (x)+19 e^{24 x} x \log (x) \log (\log (x))\right ) \log (5-x-\log (\log (x)))}{e^{5 x} \left (-5 x+x^2\right ) \log (x)+e^{5 x} x \log (x) \log (\log (x))} \, dx=e^{\left (19 \, x\right )} \log \left (-x - \log \left (\log \left (x\right )\right ) + 5\right ) + 4 \, \log \left (x + \log \left (\log \left (x\right )\right ) - 5\right ) \] Input:
integrate(((19*x*exp(24*x)*log(x)*log(log(x))+(19*x^2-95*x)*exp(24*x)*log( x))*log(-log(log(x))+5-x)+(x*exp(24*x)+4*x*exp(5*x))*log(x)+exp(24*x)+4*ex p(5*x))/(x*exp(5*x)*log(x)*log(log(x))+(x^2-5*x)*exp(5*x)*log(x)),x, algor ithm="giac")
Output:
e^(19*x)*log(-x - log(log(x)) + 5) + 4*log(x + log(log(x)) - 5)
Timed out. \[ \int \frac {4 e^{5 x}+e^{24 x}+\left (4 e^{5 x} x+e^{24 x} x\right ) \log (x)+\left (e^{24 x} \left (-95 x+19 x^2\right ) \log (x)+19 e^{24 x} x \log (x) \log (\log (x))\right ) \log (5-x-\log (\log (x)))}{e^{5 x} \left (-5 x+x^2\right ) \log (x)+e^{5 x} x \log (x) \log (\log (x))} \, dx=-\int \frac {4\,{\mathrm {e}}^{5\,x}+{\mathrm {e}}^{24\,x}-\ln \left (5-\ln \left (\ln \left (x\right )\right )-x\right )\,\left ({\mathrm {e}}^{24\,x}\,\ln \left (x\right )\,\left (95\,x-19\,x^2\right )-19\,x\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{24\,x}\,\ln \left (x\right )\right )+\ln \left (x\right )\,\left (4\,x\,{\mathrm {e}}^{5\,x}+x\,{\mathrm {e}}^{24\,x}\right )}{{\mathrm {e}}^{5\,x}\,\ln \left (x\right )\,\left (5\,x-x^2\right )-x\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{5\,x}\,\ln \left (x\right )} \,d x \] Input:
int(-(4*exp(5*x) + exp(24*x) - log(5 - log(log(x)) - x)*(exp(24*x)*log(x)* (95*x - 19*x^2) - 19*x*log(log(x))*exp(24*x)*log(x)) + log(x)*(4*x*exp(5*x ) + x*exp(24*x)))/(exp(5*x)*log(x)*(5*x - x^2) - x*log(log(x))*exp(5*x)*lo g(x)),x)
Output:
-int((4*exp(5*x) + exp(24*x) - log(5 - log(log(x)) - x)*(exp(24*x)*log(x)* (95*x - 19*x^2) - 19*x*log(log(x))*exp(24*x)*log(x)) + log(x)*(4*x*exp(5*x ) + x*exp(24*x)))/(exp(5*x)*log(x)*(5*x - x^2) - x*log(log(x))*exp(5*x)*lo g(x)), x)
Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.42 \[ \int \frac {4 e^{5 x}+e^{24 x}+\left (4 e^{5 x} x+e^{24 x} x\right ) \log (x)+\left (e^{24 x} \left (-95 x+19 x^2\right ) \log (x)+19 e^{24 x} x \log (x) \log (\log (x))\right ) \log (5-x-\log (\log (x)))}{e^{5 x} \left (-5 x+x^2\right ) \log (x)+e^{5 x} x \log (x) \log (\log (x))} \, dx=e^{19 x} \mathrm {log}\left (-\mathrm {log}\left (\mathrm {log}\left (x \right )\right )-x +5\right )+4 \,\mathrm {log}\left (\mathrm {log}\left (\mathrm {log}\left (x \right )\right )+x -5\right ) \] Input:
int(((19*x*exp(24*x)*log(x)*log(log(x))+(19*x^2-95*x)*exp(24*x)*log(x))*lo g(-log(log(x))+5-x)+(x*exp(24*x)+4*x*exp(5*x))*log(x)+exp(24*x)+4*exp(5*x) )/(x*exp(5*x)*log(x)*log(log(x))+(x^2-5*x)*exp(5*x)*log(x)),x)
Output:
e**(19*x)*log( - log(log(x)) - x + 5) + 4*log(log(log(x)) + x - 5)