Integrand size = 104, antiderivative size = 25 \[ \int \frac {e^4 \left (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)\right ) \left (-16 x-240 x^2+(-3+15 x) \log (2)+64 x \log (x)\right )}{\log ^4(2) (5 x-\log (x)) \left (-80 x^3-15 x^2 \log (2)+\left (16 x^2+3 x \log (2)\right ) \log (x)\right )} \, dx=\frac {e^4 \left (3+\frac {16 x}{\log (2)}\right )^4}{5 x-\log (x)} \] Output:
exp(ln((3+16*x/ln(2))^4)+4-ln(-ln(x)+5*x))
Time = 5.03 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.00 \[ \int \frac {e^4 \left (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)\right ) \left (-16 x-240 x^2+(-3+15 x) \log (2)+64 x \log (x)\right )}{\log ^4(2) (5 x-\log (x)) \left (-80 x^3-15 x^2 \log (2)+\left (16 x^2+3 x \log (2)\right ) \log (x)\right )} \, dx=-\frac {e^4 \left (-16 x+80 x^2+15 x \log (2)-\log (8)\right ) (16 x+\log (8))^3}{(-1+5 x) \log ^4(2) (-5 x+\log (x))} \] Input:
Integrate[(E^4*(65536*x^4 + 49152*x^3*Log[2] + 13824*x^2*Log[2]^2 + 1728*x *Log[2]^3 + 81*Log[2]^4)*(-16*x - 240*x^2 + (-3 + 15*x)*Log[2] + 64*x*Log[ x]))/(Log[2]^4*(5*x - Log[x])*(-80*x^3 - 15*x^2*Log[2] + (16*x^2 + 3*x*Log [2])*Log[x])),x]
Output:
-((E^4*(-16*x + 80*x^2 + 15*x*Log[2] - Log[8])*(16*x + Log[8])^3)/((-1 + 5 *x)*Log[2]^4*(-5*x + Log[x])))
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.
\(\displaystyle \int \frac {e^4 \left (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)\right ) \left (-240 x^2-16 x+64 x \log (x)+(15 x-3) \log (2)\right )}{\log ^4(2) (5 x-\log (x)) \left (-80 x^3-15 x^2 \log (2)+\left (16 x^2+3 x \log (2)\right ) \log (x)\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^4 \int \frac {\left (65536 x^4+49152 \log (2) x^3+13824 \log ^2(2) x^2+1728 \log ^3(2) x+81 \log ^4(2)\right ) \left (240 x^2-64 \log (x) x+16 x+3 (1-5 x) \log (2)\right )}{(5 x-\log (x)) \left (80 x^3+15 \log (2) x^2-\left (16 x^2+3 \log (2) x\right ) \log (x)\right )}dx}{\log ^4(2)}\) |
\(\Big \downarrow \) 2006 |
\(\displaystyle \frac {e^4 \int \frac {(16 x+3 \log (2))^4 \left (240 x^2-64 \log (x) x+16 x+3 (1-5 x) \log (2)\right )}{(5 x-\log (x)) \left (80 x^3+15 \log (2) x^2-\left (16 x^2+3 \log (2) x\right ) \log (x)\right )}dx}{\log ^4(2)}\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \frac {e^4 \int \frac {(16 x+\log (8))^3 \left (240 x^2-64 \log (x) x+(16-15 \log (2)) x+\log (8)\right )}{x (5 x-\log (x))^2}dx}{\log ^4(2)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {e^4 \int \left (\frac {64 (16 x+\log (8))^3}{5 x-\log (x)}+\frac {\left (-80 x^2+(16-15 \log (2)) x+\log (8)\right ) (16 x+\log (8))^3}{x (5 x-\log (x))^2}\right )dx}{\log ^4(2)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^4 \left (-327680 \int \frac {x^4}{(5 x-\log (x))^2}dx+16384 (4-5 \log (8)) \int \frac {x^3}{(5 x-\log (x))^2}dx+262144 \int \frac {x^3}{5 x-\log (x)}dx+512 (32-15 \log (8)) \log (8) \int \frac {x^2}{(5 x-\log (x))^2}dx+49152 \log (8) \int \frac {x^2}{5 x-\log (x)}dx+\log ^4(8) \int \frac {1}{x (5 x-\log (x))^2}dx+(64-5 \log (8)) \log ^3(8) \int \frac {1}{(5 x-\log (x))^2}dx+64 \log ^3(8) \int \frac {1}{5 x-\log (x)}dx+64 (24-5 \log (8)) \log ^2(8) \int \frac {x}{(5 x-\log (x))^2}dx+3072 \log ^2(8) \int \frac {x}{5 x-\log (x)}dx\right )}{\log ^4(2)}\) |
Input:
Int[(E^4*(65536*x^4 + 49152*x^3*Log[2] + 13824*x^2*Log[2]^2 + 1728*x*Log[2 ]^3 + 81*Log[2]^4)*(-16*x - 240*x^2 + (-3 + 15*x)*Log[2] + 64*x*Log[x]))/( Log[2]^4*(5*x - Log[x])*(-80*x^3 - 15*x^2*Log[2] + (16*x^2 + 3*x*Log[2])*L og[x])),x]
Output:
$Aborted
Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs. \(2(26)=52\).
Time = 28.82 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.24
method | result | size |
parallelrisch | \({\mathrm e}^{-\ln \left (-\ln \left (x \right )+5 x \right )+\ln \left (\frac {81 \ln \left (2\right )^{4}+1728 x \ln \left (2\right )^{3}+13824 x^{2} \ln \left (2\right )^{2}+49152 x^{3} \ln \left (2\right )+65536 x^{4}}{\ln \left (2\right )^{4}}\right )+4}\) | \(56\) |
norman | \(\frac {1728 \,{\mathrm e}^{4} \ln \left (2\right )^{2} x +49152 x^{3} {\mathrm e}^{4}+81 \,{\mathrm e}^{4} \ln \left (2\right )^{3}+13824 x^{2} {\mathrm e}^{4} \ln \left (2\right )+\frac {65536 \,{\mathrm e}^{4} x^{4}}{\ln \left (2\right )}}{\left (-\ln \left (x \right )+5 x \right ) \ln \left (2\right )^{3}}\) | \(61\) |
default | \(\frac {{\mathrm e}^{4} \left (-\frac {65536 x^{4}}{\ln \left (x \right )-5 x}-\frac {1728 \ln \left (2\right )^{3} x}{\ln \left (x \right )-5 x}-\frac {13824 \ln \left (2\right )^{2} x^{2}}{\ln \left (x \right )-5 x}-\frac {49152 \ln \left (2\right ) x^{3}}{\ln \left (x \right )-5 x}+\frac {81 \ln \left (2\right )^{4}}{-\ln \left (x \right )+5 x}\right )}{\ln \left (2\right )^{4}}\) | \(85\) |
Input:
int((64*x*ln(x)+(15*x-3)*ln(2)-240*x^2-16*x)*exp(-ln(-ln(x)+5*x)+ln((81*ln (2)^4+1728*x*ln(2)^3+13824*x^2*ln(2)^2+49152*x^3*ln(2)+65536*x^4)/ln(2)^4) +4)/((3*x*ln(2)+16*x^2)*ln(x)-15*x^2*ln(2)-80*x^3),x,method=_RETURNVERBOSE )
Output:
exp(-ln(-ln(x)+5*x)+ln((81*ln(2)^4+1728*x*ln(2)^3+13824*x^2*ln(2)^2+49152* x^3*ln(2)+65536*x^4)/ln(2)^4)+4)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (26) = 52\).
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.56 \[ \int \frac {e^4 \left (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)\right ) \left (-16 x-240 x^2+(-3+15 x) \log (2)+64 x \log (x)\right )}{\log ^4(2) (5 x-\log (x)) \left (-80 x^3-15 x^2 \log (2)+\left (16 x^2+3 x \log (2)\right ) \log (x)\right )} \, dx=\frac {65536 \, x^{4} e^{4} + 49152 \, x^{3} e^{4} \log \left (2\right ) + 13824 \, x^{2} e^{4} \log \left (2\right )^{2} + 1728 \, x e^{4} \log \left (2\right )^{3} + 81 \, e^{4} \log \left (2\right )^{4}}{5 \, x \log \left (2\right )^{4} - \log \left (2\right )^{4} \log \left (x\right )} \] Input:
integrate((64*x*log(x)+(15*x-3)*log(2)-240*x^2-16*x)*exp(-log(-log(x)+5*x) +log((81*log(2)^4+1728*x*log(2)^3+13824*x^2*log(2)^2+49152*x^3*log(2)+6553 6*x^4)/log(2)^4)+4)/((3*x*log(2)+16*x^2)*log(x)-15*x^2*log(2)-80*x^3),x, a lgorithm="fricas")
Output:
(65536*x^4*e^4 + 49152*x^3*e^4*log(2) + 13824*x^2*e^4*log(2)^2 + 1728*x*e^ 4*log(2)^3 + 81*e^4*log(2)^4)/(5*x*log(2)^4 - log(2)^4*log(x))
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (19) = 38\).
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.92 \[ \int \frac {e^4 \left (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)\right ) \left (-16 x-240 x^2+(-3+15 x) \log (2)+64 x \log (x)\right )}{\log ^4(2) (5 x-\log (x)) \left (-80 x^3-15 x^2 \log (2)+\left (16 x^2+3 x \log (2)\right ) \log (x)\right )} \, dx=\frac {- 65536 x^{4} e^{4} - 49152 x^{3} e^{4} \log {\left (2 \right )} - 13824 x^{2} e^{4} \log {\left (2 \right )}^{2} - 1728 x e^{4} \log {\left (2 \right )}^{3} - 81 e^{4} \log {\left (2 \right )}^{4}}{- 5 x \log {\left (2 \right )}^{4} + \log {\left (2 \right )}^{4} \log {\left (x \right )}} \] Input:
integrate((64*x*ln(x)+(15*x-3)*ln(2)-240*x**2-16*x)*exp(-ln(-ln(x)+5*x)+ln ((81*ln(2)**4+1728*x*ln(2)**3+13824*x**2*ln(2)**2+49152*x**3*ln(2)+65536*x **4)/ln(2)**4)+4)/((3*x*ln(2)+16*x**2)*ln(x)-15*x**2*ln(2)-80*x**3),x)
Output:
(-65536*x**4*exp(4) - 49152*x**3*exp(4)*log(2) - 13824*x**2*exp(4)*log(2)* *2 - 1728*x*exp(4)*log(2)**3 - 81*exp(4)*log(2)**4)/(-5*x*log(2)**4 + log( 2)**4*log(x))
Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.08 \[ \int \frac {e^4 \left (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)\right ) \left (-16 x-240 x^2+(-3+15 x) \log (2)+64 x \log (x)\right )}{\log ^4(2) (5 x-\log (x)) \left (-80 x^3-15 x^2 \log (2)+\left (16 x^2+3 x \log (2)\right ) \log (x)\right )} \, dx=\frac {{\left (65536 \, x^{4} + 49152 \, x^{3} \log \left (2\right ) + 13824 \, x^{2} \log \left (2\right )^{2} + 1728 \, x \log \left (2\right )^{3} + 81 \, \log \left (2\right )^{4}\right )} e^{4}}{{\left (5 \, x - \log \left (x\right )\right )} \log \left (2\right )^{4}} \] Input:
integrate((64*x*log(x)+(15*x-3)*log(2)-240*x^2-16*x)*exp(-log(-log(x)+5*x) +log((81*log(2)^4+1728*x*log(2)^3+13824*x^2*log(2)^2+49152*x^3*log(2)+6553 6*x^4)/log(2)^4)+4)/((3*x*log(2)+16*x^2)*log(x)-15*x^2*log(2)-80*x^3),x, a lgorithm="maxima")
Output:
(65536*x^4 + 49152*x^3*log(2) + 13824*x^2*log(2)^2 + 1728*x*log(2)^3 + 81* log(2)^4)*e^4/((5*x - log(x))*log(2)^4)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (26) = 52\).
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.56 \[ \int \frac {e^4 \left (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)\right ) \left (-16 x-240 x^2+(-3+15 x) \log (2)+64 x \log (x)\right )}{\log ^4(2) (5 x-\log (x)) \left (-80 x^3-15 x^2 \log (2)+\left (16 x^2+3 x \log (2)\right ) \log (x)\right )} \, dx=\frac {65536 \, x^{4} e^{4} + 49152 \, x^{3} e^{4} \log \left (2\right ) + 13824 \, x^{2} e^{4} \log \left (2\right )^{2} + 1728 \, x e^{4} \log \left (2\right )^{3} + 81 \, e^{4} \log \left (2\right )^{4}}{5 \, x \log \left (2\right )^{4} - \log \left (2\right )^{4} \log \left (x\right )} \] Input:
integrate((64*x*log(x)+(15*x-3)*log(2)-240*x^2-16*x)*exp(-log(-log(x)+5*x) +log((81*log(2)^4+1728*x*log(2)^3+13824*x^2*log(2)^2+49152*x^3*log(2)+6553 6*x^4)/log(2)^4)+4)/((3*x*log(2)+16*x^2)*log(x)-15*x^2*log(2)-80*x^3),x, a lgorithm="giac")
Output:
(65536*x^4*e^4 + 49152*x^3*e^4*log(2) + 13824*x^2*e^4*log(2)^2 + 1728*x*e^ 4*log(2)^3 + 81*e^4*log(2)^4)/(5*x*log(2)^4 - log(2)^4*log(x))
Time = 2.60 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^4 \left (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)\right ) \left (-16 x-240 x^2+(-3+15 x) \log (2)+64 x \log (x)\right )}{\log ^4(2) (5 x-\log (x)) \left (-80 x^3-15 x^2 \log (2)+\left (16 x^2+3 x \log (2)\right ) \log (x)\right )} \, dx=\frac {{\mathrm {e}}^4\,{\left (16\,x+\ln \left (8\right )\right )}^4}{{\ln \left (2\right )}^4\,\left (5\,x-\ln \left (x\right )\right )} \] Input:
int((exp(log((13824*x^2*log(2)^2 + 1728*x*log(2)^3 + 49152*x^3*log(2) + 81 *log(2)^4 + 65536*x^4)/log(2)^4) - log(5*x - log(x)) + 4)*(16*x - log(2)*( 15*x - 3) - 64*x*log(x) + 240*x^2))/(15*x^2*log(2) - log(x)*(3*x*log(2) + 16*x^2) + 80*x^3),x)
Output:
(exp(4)*(16*x + log(8))^4)/(log(2)^4*(5*x - log(x)))
Time = 0.21 (sec) , antiderivative size = 53, normalized size of antiderivative = 2.12 \[ \int \frac {e^4 \left (65536 x^4+49152 x^3 \log (2)+13824 x^2 \log ^2(2)+1728 x \log ^3(2)+81 \log ^4(2)\right ) \left (-16 x-240 x^2+(-3+15 x) \log (2)+64 x \log (x)\right )}{\log ^4(2) (5 x-\log (x)) \left (-80 x^3-15 x^2 \log (2)+\left (16 x^2+3 x \log (2)\right ) \log (x)\right )} \, dx=\frac {e^{4} \left (-1728 \,\mathrm {log}\left (x \right ) \mathrm {log}\left (2\right )^{3}-405 \mathrm {log}\left (2\right )^{4}-69120 \mathrm {log}\left (2\right )^{2} x^{2}-245760 \,\mathrm {log}\left (2\right ) x^{3}-327680 x^{4}\right )}{5 \mathrm {log}\left (2\right )^{4} \left (\mathrm {log}\left (x \right )-5 x \right )} \] Input:
int((64*x*log(x)+(15*x-3)*log(2)-240*x^2-16*x)*exp(-log(-log(x)+5*x)+log(( 81*log(2)^4+1728*x*log(2)^3+13824*x^2*log(2)^2+49152*x^3*log(2)+65536*x^4) /log(2)^4)+4)/((3*x*log(2)+16*x^2)*log(x)-15*x^2*log(2)-80*x^3),x)
Output:
(e**4*( - 1728*log(x)*log(2)**3 - 405*log(2)**4 - 69120*log(2)**2*x**2 - 2 45760*log(2)*x**3 - 327680*x**4))/(5*log(2)**4*(log(x) - 5*x))