Integrand size = 85, antiderivative size = 20 \[ \int \frac {e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))+e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{-320+240 \log (x)-60 \log ^2(x)+5 \log ^3(x)} \, dx=5+e^{e^x}+e^{\frac {144 x}{5 (-4+\log (x))^2}} \] Output:
5+exp(144/5*x/(ln(x)-4)^2)+exp(exp(x))
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))+e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{-320+240 \log (x)-60 \log ^2(x)+5 \log ^3(x)} \, dx=\frac {1}{5} \left (5 e^{e^x}+5 e^{\frac {144 x}{5 (-4+\log (x))^2}}\right ) \] Input:
Integrate[(E^((144*x)/(80 - 40*Log[x] + 5*Log[x]^2))*(-864 + 144*Log[x]) + E^E^x*(-320*E^x + 240*E^x*Log[x] - 60*E^x*Log[x]^2 + 5*E^x*Log[x]^3))/(-3 20 + 240*Log[x] - 60*Log[x]^2 + 5*Log[x]^3),x]
Output:
(5*E^E^x + 5*E^((144*x)/(5*(-4 + Log[x])^2)))/5
Time = 1.41 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {7292, 27, 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.
\(\displaystyle \int \frac {e^{\frac {144 x}{5 \log ^2(x)-40 \log (x)+80}} (144 \log (x)-864)+e^{e^x} \left (-320 e^x+5 e^x \log ^3(x)-60 e^x \log ^2(x)+240 e^x \log (x)\right )}{5 \log ^3(x)-60 \log ^2(x)+240 \log (x)-320} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-e^{\frac {144 x}{5 \log ^2(x)-40 \log (x)+80}} (144 \log (x)-864)-e^{e^x} \left (-320 e^x+5 e^x \log ^3(x)-60 e^x \log ^2(x)+240 e^x \log (x)\right )}{5 (4-\log (x))^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {144 e^{\frac {144 x}{5 \left (\log ^2(x)-8 \log (x)+16\right )}} (6-\log (x))+5 e^{e^x} \left (-e^x \log ^3(x)+12 e^x \log ^2(x)-48 e^x \log (x)+64 e^x\right )}{(4-\log (x))^3}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \frac {1}{5} \int \frac {-5 e^{x+e^x} (\log (x)-4)^3-144 e^{\frac {144 x}{5 (\log (x)-4)^2}} (\log (x)-6)}{(4-\log (x))^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (\frac {144 e^{\frac {144 x}{5 (\log (x)-4)^2}} (\log (x)-6)}{(\log (x)-4)^3}+5 e^{x+e^x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (5 e^{e^x}+5 e^{\frac {144 x}{5 (4-\log (x))^2}}\right )\) |
Input:
Int[(E^((144*x)/(80 - 40*Log[x] + 5*Log[x]^2))*(-864 + 144*Log[x]) + E^E^x *(-320*E^x + 240*E^x*Log[x] - 60*E^x*Log[x]^2 + 5*E^x*Log[x]^3))/(-320 + 2 40*Log[x] - 60*Log[x]^2 + 5*Log[x]^3),x]
Output:
(5*E^E^x + 5*E^((144*x)/(5*(4 - Log[x])^2)))/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 = 33.83 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75
method | result | size |
risch | \({\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{\frac {144 x}{5 \left (\ln \left (x \right )-4\right )^{2}}}\) | \(15\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{x}}+{\mathrm e}^{\frac {144 x}{5 \left (\ln \left (x \right )^{2}-8 \ln \left (x \right )+16\right )}}\) | \(21\) |
parts | \({\mathrm e}^{{\mathrm e}^{x}}+\frac {\ln \left (x \right )^{2} {\mathrm e}^{\frac {144 x}{5 \ln \left (x \right )^{2}-40 \ln \left (x \right )+80}}-8 \ln \left (x \right ) {\mathrm e}^{\frac {144 x}{5 \ln \left (x \right )^{2}-40 \ln \left (x \right )+80}}+16 \,{\mathrm e}^{\frac {144 x}{5 \ln \left (x \right )^{2}-40 \ln \left (x \right )+80}}}{\left (\ln \left (x \right )-4\right )^{2}}\) | \(78\) |
default | \(\frac {-40 \ln \left (x \right ) {\mathrm e}^{\frac {144 x}{5 \ln \left (x \right )^{2}-40 \ln \left (x \right )+80}}+5 \ln \left (x \right )^{2} {\mathrm e}^{\frac {144 x}{5 \ln \left (x \right )^{2}-40 \ln \left (x \right )+80}}+80 \,{\mathrm e}^{\frac {144 x}{5 \ln \left (x \right )^{2}-40 \ln \left (x \right )+80}}}{5 \left (\ln \left (x \right )-4\right )^{2}}+{\mathrm e}^{{\mathrm e}^{x}}\) | \(80\) |
Input:
int(((5*exp(x)*ln(x)^3-60*exp(x)*ln(x)^2+240*exp(x)*ln(x)-320*exp(x))*exp( exp(x))+(144*ln(x)-864)*exp(144*x/(5*ln(x)^2-40*ln(x)+80)))/(5*ln(x)^3-60* ln(x)^2+240*ln(x)-320),x,method=_RETURNVERBOSE)
Output:
exp(exp(x))+exp(144/5*x/(ln(x)-4)^2)
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))+e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{-320+240 \log (x)-60 \log ^2(x)+5 \log ^3(x)} \, dx=e^{\left (\frac {144 \, x}{5 \, {\left (\log \left (x\right )^{2} - 8 \, \log \left (x\right ) + 16\right )}}\right )} + e^{\left (e^{x}\right )} \] Input:
integrate(((5*exp(x)*log(x)^3-60*exp(x)*log(x)^2+240*exp(x)*log(x)-320*exp (x))*exp(exp(x))+(144*log(x)-864)*exp(144*x/(5*log(x)^2-40*log(x)+80)))/(5 *log(x)^3-60*log(x)^2+240*log(x)-320),x, algorithm="fricas")
Output:
e^(144/5*x/(log(x)^2 - 8*log(x) + 16)) + e^(e^x)
Time = 0.60 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))+e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{-320+240 \log (x)-60 \log ^2(x)+5 \log ^3(x)} \, dx=e^{\frac {144 x}{5 \log {\left (x \right )}^{2} - 40 \log {\left (x \right )} + 80}} + e^{e^{x}} \] Input:
integrate(((5*exp(x)*ln(x)**3-60*exp(x)*ln(x)**2+240*exp(x)*ln(x)-320*exp( x))*exp(exp(x))+(144*ln(x)-864)*exp(144*x/(5*ln(x)**2-40*ln(x)+80)))/(5*ln (x)**3-60*ln(x)**2+240*ln(x)-320),x)
Output:
exp(144*x/(5*log(x)**2 - 40*log(x) + 80)) + exp(exp(x))
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))+e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{-320+240 \log (x)-60 \log ^2(x)+5 \log ^3(x)} \, dx=e^{\left (\frac {144 \, x}{5 \, {\left (\log \left (x\right )^{2} - 8 \, \log \left (x\right ) + 16\right )}}\right )} + e^{\left (e^{x}\right )} \] Input:
integrate(((5*exp(x)*log(x)^3-60*exp(x)*log(x)^2+240*exp(x)*log(x)-320*exp (x))*exp(exp(x))+(144*log(x)-864)*exp(144*x/(5*log(x)^2-40*log(x)+80)))/(5 *log(x)^3-60*log(x)^2+240*log(x)-320),x, algorithm="maxima")
Output:
e^(144/5*x/(log(x)^2 - 8*log(x) + 16)) + e^(e^x)
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))+e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{-320+240 \log (x)-60 \log ^2(x)+5 \log ^3(x)} \, dx=e^{\left (\frac {144 \, x}{5 \, {\left (\log \left (x\right )^{2} - 8 \, \log \left (x\right ) + 16\right )}}\right )} + e^{\left (e^{x}\right )} \] Input:
integrate(((5*exp(x)*log(x)^3-60*exp(x)*log(x)^2+240*exp(x)*log(x)-320*exp (x))*exp(exp(x))+(144*log(x)-864)*exp(144*x/(5*log(x)^2-40*log(x)+80)))/(5 *log(x)^3-60*log(x)^2+240*log(x)-320),x, algorithm="giac")
Output:
e^(144/5*x/(log(x)^2 - 8*log(x) + 16)) + e^(e^x)
Time = 3.75 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))+e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{-320+240 \log (x)-60 \log ^2(x)+5 \log ^3(x)} \, dx={\mathrm {e}}^{{\mathrm {e}}^x}+{\mathrm {e}}^{\frac {144\,x}{5\,{\ln \left (x\right )}^2-40\,\ln \left (x\right )+80}} \] Input:
int(-(exp(exp(x))*(320*exp(x) - 240*exp(x)*log(x) + 60*exp(x)*log(x)^2 - 5 *exp(x)*log(x)^3) - exp((144*x)/(5*log(x)^2 - 40*log(x) + 80))*(144*log(x) - 864))/(240*log(x) - 60*log(x)^2 + 5*log(x)^3 - 320),x)
Output:
exp(exp(x)) + exp((144*x)/(5*log(x)^2 - 40*log(x) + 80))
Time = 0.16 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {e^{\frac {144 x}{80-40 \log (x)+5 \log ^2(x)}} (-864+144 \log (x))+e^{e^x} \left (-320 e^x+240 e^x \log (x)-60 e^x \log ^2(x)+5 e^x \log ^3(x)\right )}{-320+240 \log (x)-60 \log ^2(x)+5 \log ^3(x)} \, dx=e^{e^{x}}+e^{\frac {144 x}{5 \mathrm {log}\left (x \right )^{2}-40 \,\mathrm {log}\left (x \right )+80}} \] Input:
int(((5*exp(x)*log(x)^3-60*exp(x)*log(x)^2+240*exp(x)*log(x)-320*exp(x))*e xp(exp(x))+(144*log(x)-864)*exp(144*x/(5*log(x)^2-40*log(x)+80)))/(5*log(x )^3-60*log(x)^2+240*log(x)-320),x)
Output:
e**(e**x) + e**((144*x)/(5*log(x)**2 - 40*log(x) + 80))