Integrand size = 125, antiderivative size = 23 \[ \int \frac {e^{\frac {x}{2+\log (-18-2 x+\log (4)+3 \log (25))}} (-36-2 x+2 \log (4)+6 \log (25)+(-18-2 x+\log (4)+3 \log (25)) \log (-18-2 x+\log (4)+3 \log (25)))}{-72-8 x+4 \log (4)+12 \log (25)+(-72-8 x+4 \log (4)+12 \log (25)) \log (-18-2 x+\log (4)+3 \log (25))+(-18-2 x+\log (4)+3 \log (25)) \log ^2(-18-2 x+\log (4)+3 \log (25))} \, dx=e^{\frac {x}{2+\log (3+x+\log (4)-3 (7+x-\log (25)))}} \] Output:
exp(x/(ln(6*ln(5)+2*ln(2)-2*x-18)+2))
Time = 0.35 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70 \[ \int \frac {e^{\frac {x}{2+\log (-18-2 x+\log (4)+3 \log (25))}} (-36-2 x+2 \log (4)+6 \log (25)+(-18-2 x+\log (4)+3 \log (25)) \log (-18-2 x+\log (4)+3 \log (25)))}{-72-8 x+4 \log (4)+12 \log (25)+(-72-8 x+4 \log (4)+12 \log (25)) \log (-18-2 x+\log (4)+3 \log (25))+(-18-2 x+\log (4)+3 \log (25)) \log ^2(-18-2 x+\log (4)+3 \log (25))} \, dx=e^{\frac {x}{2+\log (-18-2 x+\log (62500))}} \] Input:
Integrate[(E^(x/(2 + Log[-18 - 2*x + Log[4] + 3*Log[25]]))*(-36 - 2*x + 2* Log[4] + 6*Log[25] + (-18 - 2*x + Log[4] + 3*Log[25])*Log[-18 - 2*x + Log[ 4] + 3*Log[25]]))/(-72 - 8*x + 4*Log[4] + 12*Log[25] + (-72 - 8*x + 4*Log[ 4] + 12*Log[25])*Log[-18 - 2*x + Log[4] + 3*Log[25]] + (-18 - 2*x + Log[4] + 3*Log[25])*Log[-18 - 2*x + Log[4] + 3*Log[25]]^2),x]
Output:
E^(x/(2 + Log[-18 - 2*x + Log[62500]]))
Time = 0.96 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.70, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {7239, 7257}
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.
\(\displaystyle \int \frac {e^{\frac {x}{\log (-2 x-18+3 \log (25)+\log (4))+2}} (-2 x+(-2 x-18+3 \log (25)+\log (4)) \log (-2 x-18+3 \log (25)+\log (4))-36+6 \log (25)+2 \log (4))}{-8 x+(-2 x-18+3 \log (25)+\log (4)) \log ^2(-2 x-18+3 \log (25)+\log (4))+(-8 x-72+12 \log (25)+4 \log (4)) \log (-2 x-18+3 \log (25)+\log (4))-72+12 \log (25)+4 \log (4)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {e^{\frac {x}{\log (-2 x-18+\log (62500))+2}} (2 (x+18-\log (62500))+(2 x+18-\log (62500)) \log (-2 x-18+\log (62500)))}{(2 x+18-\log (62500)) (\log (-2 x-18+\log (62500))+2)^2}dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle e^{\frac {x}{\log (-2 x-18+\log (62500))+2}}\) |
Input:
Int[(E^(x/(2 + Log[-18 - 2*x + Log[4] + 3*Log[25]]))*(-36 - 2*x + 2*Log[4] + 6*Log[25] + (-18 - 2*x + Log[4] + 3*Log[25])*Log[-18 - 2*x + Log[4] + 3 *Log[25]]))/(-72 - 8*x + 4*Log[4] + 12*Log[25] + (-72 - 8*x + 4*Log[4] + 1 2*Log[25])*Log[-18 - 2*x + Log[4] + 3*Log[25]] + (-18 - 2*x + Log[4] + 3*L og[25])*Log[-18 - 2*x + Log[4] + 3*Log[25]]^2),x]
Output:
E^(x/(2 + Log[-18 - 2*x + Log[62500]]))
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 1.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
risch | \({\mathrm e}^{\frac {x}{\ln \left (6 \ln \left (5\right )+2 \ln \left (2\right )-2 x -18\right )+2}}\) | \(22\) |
parallelrisch | \({\mathrm e}^{\frac {x}{\ln \left (6 \ln \left (5\right )+2 \ln \left (2\right )-2 x -18\right )+2}}\) | \(22\) |
norman | \(\frac {\ln \left (6 \ln \left (5\right )+2 \ln \left (2\right )-2 x -18\right ) {\mathrm e}^{\frac {x}{\ln \left (6 \ln \left (5\right )+2 \ln \left (2\right )-2 x -18\right )+2}}+2 \,{\mathrm e}^{\frac {x}{\ln \left (6 \ln \left (5\right )+2 \ln \left (2\right )-2 x -18\right )+2}}}{\ln \left (6 \ln \left (5\right )+2 \ln \left (2\right )-2 x -18\right )+2}\) | \(80\) |
Input:
int(((6*ln(5)+2*ln(2)-2*x-18)*ln(6*ln(5)+2*ln(2)-2*x-18)+12*ln(5)+4*ln(2)- 2*x-36)*exp(x/(ln(6*ln(5)+2*ln(2)-2*x-18)+2))/((6*ln(5)+2*ln(2)-2*x-18)*ln (6*ln(5)+2*ln(2)-2*x-18)^2+(24*ln(5)+8*ln(2)-8*x-72)*ln(6*ln(5)+2*ln(2)-2* x-18)+24*ln(5)+8*ln(2)-8*x-72),x,method=_RETURNVERBOSE)
Output:
exp(x/(ln(6*ln(5)+2*ln(2)-2*x-18)+2))
Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {x}{2+\log (-18-2 x+\log (4)+3 \log (25))}} (-36-2 x+2 \log (4)+6 \log (25)+(-18-2 x+\log (4)+3 \log (25)) \log (-18-2 x+\log (4)+3 \log (25)))}{-72-8 x+4 \log (4)+12 \log (25)+(-72-8 x+4 \log (4)+12 \log (25)) \log (-18-2 x+\log (4)+3 \log (25))+(-18-2 x+\log (4)+3 \log (25)) \log ^2(-18-2 x+\log (4)+3 \log (25))} \, dx=e^{\left (\frac {x}{\log \left (-2 \, x + 6 \, \log \left (5\right ) + 2 \, \log \left (2\right ) - 18\right ) + 2}\right )} \] Input:
integrate(((6*log(5)+2*log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)+12*log (5)+4*log(2)-2*x-36)*exp(x/(log(6*log(5)+2*log(2)-2*x-18)+2))/((6*log(5)+2 *log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)^2+(24*log(5)+8*log(2)-8*x-72 )*log(6*log(5)+2*log(2)-2*x-18)+24*log(5)+8*log(2)-8*x-72),x, algorithm="f ricas")
Output:
e^(x/(log(-2*x + 6*log(5) + 2*log(2) - 18) + 2))
Exception generated. \[ \int \frac {e^{\frac {x}{2+\log (-18-2 x+\log (4)+3 \log (25))}} (-36-2 x+2 \log (4)+6 \log (25)+(-18-2 x+\log (4)+3 \log (25)) \log (-18-2 x+\log (4)+3 \log (25)))}{-72-8 x+4 \log (4)+12 \log (25)+(-72-8 x+4 \log (4)+12 \log (25)) \log (-18-2 x+\log (4)+3 \log (25))+(-18-2 x+\log (4)+3 \log (25)) \log ^2(-18-2 x+\log (4)+3 \log (25))} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(((6*ln(5)+2*ln(2)-2*x-18)*ln(6*ln(5)+2*ln(2)-2*x-18)+12*ln(5)+4* ln(2)-2*x-36)*exp(x/(ln(6*ln(5)+2*ln(2)-2*x-18)+2))/((6*ln(5)+2*ln(2)-2*x- 18)*ln(6*ln(5)+2*ln(2)-2*x-18)**2+(24*ln(5)+8*ln(2)-8*x-72)*ln(6*ln(5)+2*l n(2)-2*x-18)+24*ln(5)+8*ln(2)-8*x-72),x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Exception generated. \[ \int \frac {e^{\frac {x}{2+\log (-18-2 x+\log (4)+3 \log (25))}} (-36-2 x+2 \log (4)+6 \log (25)+(-18-2 x+\log (4)+3 \log (25)) \log (-18-2 x+\log (4)+3 \log (25)))}{-72-8 x+4 \log (4)+12 \log (25)+(-72-8 x+4 \log (4)+12 \log (25)) \log (-18-2 x+\log (4)+3 \log (25))+(-18-2 x+\log (4)+3 \log (25)) \log ^2(-18-2 x+\log (4)+3 \log (25))} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(((6*log(5)+2*log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)+12*log (5)+4*log(2)-2*x-36)*exp(x/(log(6*log(5)+2*log(2)-2*x-18)+2))/((6*log(5)+2 *log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)^2+(24*log(5)+8*log(2)-8*x-72 )*log(6*log(5)+2*log(2)-2*x-18)+24*log(5)+8*log(2)-8*x-72),x, algorithm="m axima")
Output:
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not of the expected type LIST
Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {x}{2+\log (-18-2 x+\log (4)+3 \log (25))}} (-36-2 x+2 \log (4)+6 \log (25)+(-18-2 x+\log (4)+3 \log (25)) \log (-18-2 x+\log (4)+3 \log (25)))}{-72-8 x+4 \log (4)+12 \log (25)+(-72-8 x+4 \log (4)+12 \log (25)) \log (-18-2 x+\log (4)+3 \log (25))+(-18-2 x+\log (4)+3 \log (25)) \log ^2(-18-2 x+\log (4)+3 \log (25))} \, dx=e^{\left (\frac {x}{\log \left (-2 \, x + 6 \, \log \left (5\right ) + 2 \, \log \left (2\right ) - 18\right ) + 2}\right )} \] Input:
integrate(((6*log(5)+2*log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)+12*log (5)+4*log(2)-2*x-36)*exp(x/(log(6*log(5)+2*log(2)-2*x-18)+2))/((6*log(5)+2 *log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)^2+(24*log(5)+8*log(2)-8*x-72 )*log(6*log(5)+2*log(2)-2*x-18)+24*log(5)+8*log(2)-8*x-72),x, algorithm="g iac")
Output:
e^(x/(log(-2*x + 6*log(5) + 2*log(2) - 18) + 2))
Time = 0.46 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {e^{\frac {x}{2+\log (-18-2 x+\log (4)+3 \log (25))}} (-36-2 x+2 \log (4)+6 \log (25)+(-18-2 x+\log (4)+3 \log (25)) \log (-18-2 x+\log (4)+3 \log (25)))}{-72-8 x+4 \log (4)+12 \log (25)+(-72-8 x+4 \log (4)+12 \log (25)) \log (-18-2 x+\log (4)+3 \log (25))+(-18-2 x+\log (4)+3 \log (25)) \log ^2(-18-2 x+\log (4)+3 \log (25))} \, dx={\mathrm {e}}^{\frac {x}{\ln \left (2\,\ln \left (250\right )-2\,x-18\right )+2}} \] Input:
int((exp(x/(log(2*log(2) - 2*x + 6*log(5) - 18) + 2))*(2*x - 4*log(2) - 12 *log(5) + log(2*log(2) - 2*x + 6*log(5) - 18)*(2*x - 2*log(2) - 6*log(5) + 18) + 36))/(8*x - 8*log(2) - 24*log(5) + log(2*log(2) - 2*x + 6*log(5) - 18)*(8*x - 8*log(2) - 24*log(5) + 72) + log(2*log(2) - 2*x + 6*log(5) - 18 )^2*(2*x - 2*log(2) - 6*log(5) + 18) + 72),x)
Output:
exp(x/(log(2*log(250) - 2*x - 18) + 2))
Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {e^{\frac {x}{2+\log (-18-2 x+\log (4)+3 \log (25))}} (-36-2 x+2 \log (4)+6 \log (25)+(-18-2 x+\log (4)+3 \log (25)) \log (-18-2 x+\log (4)+3 \log (25)))}{-72-8 x+4 \log (4)+12 \log (25)+(-72-8 x+4 \log (4)+12 \log (25)) \log (-18-2 x+\log (4)+3 \log (25))+(-18-2 x+\log (4)+3 \log (25)) \log ^2(-18-2 x+\log (4)+3 \log (25))} \, dx=e^{\frac {x}{\mathrm {log}\left (6 \,\mathrm {log}\left (5\right )+2 \,\mathrm {log}\left (2\right )-2 x -18\right )+2}} \] Input:
int(((6*log(5)+2*log(2)-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)+12*log(5)+4* log(2)-2*x-36)*exp(x/(log(6*log(5)+2*log(2)-2*x-18)+2))/((6*log(5)+2*log(2 )-2*x-18)*log(6*log(5)+2*log(2)-2*x-18)^2+(24*log(5)+8*log(2)-8*x-72)*log( 6*log(5)+2*log(2)-2*x-18)+24*log(5)+8*log(2)-8*x-72),x)
Output:
e**(x/(log(6*log(5) + 2*log(2) - 2*x - 18) + 2))