Integrand size = 134, antiderivative size = 25 \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=25 x \log \left (\log \left (4 e^{2 e^{4+\frac {\log (5)}{e^5}}}-x\right )\right ) \] Output:
25*x*ln(ln(4*exp(exp(ln(5)/exp(5)+4))^2-x))
Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=25 x \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right ) \] Input:
Integrate[(-25*x + (100*E^(2*E^((4*E^5 + Log[5])/E^5)) - 25*x)*Log[4*E^(2* E^((4*E^5 + Log[5])/E^5)) - x]*Log[Log[4*E^(2*E^((4*E^5 + Log[5])/E^5)) - x]])/((4*E^(2*E^((4*E^5 + Log[5])/E^5)) - x)*Log[4*E^(2*E^((4*E^5 + Log[5] )/E^5)) - x]),x]
Output:
25*x*Log[Log[4*E^(2*5^E^(-5)*E^4) - x]]
Leaf count is larger than twice the leaf count of optimal. \(77\) vs. \(2(25)=50\).
Time = 0.75 (sec) , antiderivative size = 77, normalized size of antiderivative = 3.08, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {7239, 27, 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.
| \(\displaystyle \int \frac {\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )-25 x}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int 25 \left (\log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )-\frac {x}{\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )}\right )dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle 25 \int \left (\log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )-\frac {x}{\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 25 \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4} \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )-\left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right ) \log \left (\log \left (4 e^{2\ 5^{\frac {1}{e^5}} e^4}-x\right )\right )\right )\) |
Input:
Int[(-25*x + (100*E^(2*E^((4*E^5 + Log[5])/E^5)) - 25*x)*Log[4*E^(2*E^((4* E^5 + Log[5])/E^5)) - x]*Log[Log[4*E^(2*E^((4*E^5 + Log[5])/E^5)) - x]])/( (4*E^(2*E^((4*E^5 + Log[5])/E^5)) - x)*Log[4*E^(2*E^((4*E^5 + Log[5])/E^5) ) - x]),x]
Output:
25*(4*E^(2*5^E^(-5)*E^4)*Log[Log[4*E^(2*5^E^(-5)*E^4) - x]] - (4*E^(2*5^E^ (-5)*E^4) - x)*Log[Log[4*E^(2*5^E^(-5)*E^4) - x]])
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 = 0.43 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84
| method | result | size |
| risch | \(25 x \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,5^{{\mathrm e}^{-5}} {\mathrm e}^{4}}-x \right )\right )\) | \(21\) |
| default | \(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \left (5\right )+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\) | \(28\) |
| norman | \(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \left (5\right )+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\) | \(28\) |
| parallelrisch | \(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \left (5\right )+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\) | \(28\) |
| parts | \(25 \ln \left (\ln \left (4 \,{\mathrm e}^{2 \,{\mathrm e}^{\left (\ln \left (5\right )+4 \,{\mathrm e}^{5}\right ) {\mathrm e}^{-5}}}-x \right )\right ) x\) | \(28\) |
Input:
int(((100*exp(exp((ln(5)+4*exp(5))/exp(5)))^2-25*x)*ln(4*exp(exp((ln(5)+4* exp(5))/exp(5)))^2-x)*ln(ln(4*exp(exp((ln(5)+4*exp(5))/exp(5)))^2-x))-25*x )/(4*exp(exp((ln(5)+4*exp(5))/exp(5)))^2-x)/ln(4*exp(exp((ln(5)+4*exp(5))/ exp(5)))^2-x),x,method=_RETURNVERBOSE)
Output:
25*x*ln(ln(4*exp(2*5^exp(-5)*exp(4))-x))
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=25 \, x \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )\right ) \] Input:
integrate(((100*exp(exp((log(5)+4*exp(5))/exp(5)))^2-25*x)*log(4*exp(exp(( log(5)+4*exp(5))/exp(5)))^2-x)*log(log(4*exp(exp((log(5)+4*exp(5))/exp(5)) )^2-x))-25*x)/(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x)/log(4*exp(exp((lo g(5)+4*exp(5))/exp(5)))^2-x),x, algorithm="fricas")
Output:
25*x*log(log(-x + 4*e^(2*e^((4*e^5 + log(5))*e^(-5)))))
Exception generated. \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=\text {Exception raised: CoercionFailed} \] Input:
integrate(((100*exp(exp((ln(5)+4*exp(5))/exp(5)))**2-25*x)*ln(4*exp(exp((l n(5)+4*exp(5))/exp(5)))**2-x)*ln(ln(4*exp(exp((ln(5)+4*exp(5))/exp(5)))**2 -x))-25*x)/(4*exp(exp((ln(5)+4*exp(5))/exp(5)))**2-x)/ln(4*exp(exp((ln(5)+ 4*exp(5))/exp(5)))**2-x),x)
Output:
Exception raised: CoercionFailed >> Cannot convert x - 4*exp(2*5**exp(-5)* exp(4)) of type <class 'sympy.core.add.Add'> to QQ[x,exp(5**exp(-5)*exp(4) )]
\[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=\int { \frac {25 \, {\left ({\left (x - 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )} \log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right ) \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )\right ) + x\right )}}{{\left (x - 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )} \log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )} \,d x } \] Input:
integrate(((100*exp(exp((log(5)+4*exp(5))/exp(5)))^2-25*x)*log(4*exp(exp(( log(5)+4*exp(5))/exp(5)))^2-x)*log(log(4*exp(exp((log(5)+4*exp(5))/exp(5)) )^2-x))-25*x)/(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x)/log(4*exp(exp((lo g(5)+4*exp(5))/exp(5)))^2-x),x, algorithm="maxima")
Output:
100*(log(-x + 4*e^(2*5^e^(-5)*e^4))*log(log(-x + 4*e^(2*5^e^(-5)*e^4))) - log(-x + 4*e^(2*e^(e^(-5)*log(5) + 4)))*log(log(-x + 4*e^(2*5^e^(-5)*e^4)) ) - log(-x + 4*e^(2*5^e^(-5)*e^4)))*e^(2*e^(e^(-5)*log(5) + 4))*log(log(-x + 4*e^(2*e^(e^(-5)*log(5) + 4)))) + 100*e^(2*e^(e^(-5)*log(5) + 4))*log(x - 4*e^(2*5^e^(-5)*e^4)) + 25*integrate((x*log(-x + 4*e^(2*5^e^(-5)*e^4))* log(log(-x + 4*e^(2*5^e^(-5)*e^4))) + x)/((x - 4*e^(2*5^e^(-5)*e^4))*log(- x + 4*e^(2*5^e^(-5)*e^4))), x)
Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=25 \, x \log \left (\log \left (-x + 4 \, e^{\left (2 \, e^{\left ({\left (4 \, e^{5} + \log \left (5\right )\right )} e^{\left (-5\right )}\right )}\right )}\right )\right ) \] Input:
integrate(((100*exp(exp((log(5)+4*exp(5))/exp(5)))^2-25*x)*log(4*exp(exp(( log(5)+4*exp(5))/exp(5)))^2-x)*log(log(4*exp(exp((log(5)+4*exp(5))/exp(5)) )^2-x))-25*x)/(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x)/log(4*exp(exp((lo g(5)+4*exp(5))/exp(5)))^2-x),x, algorithm="giac")
Output:
25*x*log(log(-x + 4*e^(2*e^((4*e^5 + log(5))*e^(-5)))))
Time = 3.49 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=25\,x\,\ln \left (\ln \left (4\,{\mathrm {e}}^{2\,5^{{\mathrm {e}}^{-5}}\,{\mathrm {e}}^4}-x\right )\right ) \] Input:
int((25*x + log(4*exp(2*exp(exp(-5)*(4*exp(5) + log(5)))) - x)*log(log(4*e xp(2*exp(exp(-5)*(4*exp(5) + log(5)))) - x))*(25*x - 100*exp(2*exp(exp(-5) *(4*exp(5) + log(5))))))/(log(4*exp(2*exp(exp(-5)*(4*exp(5) + log(5)))) - x)*(x - 4*exp(2*exp(exp(-5)*(4*exp(5) + log(5)))))),x)
Output:
25*x*log(log(4*exp(2*5^exp(-5)*exp(4)) - x))
Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-25 x+\left (100 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-25 x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (\log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )\right )}{\left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right ) \log \left (4 e^{2 e^{\frac {4 e^5+\log (5)}{e^5}}}-x\right )} \, dx=25 \,\mathrm {log}\left (\mathrm {log}\left (4 e^{2 e^{\frac {\mathrm {log}\left (5\right )}{e^{5}}} e^{4}}-x \right )\right ) x \] Input:
int(((100*exp(exp((log(5)+4*exp(5))/exp(5)))^2-25*x)*log(4*exp(exp((log(5) +4*exp(5))/exp(5)))^2-x)*log(log(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x) )-25*x)/(4*exp(exp((log(5)+4*exp(5))/exp(5)))^2-x)/log(4*exp(exp((log(5)+4 *exp(5))/exp(5)))^2-x),x)
Output:
25*log(log(4*e**(2*e**(log(5)/e**5)*e**4) - x))*x