Integrand size = 119, antiderivative size = 25 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=1+\log \left (-e^4+e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \] Output:
1+ln(-exp(4)+exp(x/ln(exp(x)-5*x^2)))
Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\log \left (-e^4+e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \] Input:
Integrate[(E^(x/Log[E^x - 5*x^2])*(-(E^x*x) + 10*x^2 + (E^x - 5*x^2)*Log[E ^x - 5*x^2]))/(E^(x/Log[E^x - 5*x^2])*(E^x - 5*x^2)*Log[E^x - 5*x^2]^2 + ( -E^(4 + x) + 5*E^4*x^2)*Log[E^x - 5*x^2]^2),x]
Output:
Log[-E^4 + E^(x/Log[E^x - 5*x^2])]
Time = 1.82 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {7292, 7235}
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.
\(\displaystyle \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )-e^x x\right )}{\left (e^x-5 x^2\right ) e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \log ^2\left (e^x-5 x^2\right )+\left (5 e^4 x^2-e^{x+4}\right ) \log ^2\left (e^x-5 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-10 x^2-\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )+e^x x\right )}{\left (e^x-5 x^2\right ) \left (e^4-e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right ) \log ^2\left (e^x-5 x^2\right )}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle \log \left (e^4-e^{\frac {x}{\log \left (e^x-5 x^2\right )}}\right )\) |
Input:
Int[(E^(x/Log[E^x - 5*x^2])*(-(E^x*x) + 10*x^2 + (E^x - 5*x^2)*Log[E^x - 5 *x^2]))/(E^(x/Log[E^x - 5*x^2])*(E^x - 5*x^2)*Log[E^x - 5*x^2]^2 + (-E^(4 + x) + 5*E^4*x^2)*Log[E^x - 5*x^2]^2),x]
Output:
Log[E^4 - E^(x/Log[E^x - 5*x^2])]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Time = 32.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\ln \left (-{\mathrm e}^{4}+{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{x}-5 x^{2}\right )}}\right )\) | \(21\) |
parallelrisch | \(\ln \left (-{\mathrm e}^{4}+{\mathrm e}^{\frac {x}{\ln \left ({\mathrm e}^{x}-5 x^{2}\right )}}\right )\) | \(21\) |
Input:
int(((exp(x)-5*x^2)*ln(exp(x)-5*x^2)-exp(x)*x+10*x^2)*exp(x/ln(exp(x)-5*x^ 2))/((exp(x)-5*x^2)*ln(exp(x)-5*x^2)^2*exp(x/ln(exp(x)-5*x^2))+(-exp(4)*ex p(x)+5*x^2*exp(4))*ln(exp(x)-5*x^2)^2),x,method=_RETURNVERBOSE)
Output:
ln(-exp(4)+exp(x/ln(exp(x)-5*x^2)))
Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\log \left (-e^{4} + e^{\left (\frac {x}{\log \left (-{\left (5 \, x^{2} e^{4} - e^{\left (x + 4\right )}\right )} e^{\left (-4\right )}\right )}\right )}\right ) \] Input:
integrate(((exp(x)-5*x^2)*log(exp(x)-5*x^2)-exp(x)*x+10*x^2)*exp(x/log(exp (x)-5*x^2))/((exp(x)-5*x^2)*log(exp(x)-5*x^2)^2*exp(x/log(exp(x)-5*x^2))+( -exp(4)*exp(x)+5*x^2*exp(4))*log(exp(x)-5*x^2)^2),x, algorithm="fricas")
Output:
log(-e^4 + e^(x/log(-(5*x^2*e^4 - e^(x + 4))*e^(-4))))
Timed out. \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(((exp(x)-5*x**2)*ln(exp(x)-5*x**2)-exp(x)*x+10*x**2)*exp(x/ln(ex p(x)-5*x**2))/((exp(x)-5*x**2)*ln(exp(x)-5*x**2)**2*exp(x/ln(exp(x)-5*x**2 ))+(-exp(4)*exp(x)+5*x**2*exp(4))*ln(exp(x)-5*x**2)**2),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\log \left (-e^{4} + e^{\left (\frac {x}{\log \left (-5 \, x^{2} + e^{x}\right )}\right )}\right ) \] Input:
integrate(((exp(x)-5*x^2)*log(exp(x)-5*x^2)-exp(x)*x+10*x^2)*exp(x/log(exp (x)-5*x^2))/((exp(x)-5*x^2)*log(exp(x)-5*x^2)^2*exp(x/log(exp(x)-5*x^2))+( -exp(4)*exp(x)+5*x^2*exp(4))*log(exp(x)-5*x^2)^2),x, algorithm="maxima")
Output:
log(-e^4 + e^(x/log(-5*x^2 + e^x)))
Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\log \left (e^{4} - e^{\left (\frac {x}{\log \left (-5 \, x^{2} + e^{x}\right )}\right )}\right ) \] Input:
integrate(((exp(x)-5*x^2)*log(exp(x)-5*x^2)-exp(x)*x+10*x^2)*exp(x/log(exp (x)-5*x^2))/((exp(x)-5*x^2)*log(exp(x)-5*x^2)^2*exp(x/log(exp(x)-5*x^2))+( -exp(4)*exp(x)+5*x^2*exp(4))*log(exp(x)-5*x^2)^2),x, algorithm="giac")
Output:
log(e^4 - e^(x/log(-5*x^2 + e^x)))
Time = 1.49 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\ln \left ({\mathrm {e}}^{\frac {x}{\ln \left ({\mathrm {e}}^x-5\,x^2\right )}}-{\mathrm {e}}^4\right ) \] Input:
int((exp(x/log(exp(x) - 5*x^2))*(log(exp(x) - 5*x^2)*(exp(x) - 5*x^2) - x* exp(x) + 10*x^2))/(log(exp(x) - 5*x^2)^2*(5*x^2*exp(4) - exp(4)*exp(x)) + exp(x/log(exp(x) - 5*x^2))*log(exp(x) - 5*x^2)^2*(exp(x) - 5*x^2)),x)
Output:
log(exp(x/log(exp(x) - 5*x^2)) - exp(4))
Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (-e^x x+10 x^2+\left (e^x-5 x^2\right ) \log \left (e^x-5 x^2\right )\right )}{e^{\frac {x}{\log \left (e^x-5 x^2\right )}} \left (e^x-5 x^2\right ) \log ^2\left (e^x-5 x^2\right )+\left (-e^{4+x}+5 e^4 x^2\right ) \log ^2\left (e^x-5 x^2\right )} \, dx=\mathrm {log}\left (e^{\frac {x}{\mathrm {log}\left (e^{x}-5 x^{2}\right )}}-e^{4}\right ) \] Input:
int(((exp(x)-5*x^2)*log(exp(x)-5*x^2)-exp(x)*x+10*x^2)*exp(x/log(exp(x)-5* x^2))/((exp(x)-5*x^2)*log(exp(x)-5*x^2)^2*exp(x/log(exp(x)-5*x^2))+(-exp(4 )*exp(x)+5*x^2*exp(4))*log(exp(x)-5*x^2)^2),x)
Output:
log(e**(x/log(e**x - 5*x**2)) - e**4)