Integrand size = 91, antiderivative size = 32 \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=\log \left (\left (3-e^{5/3}-x-\frac {4 x}{\log (3)}-\log \left (-e^x+x\right )\right )^4\right ) \] Output:
ln((3-ln(x-exp(x))-x-exp(5/3)-4*x/ln(3))^4)
\[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=\int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx \] Input:
Integrate[(-16*x + (-4 - 4*x)*Log[3] + E^x*(16 + 8*Log[3]))/(-4*x^2 + (3*x - E^(5/3)*x - x^2)*Log[3] + E^x*(4*x + (-3 + E^(5/3) + x)*Log[3]) + (E^x* Log[3] - x*Log[3])*Log[-E^x + x]),x]
Output:
Integrate[(-16*x + (-4 - 4*x)*Log[3] + E^x*(16 + 8*Log[3]))/(-4*x^2 + (3*x - E^(5/3)*x - x^2)*Log[3] + E^x*(4*x + (-3 + E^(5/3) + x)*Log[3]) + (E^x* Log[3] - x*Log[3])*Log[-E^x + x]), x]
Time = 0.69 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, 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.
\(\displaystyle \int \frac {-16 x+(-4 x-4) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (-x^2-e^{5/3} x+3 x\right ) \log (3)+e^x \left (4 x+\left (x+e^{5/3}-3\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (x-e^x\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-16 x+(-4 x-4) \log (3)+e^x (16+8 \log (3))}{\left (e^x-x\right ) \left (4 x \left (1+\frac {\log (3)}{4}\right )+\log (3) \log \left (x-e^x\right )-3 \left (1-\frac {e^{5/3}}{3}\right ) \log (3)\right )}dx\) |
\(\Big \downarrow \) 7235 |
\(\displaystyle 4 \log \left (x (4+\log (3))+\log (3) \log \left (x-e^x\right )-\left (3-e^{5/3}\right ) \log (3)\right )\) |
Input:
Int[(-16*x + (-4 - 4*x)*Log[3] + E^x*(16 + 8*Log[3]))/(-4*x^2 + (3*x - E^( 5/3)*x - x^2)*Log[3] + E^x*(4*x + (-3 + E^(5/3) + x)*Log[3]) + (E^x*Log[3] - x*Log[3])*Log[-E^x + x]),x]
Output:
4*Log[-((3 - E^(5/3))*Log[3]) + x*(4 + Log[3]) + Log[3]*Log[-E^x + x]]
Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*L og[RemoveContent[y, x]], x] /; !FalseQ[q]]
Time = 0.38 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97
method | result | size |
norman | \(4 \ln \left ({\mathrm e}^{\frac {5}{3}} \ln \left (3\right )+\ln \left (x -{\mathrm e}^{x}\right ) \ln \left (3\right )+x \ln \left (3\right )-3 \ln \left (3\right )+4 x \right )\) | \(31\) |
risch | \(4 \ln \left (\ln \left (x -{\mathrm e}^{x}\right )+\frac {{\mathrm e}^{\frac {5}{3}} \ln \left (3\right )+x \ln \left (3\right )-3 \ln \left (3\right )+4 x}{\ln \left (3\right )}\right )\) | \(34\) |
parallelrisch | \(4 \ln \left (\frac {{\mathrm e}^{\frac {5}{3}} \ln \left (3\right )+\ln \left (x -{\mathrm e}^{x}\right ) \ln \left (3\right )+x \ln \left (3\right )-3 \ln \left (3\right )+4 x}{\ln \left (3\right )+4}\right )\) | \(38\) |
Input:
int(((8*ln(3)+16)*exp(x)+(-4-4*x)*ln(3)-16*x)/((ln(3)*exp(x)-x*ln(3))*ln(x -exp(x))+((exp(5/3)+x-3)*ln(3)+4*x)*exp(x)+(-x*exp(5/3)-x^2+3*x)*ln(3)-4*x ^2),x,method=_RETURNVERBOSE)
Output:
4*ln(exp(5/3)*ln(3)+ln(x-exp(x))*ln(3)+x*ln(3)-3*ln(3)+4*x)
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=4 \, \log \left ({\left (x + e^{\frac {5}{3}} - 3\right )} \log \left (3\right ) + \log \left (3\right ) \log \left (x - e^{x}\right ) + 4 \, x\right ) \] Input:
integrate(((8*log(3)+16)*exp(x)+(-4-4*x)*log(3)-16*x)/((log(3)*exp(x)-x*lo g(3))*log(x-exp(x))+((exp(5/3)+x-3)*log(3)+4*x)*exp(x)+(-x*exp(5/3)-x^2+3* x)*log(3)-4*x^2),x, algorithm="fricas")
Output:
4*log((x + e^(5/3) - 3)*log(3) + log(3)*log(x - e^x) + 4*x)
Time = 0.21 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=4 \log {\left (\frac {x \log {\left (3 \right )} + 4 x - 3 \log {\left (3 \right )} + e^{\frac {5}{3}} \log {\left (3 \right )}}{\log {\left (3 \right )}} + \log {\left (x - e^{x} \right )} \right )} \] Input:
integrate(((8*ln(3)+16)*exp(x)+(-4-4*x)*ln(3)-16*x)/((ln(3)*exp(x)-x*ln(3) )*ln(x-exp(x))+((exp(5/3)+x-3)*ln(3)+4*x)*exp(x)+(-x*exp(5/3)-x**2+3*x)*ln (3)-4*x**2),x)
Output:
4*log((x*log(3) + 4*x - 3*log(3) + exp(5/3)*log(3))/log(3) + log(x - exp(x )))
Exception generated. \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(((8*log(3)+16)*exp(x)+(-4-4*x)*log(3)-16*x)/((log(3)*exp(x)-x*lo g(3))*log(x-exp(x))+((exp(5/3)+x-3)*log(3)+4*x)*exp(x)+(-x*exp(5/3)-x^2+3* x)*log(3)-4*x^2),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=4 \, \log \left (x \log \left (3\right ) + e^{\frac {5}{3}} \log \left (3\right ) + \log \left (3\right ) \log \left (x - e^{x}\right ) + 4 \, x - 3 \, \log \left (3\right )\right ) \] Input:
integrate(((8*log(3)+16)*exp(x)+(-4-4*x)*log(3)-16*x)/((log(3)*exp(x)-x*lo g(3))*log(x-exp(x))+((exp(5/3)+x-3)*log(3)+4*x)*exp(x)+(-x*exp(5/3)-x^2+3* x)*log(3)-4*x^2),x, algorithm="giac")
Output:
4*log(x*log(3) + e^(5/3)*log(3) + log(3)*log(x - e^x) + 4*x - 3*log(3))
Time = 3.52 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=4\,\ln \left (4\,x+x\,\ln \left (3\right )+\ln \left (3\right )\,\ln \left (x-{\mathrm {e}}^x\right )+\ln \left (3\right )\,\left ({\mathrm {e}}^{5/3}-3\right )\right ) \] Input:
int((16*x + log(3)*(4*x + 4) - exp(x)*(8*log(3) + 16))/(log(3)*(x*exp(5/3) - 3*x + x^2) + log(x - exp(x))*(x*log(3) - exp(x)*log(3)) - exp(x)*(4*x + log(3)*(x + exp(5/3) - 3)) + 4*x^2),x)
Output:
4*log(4*x + x*log(3) + log(3)*log(x - exp(x)) + log(3)*(exp(5/3) - 3))
Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=4 \,\mathrm {log}\left (e^{\frac {5}{3}} \mathrm {log}\left (3\right )+\mathrm {log}\left (-e^{x}+x \right ) \mathrm {log}\left (3\right )+\mathrm {log}\left (3\right ) x -3 \,\mathrm {log}\left (3\right )+4 x \right ) \] Input:
int(((8*log(3)+16)*exp(x)+(-4-4*x)*log(3)-16*x)/((log(3)*exp(x)-x*log(3))* log(x-exp(x))+((exp(5/3)+x-3)*log(3)+4*x)*exp(x)+(-x*exp(5/3)-x^2+3*x)*log (3)-4*x^2),x)
Output:
4*log(e**(2/3)*log(3)*e + log( - e**x + x)*log(3) + log(3)*x - 3*log(3) + 4*x)