\(\int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+(3 x-e^{5/3} x-x^2) \log (3)+e^x (4 x+(-3+e^{5/3}+x) \log (3))+(e^x \log (3)-x \log (3)) \log (-e^x+x)} \, dx\) [6604]

Optimal result

Integrand size = 91, antiderivative size = 32 \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=\log \left (\left (3-e^{5/3}-x-\frac {4 x}{\log (3)}-\log \left (-e^x+x\right )\right )^4\right ) \]

[Out]

ln((3-ln(x-exp(x))-x-exp(5/3)-4*x/ln(3))^4)

Rubi [A] (verified)

Time = 0.26 (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 = {6820, 6816} \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=4 \log \left (x (4+\log (3))+\log (3) \log \left (x-e^x\right )-\left (3-e^{5/3}\right ) \log (3)\right ) \]

[In]

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]

[Out]

4*Log[-((3 - E^(5/3))*Log[3]) + x*(4 + Log[3]) + Log[3]*Log[-E^x + x]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-4 (\log (3)+x (4+\log (3)))+4 e^x (4+\log (9))}{\left (e^x-x\right ) \left (\left (-3+e^{5/3}\right ) \log (3)+x (4+\log (3))+\log (3) \log \left (-e^x+x\right )\right )} \, dx \\ & = 4 \log \left (-\left (\left (3-e^{5/3}\right ) \log (3)\right )+x (4+\log (3))+\log (3) \log \left (-e^x+x\right )\right ) \\ \end{align*}

Mathematica [F]

\[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=\int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx \]

[In]

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]

[Out]

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]

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97

method	result	size
norman	\(4 \ln \left (\ln \left (x -{\mathrm e}^{x}\right ) \ln \left (3\right )+{\mathrm e}^{\frac {5}{3}} \ln \left (3\right )+x \ln \left (3\right )-3 \ln \left (3\right )+4 x \right )\)	\(31\)
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	\(4 \ln \left (\frac {\ln \left (x -{\mathrm e}^{x}\right ) \ln \left (3\right )+{\mathrm e}^{\frac {5}{3}} \ln \left (3\right )+x \ln \left (3\right )-3 \ln \left (3\right )+4 x}{\ln \left (3\right )+4}\right )\)	\(38\)

[In]

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)

[Out]

4*ln(ln(x-exp(x))*ln(3)+exp(5/3)*ln(3)+x*ln(3)-3*ln(3)+4*x)

Fricas [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=4 \, \log \left ({\left (x + e^{\frac {5}{3}} - 3\right )} \log \left (3\right ) + \log \left (3\right ) \log \left (x - e^{x}\right ) + 4 \, x\right ) \]

[In]

integrate(((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, algorithm="fricas")

[Out]

4*log((x + e^(5/3) - 3)*log(3) + log(3)*log(x - e^x) + 4*x)

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=4 \log {\left (\frac {x \log {\left (3 \right )} + 4 x - 3 \log {\left (3 \right )} + e^{\frac {5}{3}} \log {\left (3 \right )}}{\log {\left (3 \right )}} + \log {\left (x - e^{x} \right )} \right )} \]

[In]

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)

[Out]

4*log((x*log(3) + 4*x - 3*log(3) + exp(5/3)*log(3))/log(3) + log(x - exp(x)))

Maxima [F(-2)]

Exception generated. \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(((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, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.94 \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=4 \, \log \left (x \log \left (3\right ) + e^{\frac {5}{3}} \log \left (3\right ) + \log \left (3\right ) \log \left (x - e^{x}\right ) + 4 \, x - 3 \, \log \left (3\right )\right ) \]

[In]

integrate(((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, algorithm="giac")

[Out]

4*log(x*log(3) + e^(5/3)*log(3) + log(3)*log(x - e^x) + 4*x - 3*log(3))

Mupad [B] (verification not implemented)

Time = 11.90 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {-16 x+(-4-4 x) \log (3)+e^x (16+8 \log (3))}{-4 x^2+\left (3 x-e^{5/3} x-x^2\right ) \log (3)+e^x \left (4 x+\left (-3+e^{5/3}+x\right ) \log (3)\right )+\left (e^x \log (3)-x \log (3)\right ) \log \left (-e^x+x\right )} \, dx=4\,\ln \left (4\,x+x\,\ln \left (3\right )+\ln \left (3\right )\,\ln \left (x-{\mathrm {e}}^x\right )+\ln \left (3\right )\,\left ({\mathrm {e}}^{5/3}-3\right )\right ) \]

[In]

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*l
og(3) - exp(x)*log(3)) - exp(x)*(4*x + log(3)*(x + exp(5/3) - 3)) + 4*x^2),x)

[Out]

4*log(4*x + x*log(3) + log(3)*log(x - exp(x)) + log(3)*(exp(5/3) - 3))