\(\int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 (8 x+2 x^2-2 x^3) \log (3)+e^{12} x \log ^2(3)+(-8 x-2 x^2+2 x^3-2 e^6 x \log (3)) \log (x)+x \log ^2(x)} \, dx\) [3795]

Optimal result

Integrand size = 112, antiderivative size = 23 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {-4+x}{-4-x+x^2-e^6 \log (3)+\log (x)} \]

[Out]

(x-4)/(x^2-4+ln(x)-x-exp(3)^2*ln(3))

Rubi [F]

\[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx \]

[In]

Int[(4 - 9*x + 8*x^2 - x^3 - E^6*x*Log[3] + x*Log[x])/(16*x + 8*x^2 - 7*x^3 - 2*x^4 + x^5 + E^6*(8*x + 2*x^2 -
2*x^3)*Log[3] + E^12*x*Log[3]^2 + (-8*x - 2*x^2 + 2*x^3 - 2*E^6*x*Log[3])*Log[x] + x*Log[x]^2),x]

[Out]

-5*Defer[Int][(x - x^2 + 4*(1 + (E^6*Log[3])/4) - Log[x])^(-2), x] + 4*Defer[Int][1/(x*(x - x^2 + 4*(1 + (E^6*
Log[3])/4) - Log[x])^2), x] + 9*Defer[Int][x/(x - x^2 + 4*(1 + (E^6*Log[3])/4) - Log[x])^2, x] - 2*Defer[Int][
x^2/(x - x^2 + 4*(1 + (E^6*Log[3])/4) - Log[x])^2, x] + Defer[Int][(-x + x^2 - 4*(1 + (E^6*Log[3])/4) + Log[x]
)^(-1), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4+8 x^2-x^3+x \left (-9-e^6 \log (3)\right )+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx \\ & = \int \frac {4+8 x^2-x^3+x \left (-9-e^6 \log (3)\right )+x \log (x)}{8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+x \left (16+e^{12} \log ^2(3)\right )+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx \\ & = \int \frac {4+8 x^2-x^3-x \left (9+e^6 \log (3)\right )+x \log (x)}{x \left (x-x^2+4 \left (1+\frac {1}{4} e^6 \log (3)\right )-\log (x)\right )^2} \, dx \\ & = \int \left (\frac {4-5 x+9 x^2-2 x^3}{x \left (x-x^2+4 \left (1+\frac {1}{4} e^6 \log (3)\right )-\log (x)\right )^2}+\frac {1}{-x+x^2-4 \left (1+\frac {1}{4} e^6 \log (3)\right )+\log (x)}\right ) \, dx \\ & = \int \frac {4-5 x+9 x^2-2 x^3}{x \left (x-x^2+4 \left (1+\frac {1}{4} e^6 \log (3)\right )-\log (x)\right )^2} \, dx+\int \frac {1}{-x+x^2-4 \left (1+\frac {1}{4} e^6 \log (3)\right )+\log (x)} \, dx \\ & = \int \left (-\frac {5}{\left (x-x^2+4 \left (1+\frac {1}{4} e^6 \log (3)\right )-\log (x)\right )^2}+\frac {4}{x \left (x-x^2+4 \left (1+\frac {1}{4} e^6 \log (3)\right )-\log (x)\right )^2}+\frac {9 x}{\left (x-x^2+4 \left (1+\frac {1}{4} e^6 \log (3)\right )-\log (x)\right )^2}-\frac {2 x^2}{\left (x-x^2+4 \left (1+\frac {1}{4} e^6 \log (3)\right )-\log (x)\right )^2}\right ) \, dx+\int \frac {1}{-x+x^2-4 \left (1+\frac {1}{4} e^6 \log (3)\right )+\log (x)} \, dx \\ & = -\left (2 \int \frac {x^2}{\left (x-x^2+4 \left (1+\frac {1}{4} e^6 \log (3)\right )-\log (x)\right )^2} \, dx\right )+4 \int \frac {1}{x \left (x-x^2+4 \left (1+\frac {1}{4} e^6 \log (3)\right )-\log (x)\right )^2} \, dx-5 \int \frac {1}{\left (x-x^2+4 \left (1+\frac {1}{4} e^6 \log (3)\right )-\log (x)\right )^2} \, dx+9 \int \frac {x}{\left (x-x^2+4 \left (1+\frac {1}{4} e^6 \log (3)\right )-\log (x)\right )^2} \, dx+\int \frac {1}{-x+x^2-4 \left (1+\frac {1}{4} e^6 \log (3)\right )+\log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.82 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {-4+x}{-4-x+x^2-e^6 \log (3)+\log (x)} \]

[In]

Integrate[(4 - 9*x + 8*x^2 - x^3 - E^6*x*Log[3] + x*Log[x])/(16*x + 8*x^2 - 7*x^3 - 2*x^4 + x^5 + E^6*(8*x + 2
*x^2 - 2*x^3)*Log[3] + E^12*x*Log[3]^2 + (-8*x - 2*x^2 + 2*x^3 - 2*E^6*x*Log[3])*Log[x] + x*Log[x]^2),x]

[Out]

(-4 + x)/(-4 - x + x^2 - E^6*Log[3] + Log[x])

Maple [A] (verified)

Time = 5.10 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09

method	result	size
risch	\(-\frac {x -4}{{\mathrm e}^{6} \ln \left (3\right )-x^{2}+x -\ln \left (x \right )+4}\)	\(25\)
parallelrisch	\(-\frac {x -4}{{\mathrm e}^{6} \ln \left (3\right )-x^{2}+x -\ln \left (x \right )+4}\)	\(27\)
default	\(\frac {-x +4}{{\mathrm e}^{6} \ln \left (3\right )-x^{2}+x -\ln \left (x \right )+4}\)	\(28\)
norman	\(\frac {-x +4}{{\mathrm e}^{6} \ln \left (3\right )-x^{2}+x -\ln \left (x \right )+4}\)	\(28\)

[In]

int((x*ln(x)-x*exp(3)^2*ln(3)-x^3+8*x^2-9*x+4)/(x*ln(x)^2+(-2*x*exp(3)^2*ln(3)+2*x^3-2*x^2-8*x)*ln(x)+x*exp(3)
^4*ln(3)^2+(-2*x^3+2*x^2+8*x)*exp(3)^2*ln(3)+x^5-2*x^4-7*x^3+8*x^2+16*x),x,method=_RETURNVERBOSE)

[Out]

-(x-4)/(exp(6)*ln(3)-x^2+x-ln(x)+4)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {x - 4}{x^{2} - e^{6} \log \left (3\right ) - x + \log \left (x\right ) - 4} \]

[In]

integrate((x*log(x)-x*exp(3)^2*log(3)-x^3+8*x^2-9*x+4)/(x*log(x)^2+(-2*x*exp(3)^2*log(3)+2*x^3-2*x^2-8*x)*log(
x)+x*exp(3)^4*log(3)^2+(-2*x^3+2*x^2+8*x)*exp(3)^2*log(3)+x^5-2*x^4-7*x^3+8*x^2+16*x),x, algorithm="fricas")

[Out]

(x - 4)/(x^2 - e^6*log(3) - x + log(x) - 4)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {x - 4}{x^{2} - x + \log {\left (x \right )} - e^{6} \log {\left (3 \right )} - 4} \]

[In]

integrate((x*ln(x)-x*exp(3)**2*ln(3)-x**3+8*x**2-9*x+4)/(x*ln(x)**2+(-2*x*exp(3)**2*ln(3)+2*x**3-2*x**2-8*x)*l
n(x)+x*exp(3)**4*ln(3)**2+(-2*x**3+2*x**2+8*x)*exp(3)**2*ln(3)+x**5-2*x**4-7*x**3+8*x**2+16*x),x)

[Out]

(x - 4)/(x**2 - x + log(x) - exp(6)*log(3) - 4)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {x - 4}{x^{2} - e^{6} \log \left (3\right ) - x + \log \left (x\right ) - 4} \]

[In]

integrate((x*log(x)-x*exp(3)^2*log(3)-x^3+8*x^2-9*x+4)/(x*log(x)^2+(-2*x*exp(3)^2*log(3)+2*x^3-2*x^2-8*x)*log(
x)+x*exp(3)^4*log(3)^2+(-2*x^3+2*x^2+8*x)*exp(3)^2*log(3)+x^5-2*x^4-7*x^3+8*x^2+16*x),x, algorithm="maxima")

[Out]

(x - 4)/(x^2 - e^6*log(3) - x + log(x) - 4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=\frac {x - 4}{x^{2} - e^{6} \log \left (3\right ) - x + \log \left (x\right ) - 4} \]

[In]

integrate((x*log(x)-x*exp(3)^2*log(3)-x^3+8*x^2-9*x+4)/(x*log(x)^2+(-2*x*exp(3)^2*log(3)+2*x^3-2*x^2-8*x)*log(
x)+x*exp(3)^4*log(3)^2+(-2*x^3+2*x^2+8*x)*exp(3)^2*log(3)+x^5-2*x^4-7*x^3+8*x^2+16*x),x, algorithm="giac")

[Out]

(x - 4)/(x^2 - e^6*log(3) - x + log(x) - 4)

Mupad [B] (verification not implemented)

Time = 9.68 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {4-9 x+8 x^2-x^3-e^6 x \log (3)+x \log (x)}{16 x+8 x^2-7 x^3-2 x^4+x^5+e^6 \left (8 x+2 x^2-2 x^3\right ) \log (3)+e^{12} x \log ^2(3)+\left (-8 x-2 x^2+2 x^3-2 e^6 x \log (3)\right ) \log (x)+x \log ^2(x)} \, dx=-\frac {x-4}{x-\ln \left (x\right )+{\mathrm {e}}^6\,\ln \left (3\right )-x^2+4} \]

[In]

int(-(9*x - x*log(x) - 8*x^2 + x^3 + x*exp(6)*log(3) - 4)/(16*x + x*log(x)^2 - log(x)*(8*x + 2*x^2 - 2*x^3 + 2
*x*exp(6)*log(3)) + 8*x^2 - 7*x^3 - 2*x^4 + x^5 + x*exp(12)*log(3)^2 + exp(6)*log(3)*(8*x + 2*x^2 - 2*x^3)),x)

[Out]

-(x - 4)/(x - log(x) + exp(6)*log(3) - x^2 + 4)