\(\int \frac {-1+4 x-2 x^2+(-6 x+6 x^2+8 x^3-4 x^4) \log (3)+(-1+x+(-3 x+2 x^3) \log (3)) \log (x-x^2+(3 x^2-2 x^4) \log (3))}{(x^2-x^3+(3 x^3-2 x^5) \log (3)+(-x+x^2+(-3 x^2+2 x^4) \log (3)) \log (x-x^2+(3 x^2-2 x^4) \log (3))) \log (x^2-x \log (x-x^2+(3 x^2-2 x^4) \log (3)))} \, dx\) [6738]

Optimal result

Integrand size = 172, antiderivative size = 31 \[ \int \frac {-1+4 x-2 x^2+\left (-6 x+6 x^2+8 x^3-4 x^4\right ) \log (3)+\left (-1+x+\left (-3 x+2 x^3\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )}{\left (x^2-x^3+\left (3 x^3-2 x^5\right ) \log (3)+\left (-x+x^2+\left (-3 x^2+2 x^4\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right ) \log \left (x^2-x \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right )} \, dx=\log \left (\log \left (x \left (x-\log \left (x-x \left (x+x^2 \left (-\frac {3}{x}+2 x\right ) \log (3)\right )\right )\right )\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.006, Rules used = {6816} \[ \int \frac {-1+4 x-2 x^2+\left (-6 x+6 x^2+8 x^3-4 x^4\right ) \log (3)+\left (-1+x+\left (-3 x+2 x^3\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )}{\left (x^2-x^3+\left (3 x^3-2 x^5\right ) \log (3)+\left (-x+x^2+\left (-3 x^2+2 x^4\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right ) \log \left (x^2-x \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right )} \, dx=\log \left (\log \left (x^2-x \log \left (-x^2+\left (3 x^2-2 x^4\right ) \log (3)+x\right )\right )\right ) \]

[In]

Int[(-1 + 4*x - 2*x^2 + (-6*x + 6*x^2 + 8*x^3 - 4*x^4)*Log[3] + (-1 + x + (-3*x + 2*x^3)*Log[3])*Log[x - x^2 +
(3*x^2 - 2*x^4)*Log[3]])/((x^2 - x^3 + (3*x^3 - 2*x^5)*Log[3] + (-x + x^2 + (-3*x^2 + 2*x^4)*Log[3])*Log[x -
x^2 + (3*x^2 - 2*x^4)*Log[3]])*Log[x^2 - x*Log[x - x^2 + (3*x^2 - 2*x^4)*Log[3]]]),x]

[Out]

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

Rule 6816

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

Rubi steps \begin{align*} \text {integral}& = \log \left (\log \left (x^2-x \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right )\right ) \\ \end{align*}

Mathematica [F]

\[ \int \frac {-1+4 x-2 x^2+\left (-6 x+6 x^2+8 x^3-4 x^4\right ) \log (3)+\left (-1+x+\left (-3 x+2 x^3\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )}{\left (x^2-x^3+\left (3 x^3-2 x^5\right ) \log (3)+\left (-x+x^2+\left (-3 x^2+2 x^4\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right ) \log \left (x^2-x \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right )} \, dx=\int \frac {-1+4 x-2 x^2+\left (-6 x+6 x^2+8 x^3-4 x^4\right ) \log (3)+\left (-1+x+\left (-3 x+2 x^3\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )}{\left (x^2-x^3+\left (3 x^3-2 x^5\right ) \log (3)+\left (-x+x^2+\left (-3 x^2+2 x^4\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right ) \log \left (x^2-x \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right )} \, dx \]

[In]

Integrate[(-1 + 4*x - 2*x^2 + (-6*x + 6*x^2 + 8*x^3 - 4*x^4)*Log[3] + (-1 + x + (-3*x + 2*x^3)*Log[3])*Log[x -
x^2 + (3*x^2 - 2*x^4)*Log[3]])/((x^2 - x^3 + (3*x^3 - 2*x^5)*Log[3] + (-x + x^2 + (-3*x^2 + 2*x^4)*Log[3])*Lo
g[x - x^2 + (3*x^2 - 2*x^4)*Log[3]])*Log[x^2 - x*Log[x - x^2 + (3*x^2 - 2*x^4)*Log[3]]]),x]

[Out]

Integrate[(-1 + 4*x - 2*x^2 + (-6*x + 6*x^2 + 8*x^3 - 4*x^4)*Log[3] + (-1 + x + (-3*x + 2*x^3)*Log[3])*Log[x -
x^2 + (3*x^2 - 2*x^4)*Log[3]])/((x^2 - x^3 + (3*x^3 - 2*x^5)*Log[3] + (-x + x^2 + (-3*x^2 + 2*x^4)*Log[3])*Lo
g[x - x^2 + (3*x^2 - 2*x^4)*Log[3]])*Log[x^2 - x*Log[x - x^2 + (3*x^2 - 2*x^4)*Log[3]]]), x]

Maple [A] (verified)

Time = 14.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03

method	result	size
parallelrisch	\(\ln \left (\ln \left (-x \ln \left (\left (-2 x^{4}+3 x^{2}\right ) \ln \left (3\right )-x^{2}+x \right )+x^{2}\right )\right )\)	\(32\)

[In]

int((((2*x^3-3*x)*ln(3)+x-1)*ln((-2*x^4+3*x^2)*ln(3)-x^2+x)+(-4*x^4+8*x^3+6*x^2-6*x)*ln(3)-2*x^2+4*x-1)/(((2*x
^4-3*x^2)*ln(3)+x^2-x)*ln((-2*x^4+3*x^2)*ln(3)-x^2+x)+(-2*x^5+3*x^3)*ln(3)-x^3+x^2)/ln(-x*ln((-2*x^4+3*x^2)*ln
(3)-x^2+x)+x^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {-1+4 x-2 x^2+\left (-6 x+6 x^2+8 x^3-4 x^4\right ) \log (3)+\left (-1+x+\left (-3 x+2 x^3\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )}{\left (x^2-x^3+\left (3 x^3-2 x^5\right ) \log (3)+\left (-x+x^2+\left (-3 x^2+2 x^4\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right ) \log \left (x^2-x \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right )} \, dx=\log \left (\log \left (x^{2} - x \log \left (-x^{2} - {\left (2 \, x^{4} - 3 \, x^{2}\right )} \log \left (3\right ) + x\right )\right )\right ) \]

[In]

integrate((((2*x^3-3*x)*log(3)+x-1)*log((-2*x^4+3*x^2)*log(3)-x^2+x)+(-4*x^4+8*x^3+6*x^2-6*x)*log(3)-2*x^2+4*x
-1)/(((2*x^4-3*x^2)*log(3)+x^2-x)*log((-2*x^4+3*x^2)*log(3)-x^2+x)+(-2*x^5+3*x^3)*log(3)-x^3+x^2)/log(-x*log((
-2*x^4+3*x^2)*log(3)-x^2+x)+x^2),x, algorithm="fricas")

[Out]

log(log(x^2 - x*log(-x^2 - (2*x^4 - 3*x^2)*log(3) + x)))

Sympy [A] (verification not implemented)

Time = 1.56 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {-1+4 x-2 x^2+\left (-6 x+6 x^2+8 x^3-4 x^4\right ) \log (3)+\left (-1+x+\left (-3 x+2 x^3\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )}{\left (x^2-x^3+\left (3 x^3-2 x^5\right ) \log (3)+\left (-x+x^2+\left (-3 x^2+2 x^4\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right ) \log \left (x^2-x \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right )} \, dx=\log {\left (\log {\left (x^{2} - x \log {\left (- x^{2} + x + \left (- 2 x^{4} + 3 x^{2}\right ) \log {\left (3 \right )} \right )} \right )} \right )} \]

[In]

integrate((((2*x**3-3*x)*ln(3)+x-1)*ln((-2*x**4+3*x**2)*ln(3)-x**2+x)+(-4*x**4+8*x**3+6*x**2-6*x)*ln(3)-2*x**2
+4*x-1)/(((2*x**4-3*x**2)*ln(3)+x**2-x)*ln((-2*x**4+3*x**2)*ln(3)-x**2+x)+(-2*x**5+3*x**3)*ln(3)-x**3+x**2)/ln
(-x*ln((-2*x**4+3*x**2)*ln(3)-x**2+x)+x**2),x)

[Out]

log(log(x**2 - x*log(-x**2 + x + (-2*x**4 + 3*x**2)*log(3))))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-1+4 x-2 x^2+\left (-6 x+6 x^2+8 x^3-4 x^4\right ) \log (3)+\left (-1+x+\left (-3 x+2 x^3\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )}{\left (x^2-x^3+\left (3 x^3-2 x^5\right ) \log (3)+\left (-x+x^2+\left (-3 x^2+2 x^4\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right ) \log \left (x^2-x \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right )} \, dx=\log \left (\log \left (x - \log \left (-2 \, x^{3} \log \left (3\right ) + x {\left (3 \, \log \left (3\right ) - 1\right )} + 1\right ) - \log \left (x\right )\right ) + \log \left (x\right )\right ) \]

[In]

integrate((((2*x^3-3*x)*log(3)+x-1)*log((-2*x^4+3*x^2)*log(3)-x^2+x)+(-4*x^4+8*x^3+6*x^2-6*x)*log(3)-2*x^2+4*x
-1)/(((2*x^4-3*x^2)*log(3)+x^2-x)*log((-2*x^4+3*x^2)*log(3)-x^2+x)+(-2*x^5+3*x^3)*log(3)-x^3+x^2)/log(-x*log((
-2*x^4+3*x^2)*log(3)-x^2+x)+x^2),x, algorithm="maxima")

[Out]

log(log(x - log(-2*x^3*log(3) + x*(3*log(3) - 1) + 1) - log(x)) + log(x))

Giac [F]

\[ \int \frac {-1+4 x-2 x^2+\left (-6 x+6 x^2+8 x^3-4 x^4\right ) \log (3)+\left (-1+x+\left (-3 x+2 x^3\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )}{\left (x^2-x^3+\left (3 x^3-2 x^5\right ) \log (3)+\left (-x+x^2+\left (-3 x^2+2 x^4\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right ) \log \left (x^2-x \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right )} \, dx=\int { \frac {2 \, x^{2} + 2 \, {\left (2 \, x^{4} - 4 \, x^{3} - 3 \, x^{2} + 3 \, x\right )} \log \left (3\right ) - {\left ({\left (2 \, x^{3} - 3 \, x\right )} \log \left (3\right ) + x - 1\right )} \log \left (-x^{2} - {\left (2 \, x^{4} - 3 \, x^{2}\right )} \log \left (3\right ) + x\right ) - 4 \, x + 1}{{\left (x^{3} - x^{2} + {\left (2 \, x^{5} - 3 \, x^{3}\right )} \log \left (3\right ) - {\left (x^{2} + {\left (2 \, x^{4} - 3 \, x^{2}\right )} \log \left (3\right ) - x\right )} \log \left (-x^{2} - {\left (2 \, x^{4} - 3 \, x^{2}\right )} \log \left (3\right ) + x\right )\right )} \log \left (x^{2} - x \log \left (-x^{2} - {\left (2 \, x^{4} - 3 \, x^{2}\right )} \log \left (3\right ) + x\right )\right )} \,d x } \]

[In]

integrate((((2*x^3-3*x)*log(3)+x-1)*log((-2*x^4+3*x^2)*log(3)-x^2+x)+(-4*x^4+8*x^3+6*x^2-6*x)*log(3)-2*x^2+4*x
-1)/(((2*x^4-3*x^2)*log(3)+x^2-x)*log((-2*x^4+3*x^2)*log(3)-x^2+x)+(-2*x^5+3*x^3)*log(3)-x^3+x^2)/log(-x*log((
-2*x^4+3*x^2)*log(3)-x^2+x)+x^2),x, algorithm="giac")

[Out]

integrate((2*x^2 + 2*(2*x^4 - 4*x^3 - 3*x^2 + 3*x)*log(3) - ((2*x^3 - 3*x)*log(3) + x - 1)*log(-x^2 - (2*x^4 -
3*x^2)*log(3) + x) - 4*x + 1)/((x^3 - x^2 + (2*x^5 - 3*x^3)*log(3) - (x^2 + (2*x^4 - 3*x^2)*log(3) - x)*log(-
x^2 - (2*x^4 - 3*x^2)*log(3) + x))*log(x^2 - x*log(-x^2 - (2*x^4 - 3*x^2)*log(3) + x))), x)

Mupad [B] (verification not implemented)

Time = 19.81 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {-1+4 x-2 x^2+\left (-6 x+6 x^2+8 x^3-4 x^4\right ) \log (3)+\left (-1+x+\left (-3 x+2 x^3\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )}{\left (x^2-x^3+\left (3 x^3-2 x^5\right ) \log (3)+\left (-x+x^2+\left (-3 x^2+2 x^4\right ) \log (3)\right ) \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right ) \log \left (x^2-x \log \left (x-x^2+\left (3 x^2-2 x^4\right ) \log (3)\right )\right )} \, dx=\ln \left (\ln \left (x^2-x\,\ln \left (x+\ln \left (3\right )\,\left (3\,x^2-2\,x^4\right )-x^2\right )\right )\right ) \]

[In]

int((log(3)*(6*x - 6*x^2 - 8*x^3 + 4*x^4) - 4*x + 2*x^2 + log(x + log(3)*(3*x^2 - 2*x^4) - x^2)*(log(3)*(3*x -
2*x^3) - x + 1) + 1)/(log(x^2 - x*log(x + log(3)*(3*x^2 - 2*x^4) - x^2))*(log(x + log(3)*(3*x^2 - 2*x^4) - x^
2)*(x + log(3)*(3*x^2 - 2*x^4) - x^2) - log(3)*(3*x^3 - 2*x^5) - x^2 + x^3)),x)

[Out]

log(log(x^2 - x*log(x + log(3)*(3*x^2 - 2*x^4) - x^2)))