\(\int \frac {-20+6 x+6 \log (3)}{(-20 x+4 x^2+6 x \log (3)) \log (\frac {1}{2} (10 x^2-2 x^3-3 x^2 \log (3)))+(-10 x+2 x^2+3 x \log (3)) \log (\frac {1}{2} (10 x^2-2 x^3-3 x^2 \log (3))) \log (\log (\frac {1}{2} (10 x^2-2 x^3-3 x^2 \log (3))))} \, dx\) [195]

Optimal result

Integrand size = 113, antiderivative size = 20 \[ \int \frac {-20+6 x+6 \log (3)}{\left (-20 x+4 x^2+6 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )+\left (-10 x+2 x^2+3 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right ) \log \left (\log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )\right )} \, dx=\log \left (2+\log \left (\log \left (x^2 \left (5-x-\frac {3 \log (3)}{2}\right )\right )\right )\right ) \]

[Out]

ln(ln(ln(x^2*(5-3/2*ln(3)-x)))+2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6820, 6816} \[ \int \frac {-20+6 x+6 \log (3)}{\left (-20 x+4 x^2+6 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )+\left (-10 x+2 x^2+3 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right ) \log \left (\log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )\right )} \, dx=\log \left (\log \left (\log \left (\frac {1}{2} x^2 (-2 x+10-\log (27))\right )\right )+2\right ) \]

[In]

Int[(-20 + 6*x + 6*Log[3])/((-20*x + 4*x^2 + 6*x*Log[3])*Log[(10*x^2 - 2*x^3 - 3*x^2*Log[3])/2] + (-10*x + 2*x
^2 + 3*x*Log[3])*Log[(10*x^2 - 2*x^3 - 3*x^2*Log[3])/2]*Log[Log[(10*x^2 - 2*x^3 - 3*x^2*Log[3])/2]]),x]

[Out]

Log[2 + Log[Log[(x^2*(10 - 2*x - Log[27]))/2]]]

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 {20-6 x-\log (729)}{x (10-2 x-\log (27)) \log \left (-\frac {1}{2} x^2 (-10+2 x+\log (27))\right ) \left (2+\log \left (\log \left (-\frac {1}{2} x^2 (-10+2 x+\log (27))\right )\right )\right )} \, dx \\ & = \log \left (2+\log \left (\log \left (\frac {1}{2} x^2 (10-2 x-\log (27))\right )\right )\right ) \\ \end{align*}

Mathematica [F]

\[ \int \frac {-20+6 x+6 \log (3)}{\left (-20 x+4 x^2+6 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )+\left (-10 x+2 x^2+3 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right ) \log \left (\log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )\right )} \, dx=\int \frac {-20+6 x+6 \log (3)}{\left (-20 x+4 x^2+6 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )+\left (-10 x+2 x^2+3 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right ) \log \left (\log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )\right )} \, dx \]

[In]

Integrate[(-20 + 6*x + 6*Log[3])/((-20*x + 4*x^2 + 6*x*Log[3])*Log[(10*x^2 - 2*x^3 - 3*x^2*Log[3])/2] + (-10*x
+ 2*x^2 + 3*x*Log[3])*Log[(10*x^2 - 2*x^3 - 3*x^2*Log[3])/2]*Log[Log[(10*x^2 - 2*x^3 - 3*x^2*Log[3])/2]]),x]

[Out]

Integrate[(-20 + 6*x + 6*Log[3])/((-20*x + 4*x^2 + 6*x*Log[3])*Log[(10*x^2 - 2*x^3 - 3*x^2*Log[3])/2] + (-10*x
+ 2*x^2 + 3*x*Log[3])*Log[(10*x^2 - 2*x^3 - 3*x^2*Log[3])/2]*Log[Log[(10*x^2 - 2*x^3 - 3*x^2*Log[3])/2]]), x]

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00

method	result	size
parallelrisch	\(\ln \left (\ln \left (\ln \left (-\frac {x^{2} \left (3 \ln \left (3\right )+2 x -10\right )}{2}\right )\right )+2\right )\)	\(20\)
norman	\(\ln \left (\ln \left (\ln \left (-\frac {3 x^{2} \ln \left (3\right )}{2}-x^{3}+5 x^{2}\right )\right )+2\right )\)	\(24\)
default	\(\ln \left (\ln \left (-\ln \left (2\right )+\ln \left (-3 x^{2} \ln \left (3\right )-2 x^{3}+10 x^{2}\right )\right )+2\right )\)	\(29\)

[In]

int((6*ln(3)+6*x-20)/((3*x*ln(3)+2*x^2-10*x)*ln(-3/2*x^2*ln(3)-x^3+5*x^2)*ln(ln(-3/2*x^2*ln(3)-x^3+5*x^2))+(6*
x*ln(3)+4*x^2-20*x)*ln(-3/2*x^2*ln(3)-x^3+5*x^2)),x,method=_RETURNVERBOSE)

[Out]

ln(ln(ln(-1/2*x^2*(3*ln(3)+2*x-10)))+2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-20+6 x+6 \log (3)}{\left (-20 x+4 x^2+6 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )+\left (-10 x+2 x^2+3 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right ) \log \left (\log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )\right )} \, dx=\log \left (\log \left (\log \left (-x^{3} - \frac {3}{2} \, x^{2} \log \left (3\right ) + 5 \, x^{2}\right )\right ) + 2\right ) \]

[In]

integrate((6*log(3)+6*x-20)/((3*x*log(3)+2*x^2-10*x)*log(-3/2*x^2*log(3)-x^3+5*x^2)*log(log(-3/2*x^2*log(3)-x^
3+5*x^2))+(6*x*log(3)+4*x^2-20*x)*log(-3/2*x^2*log(3)-x^3+5*x^2)),x, algorithm="fricas")

[Out]

log(log(log(-x^3 - 3/2*x^2*log(3) + 5*x^2)) + 2)

Sympy [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.20 \[ \int \frac {-20+6 x+6 \log (3)}{\left (-20 x+4 x^2+6 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )+\left (-10 x+2 x^2+3 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right ) \log \left (\log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )\right )} \, dx=\log {\left (\log {\left (\log {\left (- x^{3} - \frac {3 x^{2} \log {\left (3 \right )}}{2} + 5 x^{2} \right )} \right )} + 2 \right )} \]

[In]

integrate((6*ln(3)+6*x-20)/((3*x*ln(3)+2*x**2-10*x)*ln(-3/2*x**2*ln(3)-x**3+5*x**2)*ln(ln(-3/2*x**2*ln(3)-x**3
+5*x**2))+(6*x*ln(3)+4*x**2-20*x)*ln(-3/2*x**2*ln(3)-x**3+5*x**2)),x)

[Out]

log(log(log(-x**3 - 3*x**2*log(3)/2 + 5*x**2)) + 2)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-20+6 x+6 \log (3)}{\left (-20 x+4 x^2+6 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )+\left (-10 x+2 x^2+3 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right ) \log \left (\log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )\right )} \, dx=\log \left (\log \left (-\log \left (2\right ) + 2 \, \log \left (x\right ) + \log \left (-2 \, x - 3 \, \log \left (3\right ) + 10\right )\right ) + 2\right ) \]

[In]

integrate((6*log(3)+6*x-20)/((3*x*log(3)+2*x^2-10*x)*log(-3/2*x^2*log(3)-x^3+5*x^2)*log(log(-3/2*x^2*log(3)-x^
3+5*x^2))+(6*x*log(3)+4*x^2-20*x)*log(-3/2*x^2*log(3)-x^3+5*x^2)),x, algorithm="maxima")

[Out]

log(log(-log(2) + 2*log(x) + log(-2*x - 3*log(3) + 10)) + 2)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40 \[ \int \frac {-20+6 x+6 \log (3)}{\left (-20 x+4 x^2+6 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )+\left (-10 x+2 x^2+3 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right ) \log \left (\log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )\right )} \, dx=\log \left (\log \left (-\log \left (2\right ) + \log \left (-2 \, x^{3} - 3 \, x^{2} \log \left (3\right ) + 10 \, x^{2}\right )\right ) + 2\right ) \]

[In]

integrate((6*log(3)+6*x-20)/((3*x*log(3)+2*x^2-10*x)*log(-3/2*x^2*log(3)-x^3+5*x^2)*log(log(-3/2*x^2*log(3)-x^
3+5*x^2))+(6*x*log(3)+4*x^2-20*x)*log(-3/2*x^2*log(3)-x^3+5*x^2)),x, algorithm="giac")

[Out]

log(log(-log(2) + log(-2*x^3 - 3*x^2*log(3) + 10*x^2)) + 2)

Mupad [B] (verification not implemented)

Time = 11.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {-20+6 x+6 \log (3)}{\left (-20 x+4 x^2+6 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )+\left (-10 x+2 x^2+3 x \log (3)\right ) \log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right ) \log \left (\log \left (\frac {1}{2} \left (10 x^2-2 x^3-3 x^2 \log (3)\right )\right )\right )} \, dx=\ln \left (\ln \left (\ln \left (5\,x^2-\frac {3\,x^2\,\ln \left (3\right )}{2}-x^3\right )\right )+2\right ) \]

[In]

int((6*x + 6*log(3) - 20)/(log(5*x^2 - (3*x^2*log(3))/2 - x^3)*(6*x*log(3) - 20*x + 4*x^2) + log(log(5*x^2 - (
3*x^2*log(3))/2 - x^3))*log(5*x^2 - (3*x^2*log(3))/2 - x^3)*(3*x*log(3) - 10*x + 2*x^2)),x)

[Out]

log(log(log(5*x^2 - (3*x^2*log(3))/2 - x^3)) + 2)