\(\int \frac {1-12 x-24 x^2}{(10 x-12 x^2-12 x^3+x \log (x)) \log (10-12 x-12 x^2+\log (x)) \log (\log (10-12 x-12 x^2+\log (x)))} \, dx\) [116]

Optimal result

Integrand size = 62, antiderivative size = 13 \[ \int \frac {1-12 x-24 x^2}{\left (10 x-12 x^2-12 x^3+x \log (x)\right ) \log \left (10-12 x-12 x^2+\log (x)\right ) \log \left (\log \left (10-12 x-12 x^2+\log (x)\right )\right )} \, dx=\log (\log (\log (10-12 x (1+x)+\log (x)))) \]

[Out]

ln(ln(ln(10-12*(1+x)*x+ln(x))))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6816} \[ \int \frac {1-12 x-24 x^2}{\left (10 x-12 x^2-12 x^3+x \log (x)\right ) \log \left (10-12 x-12 x^2+\log (x)\right ) \log \left (\log \left (10-12 x-12 x^2+\log (x)\right )\right )} \, dx=\log \left (\log \left (\log \left (-12 x^2-12 x+\log (x)+10\right )\right )\right ) \]

[In]

Int[(1 - 12*x - 24*x^2)/((10*x - 12*x^2 - 12*x^3 + x*Log[x])*Log[10 - 12*x - 12*x^2 + Log[x]]*Log[Log[10 - 12*
x - 12*x^2 + Log[x]]]),x]

[Out]

Log[Log[Log[10 - 12*x - 12*x^2 + Log[x]]]]

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 (\log \left (10-12 x-12 x^2+\log (x)\right )\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {1-12 x-24 x^2}{\left (10 x-12 x^2-12 x^3+x \log (x)\right ) \log \left (10-12 x-12 x^2+\log (x)\right ) \log \left (\log \left (10-12 x-12 x^2+\log (x)\right )\right )} \, dx=\log \left (\log \left (\log \left (10-12 x-12 x^2+\log (x)\right )\right )\right ) \]

[In]

Integrate[(1 - 12*x - 24*x^2)/((10*x - 12*x^2 - 12*x^3 + x*Log[x])*Log[10 - 12*x - 12*x^2 + Log[x]]*Log[Log[10
- 12*x - 12*x^2 + Log[x]]]),x]

[Out]

Log[Log[Log[10 - 12*x - 12*x^2 + Log[x]]]]

Maple [A] (verified)

Time = 3.13 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23

method	result	size
default	\(\ln \left (\ln \left (\ln \left (\ln \left (x \right )-12 x^{2}-12 x +10\right )\right )\right )\)	\(16\)
risch	\(\ln \left (\ln \left (\ln \left (\ln \left (x \right )-12 x^{2}-12 x +10\right )\right )\right )\)	\(16\)
parallelrisch	\(\ln \left (\ln \left (\ln \left (\ln \left (x \right )-12 x^{2}-12 x +10\right )\right )\right )\)	\(16\)

[In]

int((-24*x^2-12*x+1)/(x*ln(x)-12*x^3-12*x^2+10*x)/ln(ln(x)-12*x^2-12*x+10)/ln(ln(ln(x)-12*x^2-12*x+10)),x,meth
od=_RETURNVERBOSE)

[Out]

ln(ln(ln(ln(x)-12*x^2-12*x+10)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {1-12 x-24 x^2}{\left (10 x-12 x^2-12 x^3+x \log (x)\right ) \log \left (10-12 x-12 x^2+\log (x)\right ) \log \left (\log \left (10-12 x-12 x^2+\log (x)\right )\right )} \, dx=\log \left (\log \left (\log \left (-12 \, x^{2} - 12 \, x + \log \left (x\right ) + 10\right )\right )\right ) \]

[In]

integrate((-24*x^2-12*x+1)/(x*log(x)-12*x^3-12*x^2+10*x)/log(log(x)-12*x^2-12*x+10)/log(log(log(x)-12*x^2-12*x
+10)),x, algorithm="fricas")

[Out]

log(log(log(-12*x^2 - 12*x + log(x) + 10)))

Sympy [A] (verification not implemented)

Time = 0.84 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {1-12 x-24 x^2}{\left (10 x-12 x^2-12 x^3+x \log (x)\right ) \log \left (10-12 x-12 x^2+\log (x)\right ) \log \left (\log \left (10-12 x-12 x^2+\log (x)\right )\right )} \, dx=\log {\left (\log {\left (\log {\left (- 12 x^{2} - 12 x + \log {\left (x \right )} + 10 \right )} \right )} \right )} \]

[In]

integrate((-24*x**2-12*x+1)/(x*ln(x)-12*x**3-12*x**2+10*x)/ln(ln(x)-12*x**2-12*x+10)/ln(ln(ln(x)-12*x**2-12*x+
10)),x)

[Out]

log(log(log(-12*x**2 - 12*x + log(x) + 10)))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {1-12 x-24 x^2}{\left (10 x-12 x^2-12 x^3+x \log (x)\right ) \log \left (10-12 x-12 x^2+\log (x)\right ) \log \left (\log \left (10-12 x-12 x^2+\log (x)\right )\right )} \, dx=\log \left (\log \left (\log \left (-12 \, x^{2} - 12 \, x + \log \left (x\right ) + 10\right )\right )\right ) \]

[In]

integrate((-24*x^2-12*x+1)/(x*log(x)-12*x^3-12*x^2+10*x)/log(log(x)-12*x^2-12*x+10)/log(log(log(x)-12*x^2-12*x
+10)),x, algorithm="maxima")

[Out]

log(log(log(-12*x^2 - 12*x + log(x) + 10)))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {1-12 x-24 x^2}{\left (10 x-12 x^2-12 x^3+x \log (x)\right ) \log \left (10-12 x-12 x^2+\log (x)\right ) \log \left (\log \left (10-12 x-12 x^2+\log (x)\right )\right )} \, dx=\log \left (\log \left (\log \left (-12 \, x^{2} - 12 \, x + \log \left (x\right ) + 10\right )\right )\right ) \]

[In]

integrate((-24*x^2-12*x+1)/(x*log(x)-12*x^3-12*x^2+10*x)/log(log(x)-12*x^2-12*x+10)/log(log(log(x)-12*x^2-12*x
+10)),x, algorithm="giac")

[Out]

log(log(log(-12*x^2 - 12*x + log(x) + 10)))

Mupad [B] (verification not implemented)

Time = 8.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {1-12 x-24 x^2}{\left (10 x-12 x^2-12 x^3+x \log (x)\right ) \log \left (10-12 x-12 x^2+\log (x)\right ) \log \left (\log \left (10-12 x-12 x^2+\log (x)\right )\right )} \, dx=\ln \left (\ln \left (\ln \left (\ln \left (x\right )-12\,x-12\,x^2+10\right )\right )\right ) \]

[In]

int(-(12*x + 24*x^2 - 1)/(log(log(log(x) - 12*x - 12*x^2 + 10))*log(log(x) - 12*x - 12*x^2 + 10)*(10*x + x*log
(x) - 12*x^2 - 12*x^3)),x)

[Out]

log(log(log(log(x) - 12*x - 12*x^2 + 10)))