\(\int \frac {-4 x+8 x^2+20 x^3+(-3 x-5 x^2) \log (\log (2))+(-8 x-16 x^2+(2+4 x) \log (\log (2))) \log (\frac {1}{4} (4 x-\log (\log (2))))}{4 x^3+4 x^4+(-x^2-x^3) \log (\log (2))+(-4 x^2-4 x^3+(x+x^2) \log (\log (2))) \log (\frac {1}{4} (4 x-\log (\log (2))))} \, dx\) [7576]

Optimal result

Integrand size = 124, antiderivative size = 25 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=\log \left (4 x^2 (1+x)^2 \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )\right ) \]

[Out]

ln(4*(x-ln(-1/4*ln(ln(2))+x))*x^2*(1+x)^2)

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6820, 6874, 78, 6816} \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=2 \log (x)+2 \log (x+1)+\log \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right ) \]

[In]

Int[(-4*x + 8*x^2 + 20*x^3 + (-3*x - 5*x^2)*Log[Log[2]] + (-8*x - 16*x^2 + (2 + 4*x)*Log[Log[2]])*Log[(4*x - L
og[Log[2]])/4])/(4*x^3 + 4*x^4 + (-x^2 - x^3)*Log[Log[2]] + (-4*x^2 - 4*x^3 + (x + x^2)*Log[Log[2]])*Log[(4*x
- Log[Log[2]])/4]),x]

[Out]

2*Log[x] + 2*Log[1 + x] + Log[x - Log[x - Log[Log[2]]/4]]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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]]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {x \left (-4+20 x^2+x (8-5 \log (\log (2)))-3 \log (\log (2))\right )-2 (1+2 x) (4 x-\log (\log (2))) \log \left (x-\frac {1}{4} \log (\log (2))\right )}{x (1+x) (4 x-\log (\log (2))) \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )} \, dx \\ & = \int \left (\frac {2 (1+2 x)}{x (1+x)}+\frac {-4+4 x-\log (\log (2))}{(4 x-\log (\log (2))) \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )}\right ) \, dx \\ & = 2 \int \frac {1+2 x}{x (1+x)} \, dx+\int \frac {-4+4 x-\log (\log (2))}{(4 x-\log (\log (2))) \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )} \, dx \\ & = \log \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )+2 \int \left (\frac {1}{x}+\frac {1}{1+x}\right ) \, dx \\ & = 2 \log (x)+2 \log (1+x)+\log \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=2 \log (x)+2 \log (1+x)+\log \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right ) \]

[In]

Integrate[(-4*x + 8*x^2 + 20*x^3 + (-3*x - 5*x^2)*Log[Log[2]] + (-8*x - 16*x^2 + (2 + 4*x)*Log[Log[2]])*Log[(4
*x - Log[Log[2]])/4])/(4*x^3 + 4*x^4 + (-x^2 - x^3)*Log[Log[2]] + (-4*x^2 - 4*x^3 + (x + x^2)*Log[Log[2]])*Log
[(4*x - Log[Log[2]])/4]),x]

[Out]

2*Log[x] + 2*Log[1 + x] + Log[x - Log[x - Log[Log[2]]/4]]

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92

method	result	size
risch	\(2 \ln \left (x^{2}+x \right )+\ln \left (-x +\ln \left (-\frac {\ln \left (\ln \left (2\right )\right )}{4}+x \right )\right )\)	\(23\)
norman	\(2 \ln \left (x \right )+2 \ln \left (1+x \right )+\ln \left (x -\ln \left (-\frac {\ln \left (\ln \left (2\right )\right )}{4}+x \right )\right )\)	\(25\)
parallelrisch	\(2 \ln \left (x \right )+2 \ln \left (1+x \right )+\ln \left (x -\ln \left (-\frac {\ln \left (\ln \left (2\right )\right )}{4}+x \right )\right )\)	\(25\)
derivativedivides	\(2 \ln \left (4 x \right )+\ln \left (-4 \ln \left (-\frac {\ln \left (\ln \left (2\right )\right )}{4}+x \right )+4 x \right )+2 \ln \left (4+4 x \right )\)	\(31\)
default	\(2 \ln \left (4 x \right )+\ln \left (-4 \ln \left (-\frac {\ln \left (\ln \left (2\right )\right )}{4}+x \right )+4 x \right )+2 \ln \left (4+4 x \right )\)	\(31\)

[In]

int((((4*x+2)*ln(ln(2))-16*x^2-8*x)*ln(-1/4*ln(ln(2))+x)+(-5*x^2-3*x)*ln(ln(2))+20*x^3+8*x^2-4*x)/(((x^2+x)*ln
(ln(2))-4*x^3-4*x^2)*ln(-1/4*ln(ln(2))+x)+(-x^3-x^2)*ln(ln(2))+4*x^4+4*x^3),x,method=_RETURNVERBOSE)

[Out]

2*ln(x^2+x)+ln(-x+ln(-1/4*ln(ln(2))+x))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=2 \, \log \left (x^{2} + x\right ) + \log \left (-x + \log \left (x - \frac {1}{4} \, \log \left (\log \left (2\right )\right )\right )\right ) \]

[In]

integrate((((4*x+2)*log(log(2))-16*x^2-8*x)*log(-1/4*log(log(2))+x)+(-5*x^2-3*x)*log(log(2))+20*x^3+8*x^2-4*x)
/(((x^2+x)*log(log(2))-4*x^3-4*x^2)*log(-1/4*log(log(2))+x)+(-x^3-x^2)*log(log(2))+4*x^4+4*x^3),x, algorithm="
fricas")

[Out]

2*log(x^2 + x) + log(-x + log(x - 1/4*log(log(2))))

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=\log {\left (- x + \log {\left (x - \frac {\log {\left (\log {\left (2 \right )} \right )}}{4} \right )} \right )} + 2 \log {\left (x^{2} + x \right )} \]

[In]

integrate((((4*x+2)*ln(ln(2))-16*x**2-8*x)*ln(-1/4*ln(ln(2))+x)+(-5*x**2-3*x)*ln(ln(2))+20*x**3+8*x**2-4*x)/((
(x**2+x)*ln(ln(2))-4*x**3-4*x**2)*ln(-1/4*ln(ln(2))+x)+(-x**3-x**2)*ln(ln(2))+4*x**4+4*x**3),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=2 \, \log \left (x + 1\right ) + 2 \, \log \left (x\right ) + \log \left (-x - 2 \, \log \left (2\right ) + \log \left (4 \, x - \log \left (\log \left (2\right )\right )\right )\right ) \]

[In]

integrate((((4*x+2)*log(log(2))-16*x^2-8*x)*log(-1/4*log(log(2))+x)+(-5*x^2-3*x)*log(log(2))+20*x^3+8*x^2-4*x)
/(((x^2+x)*log(log(2))-4*x^3-4*x^2)*log(-1/4*log(log(2))+x)+(-x^3-x^2)*log(log(2))+4*x^4+4*x^3),x, algorithm="
maxima")

[Out]

2*log(x + 1) + 2*log(x) + log(-x - 2*log(2) + log(4*x - log(log(2))))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.20 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=2 \, \log \left (x + 1\right ) + 2 \, \log \left (x\right ) + \log \left (-x - 2 \, \log \left (2\right ) + \log \left (4 \, x - \log \left (\log \left (2\right )\right )\right )\right ) \]

[In]

integrate((((4*x+2)*log(log(2))-16*x^2-8*x)*log(-1/4*log(log(2))+x)+(-5*x^2-3*x)*log(log(2))+20*x^3+8*x^2-4*x)
/(((x^2+x)*log(log(2))-4*x^3-4*x^2)*log(-1/4*log(log(2))+x)+(-x^3-x^2)*log(log(2))+4*x^4+4*x^3),x, algorithm="
giac")

[Out]

2*log(x + 1) + 2*log(x) + log(-x - 2*log(2) + log(4*x - log(log(2))))

Mupad [B] (verification not implemented)

Time = 13.56 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {-4 x+8 x^2+20 x^3+\left (-3 x-5 x^2\right ) \log (\log (2))+\left (-8 x-16 x^2+(2+4 x) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )}{4 x^3+4 x^4+\left (-x^2-x^3\right ) \log (\log (2))+\left (-4 x^2-4 x^3+\left (x+x^2\right ) \log (\log (2))\right ) \log \left (\frac {1}{4} (4 x-\log (\log (2)))\right )} \, dx=2\,\ln \left (x\,\left (x+1\right )\right )+\ln \left (\ln \left (x-\frac {\ln \left (\ln \left (2\right )\right )}{4}\right )-x\right ) \]

[In]

int((4*x + log(log(2))*(3*x + 5*x^2) + log(x - log(log(2))/4)*(8*x - log(log(2))*(4*x + 2) + 16*x^2) - 8*x^2 -
20*x^3)/(log(log(2))*(x^2 + x^3) + log(x - log(log(2))/4)*(4*x^2 + 4*x^3 - log(log(2))*(x + x^2)) - 4*x^3 - 4
*x^4),x)

[Out]

2*log(x*(x + 1)) + log(log(x - log(log(2))/4) - x)