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

Optimal. Leaf size=25 \[ \log \left (4 x^2 (1+x)^2 \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right )\right ) \]

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Rubi [A]  time = 1.96, antiderivative size = 26, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 4, integrand size = 124, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6688, 6742, 72, 6684} \begin {gather*} 2 \log (x)+2 \log (x+1)+\log \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 6684

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

Rule 6688

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 26, normalized size = 1.04 \begin {gather*} 2 \log (x)+2 \log (1+x)+\log \left (x-\log \left (x-\frac {1}{4} \log (\log (2))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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]]

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fricas [A]  time = 0.83, size = 22, normalized size = 0.88 \begin {gather*} 2 \, \log \left (x^{2} + x\right ) + \log \left (-x + \log \left (x - \frac {1}{4} \, \log \left (\log \relax (2)\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))))

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giac [A]  time = 0.23, size = 30, normalized size = 1.20 \begin {gather*} 2 \, \log \left (x + 1\right ) + 2 \, \log \relax (x) + \log \left (-x - 2 \, \log \relax (2) + \log \left (4 \, x - \log \left (\log \relax (2)\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))))

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maple [A]  time = 0.18, size = 23, normalized size = 0.92

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.54, size = 30, normalized size = 1.20 \begin {gather*} 2 \, \log \left (x + 1\right ) + 2 \, \log \relax (x) + \log \left (-x - 2 \, \log \relax (2) + \log \left (4 \, x - \log \left (\log \relax (2)\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))))

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mupad [B]  time = 4.93, size = 22, normalized size = 0.88 \begin {gather*} 2\,\ln \left (x\,\left (x+1\right )\right )+\ln \left (\ln \left (x-\frac {\ln \left (\ln \relax (2)\right )}{4}\right )-x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.27, size = 20, normalized size = 0.80 \begin {gather*} \log {\left (- x + \log {\left (x - \frac {\log {\left (\log {\relax (2 )} \right )}}{4} \right )} \right )} + 2 \log {\left (x^{2} + x \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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