\(\int \frac {-32-32 x+(-128-64 x-32 x^2-32 \log (2)) \log (\frac {5}{16+8 x+4 x^2+4 \log (2)})}{(-8 x-4 x^2-2 x^3-2 x \log (2)) \log (\frac {5}{16+8 x+4 x^2+4 \log (2)})+(4+2 x+x^2+\log (2)) \log (\frac {5}{16+8 x+4 x^2+4 \log (2)}) \log (\log (\frac {5}{16+8 x+4 x^2+4 \log (2)}))} \, dx\) [8401]

Optimal result

Integrand size = 132, antiderivative size = 27 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \log \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right ) \]

[Out]

16*ln(2*x-ln(ln(5/(4*ln(2)+4*x^2+8*x+16))))

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.023, Rules used = {6820, 12, 6816} \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \log \left (2 x-\log \left (\log \left (\frac {5}{4 \left (x^2+2 x+4+\log (2)\right )}\right )\right )\right ) \]

[In]

Int[(-32 - 32*x + (-128 - 64*x - 32*x^2 - 32*Log[2])*Log[5/(16 + 8*x + 4*x^2 + 4*Log[2])])/((-8*x - 4*x^2 - 2*
x^3 - 2*x*Log[2])*Log[5/(16 + 8*x + 4*x^2 + 4*Log[2])] + (4 + 2*x + x^2 + Log[2])*Log[5/(16 + 8*x + 4*x^2 + 4*
Log[2])]*Log[Log[5/(16 + 8*x + 4*x^2 + 4*Log[2])]]),x]

[Out]

16*Log[2*x - Log[Log[5/(4*(4 + 2*x + x^2 + Log[2]))]]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 {32 \left (1+x+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )}{\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right ) \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right )} \, dx \\ & = 32 \int \frac {1+x+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )}{\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right ) \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right )} \, dx \\ & = 16 \log \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \log \left (2 x-\log \left (\log \left (\frac {5}{4 \left (4+2 x+x^2+\log (2)\right )}\right )\right )\right ) \]

[In]

Integrate[(-32 - 32*x + (-128 - 64*x - 32*x^2 - 32*Log[2])*Log[5/(16 + 8*x + 4*x^2 + 4*Log[2])])/((-8*x - 4*x^
2 - 2*x^3 - 2*x*Log[2])*Log[5/(16 + 8*x + 4*x^2 + 4*Log[2])] + (4 + 2*x + x^2 + Log[2])*Log[5/(16 + 8*x + 4*x^
2 + 4*Log[2])]*Log[Log[5/(16 + 8*x + 4*x^2 + 4*Log[2])]]),x]

[Out]

16*Log[2*x - Log[Log[5/(4*(4 + 2*x + x^2 + Log[2]))]]]

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89

method	result	size
parallelrisch	\(16 \ln \left (x -\frac {\ln \left (\ln \left (\frac {5}{4 \left (\ln \left (2\right )+x^{2}+2 x +4\right )}\right )\right )}{2}\right )\)	\(24\)

[In]

int(((-32*ln(2)-32*x^2-64*x-128)*ln(5/(4*ln(2)+4*x^2+8*x+16))-32*x-32)/((ln(2)+x^2+2*x+4)*ln(5/(4*ln(2)+4*x^2+
8*x+16))*ln(ln(5/(4*ln(2)+4*x^2+8*x+16)))+(-2*x*ln(2)-2*x^3-4*x^2-8*x)*ln(5/(4*ln(2)+4*x^2+8*x+16))),x,method=
_RETURNVERBOSE)

[Out]

16*ln(x-1/2*ln(ln(5/4/(ln(2)+x^2+2*x+4))))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \, \log \left (-2 \, x + \log \left (\log \left (\frac {5}{4 \, {\left (x^{2} + 2 \, x + \log \left (2\right ) + 4\right )}}\right )\right )\right ) \]

[In]

integrate(((-32*log(2)-32*x^2-64*x-128)*log(5/(4*log(2)+4*x^2+8*x+16))-32*x-32)/((log(2)+x^2+2*x+4)*log(5/(4*l
og(2)+4*x^2+8*x+16))*log(log(5/(4*log(2)+4*x^2+8*x+16)))+(-2*x*log(2)-2*x^3-4*x^2-8*x)*log(5/(4*log(2)+4*x^2+8
*x+16))),x, algorithm="fricas")

[Out]

16*log(-2*x + log(log(5/4/(x^2 + 2*x + log(2) + 4))))

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \log {\left (- 2 x + \log {\left (\log {\left (\frac {5}{4 x^{2} + 8 x + 4 \log {\left (2 \right )} + 16} \right )} \right )} \right )} \]

[In]

integrate(((-32*ln(2)-32*x**2-64*x-128)*ln(5/(4*ln(2)+4*x**2+8*x+16))-32*x-32)/((ln(2)+x**2+2*x+4)*ln(5/(4*ln(
2)+4*x**2+8*x+16))*ln(ln(5/(4*ln(2)+4*x**2+8*x+16)))+(-2*x*ln(2)-2*x**3-4*x**2-8*x)*ln(5/(4*ln(2)+4*x**2+8*x+1
6))),x)

[Out]

16*log(-2*x + log(log(5/(4*x**2 + 8*x + 4*log(2) + 16))))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \, \log \left (-2 \, x + \log \left (\log \left (5\right ) - 2 \, \log \left (2\right ) - \log \left (x^{2} + 2 \, x + \log \left (2\right ) + 4\right )\right )\right ) \]

[In]

integrate(((-32*log(2)-32*x^2-64*x-128)*log(5/(4*log(2)+4*x^2+8*x+16))-32*x-32)/((log(2)+x^2+2*x+4)*log(5/(4*l
og(2)+4*x^2+8*x+16))*log(log(5/(4*log(2)+4*x^2+8*x+16)))+(-2*x*log(2)-2*x^3-4*x^2-8*x)*log(5/(4*log(2)+4*x^2+8
*x+16))),x, algorithm="maxima")

[Out]

16*log(-2*x + log(log(5) - 2*log(2) - log(x^2 + 2*x + log(2) + 4)))

Giac [A] (verification not implemented)

none

Time = 0.46 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=16 \, \log \left (2 \, x - \log \left (\log \left (5\right ) - 2 \, \log \left (2\right ) - \log \left (x^{2} + 2 \, x + \log \left (2\right ) + 4\right )\right )\right ) \]

[In]

integrate(((-32*log(2)-32*x^2-64*x-128)*log(5/(4*log(2)+4*x^2+8*x+16))-32*x-32)/((log(2)+x^2+2*x+4)*log(5/(4*l
og(2)+4*x^2+8*x+16))*log(log(5/(4*log(2)+4*x^2+8*x+16)))+(-2*x*log(2)-2*x^3-4*x^2-8*x)*log(5/(4*log(2)+4*x^2+8
*x+16))),x, algorithm="giac")

[Out]

16*log(2*x - log(log(5) - 2*log(2) - log(x^2 + 2*x + log(2) + 4)))

Mupad [F(-1)]

Timed out. \[ \int \frac {-32-32 x+\left (-128-64 x-32 x^2-32 \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )}{\left (-8 x-4 x^2-2 x^3-2 x \log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )+\left (4+2 x+x^2+\log (2)\right ) \log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right ) \log \left (\log \left (\frac {5}{16+8 x+4 x^2+4 \log (2)}\right )\right )} \, dx=\text {Hanged} \]

[In]

int((32*x + log(5/(8*x + 4*log(2) + 4*x^2 + 16))*(64*x + 32*log(2) + 32*x^2 + 128) + 32)/(log(5/(8*x + 4*log(2
) + 4*x^2 + 16))*(8*x + 2*x*log(2) + 4*x^2 + 2*x^3) - log(5/(8*x + 4*log(2) + 4*x^2 + 16))*log(log(5/(8*x + 4*
log(2) + 4*x^2 + 16)))*(2*x + log(2) + x^2 + 4)),x)

[Out]

\text{Hanged}