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

Optimal. Leaf size=21 \[ \log \left (\log \left (2 \left (-4-2 x-\frac {\log (x)}{4 x}+\log (\log (5))\right )\right )\right ) \]

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Rubi [A]  time = 0.12, antiderivative size = 26, normalized size of antiderivative = 1.24, number of steps used = 2, number of rules used = 2, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6, 6684} \begin {gather*} \log \left (\log \left (-\frac {8 x^2+16 x-4 x \log (\log (5))+\log (x)}{2 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 8*x^2 + Log[x])/((-16*x^2 - 8*x^3 - x*Log[x] + 4*x^2*Log[Log[5]])*Log[(-16*x - 8*x^2 - Log[x] + 4*x*
Log[Log[5]])/(2*x)]),x]

[Out]

Log[Log[-1/2*(16*x + 8*x^2 + Log[x] - 4*x*Log[Log[5]])/x]]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 6684

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1-8 x^2+\log (x)}{\left (-8 x^3-x \log (x)+x^2 (-16+4 \log (\log (5)))\right ) \log \left (\frac {-16 x-8 x^2-\log (x)+4 x \log (\log (5))}{2 x}\right )} \, dx\\ &=\log \left (\log \left (-\frac {16 x+8 x^2+\log (x)-4 x \log (\log (5))}{2 x}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.65, size = 21, normalized size = 1.00 \begin {gather*} \log \left (\log \left (-8-4 x-\frac {\log (x)}{2 x}+2 \log (\log (5))\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 8*x^2 + Log[x])/((-16*x^2 - 8*x^3 - x*Log[x] + 4*x^2*Log[Log[5]])*Log[(-16*x - 8*x^2 - Log[x]
+ 4*x*Log[Log[5]])/(2*x)]),x]

[Out]

Log[Log[-8 - 4*x - Log[x]/(2*x) + 2*Log[Log[5]]]]

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fricas [A]  time = 0.90, size = 24, normalized size = 1.14 \begin {gather*} \log \left (\log \left (-\frac {8 \, x^{2} - 4 \, x \log \left (\log \relax (5)\right ) + 16 \, x + \log \relax (x)}{2 \, x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(-1/2*(8*x^2 - 4*x*log(log(5)) + 16*x + log(x))/x))

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giac [A]  time = 0.19, size = 30, normalized size = 1.43 \begin {gather*} \log \left (-\log \relax (2) + \log \left (-8 \, x^{2} + 4 \, x \log \left (\log \relax (5)\right ) - 16 \, x - \log \relax (x)\right ) - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.13, size = 216, normalized size = 10.29

method	result	size
risch	\(\ln \left (\ln \left (x \ln \left (\ln \relax (5)\right )-2 x^{2}-4 x -\frac {\ln \relax (x )}{4}\right )-\frac {i \left (\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (-x \ln \left (\ln \relax (5)\right )+2 x^{2}+4 x +\frac {\ln \relax (x )}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-x \ln \left (\ln \relax (5)\right )+2 x^{2}+4 x +\frac {\ln \relax (x )}{4}\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (-x \ln \left (\ln \relax (5)\right )+2 x^{2}+4 x +\frac {\ln \relax (x )}{4}\right )}{x}\right )^{2}+\pi \,\mathrm {csgn}\left (i \left (-x \ln \left (\ln \relax (5)\right )+2 x^{2}+4 x +\frac {\ln \relax (x )}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-x \ln \left (\ln \relax (5)\right )+2 x^{2}+4 x +\frac {\ln \relax (x )}{4}\right )}{x}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i \left (-x \ln \left (\ln \relax (5)\right )+2 x^{2}+4 x +\frac {\ln \relax (x )}{4}\right )}{x}\right )^{3}+2 i \ln \relax (2)-2 i \ln \relax (x )\right )}{2}\right )\)	\(216\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(ln(x*ln(ln(5))-2*x^2-4*x-1/4*ln(x))-1/2*I*(Pi*csgn(I/x)*csgn(I*(-x*ln(ln(5))+2*x^2+4*x+1/4*ln(x)))*csgn(I/x
*(-x*ln(ln(5))+2*x^2+4*x+1/4*ln(x)))-Pi*csgn(I/x)*csgn(I/x*(-x*ln(ln(5))+2*x^2+4*x+1/4*ln(x)))^2+Pi*csgn(I*(-x
*ln(ln(5))+2*x^2+4*x+1/4*ln(x)))*csgn(I/x*(-x*ln(ln(5))+2*x^2+4*x+1/4*ln(x)))^2-Pi*csgn(I/x*(-x*ln(ln(5))+2*x^
2+4*x+1/4*ln(x)))^3+2*I*ln(2)-2*I*ln(x)))

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maxima [A]  time = 0.47, size = 29, normalized size = 1.38 \begin {gather*} \log \left (-\log \relax (2) + \log \left (-8 \, x^{2} + 4 \, x {\left (\log \left (\log \relax (5)\right ) - 4\right )} - \log \relax (x)\right ) - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(-log(2) + log(-8*x^2 + 4*x*(log(log(5)) - 4) - log(x)) - log(x))

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mupad [B]  time = 8.80, size = 24, normalized size = 1.14 \begin {gather*} \ln \left (\ln \left (-\frac {16\,x+\ln \relax (x)-4\,x\,\ln \left (\ln \relax (5)\right )+8\,x^2}{2\,x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x^2 - log(x) + 1)/(log(-(8*x + log(x)/2 - 2*x*log(log(5)) + 4*x^2)/x)*(x*log(x) - 4*x^2*log(log(5)) + 1
6*x^2 + 8*x^3)),x)

[Out]

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

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sympy [A]  time = 0.73, size = 26, normalized size = 1.24 \begin {gather*} \log {\left (\log {\left (\frac {- 4 x^{2} - 8 x + 2 x \log {\left (\log {\relax (5 )} \right )} - \frac {\log {\relax (x )}}{2}}{x} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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