3.59.94 \(\int \frac {(-8 x-2 x \log (\frac {1}{x})) \log (16+8 \log (\frac {1}{x})+\log ^2(\frac {1}{x}))+(8+(64+16 \log (\frac {1}{x})) \log (16+8 \log (\frac {1}{x})+\log ^2(\frac {1}{x}))) \log (\frac {\log (16+8 \log (\frac {1}{x})+\log ^2(\frac {1}{x}))}{x^4})}{(4 x^3+x^3 \log (\frac {1}{x})) \log (16+8 \log (\frac {1}{x})+\log ^2(\frac {1}{x}))+(8 x^2+2 x^2 \log (\frac {1}{x})) \log (16+8 \log (\frac {1}{x})+\log ^2(\frac {1}{x})) \log ^2(\frac {\log (16+8 \log (\frac {1}{x})+\log ^2(\frac {1}{x}))}{x^4})+(4 x+x \log (\frac {1}{x})) \log (16+8 \log (\frac {1}{x})+\log ^2(\frac {1}{x})) \log ^4(\frac {\log (16+8 \log (\frac {1}{x})+\log ^2(\frac {1}{x}))}{x^4})} \, dx\)

Optimal. Leaf size=22 \[ \frac {2}{x+\log ^2\left (\frac {\log \left (\left (4+\log \left (\frac {1}{x}\right )\right )^2\right )}{x^4}\right )} \]

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Rubi [A]  time = 0.27, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 210, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.010, Rules used = {6688, 6686} \begin {gather*} \frac {2}{\log ^2\left (\frac {\log \left (\left (\log \left (\frac {1}{x}\right )+4\right )^2\right )}{x^4}\right )+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-8*x - 2*x*Log[x^(-1)])*Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2] + (8 + (64 + 16*Log[x^(-1)])*Log[16 + 8*
Log[x^(-1)] + Log[x^(-1)]^2])*Log[Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2]/x^4])/((4*x^3 + x^3*Log[x^(-1)])*Log
[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2] + (8*x^2 + 2*x^2*Log[x^(-1)])*Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2]*Log
[Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2]/x^4]^2 + (4*x + x*Log[x^(-1)])*Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2
]*Log[Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2]/x^4]^4),x]

[Out]

2/(x + Log[Log[(4 + Log[x^(-1)])^2]/x^4]^2)

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.11, size = 22, normalized size = 1.00 \begin {gather*} \frac {2}{x+\log ^2\left (\frac {\log \left (\left (4+\log \left (\frac {1}{x}\right )\right )^2\right )}{x^4}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-8*x - 2*x*Log[x^(-1)])*Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2] + (8 + (64 + 16*Log[x^(-1)])*Log[1
6 + 8*Log[x^(-1)] + Log[x^(-1)]^2])*Log[Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2]/x^4])/((4*x^3 + x^3*Log[x^(-1)
])*Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2] + (8*x^2 + 2*x^2*Log[x^(-1)])*Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^
2]*Log[Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2]/x^4]^2 + (4*x + x*Log[x^(-1)])*Log[16 + 8*Log[x^(-1)] + Log[x^(
-1)]^2]*Log[Log[16 + 8*Log[x^(-1)] + Log[x^(-1)]^2]/x^4]^4),x]

[Out]

2/(x + Log[Log[(4 + Log[x^(-1)])^2]/x^4]^2)

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fricas [A]  time = 0.67, size = 28, normalized size = 1.27 \begin {gather*} \frac {2}{\log \left (\frac {\log \left (\log \left (\frac {1}{x}\right )^{2} + 8 \, \log \left (\frac {1}{x}\right ) + 16\right )}{x^{4}}\right )^{2} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*log(1/x)+64)*log(log(1/x)^2+8*log(1/x)+16)+8)*log(log(log(1/x)^2+8*log(1/x)+16)/x^4)+(-2*x*log
(1/x)-8*x)*log(log(1/x)^2+8*log(1/x)+16))/((x*log(1/x)+4*x)*log(log(1/x)^2+8*log(1/x)+16)*log(log(log(1/x)^2+8
*log(1/x)+16)/x^4)^4+(2*x^2*log(1/x)+8*x^2)*log(log(1/x)^2+8*log(1/x)+16)*log(log(log(1/x)^2+8*log(1/x)+16)/x^
4)^2+(x^3*log(1/x)+4*x^3)*log(log(1/x)^2+8*log(1/x)+16)),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*log(1/x)+64)*log(log(1/x)^2+8*log(1/x)+16)+8)*log(log(log(1/x)^2+8*log(1/x)+16)/x^4)+(-2*x*log
(1/x)-8*x)*log(log(1/x)^2+8*log(1/x)+16))/((x*log(1/x)+4*x)*log(log(1/x)^2+8*log(1/x)+16)*log(log(log(1/x)^2+8
*log(1/x)+16)/x^4)^4+(2*x^2*log(1/x)+8*x^2)*log(log(1/x)^2+8*log(1/x)+16)*log(log(log(1/x)^2+8*log(1/x)+16)/x^
4)^2+(x^3*log(1/x)+4*x^3)*log(log(1/x)^2+8*log(1/x)+16)),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (16 \ln \left (\frac {1}{x}\right )+64\right ) \ln \left (\ln \left (\frac {1}{x}\right )^{2}+8 \ln \left (\frac {1}{x}\right )+16\right )+8\right ) \ln \left (\frac {\ln \left (\ln \left (\frac {1}{x}\right )^{2}+8 \ln \left (\frac {1}{x}\right )+16\right )}{x^{4}}\right )+\left (-2 x \ln \left (\frac {1}{x}\right )-8 x \right ) \ln \left (\ln \left (\frac {1}{x}\right )^{2}+8 \ln \left (\frac {1}{x}\right )+16\right )}{\left (x \ln \left (\frac {1}{x}\right )+4 x \right ) \ln \left (\ln \left (\frac {1}{x}\right )^{2}+8 \ln \left (\frac {1}{x}\right )+16\right ) \ln \left (\frac {\ln \left (\ln \left (\frac {1}{x}\right )^{2}+8 \ln \left (\frac {1}{x}\right )+16\right )}{x^{4}}\right )^{4}+\left (2 x^{2} \ln \left (\frac {1}{x}\right )+8 x^{2}\right ) \ln \left (\ln \left (\frac {1}{x}\right )^{2}+8 \ln \left (\frac {1}{x}\right )+16\right ) \ln \left (\frac {\ln \left (\ln \left (\frac {1}{x}\right )^{2}+8 \ln \left (\frac {1}{x}\right )+16\right )}{x^{4}}\right )^{2}+\left (x^{3} \ln \left (\frac {1}{x}\right )+4 x^{3}\right ) \ln \left (\ln \left (\frac {1}{x}\right )^{2}+8 \ln \left (\frac {1}{x}\right )+16\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*ln(1/x)+64)*ln(ln(1/x)^2+8*ln(1/x)+16)+8)*ln(ln(ln(1/x)^2+8*ln(1/x)+16)/x^4)+(-2*x*ln(1/x)-8*x)*ln(l
n(1/x)^2+8*ln(1/x)+16))/((x*ln(1/x)+4*x)*ln(ln(1/x)^2+8*ln(1/x)+16)*ln(ln(ln(1/x)^2+8*ln(1/x)+16)/x^4)^4+(2*x^
2*ln(1/x)+8*x^2)*ln(ln(1/x)^2+8*ln(1/x)+16)*ln(ln(ln(1/x)^2+8*ln(1/x)+16)/x^4)^2+(x^3*ln(1/x)+4*x^3)*ln(ln(1/x
)^2+8*ln(1/x)+16)),x)

[Out]

int((((16*ln(1/x)+64)*ln(ln(1/x)^2+8*ln(1/x)+16)+8)*ln(ln(ln(1/x)^2+8*ln(1/x)+16)/x^4)+(-2*x*ln(1/x)-8*x)*ln(l
n(1/x)^2+8*ln(1/x)+16))/((x*ln(1/x)+4*x)*ln(ln(1/x)^2+8*ln(1/x)+16)*ln(ln(ln(1/x)^2+8*ln(1/x)+16)/x^4)^4+(2*x^
2*ln(1/x)+8*x^2)*ln(ln(1/x)^2+8*ln(1/x)+16)*ln(ln(ln(1/x)^2+8*ln(1/x)+16)/x^4)^2+(x^3*ln(1/x)+4*x^3)*ln(ln(1/x
)^2+8*ln(1/x)+16)),x)

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maxima [B]  time = 0.60, size = 45, normalized size = 2.05 \begin {gather*} \frac {2}{\log \relax (2)^{2} - 8 \, \log \relax (2) \log \relax (x) + 16 \, \log \relax (x)^{2} + 2 \, {\left (\log \relax (2) - 4 \, \log \relax (x)\right )} \log \left (\log \left (\log \relax (x) - 4\right )\right ) + \log \left (\log \left (\log \relax (x) - 4\right )\right )^{2} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*log(1/x)+64)*log(log(1/x)^2+8*log(1/x)+16)+8)*log(log(log(1/x)^2+8*log(1/x)+16)/x^4)+(-2*x*log
(1/x)-8*x)*log(log(1/x)^2+8*log(1/x)+16))/((x*log(1/x)+4*x)*log(log(1/x)^2+8*log(1/x)+16)*log(log(log(1/x)^2+8
*log(1/x)+16)/x^4)^4+(2*x^2*log(1/x)+8*x^2)*log(log(1/x)^2+8*log(1/x)+16)*log(log(log(1/x)^2+8*log(1/x)+16)/x^
4)^2+(x^3*log(1/x)+4*x^3)*log(log(1/x)^2+8*log(1/x)+16)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.15, size = 28, normalized size = 1.27 \begin {gather*} \frac {2}{{\ln \left (\frac {\ln \left ({\ln \left (\frac {1}{x}\right )}^2+8\,\ln \left (\frac {1}{x}\right )+16\right )}{x^4}\right )}^2+x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(8*log(1/x) + log(1/x)^2 + 16)*(8*x + 2*x*log(1/x)) - log(log(8*log(1/x) + log(1/x)^2 + 16)/x^4)*(log
(8*log(1/x) + log(1/x)^2 + 16)*(16*log(1/x) + 64) + 8))/(log(8*log(1/x) + log(1/x)^2 + 16)*(x^3*log(1/x) + 4*x
^3) + log(8*log(1/x) + log(1/x)^2 + 16)*log(log(8*log(1/x) + log(1/x)^2 + 16)/x^4)^2*(2*x^2*log(1/x) + 8*x^2)
+ log(8*log(1/x) + log(1/x)^2 + 16)*log(log(8*log(1/x) + log(1/x)^2 + 16)/x^4)^4*(4*x + x*log(1/x))),x)

[Out]

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

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sympy [A]  time = 0.48, size = 26, normalized size = 1.18 \begin {gather*} \frac {2}{x + \log {\left (\frac {\log {\left (\log {\left (\frac {1}{x} \right )}^{2} + 8 \log {\left (\frac {1}{x} \right )} + 16 \right )}}{x^{4}} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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