3.6.21 \(\int \frac {-5-2 x+\log (8)-2 x \log (x) \log (\log (x))}{(25 x+20 x^2+4 x^3+(-10 x-4 x^2) \log (8)+x \log ^2(8)) \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=26 \[ \frac {-2 x+\frac {x}{(5+2 x-\log (8)) \log (\log (x))}}{x} \]

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Rubi [F]  time = 1.60, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-5-2 x+\log (8)-2 x \log (x) \log (\log (x))}{\left (25 x+20 x^2+4 x^3+\left (-10 x-4 x^2\right ) \log (8)+x \log ^2(8)\right ) \log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5 - 2*x + Log[8] - 2*x*Log[x]*Log[Log[x]])/((25*x + 20*x^2 + 4*x^3 + (-10*x - 4*x^2)*Log[8] + x*Log[8]^2
)*Log[x]*Log[Log[x]]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-5-2 x+\log (8)-2 x \log (x) \log (\log (x))}{\left (20 x^2+4 x^3+\left (-10 x-4 x^2\right ) \log (8)+x \left (25+\log ^2(8)\right )\right ) \log (x) \log ^2(\log (x))} \, dx\\ &=\int \frac {-2 x-5 \left (1-\frac {3 \log (2)}{5}\right )-2 x \log (x) \log (\log (x))}{x \left (4 x^2-4 x (-5+\log (8))+(-5+\log (8))^2\right ) \log (x) \log ^2(\log (x))} \, dx\\ &=\int \frac {-2 x-5 \left (1-\frac {3 \log (2)}{5}\right )-2 x \log (x) \log (\log (x))}{x (5+2 x-\log (8))^2 \log (x) \log ^2(\log (x))} \, dx\\ &=\int \left (-\frac {1}{x (5+2 x-\log (8)) \log (x) \log ^2(\log (x))}-\frac {2}{(5+2 x-\log (8))^2 \log (\log (x))}\right ) \, dx\\ &=-\left (2 \int \frac {1}{(5+2 x-\log (8))^2 \log (\log (x))} \, dx\right )-\int \frac {1}{x (5+2 x-\log (8)) \log (x) \log ^2(\log (x))} \, dx\\ &=-\left (2 \int \frac {1}{(5+2 x-\log (8))^2 \log (\log (x))} \, dx\right )-\int \left (-\frac {1}{x (-5+\log (8)) \log (x) \log ^2(\log (x))}+\frac {2}{(5+2 x-\log (8)) (-5+\log (8)) \log (x) \log ^2(\log (x))}\right ) \, dx\\ &=-\left (2 \int \frac {1}{(5+2 x-\log (8))^2 \log (\log (x))} \, dx\right )+\frac {2 \int \frac {1}{(5+2 x-\log (8)) \log (x) \log ^2(\log (x))} \, dx}{5-\log (8)}+\frac {\int \frac {1}{x \log (x) \log ^2(\log (x))} \, dx}{-5+\log (8)}\\ &=-\left (2 \int \frac {1}{(5+2 x-\log (8))^2 \log (\log (x))} \, dx\right )+\frac {2 \int \frac {1}{(5+2 x-\log (8)) \log (x) \log ^2(\log (x))} \, dx}{5-\log (8)}+\frac {\operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,\log (x)\right )}{-5+\log (8)}\\ &=-\left (2 \int \frac {1}{(5+2 x-\log (8))^2 \log (\log (x))} \, dx\right )+\frac {2 \int \frac {1}{(5+2 x-\log (8)) \log (x) \log ^2(\log (x))} \, dx}{5-\log (8)}+\frac {\operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (\log (x))\right )}{-5+\log (8)}\\ &=\frac {1}{(5-\log (8)) \log (\log (x))}-2 \int \frac {1}{(5+2 x-\log (8))^2 \log (\log (x))} \, dx+\frac {2 \int \frac {1}{(5+2 x-\log (8)) \log (x) \log ^2(\log (x))} \, dx}{5-\log (8)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 17, normalized size = 0.65 \begin {gather*} \frac {1}{(5+2 x-\log (8)) \log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5 - 2*x + Log[8] - 2*x*Log[x]*Log[Log[x]])/((25*x + 20*x^2 + 4*x^3 + (-10*x - 4*x^2)*Log[8] + x*Lo
g[8]^2)*Log[x]*Log[Log[x]]^2),x]

[Out]

1/((5 + 2*x - Log[8])*Log[Log[x]])

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fricas [A]  time = 0.74, size = 17, normalized size = 0.65 \begin {gather*} \frac {1}{{\left (2 \, x - 3 \, \log \relax (2) + 5\right )} \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)*log(log(x))+3*log(2)-2*x-5)/(9*x*log(2)^2+3*(-4*x^2-10*x)*log(2)+4*x^3+20*x^2+25*x)/log
(x)/log(log(x))^2,x, algorithm="fricas")

[Out]

1/((2*x - 3*log(2) + 5)*log(log(x)))

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giac [A]  time = 0.29, size = 21, normalized size = 0.81 \begin {gather*} \frac {1}{2 \, x \log \left (\log \relax (x)\right ) - 3 \, \log \relax (2) \log \left (\log \relax (x)\right ) + 5 \, \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)*log(log(x))+3*log(2)-2*x-5)/(9*x*log(2)^2+3*(-4*x^2-10*x)*log(2)+4*x^3+20*x^2+25*x)/log
(x)/log(log(x))^2,x, algorithm="giac")

[Out]

1/(2*x*log(log(x)) - 3*log(2)*log(log(x)) + 5*log(log(x)))

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maple [A]  time = 0.08, size = 19, normalized size = 0.73

method	result	size
risch	\(-\frac {1}{\left (-5+3 \ln \relax (2)-2 x \right ) \ln \left (\ln \relax (x )\right )}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(x)*ln(ln(x))+3*ln(2)-2*x-5)/(9*x*ln(2)^2+3*(-4*x^2-10*x)*ln(2)+4*x^3+20*x^2+25*x)/ln(x)/ln(ln(x))
^2,x,method=_RETURNVERBOSE)

[Out]

-1/(-5+3*ln(2)-2*x)/ln(ln(x))

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maxima [A]  time = 0.95, size = 17, normalized size = 0.65 \begin {gather*} \frac {1}{{\left (2 \, x - 3 \, \log \relax (2) + 5\right )} \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(x)*log(log(x))+3*log(2)-2*x-5)/(9*x*log(2)^2+3*(-4*x^2-10*x)*log(2)+4*x^3+20*x^2+25*x)/log
(x)/log(log(x))^2,x, algorithm="maxima")

[Out]

1/((2*x - 3*log(2) + 5)*log(log(x)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {2\,x-3\,\ln \relax (2)+2\,x\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)+5}{{\ln \left (\ln \relax (x)\right )}^2\,\ln \relax (x)\,\left (25\,x-3\,\ln \relax (2)\,\left (4\,x^2+10\,x\right )+9\,x\,{\ln \relax (2)}^2+20\,x^2+4\,x^3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - 3*log(2) + 2*x*log(log(x))*log(x) + 5)/(log(log(x))^2*log(x)*(25*x - 3*log(2)*(10*x + 4*x^2) + 9*x
*log(2)^2 + 20*x^2 + 4*x^3)),x)

[Out]

int(-(2*x - 3*log(2) + 2*x*log(log(x))*log(x) + 5)/(log(log(x))^2*log(x)*(25*x - 3*log(2)*(10*x + 4*x^2) + 9*x
*log(2)^2 + 20*x^2 + 4*x^3)), x)

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sympy [A]  time = 0.31, size = 15, normalized size = 0.58 \begin {gather*} \frac {1}{\left (2 x - 3 \log {\relax (2 )} + 5\right ) \log {\left (\log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(x)*ln(ln(x))+3*ln(2)-2*x-5)/(9*x*ln(2)**2+3*(-4*x**2-10*x)*ln(2)+4*x**3+20*x**2+25*x)/ln(x)
/ln(ln(x))**2,x)

[Out]

1/((2*x - 3*log(2) + 5)*log(log(x)))

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