3.88.74 \(\int \frac {-25-10 x^2-x^4+(-20 x^2-4 x^4) \log (x) \log (\frac {1}{\log (x)})-128 x \log (x) \log ^2(\frac {1}{\log (x)})}{64 x \log (x) \log ^2(\frac {1}{\log (x)})} \, dx\)

Optimal. Leaf size=22 \[ -2 x-\frac {\left (5+x^2\right )^2}{64 \log \left (\frac {1}{\log (x)}\right )} \]

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Rubi [F]  time = 0.65, 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 {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{64 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-25 - 10*x^2 - x^4 + (-20*x^2 - 4*x^4)*Log[x]*Log[Log[x]^(-1)] - 128*x*Log[x]*Log[Log[x]^(-1)]^2)/(64*x*L
og[x]*Log[Log[x]^(-1)]^2),x]

[Out]

-2*x - 25/(64*Log[Log[x]^(-1)]) - (5*Defer[Int][x/(Log[x]*Log[Log[x]^(-1)]^2), x])/32 - Defer[Int][x^3/(Log[x]
*Log[Log[x]^(-1)]^2), x]/64 - (5*Defer[Int][x/Log[Log[x]^(-1)], x])/16 - Defer[Int][x^3/Log[Log[x]^(-1)], x]/1
6

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{64} \int \frac {-25-10 x^2-x^4+\left (-20 x^2-4 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )-128 x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx\\ &=\frac {1}{64} \int \frac {-\frac {\left (5+x^2\right )^2}{x \log (x)}-4 \log \left (\frac {1}{\log (x)}\right ) \left (5 x+x^3+32 \log \left (\frac {1}{\log (x)}\right )\right )}{\log ^2\left (\frac {1}{\log (x)}\right )} \, dx\\ &=\frac {1}{64} \int \left (-128-\frac {\left (5+x^2\right )^2}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}-\frac {4 x \left (5+x^2\right )}{\log \left (\frac {1}{\log (x)}\right )}\right ) \, dx\\ &=-2 x-\frac {1}{64} \int \frac {\left (5+x^2\right )^2}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {1}{16} \int \frac {x \left (5+x^2\right )}{\log \left (\frac {1}{\log (x)}\right )} \, dx\\ &=-2 x-\frac {1}{64} \int \left (\frac {25}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )}+\frac {10 x}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )}+\frac {x^3}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )}\right ) \, dx-\frac {1}{16} \int \left (\frac {5 x}{\log \left (\frac {1}{\log (x)}\right )}+\frac {x^3}{\log \left (\frac {1}{\log (x)}\right )}\right ) \, dx\\ &=-2 x-\frac {1}{64} \int \frac {x^3}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {1}{16} \int \frac {x^3}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{32} \int \frac {x}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{16} \int \frac {x}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {25}{64} \int \frac {1}{x \log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx\\ &=-2 x-\frac {1}{64} \int \frac {x^3}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {1}{16} \int \frac {x^3}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{32} \int \frac {x}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{16} \int \frac {x}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {25}{64} \operatorname {Subst}\left (\int \frac {1}{x \log ^2\left (\frac {1}{x}\right )} \, dx,x,\log (x)\right )\\ &=-2 x-\frac {1}{64} \int \frac {x^3}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {1}{16} \int \frac {x^3}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{32} \int \frac {x}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{16} \int \frac {x}{\log \left (\frac {1}{\log (x)}\right )} \, dx+\frac {25}{64} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log \left (\frac {1}{\log (x)}\right )\right )\\ &=-2 x-\frac {25}{64 \log \left (\frac {1}{\log (x)}\right )}-\frac {1}{64} \int \frac {x^3}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {1}{16} \int \frac {x^3}{\log \left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{32} \int \frac {x}{\log (x) \log ^2\left (\frac {1}{\log (x)}\right )} \, dx-\frac {5}{16} \int \frac {x}{\log \left (\frac {1}{\log (x)}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 22, normalized size = 1.00 \begin {gather*} -2 x-\frac {\left (5+x^2\right )^2}{64 \log \left (\frac {1}{\log (x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25 - 10*x^2 - x^4 + (-20*x^2 - 4*x^4)*Log[x]*Log[Log[x]^(-1)] - 128*x*Log[x]*Log[Log[x]^(-1)]^2)/(
64*x*Log[x]*Log[Log[x]^(-1)]^2),x]

[Out]

-2*x - (5 + x^2)^2/(64*Log[Log[x]^(-1)])

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fricas [A]  time = 0.49, size = 27, normalized size = 1.23 \begin {gather*} -\frac {x^{4} + 10 \, x^{2} + 128 \, x \log \left (\frac {1}{\log \relax (x)}\right ) + 25}{64 \, \log \left (\frac {1}{\log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-128*x*log(x)*log(1/log(x))^2+(-4*x^4-20*x^2)*log(x)*log(1/log(x))-x^4-10*x^2-25)/x/log(x)/log
(1/log(x))^2,x, algorithm="fricas")

[Out]

-1/64*(x^4 + 10*x^2 + 128*x*log(1/log(x)) + 25)/log(1/log(x))

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giac [A]  time = 0.17, size = 21, normalized size = 0.95 \begin {gather*} -2 \, x + \frac {x^{4} + 10 \, x^{2} + 25}{64 \, \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-128*x*log(x)*log(1/log(x))^2+(-4*x^4-20*x^2)*log(x)*log(1/log(x))-x^4-10*x^2-25)/x/log(x)/log
(1/log(x))^2,x, algorithm="giac")

[Out]

-2*x + 1/64*(x^4 + 10*x^2 + 25)/log(log(x))

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maple [A]  time = 0.06, size = 22, normalized size = 1.00

method	result	size
risch	\(-2 x +\frac {x^{4}+10 x^{2}+25}{64 \ln \left (\ln \relax (x )\right )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/64*(-128*x*ln(x)*ln(1/ln(x))^2+(-4*x^4-20*x^2)*ln(x)*ln(1/ln(x))-x^4-10*x^2-25)/x/ln(x)/ln(1/ln(x))^2,x,
method=_RETURNVERBOSE)

[Out]

-2*x+1/64*(x^4+10*x^2+25)/ln(ln(x))

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maxima [A]  time = 0.40, size = 27, normalized size = 1.23 \begin {gather*} -2 \, x + \frac {x^{4} + 10 \, x^{2}}{64 \, \log \left (\log \relax (x)\right )} + \frac {25}{64 \, \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-128*x*log(x)*log(1/log(x))^2+(-4*x^4-20*x^2)*log(x)*log(1/log(x))-x^4-10*x^2-25)/x/log(x)/log
(1/log(x))^2,x, algorithm="maxima")

[Out]

-2*x + 1/64*(x^4 + 10*x^2)/log(log(x)) + 25/64/log(log(x))

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mupad [B]  time = 5.60, size = 27, normalized size = 1.23 \begin {gather*} -\frac {128\,x\,\ln \left (\frac {1}{\ln \relax (x)}\right )+10\,x^2+x^4+25}{64\,\ln \left (\frac {1}{\ln \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((5*x^2)/32 + x^4/64 + (log(1/log(x))*log(x)*(20*x^2 + 4*x^4))/64 + 2*x*log(1/log(x))^2*log(x) + 25/64)/(
x*log(1/log(x))^2*log(x)),x)

[Out]

-(128*x*log(1/log(x)) + 10*x^2 + x^4 + 25)/(64*log(1/log(x)))

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sympy [A]  time = 0.25, size = 22, normalized size = 1.00 \begin {gather*} - 2 x + \frac {- x^{4} - 10 x^{2} - 25}{64 \log {\left (\frac {1}{\log {\relax (x )}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/64*(-128*x*ln(x)*ln(1/ln(x))**2+(-4*x**4-20*x**2)*ln(x)*ln(1/ln(x))-x**4-10*x**2-25)/x/ln(x)/ln(1/
ln(x))**2,x)

[Out]

-2*x + (-x**4 - 10*x**2 - 25)/(64*log(1/log(x)))

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