3.98.32 \(\int \frac {-16 x^2+(-2 e^4 x^2+6 x^4) \log (x)+(16+4 x^2 \log (x)) \log (\log ^2(x))-10 \log (x) \log ^2(\log ^2(x))}{(e^4 x^3+x^5) \log (x)-2 x^3 \log (x) \log (\log ^2(x))+x \log (x) \log ^2(\log ^2(x))} \, dx\)

Optimal. Leaf size=27 \[ \log \left (\frac {8 \left (e^4+\left (-x+\frac {\log \left (\log ^2(x)\right )}{x}\right )^2\right )^4}{x^2}\right ) \]

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Rubi [A]  time = 0.80, antiderivative size = 36, normalized size of antiderivative = 1.33, number of steps used = 4, number of rules used = 3, integrand size = 90, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6741, 6742, 6684} \begin {gather*} 4 \log \left (x^4+e^4 x^2-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )-10 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*x^2 + (-2*E^4*x^2 + 6*x^4)*Log[x] + (16 + 4*x^2*Log[x])*Log[Log[x]^2] - 10*Log[x]*Log[Log[x]^2]^2)/((
E^4*x^3 + x^5)*Log[x] - 2*x^3*Log[x]*Log[Log[x]^2] + x*Log[x]*Log[Log[x]^2]^2),x]

[Out]

-10*Log[x] + 4*Log[E^4*x^2 + x^4 - 2*x^2*Log[Log[x]^2] + Log[Log[x]^2]^2]

Rule 6684

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

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-16 x^2+\left (-2 e^4 x^2+6 x^4\right ) \log (x)+\left (16+4 x^2 \log (x)\right ) \log \left (\log ^2(x)\right )-10 \log (x) \log ^2\left (\log ^2(x)\right )}{x \log (x) \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )} \, dx\\ &=\int \left (-\frac {10}{x}+\frac {8 \left (-2 x^2+e^4 x^2 \log (x)+2 x^4 \log (x)+2 \log \left (\log ^2(x)\right )-2 x^2 \log (x) \log \left (\log ^2(x)\right )\right )}{x \log (x) \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )}\right ) \, dx\\ &=-10 \log (x)+8 \int \frac {-2 x^2+e^4 x^2 \log (x)+2 x^4 \log (x)+2 \log \left (\log ^2(x)\right )-2 x^2 \log (x) \log \left (\log ^2(x)\right )}{x \log (x) \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )} \, dx\\ &=-10 \log (x)+4 \log \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 38, normalized size = 1.41 \begin {gather*} -2 \left (5 \log (x)-2 \log \left (e^4 x^2+x^4-2 x^2 \log \left (\log ^2(x)\right )+\log ^2\left (\log ^2(x)\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x^2 + (-2*E^4*x^2 + 6*x^4)*Log[x] + (16 + 4*x^2*Log[x])*Log[Log[x]^2] - 10*Log[x]*Log[Log[x]^2]
^2)/((E^4*x^3 + x^5)*Log[x] - 2*x^3*Log[x]*Log[Log[x]^2] + x*Log[x]*Log[Log[x]^2]^2),x]

[Out]

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

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fricas [A]  time = 0.83, size = 35, normalized size = 1.30 \begin {gather*} 4 \, \log \left (x^{4} + x^{2} e^{4} - 2 \, x^{2} \log \left (\log \relax (x)^{2}\right ) + \log \left (\log \relax (x)^{2}\right )^{2}\right ) - 10 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(x)*log(log(x)^2)^2+(4*x^2*log(x)+16)*log(log(x)^2)+(-2*x^2*exp(4)+6*x^4)*log(x)-16*x^2)/(x*
log(x)*log(log(x)^2)^2-2*x^3*log(x)*log(log(x)^2)+(x^3*exp(4)+x^5)*log(x)),x, algorithm="fricas")

[Out]

4*log(x^4 + x^2*e^4 - 2*x^2*log(log(x)^2) + log(log(x)^2)^2) - 10*log(x)

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giac [A]  time = 0.31, size = 35, normalized size = 1.30 \begin {gather*} 4 \, \log \left (x^{4} + x^{2} e^{4} - 2 \, x^{2} \log \left (\log \relax (x)^{2}\right ) + \log \left (\log \relax (x)^{2}\right )^{2}\right ) - 10 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(x)*log(log(x)^2)^2+(4*x^2*log(x)+16)*log(log(x)^2)+(-2*x^2*exp(4)+6*x^4)*log(x)-16*x^2)/(x*
log(x)*log(log(x)^2)^2-2*x^3*log(x)*log(log(x)^2)+(x^3*exp(4)+x^5)*log(x)),x, algorithm="giac")

[Out]

4*log(x^4 + x^2*e^4 - 2*x^2*log(log(x)^2) + log(log(x)^2)^2) - 10*log(x)

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maple [C]  time = 0.14, size = 259, normalized size = 9.59

method	result	size
risch	\(-10 \ln \relax (x )+4 \ln \left (-\frac {\pi ^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )^{4} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}}{16}+\frac {\pi ^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )^{3} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}}{4}-\frac {3 \pi ^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{4}}{8}+\frac {\pi ^{2} \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{5}}{4}-\frac {\pi ^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{6}}{16}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )}{4}-\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}}{2}+\frac {i \pi \,x^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}}{4}+\frac {x^{4}}{4}+\frac {x^{2} {\mathrm e}^{4}}{4}+\ln \left (\ln \relax (x )\right )^{2}+\left (-\frac {i \pi \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )}{2}+i \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}-\frac {i \pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}}{2}-x^{2}\right ) \ln \left (\ln \relax (x )\right )\right )\)	\(259\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-10*ln(x)*ln(ln(x)^2)^2+(4*x^2*ln(x)+16)*ln(ln(x)^2)+(-2*x^2*exp(4)+6*x^4)*ln(x)-16*x^2)/(x*ln(x)*ln(ln(x
)^2)^2-2*x^3*ln(x)*ln(ln(x)^2)+(x^3*exp(4)+x^5)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-10*ln(x)+4*ln(-1/16*Pi^2*csgn(I*ln(x))^4*csgn(I*ln(x)^2)^2+1/4*Pi^2*csgn(I*ln(x))^3*csgn(I*ln(x)^2)^3-3/8*Pi^
2*csgn(I*ln(x))^2*csgn(I*ln(x)^2)^4+1/4*Pi^2*csgn(I*ln(x))*csgn(I*ln(x)^2)^5-1/16*Pi^2*csgn(I*ln(x)^2)^6+1/4*I
*Pi*x^2*csgn(I*ln(x))^2*csgn(I*ln(x)^2)-1/2*I*Pi*x^2*csgn(I*ln(x))*csgn(I*ln(x)^2)^2+1/4*I*Pi*x^2*csgn(I*ln(x)
^2)^3+1/4*x^4+1/4*x^2*exp(4)+ln(ln(x))^2+(-1/2*I*Pi*csgn(I*ln(x))^2*csgn(I*ln(x)^2)+I*Pi*csgn(I*ln(x))*csgn(I*
ln(x)^2)^2-1/2*I*Pi*csgn(I*ln(x)^2)^3-x^2)*ln(ln(x)))

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maxima [A]  time = 0.39, size = 34, normalized size = 1.26 \begin {gather*} 4 \, \log \left (\frac {1}{4} \, x^{4} + \frac {1}{4} \, x^{2} e^{4} - x^{2} \log \left (\log \relax (x)\right ) + \log \left (\log \relax (x)\right )^{2}\right ) - 10 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(x)*log(log(x)^2)^2+(4*x^2*log(x)+16)*log(log(x)^2)+(-2*x^2*exp(4)+6*x^4)*log(x)-16*x^2)/(x*
log(x)*log(log(x)^2)^2-2*x^3*log(x)*log(log(x)^2)+(x^3*exp(4)+x^5)*log(x)),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {10\,{\ln \left ({\ln \relax (x)}^2\right )}^2\,\ln \relax (x)+\ln \relax (x)\,\left (2\,x^2\,{\mathrm {e}}^4-6\,x^4\right )-\ln \left ({\ln \relax (x)}^2\right )\,\left (4\,x^2\,\ln \relax (x)+16\right )+16\,x^2}{\ln \relax (x)\,\left (x^5+{\mathrm {e}}^4\,x^3\right )+x\,{\ln \left ({\ln \relax (x)}^2\right )}^2\,\ln \relax (x)-2\,x^3\,\ln \left ({\ln \relax (x)}^2\right )\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*log(log(x)^2)^2*log(x) + log(x)*(2*x^2*exp(4) - 6*x^4) - log(log(x)^2)*(4*x^2*log(x) + 16) + 16*x^2)/
(log(x)*(x^3*exp(4) + x^5) + x*log(log(x)^2)^2*log(x) - 2*x^3*log(log(x)^2)*log(x)),x)

[Out]

int(-(10*log(log(x)^2)^2*log(x) + log(x)*(2*x^2*exp(4) - 6*x^4) - log(log(x)^2)*(4*x^2*log(x) + 16) + 16*x^2)/
(log(x)*(x^3*exp(4) + x^5) + x*log(log(x)^2)^2*log(x) - 2*x^3*log(log(x)^2)*log(x)), x)

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sympy [A]  time = 0.44, size = 37, normalized size = 1.37 \begin {gather*} - 10 \log {\relax (x )} + 4 \log {\left (x^{4} - 2 x^{2} \log {\left (\log {\relax (x )}^{2} \right )} + x^{2} e^{4} + \log {\left (\log {\relax (x )}^{2} \right )}^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*ln(x)*ln(ln(x)**2)**2+(4*x**2*ln(x)+16)*ln(ln(x)**2)+(-2*x**2*exp(4)+6*x**4)*ln(x)-16*x**2)/(x*
ln(x)*ln(ln(x)**2)**2-2*x**3*ln(x)*ln(ln(x)**2)+(x**3*exp(4)+x**5)*ln(x)),x)

[Out]

-10*log(x) + 4*log(x**4 - 2*x**2*log(log(x)**2) + x**2*exp(4) + log(log(x)**2)**2)

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