3.15.8 \(\int \frac {2-4 \log (x)+(4-2 x+4 x^2) \log (x) \log (\frac {\log (x)}{x^2})}{x \log (x) \log (\frac {\log (x)}{x^2})} \, dx\)

Optimal. Leaf size=30 \[ \log \left (9 e^{-2 x+2 x^2} x^4 \log ^2(5) \log ^2\left (\frac {\log (x)}{x^2}\right )\right ) \]

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Rubi [A]  time = 0.31, antiderivative size = 23, normalized size of antiderivative = 0.77, number of steps used = 5, number of rules used = 3, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {6742, 14, 6684} \begin {gather*} 2 x^2+2 \log \left (\log \left (\frac {\log (x)}{x^2}\right )\right )-2 x+4 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - 4*Log[x] + (4 - 2*x + 4*x^2)*Log[x]*Log[Log[x]/x^2])/(x*Log[x]*Log[Log[x]/x^2]),x]

[Out]

-2*x + 2*x^2 + 4*Log[x] + 2*Log[Log[Log[x]/x^2]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 6684

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

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

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Mathematica [A]  time = 0.10, size = 23, normalized size = 0.77 \begin {gather*} -2 x+2 x^2+4 \log (x)+2 \log \left (\log \left (\frac {\log (x)}{x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - 4*Log[x] + (4 - 2*x + 4*x^2)*Log[x]*Log[Log[x]/x^2])/(x*Log[x]*Log[Log[x]/x^2]),x]

[Out]

-2*x + 2*x^2 + 4*Log[x] + 2*Log[Log[Log[x]/x^2]]

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fricas [A]  time = 0.54, size = 23, normalized size = 0.77 \begin {gather*} 2 \, x^{2} - 2 \, x + 4 \, \log \relax (x) + 2 \, \log \left (\log \left (\frac {\log \relax (x)}{x^{2}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-2*x+4)*log(x)*log(log(x)/x^2)-4*log(x)+2)/x/log(x)/log(log(x)/x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.63, size = 26, normalized size = 0.87 \begin {gather*} 2 \, x^{2} - 2 \, x + 4 \, \log \relax (x) + 2 \, \log \left (2 \, \log \relax (x) - \log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-2*x+4)*log(x)*log(log(x)/x^2)-4*log(x)+2)/x/log(x)/log(log(x)/x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 24, normalized size = 0.80

method	result	size
norman	\(4 \ln \relax (x )-2 x +2 x^{2}+2 \ln \left (\ln \left (\frac {\ln \relax (x )}{x^{2}}\right )\right )\)	\(24\)
risch	\(2 x^{2}-2 x +4 \ln \relax (x )+2 \ln \left (\ln \left (\ln \relax (x )\right )+\frac {i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x^{2}}\right )+\pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x^{2}}\right )^{2}+\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x^{2}}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i \ln \relax (x )}{x^{2}}\right )^{3}+4 i \ln \relax (x )\right )}{2}\right )\)	\(152\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2-2*x+4)*ln(x)*ln(ln(x)/x^2)-4*ln(x)+2)/x/ln(x)/ln(ln(x)/x^2),x,method=_RETURNVERBOSE)

[Out]

4*ln(x)-2*x+2*x^2+2*ln(ln(ln(x)/x^2))

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maxima [A]  time = 0.44, size = 24, normalized size = 0.80 \begin {gather*} 2 \, x^{2} - 2 \, x + 4 \, \log \relax (x) + 2 \, \log \left (-2 \, \log \relax (x) + \log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2-2*x+4)*log(x)*log(log(x)/x^2)-4*log(x)+2)/x/log(x)/log(log(x)/x^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.18, size = 23, normalized size = 0.77 \begin {gather*} 4\,\ln \relax (x)-2\,x+2\,\ln \left (\ln \left (\frac {\ln \relax (x)}{x^2}\right )\right )+2\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(x)/x^2)*log(x)*(4*x^2 - 2*x + 4) - 4*log(x) + 2)/(x*log(log(x)/x^2)*log(x)),x)

[Out]

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

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sympy [A]  time = 0.31, size = 24, normalized size = 0.80 \begin {gather*} 2 x^{2} - 2 x + 4 \log {\relax (x )} + 2 \log {\left (\log {\left (\frac {\log {\relax (x )}}{x^{2}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2-2*x+4)*ln(x)*ln(ln(x)/x**2)-4*ln(x)+2)/x/ln(x)/ln(ln(x)/x**2),x)

[Out]

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

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