3.97.33 \(\int \frac {(2-6 x^3-4 \log (x \log (4))) \log (\frac {9 x^2+3 x^3-\log (x \log (4))}{x^2})}{-9 x^3-3 x^4+x \log (x \log (4))} \, dx\)

Optimal. Leaf size=18 \[ \log ^2\left (9+3 x-\frac {\log (x \log (4))}{x^2}\right ) \]

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Rubi [A]  time = 0.33, antiderivative size = 25, normalized size of antiderivative = 1.39, number of steps used = 1, number of rules used = 5, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {6741, 12, 6742, 6684, 6686} \begin {gather*} \log ^2\left (\frac {3 x^3+9 x^2-\log (x \log (4))}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2 - 6*x^3 - 4*Log[x*Log[4]])*Log[(9*x^2 + 3*x^3 - Log[x*Log[4]])/x^2])/(-9*x^3 - 3*x^4 + x*Log[x*Log[4]]
),x]

[Out]

Log[(9*x^2 + 3*x^3 - Log[x*Log[4]])/x^2]^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6684

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

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 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} &=\log ^2\left (\frac {9 x^2+3 x^3-\log (x \log (4))}{x^2}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 18, normalized size = 1.00 \begin {gather*} \log ^2\left (9+3 x-\frac {\log (x \log (4))}{x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2 - 6*x^3 - 4*Log[x*Log[4]])*Log[(9*x^2 + 3*x^3 - Log[x*Log[4]])/x^2])/(-9*x^3 - 3*x^4 + x*Log[x*L
og[4]]),x]

[Out]

Log[9 + 3*x - Log[x*Log[4]]/x^2]^2

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fricas [A]  time = 0.61, size = 26, normalized size = 1.44 \begin {gather*} \log \left (\frac {3 \, x^{3} + 9 \, x^{2} - \log \left (2 \, x \log \relax (2)\right )}{x^{2}}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(2*x*log(2))-6*x^3+2)*log((-log(2*x*log(2))+3*x^3+9*x^2)/x^2)/(x*log(2*x*log(2))-3*x^4-9*x^3)
,x, algorithm="fricas")

[Out]

log((3*x^3 + 9*x^2 - log(2*x*log(2)))/x^2)^2

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giac [B]  time = 0.21, size = 79, normalized size = 4.39 \begin {gather*} 2 \, {\left (\log \left (-3 \, x^{3} - 9 \, x^{2} + \log \relax (2) + \log \relax (x) + \log \left (\log \relax (2)\right )\right ) - 2 \, \log \relax (x)\right )} \log \left (3 \, x^{3} + 9 \, x^{2} - \log \relax (2) - \log \left (x \log \relax (2)\right )\right ) - \log \left (-3 \, x^{3} - 9 \, x^{2} + \log \relax (2) + \log \relax (x) + \log \left (\log \relax (2)\right )\right )^{2} + 4 \, \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(2*x*log(2))-6*x^3+2)*log((-log(2*x*log(2))+3*x^3+9*x^2)/x^2)/(x*log(2*x*log(2))-3*x^4-9*x^3)
,x, algorithm="giac")

[Out]

2*(log(-3*x^3 - 9*x^2 + log(2) + log(x) + log(log(2))) - 2*log(x))*log(3*x^3 + 9*x^2 - log(2) - log(x*log(2)))
- log(-3*x^3 - 9*x^2 + log(2) + log(x) + log(log(2)))^2 + 4*log(x)^2

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (-4 \ln \left (2 x \ln \relax (2)\right )-6 x^{3}+2\right ) \ln \left (\frac {-\ln \left (2 x \ln \relax (2)\right )+3 x^{3}+9 x^{2}}{x^{2}}\right )}{x \ln \left (2 x \ln \relax (2)\right )-3 x^{4}-9 x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*ln(2*x*ln(2))-6*x^3+2)*ln((-ln(2*x*ln(2))+3*x^3+9*x^2)/x^2)/(x*ln(2*x*ln(2))-3*x^4-9*x^3),x)

[Out]

int((-4*ln(2*x*ln(2))-6*x^3+2)*ln((-ln(2*x*ln(2))+3*x^3+9*x^2)/x^2)/(x*ln(2*x*ln(2))-3*x^4-9*x^3),x)

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maxima [B]  time = 0.48, size = 103, normalized size = 5.72 \begin {gather*} -\log \left (-3 \, x^{3} - 9 \, x^{2} + \log \relax (2) + \log \relax (x) + \log \left (\log \relax (2)\right )\right )^{2} + 4 \, \log \left (-3 \, x^{3} - 9 \, x^{2} + \log \relax (2) + \log \relax (x) + \log \left (\log \relax (2)\right )\right ) \log \relax (x) - 4 \, \log \relax (x)^{2} + 2 \, {\left (\log \left (-3 \, x^{3} - 9 \, x^{2} + \log \relax (2) + \log \relax (x) + \log \left (\log \relax (2)\right )\right ) - 2 \, \log \relax (x)\right )} \log \left (\frac {3 \, x^{3} + 9 \, x^{2} - \log \left (2 \, x \log \relax (2)\right )}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*log(2*x*log(2))-6*x^3+2)*log((-log(2*x*log(2))+3*x^3+9*x^2)/x^2)/(x*log(2*x*log(2))-3*x^4-9*x^3)
,x, algorithm="maxima")

[Out]

-log(-3*x^3 - 9*x^2 + log(2) + log(x) + log(log(2)))^2 + 4*log(-3*x^3 - 9*x^2 + log(2) + log(x) + log(log(2)))
*log(x) - 4*log(x)^2 + 2*(log(-3*x^3 - 9*x^2 + log(2) + log(x) + log(log(2))) - 2*log(x))*log((3*x^3 + 9*x^2 -
log(2*x*log(2)))/x^2)

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mupad [B]  time = 8.00, size = 26, normalized size = 1.44 \begin {gather*} {\ln \left (\frac {9\,x^2-\ln \left (2\,x\,\ln \relax (2)\right )+3\,x^3}{x^2}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((9*x^2 - log(2*x*log(2)) + 3*x^3)/x^2)*(4*log(2*x*log(2)) + 6*x^3 - 2))/(9*x^3 - x*log(2*x*log(2)) +
3*x^4),x)

[Out]

log((9*x^2 - log(2*x*log(2)) + 3*x^3)/x^2)^2

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sympy [A]  time = 0.39, size = 24, normalized size = 1.33 \begin {gather*} \log {\left (\frac {3 x^{3} + 9 x^{2} - \log {\left (2 x \log {\relax (2 )} \right )}}{x^{2}} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*ln(2*x*ln(2))-6*x**3+2)*ln((-ln(2*x*ln(2))+3*x**3+9*x**2)/x**2)/(x*ln(2*x*ln(2))-3*x**4-9*x**3),
x)

[Out]

log((3*x**3 + 9*x**2 - log(2*x*log(2)))/x**2)**2

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