3.92.72 \(\int \frac {\log (5 e^{-2 x^3-5 x^2 \log (\log (12 x^2))}) (-20 x-12 x^2 \log (12 x^2)-20 x \log (12 x^2) \log (\log (12 x^2)))}{\log (12 x^2)} \, dx\)

Optimal. Leaf size=25 \[ \log ^2\left (5 e^{-x^2 \left (2 x+5 \log \left (\log \left (12 x^2\right )\right )\right )}\right ) \]

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Rubi [A]  time = 0.20, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 1, number of rules used = 7, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6688, 12, 14, 2307, 2298, 2522, 6686} \begin {gather*} \log ^2\left (5 e^{-2 x^3} \log ^{-5 x^2}\left (12 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Log[5*E^(-2*x^3 - 5*x^2*Log[Log[12*x^2]])]*(-20*x - 12*x^2*Log[12*x^2] - 20*x*Log[12*x^2]*Log[Log[12*x^2]
]))/Log[12*x^2],x]

[Out]

Log[5/(E^(2*x^3)*Log[12*x^2]^(5*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 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 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2307

Int[(x_)^(m_.)/Log[(c_.)*(x_)^(n_)], x_Symbol] :> Dist[1/n, Subst[Int[1/Log[c*x], x], x, x^n], x] /; FreeQ[{c,
m, n}, x] && EqQ[m, n - 1]

Rule 2522

Int[((a_.) + Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)]*(b_.))*((e_.)*(x_))^(m_.), x_Symbol] :> Simp[((e*x)^(m + 1
)*(a + b*Log[c*Log[d*x^n]^p]))/(e*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(e*x)^m/Log[d*x^n], x], x] /; FreeQ
[{a, b, c, d, e, m, n, p}, x] && NeQ[m, -1]

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log ^2\left (5 e^{-2 x^3} \log ^{-5 x^2}\left (12 x^2\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 24, normalized size = 0.96 \begin {gather*} \log ^2\left (5 e^{-2 x^3} \log ^{-5 x^2}\left (12 x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[5*E^(-2*x^3 - 5*x^2*Log[Log[12*x^2]])]*(-20*x - 12*x^2*Log[12*x^2] - 20*x*Log[12*x^2]*Log[Log[1
2*x^2]]))/Log[12*x^2],x]

[Out]

Log[5/(E^(2*x^3)*Log[12*x^2]^(5*x^2))]^2

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fricas [B]  time = 0.64, size = 49, normalized size = 1.96 \begin {gather*} 4 \, x^{6} + 25 \, x^{4} \log \left (\log \left (12 \, x^{2}\right )\right )^{2} - 4 \, x^{3} \log \relax (5) + 10 \, {\left (2 \, x^{5} - x^{2} \log \relax (5)\right )} \log \left (\log \left (12 \, x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*log(12*x^2)*log(log(12*x^2))-12*x^2*log(12*x^2)-20*x)*log(5/exp(5*x^2*log(log(12*x^2))+2*x^3)
)/log(12*x^2),x, algorithm="fricas")

[Out]

4*x^6 + 25*x^4*log(log(12*x^2))^2 - 4*x^3*log(5) + 10*(2*x^5 - x^2*log(5))*log(log(12*x^2))

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giac [B]  time = 1.07, size = 49, normalized size = 1.96 \begin {gather*} 4 \, x^{6} + 25 \, x^{4} \log \left (\log \left (12 \, x^{2}\right )\right )^{2} - 4 \, x^{3} \log \relax (5) + 10 \, {\left (2 \, x^{5} - x^{2} \log \relax (5)\right )} \log \left (\log \left (12 \, x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*log(12*x^2)*log(log(12*x^2))-12*x^2*log(12*x^2)-20*x)*log(5/exp(5*x^2*log(log(12*x^2))+2*x^3)
)/log(12*x^2),x, algorithm="giac")

[Out]

4*x^6 + 25*x^4*log(log(12*x^2))^2 - 4*x^3*log(5) + 10*(2*x^5 - x^2*log(5))*log(log(12*x^2))

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maple [F]  time = 0.56, size = 0, normalized size = 0.00 \[\int \frac {\left (-20 x \ln \left (12 x^{2}\right ) \ln \left (\ln \left (12 x^{2}\right )\right )-12 x^{2} \ln \left (12 x^{2}\right )-20 x \right ) \ln \left (5 \,{\mathrm e}^{-5 x^{2} \ln \left (\ln \left (12 x^{2}\right )\right )-2 x^{3}}\right )}{\ln \left (12 x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-20*x*ln(12*x^2)*ln(ln(12*x^2))-12*x^2*ln(12*x^2)-20*x)*ln(5/exp(5*x^2*ln(ln(12*x^2))+2*x^3))/ln(12*x^2),
x)

[Out]

int((-20*x*ln(12*x^2)*ln(ln(12*x^2))-12*x^2*ln(12*x^2)-20*x)*ln(5/exp(5*x^2*ln(ln(12*x^2))+2*x^3))/ln(12*x^2),
x)

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maxima [B]  time = 0.50, size = 89, normalized size = 3.56 \begin {gather*} -4 \, x^{6} - 20 \, x^{5} \log \left (\log \relax (3) + 2 \, \log \relax (2) + 2 \, \log \relax (x)\right ) - 25 \, x^{4} \log \left (\log \relax (3) + 2 \, \log \relax (2) + 2 \, \log \relax (x)\right )^{2} - 2 \, {\left (2 \, x^{3} + 5 \, x^{2} \log \left (\log \relax (3) + 2 \, \log \relax (2) + 2 \, \log \relax (x)\right )\right )} \log \left (5 \, e^{\left (-2 \, x^{3} - 5 \, x^{2} \log \left (\log \left (12 \, x^{2}\right )\right )\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*log(12*x^2)*log(log(12*x^2))-12*x^2*log(12*x^2)-20*x)*log(5/exp(5*x^2*log(log(12*x^2))+2*x^3)
)/log(12*x^2),x, algorithm="maxima")

[Out]

-4*x^6 - 20*x^5*log(log(3) + 2*log(2) + 2*log(x)) - 25*x^4*log(log(3) + 2*log(2) + 2*log(x))^2 - 2*(2*x^3 + 5*
x^2*log(log(3) + 2*log(2) + 2*log(x)))*log(5*e^(-2*x^3 - 5*x^2*log(log(12*x^2))))

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mupad [B]  time = 7.48, size = 104, normalized size = 4.16 \begin {gather*} 25\,x^4\,{\ln \left (\ln \left (12\,x^2\right )\right )}^2+4\,x^6-4\,x^3\,\left (\ln \left (\frac {5}{{\ln \left (12\,x^2\right )}^{5\,x^2}}\right )+5\,x^2\,\ln \left (\ln \left (12\,x^2\right )\right )\right )+\ln \left (\ln \left (12\,x^2\right )\right )\,\left (20\,x^5-10\,x^2\,\left (\ln \left (\frac {5}{{\ln \left (12\,x^2\right )}^{5\,x^2}}\right )+5\,x^2\,\ln \left (\ln \left (12\,x^2\right )\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5*exp(- 5*x^2*log(log(12*x^2)) - 2*x^3))*(20*x + 12*x^2*log(12*x^2) + 20*x*log(log(12*x^2))*log(12*x
^2)))/log(12*x^2),x)

[Out]

25*x^4*log(log(12*x^2))^2 + 4*x^6 - 4*x^3*(log(5/log(12*x^2)^(5*x^2)) + 5*x^2*log(log(12*x^2))) + log(log(12*x
^2))*(20*x^5 - 10*x^2*(log(5/log(12*x^2)^(5*x^2)) + 5*x^2*log(log(12*x^2))))

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sympy [B]  time = 0.94, size = 49, normalized size = 1.96 \begin {gather*} 4 x^{6} + 25 x^{4} \log {\left (\log {\left (12 x^{2} \right )} \right )}^{2} - 4 x^{3} \log {\relax (5 )} + \left (20 x^{5} - 10 x^{2} \log {\relax (5 )}\right ) \log {\left (\log {\left (12 x^{2} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*x*ln(12*x**2)*ln(ln(12*x**2))-12*x**2*ln(12*x**2)-20*x)*ln(5/exp(5*x**2*ln(ln(12*x**2))+2*x**3)
)/ln(12*x**2),x)

[Out]

4*x**6 + 25*x**4*log(log(12*x**2))**2 - 4*x**3*log(5) + (20*x**5 - 10*x**2*log(5))*log(log(12*x**2))

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