3.97.40 \(\int \frac {(-10-20 x-10 \log (x)) \log (2 \log (4))}{3 x^4+6 x^3 \log (x)+3 x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=18 \[ \frac {10 \log (2 \log (4))}{3 x (x+\log (x))} \]

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Rubi [A]  time = 0.11, antiderivative size = 16, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 3, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 6688, 6687} \begin {gather*} \frac {10 \log (\log (16))}{3 x (x+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-10 - 20*x - 10*Log[x])*Log[2*Log[4]])/(3*x^4 + 6*x^3*Log[x] + 3*x^2*Log[x]^2),x]

[Out]

(10*Log[Log[16]])/(3*x*(x + Log[x]))

Rule 12

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

Rule 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, 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 (\log (16)) \int \frac {-10-20 x-10 \log (x)}{3 x^4+6 x^3 \log (x)+3 x^2 \log ^2(x)} \, dx\\ &=\log (\log (16)) \int \frac {10 (-1-2 x-\log (x))}{3 x^2 (x+\log (x))^2} \, dx\\ &=\frac {1}{3} (10 \log (\log (16))) \int \frac {-1-2 x-\log (x)}{x^2 (x+\log (x))^2} \, dx\\ &=\frac {10 \log (\log (16))}{3 x (x+\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 16, normalized size = 0.89 \begin {gather*} \frac {10 \log (\log (16))}{3 x (x+\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-10 - 20*x - 10*Log[x])*Log[2*Log[4]])/(3*x^4 + 6*x^3*Log[x] + 3*x^2*Log[x]^2),x]

[Out]

(10*Log[Log[16]])/(3*x*(x + Log[x]))

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fricas [A]  time = 0.43, size = 17, normalized size = 0.94 \begin {gather*} \frac {10 \, \log \left (4 \, \log \relax (2)\right )}{3 \, {\left (x^{2} + x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(x)-20*x-10)*log(4*log(2))/(3*x^2*log(x)^2+6*x^3*log(x)+3*x^4),x, algorithm="fricas")

[Out]

10/3*log(4*log(2))/(x^2 + x*log(x))

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giac [A]  time = 0.13, size = 17, normalized size = 0.94 \begin {gather*} \frac {10 \, \log \left (4 \, \log \relax (2)\right )}{3 \, {\left (x^{2} + x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(x)-20*x-10)*log(4*log(2))/(3*x^2*log(x)^2+6*x^3*log(x)+3*x^4),x, algorithm="giac")

[Out]

10/3*log(4*log(2))/(x^2 + x*log(x))

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maple [A]  time = 0.06, size = 20, normalized size = 1.11

method	result	size
risch	\(\frac {\frac {20 \ln \relax (2)}{3}+\frac {10 \ln \left (\ln \relax (2)\right )}{3}}{\left (x +\ln \relax (x )\right ) x}\)	\(20\)
norman	\(\frac {\frac {20 \ln \relax (2)}{3}+\frac {10 \ln \left (\ln \relax (2)\right )}{3}}{\left (x +\ln \relax (x )\right ) x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-10*ln(x)-20*x-10)*ln(4*ln(2))/(3*x^2*ln(x)^2+6*x^3*ln(x)+3*x^4),x,method=_RETURNVERBOSE)

[Out]

10/3*(2*ln(2)+ln(ln(2)))/(x+ln(x))/x

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maxima [A]  time = 0.38, size = 17, normalized size = 0.94 \begin {gather*} \frac {10 \, \log \left (4 \, \log \relax (2)\right )}{3 \, {\left (x^{2} + x \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(x)-20*x-10)*log(4*log(2))/(3*x^2*log(x)^2+6*x^3*log(x)+3*x^4),x, algorithm="maxima")

[Out]

10/3*log(4*log(2))/(x^2 + x*log(x))

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mupad [B]  time = 5.66, size = 18, normalized size = 1.00 \begin {gather*} \frac {10\,\ln \left (\ln \left (16\right )\right )}{3\,x\,\ln \relax (x)+3\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(4*log(2))*(20*x + 10*log(x) + 10))/(6*x^3*log(x) + 3*x^2*log(x)^2 + 3*x^4),x)

[Out]

(10*log(log(16)))/(3*x*log(x) + 3*x^2)

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sympy [A]  time = 0.11, size = 22, normalized size = 1.22 \begin {gather*} \frac {10 \log {\left (\log {\relax (2 )} \right )} + 20 \log {\relax (2 )}}{3 x^{2} + 3 x \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*ln(x)-20*x-10)*ln(4*ln(2))/(3*x**2*ln(x)**2+6*x**3*ln(x)+3*x**4),x)

[Out]

(10*log(log(2)) + 20*log(2))/(3*x**2 + 3*x*log(x))

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