3.32.81 \(\int \frac {960-40 \log (4 x)-60 \log ^2(4 x)}{768 x^4-96 x^4 \log ^2(4 x)+3 x^4 \log ^4(4 x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {5}{3 x^3 \left (-4+\frac {1}{4} \log ^2(4 x)\right )} \]

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Rubi [A]  time = 0.58, antiderivative size = 33, normalized size of antiderivative = 1.57, number of steps used = 14, number of rules used = 6, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {6688, 12, 6742, 2306, 2309, 2178} \begin {gather*} -\frac {5}{6 x^3 (\log (4 x)+4)}-\frac {5}{6 x^3 (4-\log (4 x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(960 - 40*Log[4*x] - 60*Log[4*x]^2)/(768*x^4 - 96*x^4*Log[4*x]^2 + 3*x^4*Log[4*x]^4),x]

[Out]

-5/(6*x^3*(4 - Log[4*x])) - 5/(6*x^3*(4 + Log[4*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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 6688

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

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 {20 \left (48-2 \log (4 x)-3 \log ^2(4 x)\right )}{3 x^4 \left (16-\log ^2(4 x)\right )^2} \, dx\\ &=\frac {20}{3} \int \frac {48-2 \log (4 x)-3 \log ^2(4 x)}{x^4 \left (16-\log ^2(4 x)\right )^2} \, dx\\ &=\frac {20}{3} \int \left (-\frac {1}{8 x^4 (-4+\log (4 x))^2}-\frac {3}{8 x^4 (-4+\log (4 x))}+\frac {1}{8 x^4 (4+\log (4 x))^2}+\frac {3}{8 x^4 (4+\log (4 x))}\right ) \, dx\\ &=-\left (\frac {5}{6} \int \frac {1}{x^4 (-4+\log (4 x))^2} \, dx\right )+\frac {5}{6} \int \frac {1}{x^4 (4+\log (4 x))^2} \, dx-\frac {5}{2} \int \frac {1}{x^4 (-4+\log (4 x))} \, dx+\frac {5}{2} \int \frac {1}{x^4 (4+\log (4 x))} \, dx\\ &=-\frac {5}{6 x^3 (4-\log (4 x))}-\frac {5}{6 x^3 (4+\log (4 x))}+\frac {5}{2} \int \frac {1}{x^4 (-4+\log (4 x))} \, dx-\frac {5}{2} \int \frac {1}{x^4 (4+\log (4 x))} \, dx-160 \operatorname {Subst}\left (\int \frac {e^{-3 x}}{-4+x} \, dx,x,\log (4 x)\right )+160 \operatorname {Subst}\left (\int \frac {e^{-3 x}}{4+x} \, dx,x,\log (4 x)\right )\\ &=-\frac {160 \text {Ei}(3 (4-\log (4 x)))}{e^{12}}+160 e^{12} \text {Ei}(-3 (4+\log (4 x)))-\frac {5}{6 x^3 (4-\log (4 x))}-\frac {5}{6 x^3 (4+\log (4 x))}+160 \operatorname {Subst}\left (\int \frac {e^{-3 x}}{-4+x} \, dx,x,\log (4 x)\right )-160 \operatorname {Subst}\left (\int \frac {e^{-3 x}}{4+x} \, dx,x,\log (4 x)\right )\\ &=-\frac {5}{6 x^3 (4-\log (4 x))}-\frac {5}{6 x^3 (4+\log (4 x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 17, normalized size = 0.81 \begin {gather*} \frac {20}{3 x^3 \left (-16+\log ^2(4 x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(960 - 40*Log[4*x] - 60*Log[4*x]^2)/(768*x^4 - 96*x^4*Log[4*x]^2 + 3*x^4*Log[4*x]^4),x]

[Out]

20/(3*x^3*(-16 + Log[4*x]^2))

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fricas [A]  time = 0.60, size = 20, normalized size = 0.95 \begin {gather*} \frac {20}{3 \, {\left (x^{3} \log \left (4 \, x\right )^{2} - 16 \, x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*log(4*x)^2-40*log(4*x)+960)/(3*x^4*log(4*x)^4-96*x^4*log(4*x)^2+768*x^4),x, algorithm="fricas")

[Out]

20/3/(x^3*log(4*x)^2 - 16*x^3)

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giac [A]  time = 0.68, size = 20, normalized size = 0.95 \begin {gather*} \frac {20}{3 \, {\left (x^{3} \log \left (4 \, x\right )^{2} - 16 \, x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*log(4*x)^2-40*log(4*x)+960)/(3*x^4*log(4*x)^4-96*x^4*log(4*x)^2+768*x^4),x, algorithm="giac")

[Out]

20/3/(x^3*log(4*x)^2 - 16*x^3)

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maple [A]  time = 0.04, size = 16, normalized size = 0.76

method	result	size
norman	\(\frac {20}{3 x^{3} \left (\ln \left (4 x \right )^{2}-16\right )}\)	\(16\)
risch	\(\frac {20}{3 x^{3} \left (\ln \left (4 x \right )^{2}-16\right )}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-60*ln(4*x)^2-40*ln(4*x)+960)/(3*x^4*ln(4*x)^4-96*x^4*ln(4*x)^2+768*x^4),x,method=_RETURNVERBOSE)

[Out]

20/3/x^3/(ln(4*x)^2-16)

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maxima [B]  time = 0.65, size = 33, normalized size = 1.57 \begin {gather*} \frac {20}{3 \, {\left (4 \, x^{3} \log \relax (2) \log \relax (x) + x^{3} \log \relax (x)^{2} + 4 \, {\left (\log \relax (2)^{2} - 4\right )} x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*log(4*x)^2-40*log(4*x)+960)/(3*x^4*log(4*x)^4-96*x^4*log(4*x)^2+768*x^4),x, algorithm="maxima")

[Out]

20/3/(4*x^3*log(2)*log(x) + x^3*log(x)^2 + 4*(log(2)^2 - 4)*x^3)

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mupad [B]  time = 1.96, size = 15, normalized size = 0.71 \begin {gather*} \frac {20}{3\,x^3\,\left ({\ln \left (4\,x\right )}^2-16\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(40*log(4*x) + 60*log(4*x)^2 - 960)/(768*x^4 - 96*x^4*log(4*x)^2 + 3*x^4*log(4*x)^4),x)

[Out]

20/(3*x^3*(log(4*x)^2 - 16))

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sympy [A]  time = 0.13, size = 17, normalized size = 0.81 \begin {gather*} \frac {20}{3 x^{3} \log {\left (4 x \right )}^{2} - 48 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-60*ln(4*x)**2-40*ln(4*x)+960)/(3*x**4*ln(4*x)**4-96*x**4*ln(4*x)**2+768*x**4),x)

[Out]

20/(3*x**3*log(4*x)**2 - 48*x**3)

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