3.51.40 \(\int \frac {-6-4 \log (\frac {x^3}{2})-3 x^3 \log ^2(\frac {x^3}{2})}{12 x^3 \log ^2(\frac {x^3}{2})} \, dx\)

Optimal. Leaf size=26 \[ \frac {1}{4} \left (-4-x+\frac {2}{3 x^2 \log \left (\frac {x^3}{2}\right )}\right ) \]

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Rubi [A]  time = 0.26, antiderivative size = 23, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 5, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {12, 6742, 2306, 2310, 2178} \begin {gather*} \frac {1}{6 x^2 \log \left (\frac {x^3}{2}\right )}-\frac {x}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-6 - 4*Log[x^3/2] - 3*x^3*Log[x^3/2]^2)/(12*x^3*Log[x^3/2]^2),x]

[Out]

-1/4*x + 1/(6*x^2*Log[x^3/2])

Rule 12

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

Rule 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 2310

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol] :> Dist[(d*x)^(m + 1)/(d*n*(c*x^n
)^((m + 1)/n)), Subst[Int[E^(((m + 1)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, 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} &=\frac {1}{12} \int \frac {-6-4 \log \left (\frac {x^3}{2}\right )-3 x^3 \log ^2\left (\frac {x^3}{2}\right )}{x^3 \log ^2\left (\frac {x^3}{2}\right )} \, dx\\ &=\frac {1}{12} \int \left (-3-\frac {6}{x^3 \log ^2\left (\frac {x^3}{2}\right )}-\frac {4}{x^3 \log \left (\frac {x^3}{2}\right )}\right ) \, dx\\ &=-\frac {x}{4}-\frac {1}{3} \int \frac {1}{x^3 \log \left (\frac {x^3}{2}\right )} \, dx-\frac {1}{2} \int \frac {1}{x^3 \log ^2\left (\frac {x^3}{2}\right )} \, dx\\ &=-\frac {x}{4}+\frac {1}{6 x^2 \log \left (\frac {x^3}{2}\right )}+\frac {1}{3} \int \frac {1}{x^3 \log \left (\frac {x^3}{2}\right )} \, dx-\frac {\left (x^3\right )^{2/3} \operatorname {Subst}\left (\int \frac {e^{-2 x/3}}{x} \, dx,x,\log \left (\frac {x^3}{2}\right )\right )}{9\ 2^{2/3} x^2}\\ &=-\frac {x}{4}-\frac {\left (x^3\right )^{2/3} \text {Ei}\left (-\frac {2}{3} \log \left (\frac {x^3}{2}\right )\right )}{9\ 2^{2/3} x^2}+\frac {1}{6 x^2 \log \left (\frac {x^3}{2}\right )}+\frac {\left (x^3\right )^{2/3} \operatorname {Subst}\left (\int \frac {e^{-2 x/3}}{x} \, dx,x,\log \left (\frac {x^3}{2}\right )\right )}{9\ 2^{2/3} x^2}\\ &=-\frac {x}{4}+\frac {1}{6 x^2 \log \left (\frac {x^3}{2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 34, normalized size = 1.31 \begin {gather*} \frac {6+x^3 \log (512)-9 x^3 \log \left (x^3\right )}{36 x^2 \log \left (\frac {x^3}{2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-6 - 4*Log[x^3/2] - 3*x^3*Log[x^3/2]^2)/(12*x^3*Log[x^3/2]^2),x]

[Out]

(6 + x^3*Log[512] - 9*x^3*Log[x^3])/(36*x^2*Log[x^3/2])

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fricas [A]  time = 0.85, size = 26, normalized size = 1.00 \begin {gather*} -\frac {3 \, x^{3} \log \left (\frac {1}{2} \, x^{3}\right ) - 2}{12 \, x^{2} \log \left (\frac {1}{2} \, x^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(-3*x^3*log(1/2*x^3)^2-4*log(1/2*x^3)-6)/x^3/log(1/2*x^3)^2,x, algorithm="fricas")

[Out]

-1/12*(3*x^3*log(1/2*x^3) - 2)/(x^2*log(1/2*x^3))

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giac [A]  time = 0.21, size = 17, normalized size = 0.65 \begin {gather*} -\frac {1}{4} \, x + \frac {1}{6 \, x^{2} \log \left (\frac {1}{2} \, x^{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(-3*x^3*log(1/2*x^3)^2-4*log(1/2*x^3)-6)/x^3/log(1/2*x^3)^2,x, algorithm="giac")

[Out]

-1/4*x + 1/6/(x^2*log(1/2*x^3))

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maple [A]  time = 0.03, size = 18, normalized size = 0.69

method	result	size
risch	\(-\frac {x}{4}+\frac {1}{6 \ln \left (\frac {x^{3}}{2}\right ) x^{2}}\)	\(18\)
norman	\(\frac {\frac {1}{6}-\frac {x^{3} \ln \left (\frac {x^{3}}{2}\right )}{4}}{x^{2} \ln \left (\frac {x^{3}}{2}\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/12*(-3*x^3*ln(1/2*x^3)^2-4*ln(1/2*x^3)-6)/x^3/ln(1/2*x^3)^2,x,method=_RETURNVERBOSE)

[Out]

-1/4*x+1/6/ln(1/2*x^3)/x^2

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maxima [A]  time = 0.46, size = 22, normalized size = 0.85 \begin {gather*} -\frac {1}{4} \, x - \frac {1}{6 \, {\left (x^{2} \log \relax (2) - 3 \, x^{2} \log \relax (x)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(-3*x^3*log(1/2*x^3)^2-4*log(1/2*x^3)-6)/x^3/log(1/2*x^3)^2,x, algorithm="maxima")

[Out]

-1/4*x - 1/6/(x^2*log(2) - 3*x^2*log(x))

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mupad [B]  time = 3.68, size = 17, normalized size = 0.65 \begin {gather*} \frac {1}{6\,x^2\,\ln \left (\frac {x^3}{2}\right )}-\frac {x}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^3/2)/3 + (x^3*log(x^3/2)^2)/4 + 1/2)/(x^3*log(x^3/2)^2),x)

[Out]

1/(6*x^2*log(x^3/2)) - x/4

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sympy [A]  time = 0.10, size = 15, normalized size = 0.58 \begin {gather*} - \frac {x}{4} + \frac {1}{6 x^{2} \log {\left (\frac {x^{3}}{2} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/12*(-3*x**3*ln(1/2*x**3)**2-4*ln(1/2*x**3)-6)/x**3/ln(1/2*x**3)**2,x)

[Out]

-x/4 + 1/(6*x**2*log(x**3/2))

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