3.60.43 \(\int \frac {-10 x^2+(10 \log (x)-10 \log ^2(x)) \log (\frac {x^2}{16})}{3 x^3 \log (\frac {x^2}{16})} \, dx\)

Optimal. Leaf size=25 \[ \frac {5}{3} \left (15+\frac {\log ^2(x)}{x^2}-\log \left (\log \left (\frac {x^2}{16}\right )\right )\right ) \]

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Rubi [A]  time = 0.17, antiderivative size = 38, normalized size of antiderivative = 1.52, number of steps used = 10, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {12, 6688, 14, 2304, 2366, 2303, 2302, 29} \begin {gather*} -\frac {5 (1-\log (x)) \log (x)}{3 x^2}+\frac {5 \log (x)}{3 x^2}-\frac {5}{3} \log \left (\log \left (\frac {x^2}{16}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*x^2 + (10*Log[x] - 10*Log[x]^2)*Log[x^2/16])/(3*x^3*Log[x^2/16]),x]

[Out]

(5*Log[x])/(3*x^2) - (5*(1 - Log[x])*Log[x])/(3*x^2) - (5*Log[Log[x^2/16]])/3

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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2303

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*(d*x)^(m + 1)*Log[c*x^n])/(
d*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
NeQ[d, 0])

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} &=\frac {1}{3} \int \frac {-10 x^2+\left (10 \log (x)-10 \log ^2(x)\right ) \log \left (\frac {x^2}{16}\right )}{x^3 \log \left (\frac {x^2}{16}\right )} \, dx\\ &=\frac {1}{3} \int \frac {10 \left (\log (x)-\log ^2(x)-\frac {x^2}{\log \left (\frac {x^2}{16}\right )}\right )}{x^3} \, dx\\ &=\frac {10}{3} \int \frac {\log (x)-\log ^2(x)-\frac {x^2}{\log \left (\frac {x^2}{16}\right )}}{x^3} \, dx\\ &=\frac {10}{3} \int \left (-\frac {(-1+\log (x)) \log (x)}{x^3}-\frac {1}{x \log \left (\frac {x^2}{16}\right )}\right ) \, dx\\ &=-\left (\frac {10}{3} \int \frac {(-1+\log (x)) \log (x)}{x^3} \, dx\right )-\frac {10}{3} \int \frac {1}{x \log \left (\frac {x^2}{16}\right )} \, dx\\ &=\frac {5 \log (x)}{6 x^2}-\frac {5 (1-\log (x)) \log (x)}{3 x^2}-\frac {5}{3} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {x^2}{16}\right )\right )+\frac {10}{3} \int \frac {1-2 \log (x)}{4 x^3} \, dx\\ &=\frac {5 \log (x)}{6 x^2}-\frac {5 (1-\log (x)) \log (x)}{3 x^2}-\frac {5}{3} \log \left (\log \left (\frac {x^2}{16}\right )\right )+\frac {5}{6} \int \frac {1-2 \log (x)}{x^3} \, dx\\ &=\frac {5 \log (x)}{3 x^2}-\frac {5 (1-\log (x)) \log (x)}{3 x^2}-\frac {5}{3} \log \left (\log \left (\frac {x^2}{16}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 25, normalized size = 1.00 \begin {gather*} \frac {5 \log ^2(x)}{3 x^2}-\frac {5}{3} \log \left (\log \left (\frac {x^2}{16}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x^2 + (10*Log[x] - 10*Log[x]^2)*Log[x^2/16])/(3*x^3*Log[x^2/16]),x]

[Out]

(5*Log[x]^2)/(3*x^2) - (5*Log[Log[x^2/16]])/3

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fricas [A]  time = 1.02, size = 26, normalized size = 1.04 \begin {gather*} -\frac {5 \, {\left (x^{2} \log \left (-4 \, \log \relax (2) + 2 \, \log \relax (x)\right ) - \log \relax (x)^{2}\right )}}{3 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-10*log(x)^2+10*log(x))*log(1/16*x^2)-10*x^2)/x^3/log(1/16*x^2),x, algorithm="fricas")

[Out]

-5/3*(x^2*log(-4*log(2) + 2*log(x)) - log(x)^2)/x^2

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giac [A]  time = 0.16, size = 22, normalized size = 0.88 \begin {gather*} \frac {5 \, \log \relax (x)^{2}}{3 \, x^{2}} - \frac {5}{3} \, \log \left (-4 \, \log \relax (2) + 2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-10*log(x)^2+10*log(x))*log(1/16*x^2)-10*x^2)/x^3/log(1/16*x^2),x, algorithm="giac")

[Out]

5/3*log(x)^2/x^2 - 5/3*log(-4*log(2) + 2*log(x))

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maple [A]  time = 0.06, size = 23, normalized size = 0.92

method	result	size
default	\(\frac {5 \ln \relax (x )^{2}}{3 x^{2}}-\frac {5 \ln \left (-4 \ln \relax (2)+\ln \left (x^{2}\right )\right )}{3}\)	\(23\)
risch	\(\frac {5 \ln \relax (x )^{2}}{3 x^{2}}-\frac {5 \ln \left (\ln \relax (x )-\frac {i \left (\pi \mathrm {csgn}\left (i x^{2}\right )^{3}+\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-8 i \ln \relax (2)\right )}{4}\right )}{3}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-10*ln(x)^2+10*ln(x))*ln(1/16*x^2)-10*x^2)/x^3/ln(1/16*x^2),x,method=_RETURNVERBOSE)

[Out]

5/3*ln(x)^2/x^2-5/3*ln(-4*ln(2)+ln(x^2))

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maxima [A]  time = 0.44, size = 39, normalized size = 1.56 \begin {gather*} \frac {5 \, {\left (2 \, \log \relax (x)^{2} + 2 \, \log \relax (x) + 1\right )}}{6 \, x^{2}} - \frac {5 \, \log \relax (x)}{3 \, x^{2}} - \frac {5}{6 \, x^{2}} - \frac {5}{3} \, \log \left (\log \left (\frac {1}{16} \, x^{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-10*log(x)^2+10*log(x))*log(1/16*x^2)-10*x^2)/x^3/log(1/16*x^2),x, algorithm="maxima")

[Out]

5/6*(2*log(x)^2 + 2*log(x) + 1)/x^2 - 5/3*log(x)/x^2 - 5/6/x^2 - 5/3*log(log(1/16*x^2))

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mupad [B]  time = 4.39, size = 19, normalized size = 0.76 \begin {gather*} \frac {5\,{\ln \relax (x)}^2}{3\,x^2}-\frac {5\,\ln \left (\ln \left (\frac {x^2}{16}\right )\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((log(x^2/16)*(10*log(x) - 10*log(x)^2))/3 - (10*x^2)/3)/(x^3*log(x^2/16)),x)

[Out]

(5*log(x)^2)/(3*x^2) - (5*log(log(x^2/16)))/3

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sympy [A]  time = 0.29, size = 24, normalized size = 0.96 \begin {gather*} - \frac {5 \log {\left (\log {\relax (x )} - 2 \log {\relax (2 )} \right )}}{3} + \frac {5 \log {\relax (x )}^{2}}{3 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-10*ln(x)**2+10*ln(x))*ln(1/16*x**2)-10*x**2)/x**3/ln(1/16*x**2),x)

[Out]

-5*log(log(x) - 2*log(2))/3 + 5*log(x)**2/(3*x**2)

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