3.15.43 \(\int \frac {-192 x^2-48 x^3+\log (2 x) (x+(-8-2 x) \log (4+x))}{(576 x^3+144 x^4) \log (2 x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {1}{3} \left (4+\frac {\log (4+x)}{48 x^2}-\log \left (\frac {\log (2 x)}{\log (5)}\right )\right ) \]

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Rubi [A]  time = 0.59, antiderivative size = 21, normalized size of antiderivative = 0.72, number of steps used = 15, number of rules used = 8, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1593, 6688, 12, 14, 44, 2302, 29, 2395} \begin {gather*} \frac {\log (x+4)}{144 x^2}-\frac {1}{3} \log (\log (2 x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-192*x^2 - 48*x^3 + Log[2*x]*(x + (-8 - 2*x)*Log[4 + x]))/((576*x^3 + 144*x^4)*Log[2*x]),x]

[Out]

Log[4 + x]/(144*x^2) - Log[Log[2*x]]/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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -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} &=\int \frac {-192 x^2-48 x^3+\log (2 x) (x+(-8-2 x) \log (4+x))}{x^3 (576+144 x) \log (2 x)} \, dx\\ &=\int \frac {\frac {x}{4+x}-\frac {48 x^2}{\log (2 x)}-2 \log (4+x)}{144 x^3} \, dx\\ &=\frac {1}{144} \int \frac {\frac {x}{4+x}-\frac {48 x^2}{\log (2 x)}-2 \log (4+x)}{x^3} \, dx\\ &=\frac {1}{144} \int \left (\frac {-192 x-48 x^2+\log (2 x)}{x^2 (4+x) \log (2 x)}-\frac {2 \log (4+x)}{x^3}\right ) \, dx\\ &=\frac {1}{144} \int \frac {-192 x-48 x^2+\log (2 x)}{x^2 (4+x) \log (2 x)} \, dx-\frac {1}{72} \int \frac {\log (4+x)}{x^3} \, dx\\ &=\frac {\log (4+x)}{144 x^2}-\frac {1}{144} \int \frac {1}{x^2 (4+x)} \, dx+\frac {1}{144} \int \frac {\frac {1}{4+x}-\frac {48 x}{\log (2 x)}}{x^2} \, dx\\ &=\frac {\log (4+x)}{144 x^2}-\frac {1}{144} \int \left (\frac {1}{4 x^2}-\frac {1}{16 x}+\frac {1}{16 (4+x)}\right ) \, dx+\frac {1}{144} \int \left (\frac {1}{x^2 (4+x)}-\frac {48}{x \log (2 x)}\right ) \, dx\\ &=\frac {1}{576 x}+\frac {\log (x)}{2304}-\frac {\log (4+x)}{2304}+\frac {\log (4+x)}{144 x^2}+\frac {1}{144} \int \frac {1}{x^2 (4+x)} \, dx-\frac {1}{3} \int \frac {1}{x \log (2 x)} \, dx\\ &=\frac {1}{576 x}+\frac {\log (x)}{2304}-\frac {\log (4+x)}{2304}+\frac {\log (4+x)}{144 x^2}+\frac {1}{144} \int \left (\frac {1}{4 x^2}-\frac {1}{16 x}+\frac {1}{16 (4+x)}\right ) \, dx-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (2 x)\right )\\ &=\frac {\log (4+x)}{144 x^2}-\frac {1}{3} \log (\log (2 x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 20, normalized size = 0.69 \begin {gather*} \frac {1}{144} \left (\frac {\log (4+x)}{x^2}-48 \log (\log (2 x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-192*x^2 - 48*x^3 + Log[2*x]*(x + (-8 - 2*x)*Log[4 + x]))/((576*x^3 + 144*x^4)*Log[2*x]),x]

[Out]

(Log[4 + x]/x^2 - 48*Log[Log[2*x]])/144

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fricas [A]  time = 0.56, size = 22, normalized size = 0.76 \begin {gather*} -\frac {48 \, x^{2} \log \left (\log \left (2 \, x\right )\right ) - \log \left (x + 4\right )}{144 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-8)*log(4+x)+x)*log(2*x)-48*x^3-192*x^2)/(144*x^4+576*x^3)/log(2*x),x, algorithm="fricas")

[Out]

-1/144*(48*x^2*log(log(2*x)) - log(x + 4))/x^2

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giac [A]  time = 0.67, size = 17, normalized size = 0.59 \begin {gather*} \frac {\log \left (x + 4\right )}{144 \, x^{2}} - \frac {1}{3} \, \log \left (\log \left (2 \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-8)*log(4+x)+x)*log(2*x)-48*x^3-192*x^2)/(144*x^4+576*x^3)/log(2*x),x, algorithm="giac")

[Out]

1/144*log(x + 4)/x^2 - 1/3*log(log(2*x))

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maple [A]  time = 0.25, size = 18, normalized size = 0.62

method	result	size
risch	\(\frac {\ln \left (4+x \right )}{144 x^{2}}-\frac {\ln \left (\ln \left (2 x \right )\right )}{3}\)	\(18\)
default	\(\frac {-\frac {1}{4}-\frac {x \ln \relax (x )}{16}}{144 x}+\frac {\ln \left (4+x \right )}{2304}-\frac {\ln \left (\ln \relax (2)+\ln \relax (x )\right )}{3}+\frac {1}{576 x}+\frac {\ln \relax (x )}{2304}-\frac {\ln \left (4+x \right ) \left (4+x \right ) \left (x -4\right )}{2304 x^{2}}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x-8)*ln(4+x)+x)*ln(2*x)-48*x^3-192*x^2)/(144*x^4+576*x^3)/ln(2*x),x,method=_RETURNVERBOSE)

[Out]

1/144*ln(4+x)/x^2-1/3*ln(ln(2*x))

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maxima [A]  time = 0.78, size = 18, normalized size = 0.62 \begin {gather*} \frac {\log \left (x + 4\right )}{144 \, x^{2}} - \frac {1}{3} \, \log \left (\log \relax (2) + \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-8)*log(4+x)+x)*log(2*x)-48*x^3-192*x^2)/(144*x^4+576*x^3)/log(2*x),x, algorithm="maxima")

[Out]

1/144*log(x + 4)/x^2 - 1/3*log(log(2) + log(x))

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mupad [B]  time = 1.20, size = 17, normalized size = 0.59 \begin {gather*} \frac {\ln \left (x+4\right )}{144\,x^2}-\frac {\ln \left (\ln \left (2\,x\right )\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(192*x^2 - log(2*x)*(x - log(x + 4)*(2*x + 8)) + 48*x^3)/(log(2*x)*(576*x^3 + 144*x^4)),x)

[Out]

log(x + 4)/(144*x^2) - log(log(2*x))/3

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sympy [A]  time = 0.37, size = 17, normalized size = 0.59 \begin {gather*} - \frac {\log {\left (\log {\left (2 x \right )} \right )}}{3} + \frac {\log {\left (x + 4 \right )}}{144 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-8)*ln(4+x)+x)*ln(2*x)-48*x**3-192*x**2)/(144*x**4+576*x**3)/ln(2*x),x)

[Out]

-log(log(2*x))/3 + log(x + 4)/(144*x**2)

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