3.34.23 \(\int \frac {-2-x+(-6-4 x) \log (\frac {3}{x})}{48 x^4+48 x^5+12 x^6} \, dx\)

Optimal. Leaf size=18 \[ \frac {\log \left (\frac {3}{x}\right )}{12 x^3 (2+x)} \]

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Rubi [B]  time = 0.26, antiderivative size = 62, normalized size of antiderivative = 3.44, number of steps used = 14, number of rules used = 9, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1594, 27, 12, 6742, 44, 2357, 2304, 2314, 31} \begin {gather*} \frac {\log \left (\frac {3}{x}\right )}{24 x^3}-\frac {\log \left (\frac {3}{x}\right )}{48 x^2}+\frac {\log \left (\frac {3}{x}\right )}{96 x}+\frac {x \log \left (\frac {3}{x}\right )}{192 (x+2)}+\frac {\log (x)}{192} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - x + (-6 - 4*x)*Log[3/x])/(48*x^4 + 48*x^5 + 12*x^6),x]

[Out]

Log[3/x]/(24*x^3) - Log[3/x]/(48*x^2) + Log[3/x]/(96*x) + (x*Log[3/x])/(192*(2 + x)) + Log[x]/192

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, 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 1594

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

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 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

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 {-2-x+(-6-4 x) \log \left (\frac {3}{x}\right )}{x^4 \left (48+48 x+12 x^2\right )} \, dx\\ &=\int \frac {-2-x+(-6-4 x) \log \left (\frac {3}{x}\right )}{12 x^4 (2+x)^2} \, dx\\ &=\frac {1}{12} \int \frac {-2-x+(-6-4 x) \log \left (\frac {3}{x}\right )}{x^4 (2+x)^2} \, dx\\ &=\frac {1}{12} \int \left (-\frac {1}{x^4 (2+x)}-\frac {2 (3+2 x) \log \left (\frac {3}{x}\right )}{x^4 (2+x)^2}\right ) \, dx\\ &=-\left (\frac {1}{12} \int \frac {1}{x^4 (2+x)} \, dx\right )-\frac {1}{6} \int \frac {(3+2 x) \log \left (\frac {3}{x}\right )}{x^4 (2+x)^2} \, dx\\ &=-\left (\frac {1}{12} \int \left (\frac {1}{2 x^4}-\frac {1}{4 x^3}+\frac {1}{8 x^2}-\frac {1}{16 x}+\frac {1}{16 (2+x)}\right ) \, dx\right )-\frac {1}{6} \int \left (\frac {3 \log \left (\frac {3}{x}\right )}{4 x^4}-\frac {\log \left (\frac {3}{x}\right )}{4 x^3}+\frac {\log \left (\frac {3}{x}\right )}{16 x^2}-\frac {\log \left (\frac {3}{x}\right )}{16 (2+x)^2}\right ) \, dx\\ &=\frac {1}{72 x^3}-\frac {1}{96 x^2}+\frac {1}{96 x}+\frac {\log (x)}{192}-\frac {1}{192} \log (2+x)-\frac {1}{96} \int \frac {\log \left (\frac {3}{x}\right )}{x^2} \, dx+\frac {1}{96} \int \frac {\log \left (\frac {3}{x}\right )}{(2+x)^2} \, dx+\frac {1}{24} \int \frac {\log \left (\frac {3}{x}\right )}{x^3} \, dx-\frac {1}{8} \int \frac {\log \left (\frac {3}{x}\right )}{x^4} \, dx\\ &=\frac {\log \left (\frac {3}{x}\right )}{24 x^3}-\frac {\log \left (\frac {3}{x}\right )}{48 x^2}+\frac {\log \left (\frac {3}{x}\right )}{96 x}+\frac {x \log \left (\frac {3}{x}\right )}{192 (2+x)}+\frac {\log (x)}{192}-\frac {1}{192} \log (2+x)+\frac {1}{192} \int \frac {1}{2+x} \, dx\\ &=\frac {\log \left (\frac {3}{x}\right )}{24 x^3}-\frac {\log \left (\frac {3}{x}\right )}{48 x^2}+\frac {\log \left (\frac {3}{x}\right )}{96 x}+\frac {x \log \left (\frac {3}{x}\right )}{192 (2+x)}+\frac {\log (x)}{192}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 21, normalized size = 1.17 \begin {gather*} \frac {\log (6561)+8 \log \left (\frac {1}{x}\right )}{96 x^3 (2+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - x + (-6 - 4*x)*Log[3/x])/(48*x^4 + 48*x^5 + 12*x^6),x]

[Out]

(Log[6561] + 8*Log[x^(-1)])/(96*x^3*(2 + x))

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fricas [A]  time = 0.57, size = 19, normalized size = 1.06 \begin {gather*} \frac {\log \left (\frac {3}{x}\right )}{12 \, {\left (x^{4} + 2 \, x^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-6)*log(3/x)-x-2)/(12*x^6+48*x^5+48*x^4),x, algorithm="fricas")

[Out]

1/12*log(3/x)/(x^4 + 2*x^3)

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giac [B]  time = 0.19, size = 42, normalized size = 2.33 \begin {gather*} \frac {1}{192} \, {\left (\frac {2}{x} + \frac {1}{\frac {2}{x} + 1} - \frac {4}{x^{2}} + \frac {8}{x^{3}}\right )} \log \left (\frac {3}{x}\right ) - \frac {1}{192} \, \log \left (\frac {3}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-6)*log(3/x)-x-2)/(12*x^6+48*x^5+48*x^4),x, algorithm="giac")

[Out]

1/192*(2/x + 1/(2/x + 1) - 4/x^2 + 8/x^3)*log(3/x) - 1/192*log(3/x)

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maple [A]  time = 0.04, size = 17, normalized size = 0.94

method	result	size
norman	\(\frac {\ln \left (\frac {3}{x}\right )}{12 x^{3} \left (2+x \right )}\)	\(17\)
risch	\(\frac {\ln \left (\frac {3}{x}\right )}{12 x^{3} \left (2+x \right )}\)	\(17\)
derivativedivides	\(\frac {\ln \left (\frac {3}{x}\right )}{24 x^{3}}-\frac {\ln \left (\frac {3}{x}\right )}{48 x^{2}}+\frac {\ln \left (\frac {3}{x}\right )}{96 x}-\frac {\ln \left (\frac {3}{x}\right )}{32 x \left (\frac {6}{x}+3\right )}\)	\(55\)
default	\(\frac {\ln \left (\frac {3}{x}\right )}{24 x^{3}}-\frac {\ln \left (\frac {3}{x}\right )}{48 x^{2}}+\frac {\ln \left (\frac {3}{x}\right )}{96 x}-\frac {\ln \left (\frac {3}{x}\right )}{32 x \left (\frac {6}{x}+3\right )}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x-6)*ln(3/x)-x-2)/(12*x^6+48*x^5+48*x^4),x,method=_RETURNVERBOSE)

[Out]

1/12*ln(3/x)/x^3/(2+x)

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maxima [B]  time = 0.90, size = 79, normalized size = 4.39 \begin {gather*} -\frac {6 \, x^{3} + 6 \, x^{2} + 3 \, {\left (x^{4} + 2 \, x^{3} + 4\right )} \log \relax (x) - 4 \, x - 12 \, \log \relax (3) + 4}{144 \, {\left (x^{4} + 2 \, x^{3}\right )}} + \frac {3 \, x^{3} + 3 \, x^{2} - 2 \, x + 2}{72 \, {\left (x^{4} + 2 \, x^{3}\right )}} + \frac {1}{48} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-6)*log(3/x)-x-2)/(12*x^6+48*x^5+48*x^4),x, algorithm="maxima")

[Out]

-1/144*(6*x^3 + 6*x^2 + 3*(x^4 + 2*x^3 + 4)*log(x) - 4*x - 12*log(3) + 4)/(x^4 + 2*x^3) + 1/72*(3*x^3 + 3*x^2
- 2*x + 2)/(x^4 + 2*x^3) + 1/48*log(x)

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mupad [B]  time = 2.04, size = 22, normalized size = 1.22 \begin {gather*} \frac {x\,\left (\ln \left (\frac {1}{x}\right )+\ln \relax (3)\right )}{12\,\left (x^5+2\,x^4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x + log(3/x)*(4*x + 6) + 2)/(48*x^4 + 48*x^5 + 12*x^6),x)

[Out]

(x*(log(1/x) + log(3)))/(12*(2*x^4 + x^5))

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sympy [A]  time = 0.16, size = 14, normalized size = 0.78 \begin {gather*} \frac {\log {\left (\frac {3}{x} \right )}}{12 x^{4} + 24 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x-6)*ln(3/x)-x-2)/(12*x**6+48*x**5+48*x**4),x)

[Out]

log(3/x)/(12*x**4 + 24*x**3)

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