∫ (6−x) log(4)+(3−x) log(4) log(−3+x)_________________________

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

Optimal. Leaf size=12 \[ \frac {\log (4) (2+\log (-3+x))}{x} \]

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Rubi [A] time = 0.36, antiderivative size = 18, normalized size of antiderivative = 1.50, number of steps used = 12, number of rules used = 10, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {1593, 6741, 12, 6688, 6742, 77, 2395, 36, 31, 29} \begin {gather*} \frac {\log (4) \log (x-3)}{x}+\frac {2 \log (4)}{x} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*Log[4])/x + (Log[4]*Log[-3 + x])/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 29

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

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -

Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran

d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[

n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1

, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 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]]

Rule 6741

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

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

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Mathematica [A] time = 0.03, size = 15, normalized size = 1.25 \begin {gather*} -\frac {\log (4) (-2-\log (-3+x))}{x} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-((Log[4]*(-2 - Log[-3 + x]))/x)

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fricas [A] time = 0.65, size = 17, normalized size = 1.42 \begin {gather*} \frac {2 \, {\left (\log \relax (2) \log \left (x - 3\right ) + 2 \, \log \relax (2)\right )}}{x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A] time = 0.24, size = 19, normalized size = 1.58 \begin {gather*} \frac {2 \, \log \relax (2) \log \left (x - 3\right )}{x} + \frac {4 \, \log \relax (2)}{x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A] time = 0.07, size = 18, normalized size = 1.50


method	result	size

norman	\(\frac {2 \ln \relax (2) \ln \left (x -3\right )+4 \ln \relax (2)}{x}\)	\(18\)
risch	\(\frac {2 \ln \relax (2) \ln \left (x -3\right )}{x}+\frac {4 \ln \relax (2)}{x}\)	\(20\)
derivativedivides	\(2 \ln \relax (2) \left (-\frac {\ln \left (x -3\right ) \left (x -3\right )}{3 x}+\frac {2}{x}+\frac {\ln \left (x -3\right )}{3}\right )\)	\(29\)
default	\(2 \ln \relax (2) \left (-\frac {\ln \left (x -3\right ) \left (x -3\right )}{3 x}+\frac {2}{x}+\frac {\ln \left (x -3\right )}{3}\right )\)	\(29\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B] time = 0.51, size = 44, normalized size = 3.67 \begin {gather*} \frac {4}{3} \, {\left (\frac {3}{x} + \log \left (x - 3\right ) - \log \relax (x)\right )} \log \relax (2) + \frac {4}{3} \, \log \relax (2) \log \relax (x) - \frac {2 \, {\left (2 \, x \log \relax (2) - 3 \, \log \relax (2)\right )} \log \left (x - 3\right )}{3 \, x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B] time = 0.15, size = 54, normalized size = 4.50 \begin {gather*} \frac {36\,\ln \relax (2)+18\,\ln \left (x-3\right )\,\ln \relax (2)-x\,\left (24\,\ln \relax (2)+12\,\ln \left (x-3\right )\,\ln \relax (2)\right )+x^2\,\left (4\,\ln \relax (2)+\ln \left (x-3\right )\,\ln \relax (4)\right )}{x\,{\left (x-3\right )}^2} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((2*log(2)*(x - 6) + 2*log(x - 3)*log(2)*(x - 3))/(3*x^2 - x^3),x)

[Out]

(36*log(2) + 18*log(x - 3)*log(2) - x*(24*log(2) + 12*log(x - 3)*log(2)) + x^2*(4*log(2) + log(x - 3)*log(4)))

/(x*(x - 3)^2)

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sympy [A] time = 0.15, size = 17, normalized size = 1.42 \begin {gather*} \frac {2 \log {\relax (2 )} \log {\left (x - 3 \right )}}{x} + \frac {4 \log {\relax (2 )}}{x} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*(3-x)*ln(2)*ln(x-3)+2*(-x+6)*ln(2))/(x**3-3*x**2),x)

[Out]

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

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