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

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} $$\begin{array}{c}\frac{\log (4)\log (x-3)}{x}+\frac{2\log (4)}{x}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{(6-x)\log (4)+(3-x)\log (4)\log (-3+x)}{(-3+x)x^2}dx\\ & =\int \frac{\log (4)(-6+x-3\log (-3+x)+x\log (-3+x))}{(3-x)x^2}dx\\ & =\log (4)\int \frac{-6+x-3\log (-3+x)+x\log (-3+x)}{(3-x)x^2}dx\\ & =\log (4)\int \frac{-6+x+(-3+x)\log (-3+x)}{(3-x)x^2}dx\\ & =\log (4)\int (\frac{6-x}{(-3+x)x^2}-\frac{\log (-3+x)}{x^2})dx\\ & =\log (4)\int \frac{6-x}{(-3+x)x^2}dx-\log (4)\int \frac{\log (-3+x)}{x^2}dx\\ & =\frac{\log (4)\log (-3+x)}{x}+\log (4)\int (\frac{1}{3(-3+x)}-\frac{2}{x^2}-\frac{1}{3x})dx-\log (4)\int \frac{1}{(-3+x)x}dx\\ & =\frac{2\log (4)}{x}+\frac{1}{3}\log (4)\log (3-x)+\frac{\log (4)\log (-3+x)}{x}-\frac{1}{3}\log (4)\log (x)-\frac{1}{3}\log (4)\int \frac{1}{-3+x}dx+\frac{1}{3}\log (4)\int \frac{1}{x}dx\\ & =\frac{2\log (4)}{x}+\frac{\log (4)\log (-3+x)}{x}\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{2(\log (2)\log (x-3)+2\log (2))}{x}\end{array}$$

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 $$\begin{array}{c}\frac{2\log (2)\log (x-3)}{x}+\frac{4\log (2)}{x}\end{array}$$

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	$\frac{2\ln (2)\ln (x-3)+4\ln (2)}{x}$	$18$
risch	$\frac{2\ln (2)\ln (x-3)}{x}+\frac{4\ln (2)}{x}$	$20$
derivativedivides	$2\ln (2)(-\frac{\ln (x-3)(x-3)}{3x}+\frac{2}{x}+\frac{\ln (x-3)}{3})$	$29$
default	$2\ln (2)(-\frac{\ln (x-3)(x-3)}{3x}+\frac{2}{x}+\frac{\ln (x-3)}{3})$	$29$

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 $$\begin{array}{c}\frac{4}{3}(\frac{3}{x}+\log (x-3)-\log (x))\log (2)+\frac{4}{3}\log (2)\log (x)-\frac{2(2x\log (2)-3\log (2))\log (x-3)}{3x}\end{array}$$

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 $$\begin{array}{c}\frac{36\ln (2)+18\ln (x-3)\ln (2)-x(24\ln (2)+12\ln (x-3)\ln (2))+x^2(4\ln (2)+\ln (x-3)\ln (4))}{x{(x-3)}^2}\end{array}$$

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 $$\begin{array}{c}\frac{2\log (2)\log (x-3)}{x}+\frac{4\log (2)}{x}\end{array}$$

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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