3.23.86 \(\int \frac {24-39 x+46 x^2-27 x^3+(-3+5 x-5 x^2+3 x^3) \log (\frac {3-2 x+3 x^2}{-3 x+3 x^2})}{-9 x^2+15 x^3-15 x^4+9 x^5} \, dx\)

Optimal. Leaf size=30 \[ \frac {3-\frac {1}{3} \log \left (1-\frac {3+x}{3 (1-x) x}\right )}{x} \]

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Rubi [A]  time = 0.91, antiderivative size = 38, normalized size of antiderivative = 1.27, number of steps used = 35, number of rules used = 12, integrand size = 78, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6741, 12, 6742, 893, 634, 618, 204, 628, 2058, 2074, 2525, 6728} \begin {gather*} \frac {3}{x}-\frac {\log \left (-\frac {3 x^2-2 x+3}{3 (1-x) x}\right )}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(24 - 39*x + 46*x^2 - 27*x^3 + (-3 + 5*x - 5*x^2 + 3*x^3)*Log[(3 - 2*x + 3*x^2)/(-3*x + 3*x^2)])/(-9*x^2 +
15*x^3 - 15*x^4 + 9*x^5),x]

[Out]

3/x - Log[-1/3*(3 - 2*x + 3*x^2)/((1 - x)*x)]/(3*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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 2058

Int[(P_)^(p_), x_Symbol] :> With[{u = Factor[P]}, Int[ExpandIntegrand[u^p, x], x] /;  !SumQ[NonfreeFactors[u,
x]]] /; PolyQ[P, x] && ILtQ[p, 0]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rule 2525

Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m
+ 1)*(a + b*Log[c*RFx^p])^n)/(e*(m + 1)), x] - Dist[(b*n*p)/(e*(m + 1)), Int[SimplifyIntegrand[((d + e*x)^(m +
1)*(a + b*Log[c*RFx^p])^(n - 1)*D[RFx, x])/RFx, x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunc
tionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

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 {-24+39 x-46 x^2+27 x^3-\left (-3+5 x-5 x^2+3 x^3\right ) \log \left (\frac {3-2 x+3 x^2}{-3 x+3 x^2}\right )}{3 x^2 \left (3-5 x+5 x^2-3 x^3\right )} \, dx\\ &=\frac {1}{3} \int \frac {-24+39 x-46 x^2+27 x^3-\left (-3+5 x-5 x^2+3 x^3\right ) \log \left (\frac {3-2 x+3 x^2}{-3 x+3 x^2}\right )}{x^2 \left (3-5 x+5 x^2-3 x^3\right )} \, dx\\ &=\frac {1}{3} \int \left (\frac {24}{(-1+x) x^2 \left (3-2 x+3 x^2\right )}+\frac {46}{-3+5 x-5 x^2+3 x^3}-\frac {39}{x \left (-3+5 x-5 x^2+3 x^3\right )}-\frac {27 x}{-3+5 x-5 x^2+3 x^3}+\frac {\log \left (\frac {3-2 x+3 x^2}{x (-3+3 x)}\right )}{x^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {\log \left (\frac {3-2 x+3 x^2}{x (-3+3 x)}\right )}{x^2} \, dx+8 \int \frac {1}{(-1+x) x^2 \left (3-2 x+3 x^2\right )} \, dx-9 \int \frac {x}{-3+5 x-5 x^2+3 x^3} \, dx-13 \int \frac {1}{x \left (-3+5 x-5 x^2+3 x^3\right )} \, dx+\frac {46}{3} \int \frac {1}{-3+5 x-5 x^2+3 x^3} \, dx\\ &=-\frac {\log \left (-\frac {3-2 x+3 x^2}{3 (1-x) x}\right )}{3 x}+\frac {1}{3} \int \frac {-3+6 x+x^2}{(1-x) x^2 \left (3-2 x+3 x^2\right )} \, dx+8 \int \left (\frac {1}{4 (-1+x)}-\frac {1}{3 x^2}-\frac {5}{9 x}+\frac {-13+33 x}{36 \left (3-2 x+3 x^2\right )}\right ) \, dx-9 \int \left (\frac {1}{4 (-1+x)}-\frac {3 (-1+x)}{4 \left (3-2 x+3 x^2\right )}\right ) \, dx-13 \int \left (\frac {1}{4 (-1+x)}-\frac {1}{3 x}+\frac {-11+3 x}{12 \left (3-2 x+3 x^2\right )}\right ) \, dx+\frac {46}{3} \int \left (\frac {1}{4 (-1+x)}+\frac {-1-3 x}{4 \left (3-2 x+3 x^2\right )}\right ) \, dx\\ &=\frac {8}{3 x}+\frac {1}{3} \log (1-x)-\frac {\log (x)}{9}-\frac {\log \left (-\frac {3-2 x+3 x^2}{3 (1-x) x}\right )}{3 x}+\frac {2}{9} \int \frac {-13+33 x}{3-2 x+3 x^2} \, dx+\frac {1}{3} \int \left (\frac {1}{1-x}-\frac {1}{x^2}+\frac {1}{3 x}+\frac {2 (7+3 x)}{3 \left (3-2 x+3 x^2\right )}\right ) \, dx-\frac {13}{12} \int \frac {-11+3 x}{3-2 x+3 x^2} \, dx+\frac {23}{6} \int \frac {-1-3 x}{3-2 x+3 x^2} \, dx+\frac {27}{4} \int \frac {-1+x}{3-2 x+3 x^2} \, dx\\ &=\frac {3}{x}-\frac {\log \left (-\frac {3-2 x+3 x^2}{3 (1-x) x}\right )}{3 x}+\frac {2}{9} \int \frac {7+3 x}{3-2 x+3 x^2} \, dx-\frac {4}{9} \int \frac {1}{3-2 x+3 x^2} \, dx-\frac {13}{24} \int \frac {-2+6 x}{3-2 x+3 x^2} \, dx+\frac {9}{8} \int \frac {-2+6 x}{3-2 x+3 x^2} \, dx+\frac {11}{9} \int \frac {-2+6 x}{3-2 x+3 x^2} \, dx-\frac {23}{12} \int \frac {-2+6 x}{3-2 x+3 x^2} \, dx-\frac {9}{2} \int \frac {1}{3-2 x+3 x^2} \, dx-\frac {23}{3} \int \frac {1}{3-2 x+3 x^2} \, dx+\frac {65}{6} \int \frac {1}{3-2 x+3 x^2} \, dx\\ &=\frac {3}{x}-\frac {1}{9} \log \left (3-2 x+3 x^2\right )-\frac {\log \left (-\frac {3-2 x+3 x^2}{3 (1-x) x}\right )}{3 x}+\frac {1}{9} \int \frac {-2+6 x}{3-2 x+3 x^2} \, dx+\frac {8}{9} \operatorname {Subst}\left (\int \frac {1}{-32-x^2} \, dx,x,-2+6 x\right )+\frac {16}{9} \int \frac {1}{3-2 x+3 x^2} \, dx+9 \operatorname {Subst}\left (\int \frac {1}{-32-x^2} \, dx,x,-2+6 x\right )+\frac {46}{3} \operatorname {Subst}\left (\int \frac {1}{-32-x^2} \, dx,x,-2+6 x\right )-\frac {65}{3} \operatorname {Subst}\left (\int \frac {1}{-32-x^2} \, dx,x,-2+6 x\right )\\ &=\frac {3}{x}+\frac {4}{9} \sqrt {2} \tan ^{-1}\left (\frac {1-3 x}{2 \sqrt {2}}\right )-\frac {\log \left (-\frac {3-2 x+3 x^2}{3 (1-x) x}\right )}{3 x}-\frac {32}{9} \operatorname {Subst}\left (\int \frac {1}{-32-x^2} \, dx,x,-2+6 x\right )\\ &=\frac {3}{x}-\frac {\log \left (-\frac {3-2 x+3 x^2}{3 (1-x) x}\right )}{3 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 38, normalized size = 1.27 \begin {gather*} \frac {1}{3} \left (\frac {9}{x}-\frac {\log \left (\frac {3-2 x+3 x^2}{-3 x+3 x^2}\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(24 - 39*x + 46*x^2 - 27*x^3 + (-3 + 5*x - 5*x^2 + 3*x^3)*Log[(3 - 2*x + 3*x^2)/(-3*x + 3*x^2)])/(-9
*x^2 + 15*x^3 - 15*x^4 + 9*x^5),x]

[Out]

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

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fricas [A]  time = 0.77, size = 29, normalized size = 0.97 \begin {gather*} -\frac {\log \left (\frac {3 \, x^{2} - 2 \, x + 3}{3 \, {\left (x^{2} - x\right )}}\right ) - 9}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3-5*x^2+5*x-3)*log((3*x^2-2*x+3)/(3*x^2-3*x))-27*x^3+46*x^2-39*x+24)/(9*x^5-15*x^4+15*x^3-9*x^
2),x, algorithm="fricas")

[Out]

-1/3*(log(1/3*(3*x^2 - 2*x + 3)/(x^2 - x)) - 9)/x

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giac [A]  time = 0.36, size = 33, normalized size = 1.10 \begin {gather*} -\frac {\log \left (\frac {3 \, x^{2} - 2 \, x + 3}{3 \, {\left (x^{2} - x\right )}}\right )}{3 \, x} + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3-5*x^2+5*x-3)*log((3*x^2-2*x+3)/(3*x^2-3*x))-27*x^3+46*x^2-39*x+24)/(9*x^5-15*x^4+15*x^3-9*x^
2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 32, normalized size = 1.07

method	result	size
norman	\(\frac {3-\frac {\ln \left (\frac {3 x^{2}-2 x +3}{3 x^{2}-3 x}\right )}{3}}{x}\)	\(32\)
risch	\(-\frac {\ln \left (\frac {3 x^{2}-2 x +3}{3 x^{2}-3 x}\right )}{3 x}+\frac {3}{x}\)	\(35\)
default	\(-\frac {\ln \left (\frac {3 x^{2}-2 x +3}{x \left (x -1\right )}\right )}{3 x}+\frac {3}{x}+\frac {\ln \relax (3)}{3 x}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^3-5*x^2+5*x-3)*ln((3*x^2-2*x+3)/(3*x^2-3*x))-27*x^3+46*x^2-39*x+24)/(9*x^5-15*x^4+15*x^3-9*x^2),x,me
thod=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 1.07, size = 70, normalized size = 2.33 \begin {gather*} \frac {{\left (x - 3\right )} \log \left (3 \, x^{2} - 2 \, x + 3\right ) - 3 \, {\left (x - 1\right )} \log \left (x - 1\right ) + {\left (x + 3\right )} \log \relax (x) + 3 \, \log \relax (3) + 3}{9 \, x} + \frac {8}{3 \, x} - \frac {1}{9} \, \log \left (3 \, x^{2} - 2 \, x + 3\right ) + \frac {1}{3} \, \log \left (x - 1\right ) - \frac {1}{9} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^3-5*x^2+5*x-3)*log((3*x^2-2*x+3)/(3*x^2-3*x))-27*x^3+46*x^2-39*x+24)/(9*x^5-15*x^4+15*x^3-9*x^
2),x, algorithm="maxima")

[Out]

1/9*((x - 3)*log(3*x^2 - 2*x + 3) - 3*(x - 1)*log(x - 1) + (x + 3)*log(x) + 3*log(3) + 3)/x + 8/3/x - 1/9*log(
3*x^2 - 2*x + 3) + 1/3*log(x - 1) - 1/9*log(x)

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mupad [B]  time = 1.53, size = 31, normalized size = 1.03 \begin {gather*} -\frac {\ln \left (-\frac {3\,x^2-2\,x+3}{3\,x-3\,x^2}\right )-9}{3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(-(3*x^2 - 2*x + 3)/(3*x - 3*x^2))*(5*x - 5*x^2 + 3*x^3 - 3) - 39*x + 46*x^2 - 27*x^3 + 24)/(9*x^2 -
15*x^3 + 15*x^4 - 9*x^5),x)

[Out]

-(log(-(3*x^2 - 2*x + 3)/(3*x - 3*x^2)) - 9)/(3*x)

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sympy [A]  time = 0.22, size = 26, normalized size = 0.87 \begin {gather*} - \frac {\log {\left (\frac {3 x^{2} - 2 x + 3}{3 x^{2} - 3 x} \right )}}{3 x} + \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**3-5*x**2+5*x-3)*ln((3*x**2-2*x+3)/(3*x**2-3*x))-27*x**3+46*x**2-39*x+24)/(9*x**5-15*x**4+15*x
**3-9*x**2),x)

[Out]

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

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