3.9.37 \(\int \frac {84+198 x+152 x^2+60 x^3+10 x^4+(72+132 x+96 x^2+32 x^3+4 x^4) \log (\frac {12+12 x+4 x^2}{12+16 x+7 x^2+x^3})}{18-3 x-24 x^2-7 x^3+9 x^4+6 x^5+x^6} \, dx\)

Optimal. Leaf size=32 \[ \frac {2+4 \left (x+\log \left (\frac {3+\frac {x^2}{(2+x)^2}}{3+x}\right )\right )}{1-x} \]

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Rubi [A]  time = 5.39, antiderivative size = 40, normalized size of antiderivative = 1.25, number of steps used = 40, number of rules used = 8, integrand size = 99, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {6741, 6742, 6728, 634, 618, 204, 628, 2525} \begin {gather*} \frac {4 \log \left (\frac {4 \left (x^2+3 x+3\right )}{(x+2)^2 (x+3)}\right )}{1-x}+\frac {6}{1-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(84 + 198*x + 152*x^2 + 60*x^3 + 10*x^4 + (72 + 132*x + 96*x^2 + 32*x^3 + 4*x^4)*Log[(12 + 12*x + 4*x^2)/(
12 + 16*x + 7*x^2 + x^3)])/(18 - 3*x - 24*x^2 - 7*x^3 + 9*x^4 + 6*x^5 + x^6),x]

[Out]

6/(1 - x) + (4*Log[(4*(3 + 3*x + x^2))/((2 + x)^2*(3 + x))])/(1 - 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 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 {84+198 x+152 x^2+60 x^3+10 x^4+\left (72+132 x+96 x^2+32 x^3+4 x^4\right ) \log \left (\frac {12+12 x+4 x^2}{12+16 x+7 x^2+x^3}\right )}{(1-x)^2 \left (18+33 x+24 x^2+8 x^3+x^4\right )} \, dx\\ &=\int \left (\frac {84}{(-1+x)^2 (2+x) (3+x) \left (3+3 x+x^2\right )}+\frac {198 x}{(-1+x)^2 (2+x) (3+x) \left (3+3 x+x^2\right )}+\frac {152 x^2}{(-1+x)^2 (2+x) (3+x) \left (3+3 x+x^2\right )}+\frac {60 x^3}{(-1+x)^2 (2+x) (3+x) \left (3+3 x+x^2\right )}+\frac {10 x^4}{(-1+x)^2 (2+x) (3+x) \left (3+3 x+x^2\right )}+\frac {4 \log \left (\frac {4 \left (3+3 x+x^2\right )}{(2+x)^2 (3+x)}\right )}{(-1+x)^2}\right ) \, dx\\ &=4 \int \frac {\log \left (\frac {4 \left (3+3 x+x^2\right )}{(2+x)^2 (3+x)}\right )}{(-1+x)^2} \, dx+10 \int \frac {x^4}{(-1+x)^2 (2+x) (3+x) \left (3+3 x+x^2\right )} \, dx+60 \int \frac {x^3}{(-1+x)^2 (2+x) (3+x) \left (3+3 x+x^2\right )} \, dx+84 \int \frac {1}{(-1+x)^2 (2+x) (3+x) \left (3+3 x+x^2\right )} \, dx+152 \int \frac {x^2}{(-1+x)^2 (2+x) (3+x) \left (3+3 x+x^2\right )} \, dx+198 \int \frac {x}{(-1+x)^2 (2+x) (3+x) \left (3+3 x+x^2\right )} \, dx\\ &=\frac {4 \log \left (\frac {4 \left (3+3 x+x^2\right )}{(2+x)^2 (3+x)}\right )}{1-x}+4 \int \frac {6+6 x+4 x^2+x^3}{(1-x) (2+x) (3+x) \left (3+3 x+x^2\right )} \, dx+10 \int \left (\frac {1}{84 (-1+x)^2}+\frac {227}{7056 (-1+x)}+\frac {16}{9 (2+x)}-\frac {27}{16 (3+x)}-\frac {3 (15+2 x)}{49 \left (3+3 x+x^2\right )}\right ) \, dx+60 \int \left (\frac {1}{84 (-1+x)^2}+\frac {143}{7056 (-1+x)}-\frac {8}{9 (2+x)}+\frac {9}{16 (3+x)}+\frac {3 (13+5 x)}{49 \left (3+3 x+x^2\right )}\right ) \, dx+84 \int \left (\frac {1}{84 (-1+x)^2}-\frac {109}{7056 (-1+x)}+\frac {1}{9 (2+x)}-\frac {1}{48 (3+x)}+\frac {-9-11 x}{147 \left (3+3 x+x^2\right )}\right ) \, dx+152 \int \left (\frac {1}{84 (-1+x)^2}+\frac {59}{7056 (-1+x)}+\frac {4}{9 (2+x)}-\frac {3}{16 (3+x)}+\frac {-24-13 x}{49 \left (3+3 x+x^2\right )}\right ) \, dx+198 \int \left (\frac {1}{84 (-1+x)^2}-\frac {25}{7056 (-1+x)}-\frac {2}{9 (2+x)}+\frac {1}{16 (3+x)}+\frac {11+8 x}{49 \left (3+3 x+x^2\right )}\right ) \, dx\\ &=\frac {6}{1-x}+\frac {17}{21} \log (1-x)-\frac {8}{3} \log (2+x)-\log (3+x)+\frac {4 \log \left (\frac {4 \left (3+3 x+x^2\right )}{(2+x)^2 (3+x)}\right )}{1-x}+\frac {4}{7} \int \frac {-9-11 x}{3+3 x+x^2} \, dx-\frac {30}{49} \int \frac {15+2 x}{3+3 x+x^2} \, dx+\frac {152}{49} \int \frac {-24-13 x}{3+3 x+x^2} \, dx+\frac {180}{49} \int \frac {13+5 x}{3+3 x+x^2} \, dx+4 \int \left (-\frac {17}{84 (-1+x)}+\frac {2}{3 (2+x)}+\frac {1}{4 (3+x)}+\frac {-6-5 x}{7 \left (3+3 x+x^2\right )}\right ) \, dx+\frac {198}{49} \int \frac {11+8 x}{3+3 x+x^2} \, dx\\ &=\frac {6}{1-x}+\frac {4 \log \left (\frac {4 \left (3+3 x+x^2\right )}{(2+x)^2 (3+x)}\right )}{1-x}+\frac {4}{7} \int \frac {-6-5 x}{3+3 x+x^2} \, dx-\frac {30}{49} \int \frac {3+2 x}{3+3 x+x^2} \, dx-\frac {22}{7} \int \frac {3+2 x}{3+3 x+x^2} \, dx-\frac {198}{49} \int \frac {1}{3+3 x+x^2} \, dx+\frac {30}{7} \int \frac {1}{3+3 x+x^2} \, dx-\frac {360}{49} \int \frac {1}{3+3 x+x^2} \, dx+\frac {450}{49} \int \frac {3+2 x}{3+3 x+x^2} \, dx-\frac {684}{49} \int \frac {1}{3+3 x+x^2} \, dx+\frac {792}{49} \int \frac {3+2 x}{3+3 x+x^2} \, dx-\frac {988}{49} \int \frac {3+2 x}{3+3 x+x^2} \, dx+\frac {990}{49} \int \frac {1}{3+3 x+x^2} \, dx\\ &=\frac {6}{1-x}+\frac {10}{7} \log \left (3+3 x+x^2\right )+\frac {4 \log \left (\frac {4 \left (3+3 x+x^2\right )}{(2+x)^2 (3+x)}\right )}{1-x}+\frac {6}{7} \int \frac {1}{3+3 x+x^2} \, dx-\frac {10}{7} \int \frac {3+2 x}{3+3 x+x^2} \, dx+\frac {396}{49} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,3+2 x\right )-\frac {60}{7} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,3+2 x\right )+\frac {720}{49} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,3+2 x\right )+\frac {1368}{49} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,3+2 x\right )-\frac {1980}{49} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,3+2 x\right )\\ &=\frac {6}{1-x}-\frac {4}{7} \sqrt {3} \tan ^{-1}\left (\frac {3+2 x}{\sqrt {3}}\right )+\frac {4 \log \left (\frac {4 \left (3+3 x+x^2\right )}{(2+x)^2 (3+x)}\right )}{1-x}-\frac {12}{7} \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,3+2 x\right )\\ &=\frac {6}{1-x}+\frac {4 \log \left (\frac {4 \left (3+3 x+x^2\right )}{(2+x)^2 (3+x)}\right )}{1-x}\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 7.48, size = 8222, normalized size = 256.94 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(84 + 198*x + 152*x^2 + 60*x^3 + 10*x^4 + (72 + 132*x + 96*x^2 + 32*x^3 + 4*x^4)*Log[(12 + 12*x + 4*
x^2)/(12 + 16*x + 7*x^2 + x^3)])/(18 - 3*x - 24*x^2 - 7*x^3 + 9*x^4 + 6*x^5 + x^6),x]

[Out]

Result too large to show

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fricas [A]  time = 1.35, size = 37, normalized size = 1.16 \begin {gather*} -\frac {2 \, {\left (2 \, \log \left (\frac {4 \, {\left (x^{2} + 3 \, x + 3\right )}}{x^{3} + 7 \, x^{2} + 16 \, x + 12}\right ) + 3\right )}}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4+32*x^3+96*x^2+132*x+72)*log((4*x^2+12*x+12)/(x^3+7*x^2+16*x+12))+10*x^4+60*x^3+152*x^2+198*x
+84)/(x^6+6*x^5+9*x^4-7*x^3-24*x^2-3*x+18),x, algorithm="fricas")

[Out]

-2*(2*log(4*(x^2 + 3*x + 3)/(x^3 + 7*x^2 + 16*x + 12)) + 3)/(x - 1)

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giac [A]  time = 1.28, size = 41, normalized size = 1.28 \begin {gather*} -\frac {4 \, \log \left (\frac {4 \, {\left (x^{2} + 3 \, x + 3\right )}}{x^{3} + 7 \, x^{2} + 16 \, x + 12}\right )}{x - 1} - \frac {6}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4+32*x^3+96*x^2+132*x+72)*log((4*x^2+12*x+12)/(x^3+7*x^2+16*x+12))+10*x^4+60*x^3+152*x^2+198*x
+84)/(x^6+6*x^5+9*x^4-7*x^3-24*x^2-3*x+18),x, algorithm="giac")

[Out]

-4*log(4*(x^2 + 3*x + 3)/(x^3 + 7*x^2 + 16*x + 12))/(x - 1) - 6/(x - 1)

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maple [A]  time = 0.11, size = 38, normalized size = 1.19

method	result	size
norman	\(\frac {-4 \ln \left (\frac {4 x^{2}+12 x +12}{x^{3}+7 x^{2}+16 x +12}\right )-6}{x -1}\)	\(38\)
risch	\(-\frac {4 \ln \left (\frac {4 x^{2}+12 x +12}{x^{3}+7 x^{2}+16 x +12}\right )}{x -1}-\frac {6}{x -1}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^4+32*x^3+96*x^2+132*x+72)*ln((4*x^2+12*x+12)/(x^3+7*x^2+16*x+12))+10*x^4+60*x^3+152*x^2+198*x+84)/(x
^6+6*x^5+9*x^4-7*x^3-24*x^2-3*x+18),x,method=_RETURNVERBOSE)

[Out]

(-4*ln((4*x^2+12*x+12)/(x^3+7*x^2+16*x+12))-6)/(x-1)

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maxima [B]  time = 2.61, size = 77, normalized size = 2.41 \begin {gather*} -\frac {6 \, {\left (5 \, x + 9\right )} \log \left (x^{2} + 3 \, x + 3\right ) - 21 \, {\left (x + 3\right )} \log \left (x + 3\right ) - 56 \, {\left (x + 2\right )} \log \left (x + 2\right ) + 168 \, \log \relax (2)}{21 \, {\left (x - 1\right )}} - \frac {6}{x - 1} + \frac {10}{7} \, \log \left (x^{2} + 3 \, x + 3\right ) - \log \left (x + 3\right ) - \frac {8}{3} \, \log \left (x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^4+32*x^3+96*x^2+132*x+72)*log((4*x^2+12*x+12)/(x^3+7*x^2+16*x+12))+10*x^4+60*x^3+152*x^2+198*x
+84)/(x^6+6*x^5+9*x^4-7*x^3-24*x^2-3*x+18),x, algorithm="maxima")

[Out]

-1/21*(6*(5*x + 9)*log(x^2 + 3*x + 3) - 21*(x + 3)*log(x + 3) - 56*(x + 2)*log(x + 2) + 168*log(2))/(x - 1) -
6/(x - 1) + 10/7*log(x^2 + 3*x + 3) - log(x + 3) - 8/3*log(x + 2)

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mupad [B]  time = 0.98, size = 42, normalized size = 1.31 \begin {gather*} -\frac {6}{x-1}-\frac {4\,\ln \left (\frac {4\,x^2+12\,x+12}{x^3+7\,x^2+16\,x+12}\right )}{x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((198*x + log((12*x + 4*x^2 + 12)/(16*x + 7*x^2 + x^3 + 12))*(132*x + 96*x^2 + 32*x^3 + 4*x^4 + 72) + 152*x
^2 + 60*x^3 + 10*x^4 + 84)/(9*x^4 - 24*x^2 - 7*x^3 - 3*x + 6*x^5 + x^6 + 18),x)

[Out]

- 6/(x - 1) - (4*log((12*x + 4*x^2 + 12)/(16*x + 7*x^2 + x^3 + 12)))/(x - 1)

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sympy [A]  time = 0.27, size = 36, normalized size = 1.12 \begin {gather*} - \frac {4 \log {\left (\frac {4 x^{2} + 12 x + 12}{x^{3} + 7 x^{2} + 16 x + 12} \right )}}{x - 1} - \frac {6}{x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**4+32*x**3+96*x**2+132*x+72)*ln((4*x**2+12*x+12)/(x**3+7*x**2+16*x+12))+10*x**4+60*x**3+152*x*
*2+198*x+84)/(x**6+6*x**5+9*x**4-7*x**3-24*x**2-3*x+18),x)

[Out]

-4*log((4*x**2 + 12*x + 12)/(x**3 + 7*x**2 + 16*x + 12))/(x - 1) - 6/(x - 1)

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