3.10.61 \(\int \frac {1-4 x+5 x^2-2 x^3+2 x^4+(-10 x+3 x^2) \log (6)}{x^2-4 x^3+4 x^4+(-2 x+4 x^2) \log (6)+\log ^2(6)} \, dx\)

Optimal. Leaf size=26 \[ \frac {1-(5-x) x^2}{-x+2 x^2+\log (6)} \]

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Rubi [B]  time = 0.14, antiderivative size = 64, normalized size of antiderivative = 2.46, number of steps used = 5, number of rules used = 5, integrand size = 65, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1680, 12, 1814, 21, 8} \begin {gather*} \frac {x}{2}-\frac {(1-8 \log (6)) (7+34 \log (6))-4 \left (x-\frac {1}{4}\right ) \left (9-16 \log ^2(6)-70 \log (6)\right )}{16 (1-8 \log (6)) \left (-2 x^2+x-\log (6)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 4*x + 5*x^2 - 2*x^3 + 2*x^4 + (-10*x + 3*x^2)*Log[6])/(x^2 - 4*x^3 + 4*x^4 + (-2*x + 4*x^2)*Log[6] +
Log[6]^2),x]

[Out]

x/2 - ((1 - 8*Log[6])*(7 + 34*Log[6]) - 4*(-1/4 + x)*(9 - 70*Log[6] - 16*Log[6]^2))/(16*(1 - 8*Log[6])*(x - 2*
x^2 - Log[6]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\operatorname {Subst}\left (\int \frac {256 x^4+37 (1-8 \log (6))+32 x^2 (17+12 \log (6))-32 x (7+34 \log (6))}{2 \left (1-16 x^2-8 \log (6)\right )^2} \, dx,x,-\frac {1}{4}+x\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {256 x^4+37 (1-8 \log (6))+32 x^2 (17+12 \log (6))-32 x (7+34 \log (6))}{\left (1-16 x^2-8 \log (6)\right )^2} \, dx,x,-\frac {1}{4}+x\right )\\ &=-\frac {(1-8 \log (6)) (7+34 \log (6))+(1-4 x) \left (9-70 \log (6)-16 \log ^2(6)\right )}{16 (1-8 \log (6)) \left (x-2 x^2-\log (6)\right )}-\frac {\operatorname {Subst}\left (\int \frac {32 x^2 (1-8 \log (6))-2 (1-8 \log (6))^2}{1-16 x^2-8 \log (6)} \, dx,x,-\frac {1}{4}+x\right )}{4 (1-8 \log (6))}\\ &=-\frac {(1-8 \log (6)) (7+34 \log (6))+(1-4 x) \left (9-70 \log (6)-16 \log ^2(6)\right )}{16 (1-8 \log (6)) \left (x-2 x^2-\log (6)\right )}+\frac {1}{2} \operatorname {Subst}\left (\int 1 \, dx,x,-\frac {1}{4}+x\right )\\ &=\frac {x}{2}-\frac {(1-8 \log (6)) (7+34 \log (6))+(1-4 x) \left (9-70 \log (6)-16 \log ^2(6)\right )}{16 (1-8 \log (6)) \left (x-2 x^2-\log (6)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.06, size = 67, normalized size = 2.58 \begin {gather*} \frac {1}{4} \left (2 x+\frac {-4+78 \log ^2(6)+\log (6) (23-2 \log (216))+x \left (9+8 \log ^2(6)+\log (46656)-4 \log (6) (19+\log (46656))\right )}{\left (-x+2 x^2+\log (6)\right ) (-1+8 \log (6))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 4*x + 5*x^2 - 2*x^3 + 2*x^4 + (-10*x + 3*x^2)*Log[6])/(x^2 - 4*x^3 + 4*x^4 + (-2*x + 4*x^2)*Log
[6] + Log[6]^2),x]

[Out]

(2*x + (-4 + 78*Log[6]^2 + Log[6]*(23 - 2*Log[216]) + x*(9 + 8*Log[6]^2 + Log[46656] - 4*Log[6]*(19 + Log[4665
6])))/((-x + 2*x^2 + Log[6])*(-1 + 8*Log[6])))/4

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fricas [A]  time = 1.00, size = 34, normalized size = 1.31 \begin {gather*} \frac {4 \, x^{3} - 2 \, x^{2} - 9 \, x + 9 \, \log \relax (6) + 4}{4 \, {\left (2 \, x^{2} - x + \log \relax (6)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-10*x)*log(6)+2*x^4-2*x^3+5*x^2-4*x+1)/(log(6)^2+(4*x^2-2*x)*log(6)+4*x^4-4*x^3+x^2),x, algor
ithm="fricas")

[Out]

1/4*(4*x^3 - 2*x^2 - 9*x + 9*log(6) + 4)/(2*x^2 - x + log(6))

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giac [A]  time = 0.36, size = 33, normalized size = 1.27 \begin {gather*} \frac {1}{2} \, x - \frac {2 \, x \log \relax (6) + 9 \, x - 9 \, \log \relax (6) - 4}{4 \, {\left (2 \, x^{2} - x + \log \relax (6)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-10*x)*log(6)+2*x^4-2*x^3+5*x^2-4*x+1)/(log(6)^2+(4*x^2-2*x)*log(6)+4*x^4-4*x^3+x^2),x, algor
ithm="giac")

[Out]

1/2*x - 1/4*(2*x*log(6) + 9*x - 9*log(6) - 4)/(2*x^2 - x + log(6))

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maple [A]  time = 0.06, size = 27, normalized size = 1.04

method	result	size
norman	\(\frac {x^{3}+1-\frac {5 x}{2}+\frac {5 \ln \relax (6)}{2}}{\ln \relax (6)+2 x^{2}-x}\)	\(27\)
gosper	\(\frac {2 x^{3}+5 \ln \relax (6)-5 x +2}{2 \ln \relax (6)+4 x^{2}-2 x}\)	\(30\)
default	\(\frac {x}{2}-\frac {\left (\ln \relax (6)+\frac {9}{2}\right ) x -2-\frac {9 \ln \relax (6)}{2}}{2 \left (\ln \relax (6)+2 x^{2}-x \right )}\)	\(32\)
risch	\(\frac {x}{2}+\frac {\frac {\left (-\frac {9}{2}-\ln \relax (2)-\ln \relax (3)\right ) x}{2}+\frac {9 \ln \relax (3)}{4}+\frac {9 \ln \relax (2)}{4}+1}{\ln \relax (2)+\ln \relax (3)+2 x^{2}-x}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^2-10*x)*ln(6)+2*x^4-2*x^3+5*x^2-4*x+1)/(ln(6)^2+(4*x^2-2*x)*ln(6)+4*x^4-4*x^3+x^2),x,method=_RETURNV
ERBOSE)

[Out]

(x^3+1-5/2*x+5/2*ln(6))/(ln(6)+2*x^2-x)

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maxima [A]  time = 0.73, size = 33, normalized size = 1.27 \begin {gather*} \frac {1}{2} \, x - \frac {x {\left (2 \, \log \relax (6) + 9\right )} - 9 \, \log \relax (6) - 4}{4 \, {\left (2 \, x^{2} - x + \log \relax (6)\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2-10*x)*log(6)+2*x^4-2*x^3+5*x^2-4*x+1)/(log(6)^2+(4*x^2-2*x)*log(6)+4*x^4-4*x^3+x^2),x, algor
ithm="maxima")

[Out]

1/2*x - 1/4*(x*(2*log(6) + 9) - 9*log(6) - 4)/(2*x^2 - x + log(6))

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mupad [B]  time = 0.12, size = 33, normalized size = 1.27 \begin {gather*} \frac {x}{2}+\frac {\frac {9\,\ln \relax (6)}{2}-x\,\left (\ln \relax (6)+\frac {9}{2}\right )+2}{4\,x^2-2\,x+2\,\ln \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x + log(6)*(10*x - 3*x^2) - 5*x^2 + 2*x^3 - 2*x^4 - 1)/(log(6)^2 - log(6)*(2*x - 4*x^2) + x^2 - 4*x^3
+ 4*x^4),x)

[Out]

x/2 + ((9*log(6))/2 - x*(log(6) + 9/2) + 2)/(2*log(6) - 2*x + 4*x^2)

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sympy [A]  time = 0.59, size = 32, normalized size = 1.23 \begin {gather*} \frac {x}{2} + \frac {x \left (-9 - 2 \log {\relax (6 )}\right ) + 4 + 9 \log {\relax (6 )}}{8 x^{2} - 4 x + 4 \log {\relax (6 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**2-10*x)*ln(6)+2*x**4-2*x**3+5*x**2-4*x+1)/(ln(6)**2+(4*x**2-2*x)*ln(6)+4*x**4-4*x**3+x**2),x)

[Out]

x/2 + (x*(-9 - 2*log(6)) + 4 + 9*log(6))/(8*x**2 - 4*x + 4*log(6))

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