3.2.56 \(\int \frac {10602576-29160 x-2916 x^2+23328 \log (\log (12))+2916 \log ^2(\log (12))}{1-1454 x+528531 x^2-1454 x^3+x^4+(-4+2904 x+2904 x^2-4 x^3) \log (\log (12))+(6-1446 x+6 x^2) \log ^2(\log (12))+(-4-4 x) \log ^3(\log (12))+\log ^4(\log (12))} \, dx\)

Optimal. Leaf size=25 \[ \frac {4 (5+x)}{-x+\frac {1}{729} (1+x-\log (\log (12)))^2} \]

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Rubi [A]  time = 0.13, antiderivative size = 34, normalized size of antiderivative = 1.36, number of steps used = 4, number of rules used = 4, integrand size = 94, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1680, 12, 1814, 8} \begin {gather*} \frac {11664 (x+5)}{4 \left (x+\frac {1}{4} (-1454-4 \log (\log (12)))\right )^2-729 (725+4 \log (\log (12)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10602576 - 29160*x - 2916*x^2 + 23328*Log[Log[12]] + 2916*Log[Log[12]]^2)/(1 - 1454*x + 528531*x^2 - 1454
*x^3 + x^4 + (-4 + 2904*x + 2904*x^2 - 4*x^3)*Log[Log[12]] + (6 - 1446*x + 6*x^2)*Log[Log[12]]^2 + (-4 - 4*x)*
Log[Log[12]]^3 + Log[Log[12]]^4),x]

[Out]

(11664*(5 + x))/(4*(x + (-1454 - 4*Log[Log[12]])/4)^2 - 729*(725 + 4*Log[Log[12]]))

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 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 {11664 \left (-4 x^2-4 x (737+2 \log (\log (12)))-729 (725+4 \log (\log (12)))\right )}{\left (4 x^2-729 (725+4 \log (\log (12)))\right )^2} \, dx,x,x+\frac {1}{4} (-1454-4 \log (\log (12)))\right )\\ &=11664 \operatorname {Subst}\left (\int \frac {-4 x^2-4 x (737+2 \log (\log (12)))-729 (725+4 \log (\log (12)))}{\left (4 x^2-729 (725+4 \log (\log (12)))\right )^2} \, dx,x,x+\frac {1}{4} (-1454-4 \log (\log (12)))\right )\\ &=\frac {11664 (5+x)}{(-727+2 x-2 \log (\log (12)))^2-729 (725+4 \log (\log (12)))}+\frac {8 \operatorname {Subst}\left (\int 0 \, dx,x,x+\frac {1}{4} (-1454-4 \log (\log (12)))\right )}{725+4 \log (\log (12))}\\ &=\frac {11664 (5+x)}{(-727+2 x-2 \log (\log (12)))^2-729 (725+4 \log (\log (12)))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 28, normalized size = 1.12 \begin {gather*} \frac {2916 (5+x)}{x^2+(-1+\log (\log (12)))^2-x (727+2 \log (\log (12)))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10602576 - 29160*x - 2916*x^2 + 23328*Log[Log[12]] + 2916*Log[Log[12]]^2)/(1 - 1454*x + 528531*x^2
- 1454*x^3 + x^4 + (-4 + 2904*x + 2904*x^2 - 4*x^3)*Log[Log[12]] + (6 - 1446*x + 6*x^2)*Log[Log[12]]^2 + (-4 -
4*x)*Log[Log[12]]^3 + Log[Log[12]]^4),x]

[Out]

(2916*(5 + x))/(x^2 + (-1 + Log[Log[12]])^2 - x*(727 + 2*Log[Log[12]]))

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fricas [A]  time = 0.69, size = 28, normalized size = 1.12 \begin {gather*} \frac {2916 \, {\left (x + 5\right )}}{x^{2} - 2 \, {\left (x + 1\right )} \log \left (\log \left (12\right )\right ) + \log \left (\log \left (12\right )\right )^{2} - 727 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2916*log(log(12))^2+23328*log(log(12))-2916*x^2-29160*x+10602576)/(log(log(12))^4+(-4*x-4)*log(log(
12))^3+(6*x^2-1446*x+6)*log(log(12))^2+(-4*x^3+2904*x^2+2904*x-4)*log(log(12))+x^4-1454*x^3+528531*x^2-1454*x+
1),x, algorithm="fricas")

[Out]

2916*(x + 5)/(x^2 - 2*(x + 1)*log(log(12)) + log(log(12))^2 - 727*x + 1)

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giac [A]  time = 0.33, size = 31, normalized size = 1.24 \begin {gather*} \frac {2916 \, {\left (x + 5\right )}}{x^{2} - 2 \, x \log \left (\log \left (12\right )\right ) + \log \left (\log \left (12\right )\right )^{2} - 727 \, x - 2 \, \log \left (\log \left (12\right )\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2916*log(log(12))^2+23328*log(log(12))-2916*x^2-29160*x+10602576)/(log(log(12))^4+(-4*x-4)*log(log(
12))^3+(6*x^2-1446*x+6)*log(log(12))^2+(-4*x^3+2904*x^2+2904*x-4)*log(log(12))+x^4-1454*x^3+528531*x^2-1454*x+
1),x, algorithm="giac")

[Out]

2916*(x + 5)/(x^2 - 2*x*log(log(12)) + log(log(12))^2 - 727*x - 2*log(log(12)) + 1)

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maple [A]  time = 0.10, size = 32, normalized size = 1.28

method	result	size
gosper	\(\frac {2916 x +14580}{\ln \left (\ln \left (12\right )\right )^{2}-2 x \ln \left (\ln \left (12\right )\right )+x^{2}-2 \ln \left (\ln \left (12\right )\right )-727 x +1}\)	\(32\)
default	\(\frac {2916 x +14580}{\ln \left (\ln \left (12\right )\right )^{2}-2 x \ln \left (\ln \left (12\right )\right )+x^{2}-2 \ln \left (\ln \left (12\right )\right )-727 x +1}\)	\(32\)
norman	\(\frac {2916 x +14580}{\ln \left (\ln \left (12\right )\right )^{2}-2 x \ln \left (\ln \left (12\right )\right )+x^{2}-2 \ln \left (\ln \left (12\right )\right )-727 x +1}\)	\(33\)
risch	\(\frac {2916 x +14580}{\ln \left (\ln \relax (3)+2 \ln \relax (2)\right )^{2}-2 x \ln \left (\ln \relax (3)+2 \ln \relax (2)\right )+x^{2}-2 \ln \left (\ln \relax (3)+2 \ln \relax (2)\right )-727 x +1}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2916*ln(ln(12))^2+23328*ln(ln(12))-2916*x^2-29160*x+10602576)/(ln(ln(12))^4+(-4*x-4)*ln(ln(12))^3+(6*x^2-
1446*x+6)*ln(ln(12))^2+(-4*x^3+2904*x^2+2904*x-4)*ln(ln(12))+x^4-1454*x^3+528531*x^2-1454*x+1),x,method=_RETUR
NVERBOSE)

[Out]

2916*(5+x)/(ln(ln(12))^2-2*x*ln(ln(12))+x^2-2*ln(ln(12))-727*x+1)

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maxima [A]  time = 0.35, size = 32, normalized size = 1.28 \begin {gather*} \frac {2916 \, {\left (x + 5\right )}}{x^{2} - x {\left (2 \, \log \left (\log \left (12\right )\right ) + 727\right )} + \log \left (\log \left (12\right )\right )^{2} - 2 \, \log \left (\log \left (12\right )\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2916*log(log(12))^2+23328*log(log(12))-2916*x^2-29160*x+10602576)/(log(log(12))^4+(-4*x-4)*log(log(
12))^3+(6*x^2-1446*x+6)*log(log(12))^2+(-4*x^3+2904*x^2+2904*x-4)*log(log(12))+x^4-1454*x^3+528531*x^2-1454*x+
1),x, algorithm="maxima")

[Out]

2916*(x + 5)/(x^2 - x*(2*log(log(12)) + 727) + log(log(12))^2 - 2*log(log(12)) + 1)

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mupad [B]  time = 0.16, size = 33, normalized size = 1.32 \begin {gather*} \frac {2916\,x+14580}{x^2+\left (-2\,\ln \left (\ln \left (12\right )\right )-727\right )\,x-2\,\ln \left (\ln \left (12\right )\right )+{\ln \left (\ln \left (12\right )\right )}^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((23328*log(log(12)) - 29160*x + 2916*log(log(12))^2 - 2916*x^2 + 10602576)/(log(log(12))^2*(6*x^2 - 1446*x
+ 6) - 1454*x + log(log(12))^4 - log(log(12))^3*(4*x + 4) + 528531*x^2 - 1454*x^3 + x^4 + log(log(12))*(2904*
x + 2904*x^2 - 4*x^3 - 4) + 1),x)

[Out]

(2916*x + 14580)/(log(log(12))^2 - 2*log(log(12)) - x*(2*log(log(12)) + 727) + x^2 + 1)

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sympy [A]  time = 0.66, size = 37, normalized size = 1.48 \begin {gather*} - \frac {- 2916 x - 14580}{x^{2} + x \left (-727 - 2 \log {\left (\log {\left (12 \right )} \right )}\right ) - 2 \log {\left (\log {\left (12 \right )} \right )} + \log {\left (\log {\left (12 \right )} \right )}^{2} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2916*ln(ln(12))**2+23328*ln(ln(12))-2916*x**2-29160*x+10602576)/(ln(ln(12))**4+(-4*x-4)*ln(ln(12))*
*3+(6*x**2-1446*x+6)*ln(ln(12))**2+(-4*x**3+2904*x**2+2904*x-4)*ln(ln(12))+x**4-1454*x**3+528531*x**2-1454*x+1
),x)

[Out]

-(-2916*x - 14580)/(x**2 + x*(-727 - 2*log(log(12))) - 2*log(log(12)) + log(log(12))**2 + 1)

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