3.12.86 \(\int \frac {(12 x^2-256 x^3+112 x^4) \log (\log (4))}{1-64 x+1080 x^2-1800 x^3+1040 x^4-224 x^5+16 x^6} \, dx\)

Optimal. Leaf size=27 \[ \frac {x^2 \log (\log (4))}{(2-x) (-4+x)+\frac {1}{4 x}+x} \]

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Rubi [B]  time = 0.35, antiderivative size = 70, normalized size of antiderivative = 2.59, number of steps used = 8, number of rules used = 5, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.096, Rules used = {12, 1594, 6688, 2102, 1588} \begin {gather*} \frac {28 x^2 \log (\log (4))}{-4 x^3+28 x^2-32 x+1}-\frac {32 x \log (\log (4))}{-4 x^3+28 x^2-32 x+1}+\frac {\log (\log (4))}{-4 x^3+28 x^2-32 x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((12*x^2 - 256*x^3 + 112*x^4)*Log[Log[4]])/(1 - 64*x + 1080*x^2 - 1800*x^3 + 1040*x^4 - 224*x^5 + 16*x^6),
x]

[Out]

Log[Log[4]]/(1 - 32*x + 28*x^2 - 4*x^3) - (32*x*Log[Log[4]])/(1 - 32*x + 28*x^2 - 4*x^3) + (28*x^2*Log[Log[4]]
)/(1 - 32*x + 28*x^2 - 4*x^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1588

Int[(Pp_)*(Qq_)^(m_.), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[(Coeff[Pp, x, p]*x^(p - q
+ 1)*Qq^(m + 1))/((p + m*q + 1)*Coeff[Qq, x, q]), x] /; NeQ[p + m*q + 1, 0] && EqQ[(p + m*q + 1)*Coeff[Qq, x,
q]*Pp, Coeff[Pp, x, p]*x^(p - q)*((p - q + 1)*Qq + (m + 1)*x*D[Qq, x])]] /; FreeQ[m, x] && PolyQ[Pp, x] && Pol
yQ[Qq, x] && NeQ[m, -1]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 2102

Int[(Pm_)*(Qn_)^(p_.), x_Symbol] :> With[{m = Expon[Pm, x], n = Expon[Qn, x]}, Simp[(Coeff[Pm, x, m]*x^(m - n
+ 1)*Qn^(p + 1))/((m + n*p + 1)*Coeff[Qn, x, n]), x] + Dist[1/((m + n*p + 1)*Coeff[Qn, x, n]), Int[ExpandToSum
[(m + n*p + 1)*Coeff[Qn, x, n]*Pm - Coeff[Pm, x, m]*x^(m - n)*((m - n + 1)*Qn + (p + 1)*x*D[Qn, x]), x]*Qn^p,
x], x] /; LtQ[1, n, m + 1] && m + n*p + 1 < 0] /; FreeQ[p, x] && PolyQ[Pm, x] && PolyQ[Qn, x] && LtQ[p, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log (\log (4)) \int \frac {12 x^2-256 x^3+112 x^4}{1-64 x+1080 x^2-1800 x^3+1040 x^4-224 x^5+16 x^6} \, dx\\ &=\log (\log (4)) \int \frac {x^2 \left (12-256 x+112 x^2\right )}{1-64 x+1080 x^2-1800 x^3+1040 x^4-224 x^5+16 x^6} \, dx\\ &=\log (\log (4)) \int \frac {4 x^2 \left (3-64 x+28 x^2\right )}{\left (1-32 x+28 x^2-4 x^3\right )^2} \, dx\\ &=(4 \log (\log (4))) \int \frac {x^2 \left (3-64 x+28 x^2\right )}{\left (1-32 x+28 x^2-4 x^3\right )^2} \, dx\\ &=\frac {28 x^2 \log (\log (4))}{1-32 x+28 x^2-4 x^3}+\log (\log (4)) \int \frac {-56 x+908 x^2-256 x^3}{\left (1-32 x+28 x^2-4 x^3\right )^2} \, dx\\ &=\frac {28 x^2 \log (\log (4))}{1-32 x+28 x^2-4 x^3}+\log (\log (4)) \int \frac {x \left (-56+908 x-256 x^2\right )}{\left (1-32 x+28 x^2-4 x^3\right )^2} \, dx\\ &=-\frac {32 x \log (\log (4))}{1-32 x+28 x^2-4 x^3}+\frac {28 x^2 \log (\log (4))}{1-32 x+28 x^2-4 x^3}+\frac {1}{8} \log (\log (4)) \int \frac {256-448 x+96 x^2}{\left (1-32 x+28 x^2-4 x^3\right )^2} \, dx\\ &=\frac {\log (\log (4))}{1-32 x+28 x^2-4 x^3}-\frac {32 x \log (\log (4))}{1-32 x+28 x^2-4 x^3}+\frac {28 x^2 \log (\log (4))}{1-32 x+28 x^2-4 x^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 32, normalized size = 1.19 \begin {gather*} \frac {4 \left (1-32 x+28 x^2\right ) \log (\log (4))}{4-128 x+112 x^2-16 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((12*x^2 - 256*x^3 + 112*x^4)*Log[Log[4]])/(1 - 64*x + 1080*x^2 - 1800*x^3 + 1040*x^4 - 224*x^5 + 16
*x^6),x]

[Out]

(4*(1 - 32*x + 28*x^2)*Log[Log[4]])/(4 - 128*x + 112*x^2 - 16*x^3)

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fricas [A]  time = 0.78, size = 34, normalized size = 1.26 \begin {gather*} -\frac {{\left (28 \, x^{2} - 32 \, x + 1\right )} \log \left (2 \, \log \relax (2)\right )}{4 \, x^{3} - 28 \, x^{2} + 32 \, x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((112*x^4-256*x^3+12*x^2)*log(2*log(2))/(16*x^6-224*x^5+1040*x^4-1800*x^3+1080*x^2-64*x+1),x, algorit
hm="fricas")

[Out]

-(28*x^2 - 32*x + 1)*log(2*log(2))/(4*x^3 - 28*x^2 + 32*x - 1)

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giac [A]  time = 0.31, size = 34, normalized size = 1.26 \begin {gather*} -\frac {{\left (28 \, x^{2} - 32 \, x + 1\right )} \log \left (2 \, \log \relax (2)\right )}{4 \, x^{3} - 28 \, x^{2} + 32 \, x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((112*x^4-256*x^3+12*x^2)*log(2*log(2))/(16*x^6-224*x^5+1040*x^4-1800*x^3+1080*x^2-64*x+1),x, algorit
hm="giac")

[Out]

-(28*x^2 - 32*x + 1)*log(2*log(2))/(4*x^3 - 28*x^2 + 32*x - 1)

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maple [A]  time = 0.04, size = 32, normalized size = 1.19

method	result	size
norman	\(\frac {\left (-4 \ln \left (\ln \relax (2)\right )-4 \ln \relax (2)\right ) x^{3}}{4 x^{3}-28 x^{2}+32 x -1}\)	\(32\)
default	\(\frac {4 \ln \left (2 \ln \relax (2)\right ) \left (-\frac {7}{4} x^{2}+2 x -\frac {1}{16}\right )}{x^{3}-7 x^{2}+8 x -\frac {1}{4}}\)	\(33\)
risch	\(\frac {\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) \left (-7 x^{2}+8 x -\frac {1}{4}\right )}{x^{3}-7 x^{2}+8 x -\frac {1}{4}}\)	\(33\)
gosper	\(-\frac {\left (28 x^{2}-32 x +1\right ) \ln \left (2 \ln \relax (2)\right )}{4 x^{3}-28 x^{2}+32 x -1}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((112*x^4-256*x^3+12*x^2)*ln(2*ln(2))/(16*x^6-224*x^5+1040*x^4-1800*x^3+1080*x^2-64*x+1),x,method=_RETURNVE
RBOSE)

[Out]

(-4*ln(ln(2))-4*ln(2))*x^3/(4*x^3-28*x^2+32*x-1)

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maxima [A]  time = 0.44, size = 34, normalized size = 1.26 \begin {gather*} -\frac {{\left (28 \, x^{2} - 32 \, x + 1\right )} \log \left (2 \, \log \relax (2)\right )}{4 \, x^{3} - 28 \, x^{2} + 32 \, x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((112*x^4-256*x^3+12*x^2)*log(2*log(2))/(16*x^6-224*x^5+1040*x^4-1800*x^3+1080*x^2-64*x+1),x, algorit
hm="maxima")

[Out]

-(28*x^2 - 32*x + 1)*log(2*log(2))/(4*x^3 - 28*x^2 + 32*x - 1)

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mupad [B]  time = 0.09, size = 34, normalized size = 1.26 \begin {gather*} -\frac {\ln \left (2\,\ln \relax (2)\right )\,\left (28\,x^2-32\,x+1\right )}{4\,x^3-28\,x^2+32\,x-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2*log(2))*(12*x^2 - 256*x^3 + 112*x^4))/(1080*x^2 - 64*x - 1800*x^3 + 1040*x^4 - 224*x^5 + 16*x^6 + 1
),x)

[Out]

-(log(2*log(2))*(28*x^2 - 32*x + 1))/(32*x - 28*x^2 + 4*x^3 - 1)

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sympy [B]  time = 1.06, size = 53, normalized size = 1.96 \begin {gather*} \frac {x^{2} \left (- 28 \log {\relax (2 )} - 28 \log {\left (\log {\relax (2 )} \right )}\right ) + x \left (32 \log {\left (\log {\relax (2 )} \right )} + 32 \log {\relax (2 )}\right ) - \log {\relax (2 )} - \log {\left (\log {\relax (2 )} \right )}}{4 x^{3} - 28 x^{2} + 32 x - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((112*x**4-256*x**3+12*x**2)*ln(2*ln(2))/(16*x**6-224*x**5+1040*x**4-1800*x**3+1080*x**2-64*x+1),x)

[Out]

(x**2*(-28*log(2) - 28*log(log(2))) + x*(32*log(log(2)) + 32*log(2)) - log(2) - log(log(2)))/(4*x**3 - 28*x**2
+ 32*x - 1)

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