Integrand size = 52, antiderivative size = 27 \[ \int \frac {\left (12 x^2-256 x^3+112 x^4\right ) \log (\log (4))}{1-64 x+1080 x^2-1800 x^3+1040 x^4-224 x^5+16 x^6} \, dx=\frac {x^2 \log (\log (4))}{(2-x) (-4+x)+\frac {1}{4 x}+x} \] Output:
x^2*ln(2*ln(2))/(x+1/4/x+(-4+x)*(2-x))
Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {\left (12 x^2-256 x^3+112 x^4\right ) \log (\log (4))}{1-64 x+1080 x^2-1800 x^3+1040 x^4-224 x^5+16 x^6} \, dx=\frac {4 \left (1-32 x+28 x^2\right ) \log (\log (4))}{4-128 x+112 x^2-16 x^3} \] Input:
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]
Output:
(4*(1 - 32*x + 28*x^2)*Log[Log[4]])/(4 - 128*x + 112*x^2 - 16*x^3)
Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(27)=54\).
Time = 10.62 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.56, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {27, 27, 2028, 2462, 7239, 2527, 27, 2028, 2527, 27, 2021}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (112 x^4-256 x^3+12 x^2\right ) \log (\log (4))}{16 x^6-224 x^5+1040 x^4-1800 x^3+1080 x^2-64 x+1} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \log (\log (4)) \int \frac {4 \left (28 x^4-64 x^3+3 x^2\right )}{16 x^6-224 x^5+1040 x^4-1800 x^3+1080 x^2-64 x+1}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \log (\log (4)) \int \frac {28 x^4-64 x^3+3 x^2}{16 x^6-224 x^5+1040 x^4-1800 x^3+1080 x^2-64 x+1}dx\) |
\(\Big \downarrow \) 2028 |
\(\displaystyle 4 \log (\log (4)) \int \frac {x^2 \left (28 x^2-64 x+3\right )}{16 x^6-224 x^5+1040 x^4-1800 x^3+1080 x^2-64 x+1}dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle 4 \log (\log (4)) \int \left (\frac {7 x+33}{4 x^3-28 x^2+32 x-1}+\frac {703 x^2-1049 x+33}{\left (4 x^3-28 x^2+32 x-1\right )^2}\right )dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle 4 \log (\log (4)) \int \frac {x^2 \left (28 x^2-64 x+3\right )}{\left (-4 x^3+28 x^2-32 x+1\right )^2}dx\) |
\(\Big \downarrow \) 2527 |
\(\displaystyle 4 \log (\log (4)) \left (\frac {1}{4} \int -\frac {4 \left (64 x^3-227 x^2+14 x\right )}{\left (-4 x^3+28 x^2-32 x+1\right )^2}dx+\frac {7 x^2}{-4 x^3+28 x^2-32 x+1}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \log (\log (4)) \left (\frac {7 x^2}{-4 x^3+28 x^2-32 x+1}-\int \frac {64 x^3-227 x^2+14 x}{\left (-4 x^3+28 x^2-32 x+1\right )^2}dx\right )\) |
\(\Big \downarrow \) 2028 |
\(\displaystyle 4 \log (\log (4)) \left (\frac {7 x^2}{-4 x^3+28 x^2-32 x+1}-\int \frac {x \left (64 x^2-227 x+14\right )}{\left (-4 x^3+28 x^2-32 x+1\right )^2}dx\right )\) |
\(\Big \downarrow \) 2527 |
\(\displaystyle 4 \log (\log (4)) \left (-\frac {1}{8} \int -\frac {8 \left (3 x^2-14 x+8\right )}{\left (-4 x^3+28 x^2-32 x+1\right )^2}dx+\frac {7 x^2}{-4 x^3+28 x^2-32 x+1}-\frac {8 x}{-4 x^3+28 x^2-32 x+1}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \log (\log (4)) \left (\int \frac {3 x^2-14 x+8}{\left (-4 x^3+28 x^2-32 x+1\right )^2}dx+\frac {7 x^2}{-4 x^3+28 x^2-32 x+1}-\frac {8 x}{-4 x^3+28 x^2-32 x+1}\right )\) |
\(\Big \downarrow \) 2021 |
\(\displaystyle 4 \left (\frac {7 x^2}{-4 x^3+28 x^2-32 x+1}-\frac {8 x}{-4 x^3+28 x^2-32 x+1}+\frac {1}{4 \left (-4 x^3+28 x^2-32 x+1\right )}\right ) \log (\log (4))\) |
Input:
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]
Output:
4*(1/(4*(1 - 32*x + 28*x^2 - 4*x^3)) - (8*x)/(1 - 32*x + 28*x^2 - 4*x^3) + (7*x^2)/(1 - 32*x + 28*x^2 - 4*x^3))*Log[Log[4]]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])]] /; Free Q[m, x] && PolyQ[Pp, x] && PolyQ[Qq, x] && NeQ[m, -1]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.) + (c_.)*(x_)^(t_.))^(p_.), x_Symbol] :> Int[x^(p*r)*(a + b*x^(s - r) + c*x^(t - r))^p*Fx, x] /; FreeQ[ {a, b, c, r, s, t}, x] && IntegerQ[p] && PosQ[s - r] && PosQ[t - r] && !(E qQ[p, 1] && EqQ[u, 1])
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u*Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ [Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0 ] && RationalFunctionQ[u, x]
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] + Simp[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]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19
method | result | size |
norman | \(\frac {\left (-4 \ln \left (\ln \left (2\right )\right )-4 \ln \left (2\right )\right ) x^{3}}{4 x^{3}-28 x^{2}+32 x -1}\) | \(32\) |
default | \(\frac {4 \ln \left (2 \ln \left (2\right )\right ) \left (-\frac {7}{4} x^{2}+2 x -\frac {1}{16}\right )}{x^{3}-7 x^{2}+8 x -\frac {1}{4}}\) | \(33\) |
risch | \(\frac {\left (\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )\right ) \left (-7 x^{2}+8 x -\frac {1}{4}\right )}{x^{3}-7 x^{2}+8 x -\frac {1}{4}}\) | \(33\) |
gosper | \(-\frac {\left (28 x^{2}-32 x +1\right ) \ln \left (2 \ln \left (2\right )\right )}{4 x^{3}-28 x^{2}+32 x -1}\) | \(35\) |
parallelrisch | \(\frac {\ln \left (2 \ln \left (2\right )\right ) \left (-112 x^{2}+128 x -4\right )}{16 x^{3}-112 x^{2}+128 x -4}\) | \(35\) |
orering | \(-\frac {x \left (4 x^{3}-28 x^{2}+32 x -1\right ) \left (112 x^{4}-256 x^{3}+12 x^{2}\right ) \ln \left (2 \ln \left (2\right )\right )}{\left (28 x^{2}-64 x +3\right ) \left (16 x^{6}-224 x^{5}+1040 x^{4}-1800 x^{3}+1080 x^{2}-64 x +1\right )}\) | \(84\) |
Input:
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=_RETURNVERBOSE)
Output:
(-4*ln(ln(2))-4*ln(2))*x^3/(4*x^3-28*x^2+32*x-1)
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\left (12 x^2-256 x^3+112 x^4\right ) \log (\log (4))}{1-64 x+1080 x^2-1800 x^3+1040 x^4-224 x^5+16 x^6} \, dx=-\frac {{\left (28 \, x^{2} - 32 \, x + 1\right )} \log \left (2 \, \log \left (2\right )\right )}{4 \, x^{3} - 28 \, x^{2} + 32 \, x - 1} \] Input:
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, algorithm="fricas")
Output:
-(28*x^2 - 32*x + 1)*log(2*log(2))/(4*x^3 - 28*x^2 + 32*x - 1)
Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (22) = 44\).
Time = 0.67 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {\left (12 x^2-256 x^3+112 x^4\right ) \log (\log (4))}{1-64 x+1080 x^2-1800 x^3+1040 x^4-224 x^5+16 x^6} \, dx=\frac {x^{2} \left (- 28 \log {\left (2 \right )} - 28 \log {\left (\log {\left (2 \right )} \right )}\right ) + x \left (32 \log {\left (\log {\left (2 \right )} \right )} + 32 \log {\left (2 \right )}\right ) - \log {\left (2 \right )} - \log {\left (\log {\left (2 \right )} \right )}}{4 x^{3} - 28 x^{2} + 32 x - 1} \] Input:
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)
Output:
(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)
Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\left (12 x^2-256 x^3+112 x^4\right ) \log (\log (4))}{1-64 x+1080 x^2-1800 x^3+1040 x^4-224 x^5+16 x^6} \, dx=-\frac {{\left (28 \, x^{2} - 32 \, x + 1\right )} \log \left (2 \, \log \left (2\right )\right )}{4 \, x^{3} - 28 \, x^{2} + 32 \, x - 1} \] Input:
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, algorithm="maxima")
Output:
-(28*x^2 - 32*x + 1)*log(2*log(2))/(4*x^3 - 28*x^2 + 32*x - 1)
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\left (12 x^2-256 x^3+112 x^4\right ) \log (\log (4))}{1-64 x+1080 x^2-1800 x^3+1040 x^4-224 x^5+16 x^6} \, dx=-\frac {{\left (28 \, x^{2} - 32 \, x + 1\right )} \log \left (2 \, \log \left (2\right )\right )}{4 \, x^{3} - 28 \, x^{2} + 32 \, x - 1} \] Input:
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, algorithm="giac")
Output:
-(28*x^2 - 32*x + 1)*log(2*log(2))/(4*x^3 - 28*x^2 + 32*x - 1)
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\left (12 x^2-256 x^3+112 x^4\right ) \log (\log (4))}{1-64 x+1080 x^2-1800 x^3+1040 x^4-224 x^5+16 x^6} \, dx=-\frac {\ln \left (2\,\ln \left (2\right )\right )\,\left (28\,x^2-32\,x+1\right )}{4\,x^3-28\,x^2+32\,x-1} \] Input:
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)
Output:
-(log(2*log(2))*(28*x^2 - 32*x + 1))/(32*x - 28*x^2 + 4*x^3 - 1)
Time = 0.24 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (12 x^2-256 x^3+112 x^4\right ) \log (\log (4))}{1-64 x+1080 x^2-1800 x^3+1040 x^4-224 x^5+16 x^6} \, dx=-\frac {4 \,\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right ) x^{3}}{4 x^{3}-28 x^{2}+32 x -1} \] Input:
int((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)
Output:
( - 4*log(2*log(2))*x**3)/(4*x**3 - 28*x**2 + 32*x - 1)