Integrand size = 55, antiderivative size = 23 \[ \int \frac {-8-36 x+8 x \log (\log (2))}{80 x^2+360 x^3+405 x^4+\left (-80 x^3-180 x^4\right ) \log (\log (2))+20 x^4 \log ^2(\log (2))} \, dx=\frac {1}{5 x^2 \left (\frac {9}{2}+\frac {2}{x}-\log (\log (2))\right )} \] Output:
1/5/x^2/(2/x+9/2-ln(ln(2)))
Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-8-36 x+8 x \log (\log (2))}{80 x^2+360 x^3+405 x^4+\left (-80 x^3-180 x^4\right ) \log (\log (2))+20 x^4 \log ^2(\log (2))} \, dx=\frac {4}{5 \left (8 x+18 x^2-4 x^2 \log (\log (2))\right )} \] Input:
Integrate[(-8 - 36*x + 8*x*Log[Log[2]])/(80*x^2 + 360*x^3 + 405*x^4 + (-80 *x^3 - 180*x^4)*Log[Log[2]] + 20*x^4*Log[Log[2]]^2),x]
Output:
4/(5*(8*x + 18*x^2 - 4*x^2*Log[Log[2]]))
Time = 0.21 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {6, 6, 2026, 1184, 27, 83}
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.
\(\displaystyle \int \frac {-36 x+8 x \log (\log (2))-8}{405 x^4+20 x^4 \log ^2(\log (2))+360 x^3+80 x^2+\left (-180 x^4-80 x^3\right ) \log (\log (2))} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x (8 \log (\log (2))-36)-8}{405 x^4+20 x^4 \log ^2(\log (2))+360 x^3+80 x^2+\left (-180 x^4-80 x^3\right ) \log (\log (2))}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {x (8 \log (\log (2))-36)-8}{x^4 \left (405+20 \log ^2(\log (2))\right )+360 x^3+80 x^2+\left (-180 x^4-80 x^3\right ) \log (\log (2))}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x (8 \log (\log (2))-36)-8}{x^2 \left (5 x^2 (9-2 \log (\log (2)))^2+40 x (9-2 \log (\log (2)))+80\right )}dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle 5 (9-2 \log (\log (2)))^2 \int -\frac {4 ((9-2 \log (\log (2))) x+2)}{25 x^2 ((9-2 \log (\log (2))) x+4)^2 (9-2 \log (\log (2)))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4}{5} \int \frac {(9-2 \log (\log (2))) x+2}{x^2 ((9-2 \log (\log (2))) x+4)^2}dx\) |
\(\Big \downarrow \) 83 |
\(\displaystyle \frac {2}{5 x (x (9-2 \log (\log (2)))+4)}\) |
Input:
Int[(-8 - 36*x + 8*x*Log[Log[2]])/(80*x^2 + 360*x^3 + 405*x^4 + (-80*x^3 - 180*x^4)*Log[Log[2]] + 20*x^4*Log[Log[2]]^2),x]
Output:
2/(5*x*(4 + x*(9 - 2*Log[Log[2]])))
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && EqQ[a*d*f *(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83
method | result | size |
gosper | \(-\frac {2}{5 x \left (2 x \ln \left (\ln \left (2\right )\right )-9 x -4\right )}\) | \(19\) |
norman | \(-\frac {2}{5 x \left (2 x \ln \left (\ln \left (2\right )\right )-9 x -4\right )}\) | \(19\) |
risch | \(-\frac {2}{5 x \left (2 x \ln \left (\ln \left (2\right )\right )-9 x -4\right )}\) | \(19\) |
parallelrisch | \(-\frac {2}{5 x \left (2 x \ln \left (\ln \left (2\right )\right )-9 x -4\right )}\) | \(19\) |
default | \(\frac {1}{10 x}-\frac {\left (2 \ln \left (\ln \left (2\right )\right )-9\right )^{2}}{10 \left (-2 \ln \left (\ln \left (2\right )\right )+9\right ) \left (-2 x \ln \left (\ln \left (2\right )\right )+9 x +4\right )}\) | \(40\) |
orering | \(-\frac {x \left (2 x \ln \left (\ln \left (2\right )\right )-9 x -4\right ) \left (8 x \ln \left (\ln \left (2\right )\right )-36 x -8\right )}{2 \left (2 x \ln \left (\ln \left (2\right )\right )-9 x -2\right ) \left (20 x^{4} \ln \left (\ln \left (2\right )\right )^{2}+\left (-180 x^{4}-80 x^{3}\right ) \ln \left (\ln \left (2\right )\right )+405 x^{4}+360 x^{3}+80 x^{2}\right )}\) | \(82\) |
Input:
int((8*x*ln(ln(2))-36*x-8)/(20*x^4*ln(ln(2))^2+(-180*x^4-80*x^3)*ln(ln(2)) +405*x^4+360*x^3+80*x^2),x,method=_RETURNVERBOSE)
Output:
-2/5/x/(2*x*ln(ln(2))-9*x-4)
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-8-36 x+8 x \log (\log (2))}{80 x^2+360 x^3+405 x^4+\left (-80 x^3-180 x^4\right ) \log (\log (2))+20 x^4 \log ^2(\log (2))} \, dx=-\frac {2}{5 \, {\left (2 \, x^{2} \log \left (\log \left (2\right )\right ) - 9 \, x^{2} - 4 \, x\right )}} \] Input:
integrate((8*x*log(log(2))-36*x-8)/(20*x^4*log(log(2))^2+(-180*x^4-80*x^3) *log(log(2))+405*x^4+360*x^3+80*x^2),x, algorithm="fricas")
Output:
-2/5/(2*x^2*log(log(2)) - 9*x^2 - 4*x)
Time = 0.34 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {-8-36 x+8 x \log (\log (2))}{80 x^2+360 x^3+405 x^4+\left (-80 x^3-180 x^4\right ) \log (\log (2))+20 x^4 \log ^2(\log (2))} \, dx=- \frac {2}{x^{2} \left (-45 + 10 \log {\left (\log {\left (2 \right )} \right )}\right ) - 20 x} \] Input:
integrate((8*x*ln(ln(2))-36*x-8)/(20*x**4*ln(ln(2))**2+(-180*x**4-80*x**3) *ln(ln(2))+405*x**4+360*x**3+80*x**2),x)
Output:
-2/(x**2*(-45 + 10*log(log(2))) - 20*x)
Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-8-36 x+8 x \log (\log (2))}{80 x^2+360 x^3+405 x^4+\left (-80 x^3-180 x^4\right ) \log (\log (2))+20 x^4 \log ^2(\log (2))} \, dx=-\frac {2}{5 \, {\left (x^{2} {\left (2 \, \log \left (\log \left (2\right )\right ) - 9\right )} - 4 \, x\right )}} \] Input:
integrate((8*x*log(log(2))-36*x-8)/(20*x^4*log(log(2))^2+(-180*x^4-80*x^3) *log(log(2))+405*x^4+360*x^3+80*x^2),x, algorithm="maxima")
Output:
-2/5/(x^2*(2*log(log(2)) - 9) - 4*x)
Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-8-36 x+8 x \log (\log (2))}{80 x^2+360 x^3+405 x^4+\left (-80 x^3-180 x^4\right ) \log (\log (2))+20 x^4 \log ^2(\log (2))} \, dx=-\frac {2}{5 \, {\left (2 \, x^{2} \log \left (\log \left (2\right )\right ) - 9 \, x^{2} - 4 \, x\right )}} \] Input:
integrate((8*x*log(log(2))-36*x-8)/(20*x^4*log(log(2))^2+(-180*x^4-80*x^3) *log(log(2))+405*x^4+360*x^3+80*x^2),x, algorithm="giac")
Output:
-2/5/(2*x^2*log(log(2)) - 9*x^2 - 4*x)
Time = 0.14 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-8-36 x+8 x \log (\log (2))}{80 x^2+360 x^3+405 x^4+\left (-80 x^3-180 x^4\right ) \log (\log (2))+20 x^4 \log ^2(\log (2))} \, dx=\frac {2}{20\,x-x^2\,\left (10\,\ln \left (\ln \left (2\right )\right )-45\right )} \] Input:
int(-(36*x - 8*x*log(log(2)) + 8)/(20*x^4*log(log(2))^2 - log(log(2))*(80* x^3 + 180*x^4) + 80*x^2 + 360*x^3 + 405*x^4),x)
Output:
2/(20*x - x^2*(10*log(log(2)) - 45))
Time = 0.20 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {-8-36 x+8 x \log (\log (2))}{80 x^2+360 x^3+405 x^4+\left (-80 x^3-180 x^4\right ) \log (\log (2))+20 x^4 \log ^2(\log (2))} \, dx=-\frac {2}{5 x \left (2 \,\mathrm {log}\left (\mathrm {log}\left (2\right )\right ) x -9 x -4\right )} \] Input:
int((8*x*log(log(2))-36*x-8)/(20*x^4*log(log(2))^2+(-180*x^4-80*x^3)*log(l og(2))+405*x^4+360*x^3+80*x^2),x)
Output:
( - 2)/(5*x*(2*log(log(2))*x - 9*x - 4))