Integrand size = 108, antiderivative size = 25 \[ \int \frac {4096-256 x-64 x^2}{4 x^3+x^4+\left (128 x^2+32 x^3\right ) \log (x)+\left (1024 x+256 x^2\right ) \log ^2(x)+\left (-128 x^2-32 x^3+\left (-2048 x-512 x^2\right ) \log (x)\right ) \log \left (\frac {5 x^2}{4+x}\right )+\left (1024 x+256 x^2\right ) \log ^2\left (\frac {5 x^2}{4+x}\right )} \, dx=\frac {4}{\frac {x}{16}+\log (x)-\log \left (\frac {5 x^2}{4+x}\right )} \] Output:
4/(ln(x)-ln(5*x^2/(4+x))+1/16*x)
Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {4096-256 x-64 x^2}{4 x^3+x^4+\left (128 x^2+32 x^3\right ) \log (x)+\left (1024 x+256 x^2\right ) \log ^2(x)+\left (-128 x^2-32 x^3+\left (-2048 x-512 x^2\right ) \log (x)\right ) \log \left (\frac {5 x^2}{4+x}\right )+\left (1024 x+256 x^2\right ) \log ^2\left (\frac {5 x^2}{4+x}\right )} \, dx=\frac {64}{x+16 \log (x)-16 \log \left (\frac {5 x^2}{4+x}\right )} \] Input:
Integrate[(4096 - 256*x - 64*x^2)/(4*x^3 + x^4 + (128*x^2 + 32*x^3)*Log[x] + (1024*x + 256*x^2)*Log[x]^2 + (-128*x^2 - 32*x^3 + (-2048*x - 512*x^2)* Log[x])*Log[(5*x^2)/(4 + x)] + (1024*x + 256*x^2)*Log[(5*x^2)/(4 + x)]^2), x]
Output:
64/(x + 16*Log[x] - 16*Log[(5*x^2)/(4 + x)])
Time = 0.48 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {7239, 27, 7237}
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.
\(\displaystyle \int \frac {-64 x^2-256 x+4096}{x^4+4 x^3+\left (256 x^2+1024 x\right ) \log ^2(x)+\left (256 x^2+1024 x\right ) \log ^2\left (\frac {5 x^2}{x+4}\right )+\left (32 x^3+128 x^2\right ) \log (x)+\left (-32 x^3-128 x^2+\left (-512 x^2-2048 x\right ) \log (x)\right ) \log \left (\frac {5 x^2}{x+4}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {64 \left (-x^2-4 x+64\right )}{x (x+4) \left (-16 \log \left (\frac {5 x^2}{x+4}\right )+x+16 \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 64 \int \frac {-x^2-4 x+64}{x (x+4) \left (x+16 \log (x)-16 \log \left (\frac {5 x^2}{x+4}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \frac {64}{-16 \log \left (\frac {5 x^2}{x+4}\right )+x+16 \log (x)}\) |
Input:
Int[(4096 - 256*x - 64*x^2)/(4*x^3 + x^4 + (128*x^2 + 32*x^3)*Log[x] + (10 24*x + 256*x^2)*Log[x]^2 + (-128*x^2 - 32*x^3 + (-2048*x - 512*x^2)*Log[x] )*Log[(5*x^2)/(4 + x)] + (1024*x + 256*x^2)*Log[(5*x^2)/(4 + x)]^2),x]
Output:
64/(x + 16*Log[x] - 16*Log[(5*x^2)/(4 + x)])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.89 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96
method | result | size |
parallelrisch | \(\frac {64}{x +16 \ln \left (x \right )-16 \ln \left (\frac {5 x^{2}}{4+x}\right )}\) | \(24\) |
default | \(-\frac {64 i}{8 \pi \,\operatorname {csgn}\left (\frac {i}{4+x}\right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{4+x}\right )-8 \pi \,\operatorname {csgn}\left (\frac {i}{4+x}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{4+x}\right )^{2}+8 \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-16 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+8 \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-8 \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{4+x}\right )^{2}+8 \pi \operatorname {csgn}\left (\frac {i x^{2}}{4+x}\right )^{3}+16 i \ln \left (5\right )+16 i \ln \left (x \right )-i x -16 i \ln \left (4+x \right )}\) | \(172\) |
risch | \(\frac {64 i}{8 \pi \,\operatorname {csgn}\left (\frac {i}{4+x}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{4+x}\right )^{2}-8 \pi \,\operatorname {csgn}\left (\frac {i}{4+x}\right ) \operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{4+x}\right )-8 \pi \operatorname {csgn}\left (\frac {i x^{2}}{4+x}\right )^{3}+8 \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (\frac {i x^{2}}{4+x}\right )^{2}-8 \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )+16 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}-8 \pi \operatorname {csgn}\left (i x^{2}\right )^{3}-16 i \ln \left (5\right )+i x -16 i \ln \left (x \right )+16 i \ln \left (4+x \right )}\) | \(172\) |
Input:
int((-64*x^2-256*x+4096)/((256*x^2+1024*x)*ln(5*x^2/(4+x))^2+((-512*x^2-20 48*x)*ln(x)-32*x^3-128*x^2)*ln(5*x^2/(4+x))+(256*x^2+1024*x)*ln(x)^2+(32*x ^3+128*x^2)*ln(x)+x^4+4*x^3),x,method=_RETURNVERBOSE)
Output:
64/(x+16*ln(x)-16*ln(5*x^2/(4+x)))
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {4096-256 x-64 x^2}{4 x^3+x^4+\left (128 x^2+32 x^3\right ) \log (x)+\left (1024 x+256 x^2\right ) \log ^2(x)+\left (-128 x^2-32 x^3+\left (-2048 x-512 x^2\right ) \log (x)\right ) \log \left (\frac {5 x^2}{4+x}\right )+\left (1024 x+256 x^2\right ) \log ^2\left (\frac {5 x^2}{4+x}\right )} \, dx=\frac {64}{x + 16 \, \log \left (x\right ) - 16 \, \log \left (\frac {5 \, x^{2}}{x + 4}\right )} \] Input:
integrate((-64*x^2-256*x+4096)/((256*x^2+1024*x)*log(5*x^2/(4+x))^2+((-512 *x^2-2048*x)*log(x)-32*x^3-128*x^2)*log(5*x^2/(4+x))+(256*x^2+1024*x)*log( x)^2+(32*x^3+128*x^2)*log(x)+x^4+4*x^3),x, algorithm="fricas")
Output:
64/(x + 16*log(x) - 16*log(5*x^2/(x + 4)))
Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {4096-256 x-64 x^2}{4 x^3+x^4+\left (128 x^2+32 x^3\right ) \log (x)+\left (1024 x+256 x^2\right ) \log ^2(x)+\left (-128 x^2-32 x^3+\left (-2048 x-512 x^2\right ) \log (x)\right ) \log \left (\frac {5 x^2}{4+x}\right )+\left (1024 x+256 x^2\right ) \log ^2\left (\frac {5 x^2}{4+x}\right )} \, dx=- \frac {4}{- \frac {x}{16} - \log {\left (x \right )} + \log {\left (\frac {5 x^{2}}{x + 4} \right )}} \] Input:
integrate((-64*x**2-256*x+4096)/((256*x**2+1024*x)*ln(5*x**2/(4+x))**2+((- 512*x**2-2048*x)*ln(x)-32*x**3-128*x**2)*ln(5*x**2/(4+x))+(256*x**2+1024*x )*ln(x)**2+(32*x**3+128*x**2)*ln(x)+x**4+4*x**3),x)
Output:
-4/(-x/16 - log(x) + log(5*x**2/(x + 4)))
Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {4096-256 x-64 x^2}{4 x^3+x^4+\left (128 x^2+32 x^3\right ) \log (x)+\left (1024 x+256 x^2\right ) \log ^2(x)+\left (-128 x^2-32 x^3+\left (-2048 x-512 x^2\right ) \log (x)\right ) \log \left (\frac {5 x^2}{4+x}\right )+\left (1024 x+256 x^2\right ) \log ^2\left (\frac {5 x^2}{4+x}\right )} \, dx=\frac {64}{x - 16 \, \log \left (5\right ) + 16 \, \log \left (x + 4\right ) - 16 \, \log \left (x\right )} \] Input:
integrate((-64*x^2-256*x+4096)/((256*x^2+1024*x)*log(5*x^2/(4+x))^2+((-512 *x^2-2048*x)*log(x)-32*x^3-128*x^2)*log(5*x^2/(4+x))+(256*x^2+1024*x)*log( x)^2+(32*x^3+128*x^2)*log(x)+x^4+4*x^3),x, algorithm="maxima")
Output:
64/(x - 16*log(5) + 16*log(x + 4) - 16*log(x))
Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {4096-256 x-64 x^2}{4 x^3+x^4+\left (128 x^2+32 x^3\right ) \log (x)+\left (1024 x+256 x^2\right ) \log ^2(x)+\left (-128 x^2-32 x^3+\left (-2048 x-512 x^2\right ) \log (x)\right ) \log \left (\frac {5 x^2}{4+x}\right )+\left (1024 x+256 x^2\right ) \log ^2\left (\frac {5 x^2}{4+x}\right )} \, dx=\frac {64}{x - 16 \, \log \left (5\right ) + 16 \, \log \left (x + 4\right ) - 16 \, \log \left (x\right )} \] Input:
integrate((-64*x^2-256*x+4096)/((256*x^2+1024*x)*log(5*x^2/(4+x))^2+((-512 *x^2-2048*x)*log(x)-32*x^3-128*x^2)*log(5*x^2/(4+x))+(256*x^2+1024*x)*log( x)^2+(32*x^3+128*x^2)*log(x)+x^4+4*x^3),x, algorithm="giac")
Output:
64/(x - 16*log(5) + 16*log(x + 4) - 16*log(x))
Time = 4.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {4096-256 x-64 x^2}{4 x^3+x^4+\left (128 x^2+32 x^3\right ) \log (x)+\left (1024 x+256 x^2\right ) \log ^2(x)+\left (-128 x^2-32 x^3+\left (-2048 x-512 x^2\right ) \log (x)\right ) \log \left (\frac {5 x^2}{4+x}\right )+\left (1024 x+256 x^2\right ) \log ^2\left (\frac {5 x^2}{4+x}\right )} \, dx=\frac {64}{x-16\,\ln \left (\frac {5\,x^2}{x+4}\right )+16\,\ln \left (x\right )} \] Input:
int(-(256*x + 64*x^2 - 4096)/(log((5*x^2)/(x + 4))^2*(1024*x + 256*x^2) + log(x)*(128*x^2 + 32*x^3) + log(x)^2*(1024*x + 256*x^2) - log((5*x^2)/(x + 4))*(log(x)*(2048*x + 512*x^2) + 128*x^2 + 32*x^3) + 4*x^3 + x^4),x)
Output:
64/(x - 16*log((5*x^2)/(x + 4)) + 16*log(x))
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {4096-256 x-64 x^2}{4 x^3+x^4+\left (128 x^2+32 x^3\right ) \log (x)+\left (1024 x+256 x^2\right ) \log ^2(x)+\left (-128 x^2-32 x^3+\left (-2048 x-512 x^2\right ) \log (x)\right ) \log \left (\frac {5 x^2}{4+x}\right )+\left (1024 x+256 x^2\right ) \log ^2\left (\frac {5 x^2}{4+x}\right )} \, dx=-\frac {64}{16 \,\mathrm {log}\left (\frac {5 x^{2}}{x +4}\right )-16 \,\mathrm {log}\left (x \right )-x} \] Input:
int((-64*x^2-256*x+4096)/((256*x^2+1024*x)*log(5*x^2/(4+x))^2+((-512*x^2-2 048*x)*log(x)-32*x^3-128*x^2)*log(5*x^2/(4+x))+(256*x^2+1024*x)*log(x)^2+( 32*x^3+128*x^2)*log(x)+x^4+4*x^3),x)
Output:
( - 64)/(16*log((5*x**2)/(x + 4)) - 16*log(x) - x)