3.50.4 \(\int \frac {4096-256 x-64 x^2}{4 x^3+x^4+(128 x^2+32 x^3) \log (x)+(1024 x+256 x^2) \log ^2(x)+(-128 x^2-32 x^3+(-2048 x-512 x^2) \log (x)) \log (\frac {5 x^2}{4+x})+(1024 x+256 x^2) \log ^2(\frac {5 x^2}{4+x})} \, dx\)

Optimal. Leaf size=25 \[ \frac {4}{\frac {x}{16}+\log (x)-\log \left (\frac {5 x^2}{4+x}\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {6688, 12, 6686} \begin {gather*} \frac {64}{-16 \log \left (\frac {5 x^2}{x+4}\right )+x+16 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(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]

[Out]

64/(x + 16*Log[x] - 16*Log[(5*x^2)/(4 + 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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -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} &=\int \frac {64 \left (64-4 x-x^2\right )}{x (4+x) \left (x+16 \log (x)-16 \log \left (\frac {5 x^2}{4+x}\right )\right )^2} \, dx\\ &=64 \int \frac {64-4 x-x^2}{x (4+x) \left (x+16 \log (x)-16 \log \left (\frac {5 x^2}{4+x}\right )\right )^2} \, dx\\ &=\frac {64}{x+16 \log (x)-16 \log \left (\frac {5 x^2}{4+x}\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 23, normalized size = 0.92 \begin {gather*} \frac {64}{x+16 \log (x)-16 \log \left (\frac {5 x^2}{4+x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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 + (-1
28*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]

[Out]

64/(x + 16*Log[x] - 16*Log[(5*x^2)/(4 + x)])

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 23, normalized size = 0.92 \begin {gather*} \frac {64}{x + 16 \, \log \relax (x) - 16 \, \log \left (\frac {5 \, x^{2}}{x + 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

64/(x + 16*log(x) - 16*log(5*x^2/(x + 4)))

________________________________________________________________________________________

giac [A]  time = 0.30, size = 20, normalized size = 0.80 \begin {gather*} \frac {64}{x - 16 \, \log \relax (5) + 16 \, \log \left (x + 4\right ) - 16 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

64/(x - 16*log(5) + 16*log(x + 4) - 16*log(x))

________________________________________________________________________________________

maple [C]  time = 0.12, size = 172, normalized size = 6.88

method	result	size
risch	\(-\frac {64 i}{8 \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-16 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+8 \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+8 \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i}{4+x}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{4+x}\right )-8 \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{4+x}\right )^{2}-8 \pi \,\mathrm {csgn}\left (\frac {i}{4+x}\right ) \mathrm {csgn}\left (\frac {i x^{2}}{4+x}\right )^{2}+8 \pi \mathrm {csgn}\left (\frac {i x^{2}}{4+x}\right )^{3}+16 i \ln \relax (5)-i x +16 i \ln \relax (x )-16 i \ln \left (4+x \right )}\)	\(172\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-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,method=_RETURNVERBOSE)

[Out]

-64*I/(8*Pi*csgn(I*x)^2*csgn(I*x^2)-16*Pi*csgn(I*x)*csgn(I*x^2)^2+8*Pi*csgn(I*x^2)^3+8*Pi*csgn(I*x^2)*csgn(I/(
4+x))*csgn(I*x^2/(4+x))-8*Pi*csgn(I*x^2)*csgn(I*x^2/(4+x))^2-8*Pi*csgn(I/(4+x))*csgn(I*x^2/(4+x))^2+8*Pi*csgn(
I*x^2/(4+x))^3+16*I*ln(5)-I*x+16*I*ln(x)-16*I*ln(4+x))

________________________________________________________________________________________

maxima [A]  time = 0.48, size = 20, normalized size = 0.80 \begin {gather*} \frac {64}{x - 16 \, \log \relax (5) + 16 \, \log \left (x + 4\right ) - 16 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

64/(x - 16*log(5) + 16*log(x + 4) - 16*log(x))

________________________________________________________________________________________

mupad [B]  time = 3.77, size = 23, normalized size = 0.92 \begin {gather*} \frac {64}{x-16\,\ln \left (\frac {5\,x^2}{x+4}\right )+16\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

64/(x - 16*log((5*x^2)/(x + 4)) + 16*log(x))

________________________________________________________________________________________

sympy [A]  time = 0.38, size = 19, normalized size = 0.76 \begin {gather*} - \frac {4}{- \frac {x}{16} - \log {\relax (x )} + \log {\left (\frac {5 x^{2}}{x + 4} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

-4/(-x/16 - log(x) + log(5*x**2/(x + 4)))

________________________________________________________________________________________