\(\int \frac {-12 x^3-3 x^7+12 x^3 \log (x^2)}{(256+128 x^4+16 x^8) \log ^3(x^2)} \, dx\) [7906]

Optimal result

Integrand size = 41, antiderivative size = 24 \[ \int \frac {-12 x^3-3 x^7+12 x^3 \log \left (x^2\right )}{\left (256+128 x^4+16 x^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {3 x^2}{64 \left (\frac {4}{x^2}+x^2\right ) \log ^2\left (x^2\right )} \]

[Out]

3/64*x^2/ln(x^2)^2/(x^2+4/x^2)

Rubi [F]

\[ \int \frac {-12 x^3-3 x^7+12 x^3 \log \left (x^2\right )}{\left (256+128 x^4+16 x^8\right ) \log ^3\left (x^2\right )} \, dx=\int \frac {-12 x^3-3 x^7+12 x^3 \log \left (x^2\right )}{\left (256+128 x^4+16 x^8\right ) \log ^3\left (x^2\right )} \, dx \]

[In]

Int[(-12*x^3 - 3*x^7 + 12*x^3*Log[x^2])/((256 + 128*x^4 + 16*x^8)*Log[x^2]^3),x]

[Out]

(-3*Defer[Subst][Defer[Int][x/((4 + x^2)*Log[x]^3), x], x, x^2])/32 + (3*Defer[Subst][Defer[Int][x/((4 + x^2)^
2*Log[x]^2), x], x, x^2])/8

Rubi steps \begin{align*} \text {integral}& = 16 \int \frac {-12 x^3-3 x^7+12 x^3 \log \left (x^2\right )}{\left (64+16 x^4\right )^2 \log ^3\left (x^2\right )} \, dx \\ & = 16 \int \frac {3 x^3 \left (-4-x^4+4 \log \left (x^2\right )\right )}{\left (64+16 x^4\right )^2 \log ^3\left (x^2\right )} \, dx \\ & = 48 \int \frac {x^3 \left (-4-x^4+4 \log \left (x^2\right )\right )}{\left (64+16 x^4\right )^2 \log ^3\left (x^2\right )} \, dx \\ & = 24 \text {Subst}\left (\int -\frac {x \left (4+x^2-4 \log (x)\right )}{256 \left (4+x^2\right )^2 \log ^3(x)} \, dx,x,x^2\right ) \\ & = -\left (\frac {3}{32} \text {Subst}\left (\int \frac {x \left (4+x^2-4 \log (x)\right )}{\left (4+x^2\right )^2 \log ^3(x)} \, dx,x,x^2\right )\right ) \\ & = -\left (\frac {3}{32} \text {Subst}\left (\int \left (\frac {x}{\left (4+x^2\right ) \log ^3(x)}-\frac {4 x}{\left (4+x^2\right )^2 \log ^2(x)}\right ) \, dx,x,x^2\right )\right ) \\ & = -\left (\frac {3}{32} \text {Subst}\left (\int \frac {x}{\left (4+x^2\right ) \log ^3(x)} \, dx,x,x^2\right )\right )+\frac {3}{8} \text {Subst}\left (\int \frac {x}{\left (4+x^2\right )^2 \log ^2(x)} \, dx,x,x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-12 x^3-3 x^7+12 x^3 \log \left (x^2\right )}{\left (256+128 x^4+16 x^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {3 x^4}{64 \left (4+x^4\right ) \log ^2\left (x^2\right )} \]

[In]

Integrate[(-12*x^3 - 3*x^7 + 12*x^3*Log[x^2])/((256 + 128*x^4 + 16*x^8)*Log[x^2]^3),x]

[Out]

(3*x^4)/(64*(4 + x^4)*Log[x^2]^2)

Maple [A] (verified)

Time = 1.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {3 x^{4}}{64 \left (x^{4}+4\right ) \ln \left (x^{2}\right )^{2}}\)	\(19\)
parallelrisch	\(\frac {3 x^{4}}{64 \left (x^{4}+4\right ) \ln \left (x^{2}\right )^{2}}\)	\(19\)
default	\(\frac {3}{64 \ln \left (x^{2}\right )^{2}}-\frac {3}{16 \left (x^{4}+4\right ) \ln \left (x^{2}\right )^{2}}\)	\(25\)
parts	\(\frac {3}{64 \ln \left (x^{2}\right )^{2}}-\frac {3}{16 \left (x^{4}+4\right ) \ln \left (x^{2}\right )^{2}}\)	\(25\)

[In]

int((12*x^3*ln(x^2)-3*x^7-12*x^3)/(16*x^8+128*x^4+256)/ln(x^2)^3,x,method=_RETURNVERBOSE)

[Out]

3/64*x^4/(x^4+4)/ln(x^2)^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-12 x^3-3 x^7+12 x^3 \log \left (x^2\right )}{\left (256+128 x^4+16 x^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {3 \, x^{4}}{64 \, {\left (x^{4} + 4\right )} \log \left (x^{2}\right )^{2}} \]

[In]

integrate((12*x^3*log(x^2)-3*x^7-12*x^3)/(16*x^8+128*x^4+256)/log(x^2)^3,x, algorithm="fricas")

[Out]

3/64*x^4/((x^4 + 4)*log(x^2)^2)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-12 x^3-3 x^7+12 x^3 \log \left (x^2\right )}{\left (256+128 x^4+16 x^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {3 x^{4}}{\left (64 x^{4} + 256\right ) \log {\left (x^{2} \right )}^{2}} \]

[In]

integrate((12*x**3*ln(x**2)-3*x**7-12*x**3)/(16*x**8+128*x**4+256)/ln(x**2)**3,x)

[Out]

3*x**4/((64*x**4 + 256)*log(x**2)**2)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-12 x^3-3 x^7+12 x^3 \log \left (x^2\right )}{\left (256+128 x^4+16 x^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {3 \, x^{4}}{256 \, {\left (x^{4} + 4\right )} \log \left (x\right )^{2}} \]

[In]

integrate((12*x^3*log(x^2)-3*x^7-12*x^3)/(16*x^8+128*x^4+256)/log(x^2)^3,x, algorithm="maxima")

[Out]

3/256*x^4/((x^4 + 4)*log(x)^2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-12 x^3-3 x^7+12 x^3 \log \left (x^2\right )}{\left (256+128 x^4+16 x^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {3 \, x^{4}}{64 \, {\left (x^{4} \log \left (x^{2}\right )^{2} + 4 \, \log \left (x^{2}\right )^{2}\right )}} \]

[In]

integrate((12*x^3*log(x^2)-3*x^7-12*x^3)/(16*x^8+128*x^4+256)/log(x^2)^3,x, algorithm="giac")

[Out]

3/64*x^4/(x^4*log(x^2)^2 + 4*log(x^2)^2)

Mupad [B] (verification not implemented)

Time = 13.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {-12 x^3-3 x^7+12 x^3 \log \left (x^2\right )}{\left (256+128 x^4+16 x^8\right ) \log ^3\left (x^2\right )} \, dx=\frac {3\,x^4}{64\,{\ln \left (x^2\right )}^2\,\left (x^4+4\right )} \]

[In]

int(-(12*x^3 - 12*x^3*log(x^2) + 3*x^7)/(log(x^2)^3*(128*x^4 + 16*x^8 + 256)),x)

[Out]

(3*x^4)/(64*log(x^2)^2*(x^4 + 4))