\(\int \frac {256+512 x+192 x^2}{1+256 x+256 x^2+64 x^3} \, dx\) [7750]

Optimal result

Integrand size = 28, antiderivative size = 11 \[ \int \frac {256+512 x+192 x^2}{1+256 x+256 x^2+64 x^3} \, dx=\log \left (4+256 x (2+x)^2\right ) \]

[Out]

ln(4+256*x*(2+x)^2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.45, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {1601} \[ \int \frac {256+512 x+192 x^2}{1+256 x+256 x^2+64 x^3} \, dx=\log \left (64 x^3+256 x^2+256 x+1\right ) \]

[In]

Int[(256 + 512*x + 192*x^2)/(1 + 256*x + 256*x^2 + 64*x^3),x]

[Out]

Log[1 + 256*x + 256*x^2 + 64*x^3]

Rule 1601

Int[(Pp_)/(Qq_), x_Symbol] :> With[{p = Expon[Pp, x], q = Expon[Qq, x]}, Simp[Coeff[Pp, x, p]*(Log[RemoveConte
nt[Qq, x]]/(q*Coeff[Qq, x, q])), x] /; EqQ[p, q - 1] && EqQ[Pp, Simplify[(Coeff[Pp, x, p]/(q*Coeff[Qq, x, q]))
*D[Qq, x]]]] /; PolyQ[Pp, x] && PolyQ[Qq, x]

Rubi steps \begin{align*} \text {integral}& = \log \left (1+256 x+256 x^2+64 x^3\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.45 \[ \int \frac {256+512 x+192 x^2}{1+256 x+256 x^2+64 x^3} \, dx=\log \left (1+256 x+256 x^2+64 x^3\right ) \]

[In]

Integrate[(256 + 512*x + 192*x^2)/(1 + 256*x + 256*x^2 + 64*x^3),x]

[Out]

Log[1 + 256*x + 256*x^2 + 64*x^3]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.36

method	result	size
parallelrisch	\(\ln \left (x^{3}+4 x^{2}+4 x +\frac {1}{64}\right )\)	\(15\)
derivativedivides	\(\ln \left (64 x^{3}+256 x^{2}+256 x +1\right )\)	\(17\)
default	\(\ln \left (64 x^{3}+256 x^{2}+256 x +1\right )\)	\(17\)
norman	\(\ln \left (64 x^{3}+256 x^{2}+256 x +1\right )\)	\(17\)
risch	\(\ln \left (64 x^{3}+256 x^{2}+256 x +1\right )\)	\(17\)

[In]

int((192*x^2+512*x+256)/(64*x^3+256*x^2+256*x+1),x,method=_RETURNVERBOSE)

[Out]

ln(x^3+4*x^2+4*x+1/64)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.45 \[ \int \frac {256+512 x+192 x^2}{1+256 x+256 x^2+64 x^3} \, dx=\log \left (64 \, x^{3} + 256 \, x^{2} + 256 \, x + 1\right ) \]

[In]

integrate((192*x^2+512*x+256)/(64*x^3+256*x^2+256*x+1),x, algorithm="fricas")

[Out]

log(64*x^3 + 256*x^2 + 256*x + 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.36 \[ \int \frac {256+512 x+192 x^2}{1+256 x+256 x^2+64 x^3} \, dx=\log {\left (64 x^{3} + 256 x^{2} + 256 x + 1 \right )} \]

[In]

integrate((192*x**2+512*x+256)/(64*x**3+256*x**2+256*x+1),x)

[Out]

log(64*x**3 + 256*x**2 + 256*x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.45 \[ \int \frac {256+512 x+192 x^2}{1+256 x+256 x^2+64 x^3} \, dx=\log \left (64 \, x^{3} + 256 \, x^{2} + 256 \, x + 1\right ) \]

[In]

integrate((192*x^2+512*x+256)/(64*x^3+256*x^2+256*x+1),x, algorithm="maxima")

[Out]

log(64*x^3 + 256*x^2 + 256*x + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.55 \[ \int \frac {256+512 x+192 x^2}{1+256 x+256 x^2+64 x^3} \, dx=\log \left ({\left | 64 \, x^{3} + 256 \, x^{2} + 256 \, x + 1 \right |}\right ) \]

[In]

integrate((192*x^2+512*x+256)/(64*x^3+256*x^2+256*x+1),x, algorithm="giac")

[Out]

log(abs(64*x^3 + 256*x^2 + 256*x + 1))

Mupad [B] (verification not implemented)

Time = 12.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.27 \[ \int \frac {256+512 x+192 x^2}{1+256 x+256 x^2+64 x^3} \, dx=\ln \left (x^3+4\,x^2+4\,x+\frac {1}{64}\right ) \]

[In]

int((512*x + 192*x^2 + 256)/(256*x + 256*x^2 + 64*x^3 + 1),x)

[Out]

log(4*x + 4*x^2 + x^3 + 1/64)