\(\int \frac {x^3}{1-2 x^4+x^8} \, dx\) [294]

Optimal result

Integrand size = 16, antiderivative size = 13 \[ \int \frac {x^3}{1-2 x^4+x^8} \, dx=\frac {1}{4 \left (1-x^4\right )} \]

[Out]

1/4/(-x^4+1)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 267} \[ \int \frac {x^3}{1-2 x^4+x^8} \, dx=\frac {1}{4 \left (1-x^4\right )} \]

[In]

Int[x^3/(1 - 2*x^4 + x^8),x]

[Out]

1/(4*(1 - x^4))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 267

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^3}{\left (-1+x^4\right )^2} \, dx \\ & = \frac {1}{4 \left (1-x^4\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {x^3}{1-2 x^4+x^8} \, dx=-\frac {1}{4 \left (-1+x^4\right )} \]

[In]

Integrate[x^3/(1 - 2*x^4 + x^8),x]

[Out]

-1/4*1/(-1 + x^4)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method	result	size
gosper	\(-\frac {1}{4 \left (x^{4}-1\right )}\)	\(10\)
default	\(-\frac {1}{4 \left (x^{4}-1\right )}\)	\(10\)
norman	\(-\frac {1}{4 \left (x^{4}-1\right )}\)	\(10\)
risch	\(-\frac {1}{4 \left (x^{4}-1\right )}\)	\(10\)
parallelrisch	\(-\frac {1}{4 \left (x^{4}-1\right )}\)	\(10\)

[In]

int(x^3/(x^8-2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

-1/4/(x^4-1)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{1-2 x^4+x^8} \, dx=-\frac {1}{4 \, {\left (x^{4} - 1\right )}} \]

[In]

integrate(x^3/(x^8-2*x^4+1),x, algorithm="fricas")

[Out]

-1/4/(x^4 - 1)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.62 \[ \int \frac {x^3}{1-2 x^4+x^8} \, dx=- \frac {1}{4 x^{4} - 4} \]

[In]

integrate(x**3/(x**8-2*x**4+1),x)

[Out]

-1/(4*x**4 - 4)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{1-2 x^4+x^8} \, dx=-\frac {1}{4 \, {\left (x^{4} - 1\right )}} \]

[In]

integrate(x^3/(x^8-2*x^4+1),x, algorithm="maxima")

[Out]

-1/4/(x^4 - 1)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {x^3}{1-2 x^4+x^8} \, dx=-\frac {1}{4 \, {\left (x^{4} - 1\right )}} \]

[In]

integrate(x^3/(x^8-2*x^4+1),x, algorithm="giac")

[Out]

-1/4/(x^4 - 1)

Mupad [B] (verification not implemented)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {x^3}{1-2 x^4+x^8} \, dx=-\frac {1}{4\,\left (x^4-1\right )} \]

[In]

int(x^3/(x^8 - 2*x^4 + 1),x)

[Out]

-1/(4*(x^4 - 1))