\(\int \frac {x^8}{1-2 x^4+x^8} \, dx\) [300]

Optimal result

Integrand size = 16, antiderivative size = 34 \[ \int \frac {x^8}{1-2 x^4+x^8} \, dx=\frac {5 x}{4}+\frac {x^5}{4 \left (1-x^4\right )}-\frac {5 \arctan (x)}{8}-\frac {5 \text {arctanh}(x)}{8} \]

[Out]

5/4*x+1/4*x^5/(-x^4+1)-5/8*arctan(x)-5/8*arctanh(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {28, 294, 327, 218, 212, 209} \[ \int \frac {x^8}{1-2 x^4+x^8} \, dx=-\frac {5 \arctan (x)}{8}-\frac {5 \text {arctanh}(x)}{8}+\frac {x^5}{4 \left (1-x^4\right )}+\frac {5 x}{4} \]

[In]

Int[x^8/(1 - 2*x^4 + x^8),x]

[Out]

(5*x)/4 + x^5/(4*(1 - x^4)) - (5*ArcTan[x])/8 - (5*ArcTanh[x])/8

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {x^8}{\left (-1+x^4\right )^2} \, dx \\ & = \frac {x^5}{4 \left (1-x^4\right )}+\frac {5}{4} \int \frac {x^4}{-1+x^4} \, dx \\ & = \frac {5 x}{4}+\frac {x^5}{4 \left (1-x^4\right )}+\frac {5}{4} \int \frac {1}{-1+x^4} \, dx \\ & = \frac {5 x}{4}+\frac {x^5}{4 \left (1-x^4\right )}-\frac {5}{8} \int \frac {1}{1-x^2} \, dx-\frac {5}{8} \int \frac {1}{1+x^2} \, dx \\ & = \frac {5 x}{4}+\frac {x^5}{4 \left (1-x^4\right )}-\frac {5}{8} \tan ^{-1}(x)-\frac {5}{8} \tanh ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {x^8}{1-2 x^4+x^8} \, dx=x-\frac {x}{4 \left (-1+x^4\right )}-\frac {5 \arctan (x)}{8}+\frac {5}{16} \log (1-x)-\frac {5}{16} \log (1+x) \]

[In]

Integrate[x^8/(1 - 2*x^4 + x^8),x]

[Out]

x - x/(4*(-1 + x^4)) - (5*ArcTan[x])/8 + (5*Log[1 - x])/16 - (5*Log[1 + x])/16

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85

method	result	size
risch	\(x -\frac {x}{4 \left (x^{4}-1\right )}+\frac {5 \ln \left (x -1\right )}{16}-\frac {5 \arctan \left (x \right )}{8}-\frac {5 \ln \left (x +1\right )}{16}\)	\(29\)
default	\(x -\frac {1}{16 \left (x +1\right )}-\frac {5 \ln \left (x +1\right )}{16}+\frac {x}{8 x^{2}+8}-\frac {5 \arctan \left (x \right )}{8}-\frac {1}{16 \left (x -1\right )}+\frac {5 \ln \left (x -1\right )}{16}\)	\(43\)
parallelrisch	\(\frac {5 i \ln \left (x -i\right ) x^{4}-5 i \ln \left (x +i\right ) x^{4}+5 \ln \left (x -1\right ) x^{4}-5 \ln \left (x +1\right ) x^{4}+16 x^{5}-5 i \ln \left (x -i\right )+5 i \ln \left (x +i\right )-5 \ln \left (x -1\right )+5 \ln \left (x +1\right )-20 x}{16 x^{4}-16}\)	\(87\)

[In]

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

[Out]

x-1/4*x/(x^4-1)+5/16*ln(x-1)-5/8*arctan(x)-5/16*ln(x+1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 49 vs. \(2 (24) = 48\).

Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.44 \[ \int \frac {x^8}{1-2 x^4+x^8} \, dx=\frac {16 \, x^{5} - 10 \, {\left (x^{4} - 1\right )} \arctan \left (x\right ) - 5 \, {\left (x^{4} - 1\right )} \log \left (x + 1\right ) + 5 \, {\left (x^{4} - 1\right )} \log \left (x - 1\right ) - 20 \, x}{16 \, {\left (x^{4} - 1\right )}} \]

[In]

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

[Out]

1/16*(16*x^5 - 10*(x^4 - 1)*arctan(x) - 5*(x^4 - 1)*log(x + 1) + 5*(x^4 - 1)*log(x - 1) - 20*x)/(x^4 - 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {x^8}{1-2 x^4+x^8} \, dx=x - \frac {x}{4 x^{4} - 4} + \frac {5 \log {\left (x - 1 \right )}}{16} - \frac {5 \log {\left (x + 1 \right )}}{16} - \frac {5 \operatorname {atan}{\left (x \right )}}{8} \]

[In]

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

[Out]

x - x/(4*x**4 - 4) + 5*log(x - 1)/16 - 5*log(x + 1)/16 - 5*atan(x)/8

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {x^8}{1-2 x^4+x^8} \, dx=x - \frac {x}{4 \, {\left (x^{4} - 1\right )}} - \frac {5}{8} \, \arctan \left (x\right ) - \frac {5}{16} \, \log \left (x + 1\right ) + \frac {5}{16} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

x - 1/4*x/(x^4 - 1) - 5/8*arctan(x) - 5/16*log(x + 1) + 5/16*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {x^8}{1-2 x^4+x^8} \, dx=x - \frac {x}{4 \, {\left (x^{4} - 1\right )}} - \frac {5}{8} \, \arctan \left (x\right ) - \frac {5}{16} \, \log \left ({\left | x + 1 \right |}\right ) + \frac {5}{16} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

x - 1/4*x/(x^4 - 1) - 5/8*arctan(x) - 5/16*log(abs(x + 1)) + 5/16*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 8.37 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {x^8}{1-2 x^4+x^8} \, dx=x-\frac {5\,\mathrm {atan}\left (x\right )}{8}-\frac {x}{4\,\left (x^4-1\right )}+\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{8} \]

[In]

int(x^8/(x^8 - 2*x^4 + 1),x)

[Out]

x + (atan(x*1i)*5i)/8 - (5*atan(x))/8 - x/(4*(x^4 - 1))