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

Optimal. Leaf size=34 \[ \frac{x^5}{4 \left (1-x^4\right )}+\frac{5 x}{4}-\frac{5}{8} \tan ^{-1}(x)-\frac{5}{8} \tanh ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0090223, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {28, 288, 321, 212, 206, 203} \[ \frac{x^5}{4 \left (1-x^4\right )}+\frac{5 x}{4}-\frac{5}{8} \tan ^{-1}(x)-\frac{5}{8} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[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 288

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 321

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]

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 206

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

Rule 203

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

Rubi steps

\begin{align*} \int \frac{x^8}{1-2 x^4+x^8} \, dx &=\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]  time = 0.0157707, size = 38, normalized size = 1.12 \[ -\frac{x}{4 \left (x^4-1\right )}+x+\frac{5}{16} \log (1-x)-\frac{5}{16} \log (x+1)-\frac{5}{8} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[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

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Maple [A]  time = 0.013, size = 43, normalized size = 1.3 \begin{align*} x+{\frac{x}{8\,{x}^{2}+8}}-{\frac{5\,\arctan \left ( x \right ) }{8}}-{\frac{1}{16+16\,x}}-{\frac{5\,\ln \left ( 1+x \right ) }{16}}-{\frac{1}{16\,x-16}}+{\frac{5\,\ln \left ( x-1 \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+1/8*x/(x^2+1)-5/8*arctan(x)-1/16/(1+x)-5/16*ln(1+x)-1/16/(x-1)+5/16*ln(x-1)

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Maxima [A]  time = 1.47366, size = 38, normalized size = 1.12 \begin{align*} 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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 1.56054, size = 144, normalized size = 4.24 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.159471, size = 32, normalized size = 0.94 \begin{align*} 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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.13069, size = 41, normalized size = 1.21 \begin{align*} 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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))