3.301 \(\int \frac{x^6}{1-2 x^4+x^8} \, dx\)

Optimal. Leaf size=29 \[ \frac{x^3}{4 \left (1-x^4\right )}+\frac{3}{8} \tan ^{-1}(x)-\frac{3}{8} \tanh ^{-1}(x) \]

[Out]

x^3/(4*(1 - x^4)) + (3*ArcTan[x])/8 - (3*ArcTanh[x])/8

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Rubi [A]  time = 0.0077958, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {28, 288, 298, 203, 206} \[ \frac{x^3}{4 \left (1-x^4\right )}+\frac{3}{8} \tan ^{-1}(x)-\frac{3}{8} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[x^6/(1 - 2*x^4 + x^8),x]

[Out]

x^3/(4*(1 - x^4)) + (3*ArcTan[x])/8 - (3*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 298

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

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])

Rubi steps

\begin{align*} \int \frac{x^6}{1-2 x^4+x^8} \, dx &=\int \frac{x^6}{\left (-1+x^4\right )^2} \, dx\\ &=\frac{x^3}{4 \left (1-x^4\right )}+\frac{3}{4} \int \frac{x^2}{-1+x^4} \, dx\\ &=\frac{x^3}{4 \left (1-x^4\right )}-\frac{3}{8} \int \frac{1}{1-x^2} \, dx+\frac{3}{8} \int \frac{1}{1+x^2} \, dx\\ &=\frac{x^3}{4 \left (1-x^4\right )}+\frac{3}{8} \tan ^{-1}(x)-\frac{3}{8} \tanh ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0162542, size = 35, normalized size = 1.21 \[ \frac{1}{16} \left (-\frac{4 x^3}{x^4-1}+3 \log (1-x)-3 \log (x+1)+6 \tan ^{-1}(x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(1 - 2*x^4 + x^8),x]

[Out]

((-4*x^3)/(-1 + x^4) + 6*ArcTan[x] + 3*Log[1 - x] - 3*Log[1 + x])/16

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Maple [A]  time = 0.011, size = 42, normalized size = 1.5 \begin{align*} -{\frac{x}{8\,{x}^{2}+8}}+{\frac{3\,\arctan \left ( x \right ) }{8}}-{\frac{1}{16+16\,x}}-{\frac{3\,\ln \left ( 1+x \right ) }{16}}-{\frac{1}{16\,x-16}}+{\frac{3\,\ln \left ( x-1 \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(x^8-2*x^4+1),x)

[Out]

-1/8*x/(x^2+1)+3/8*arctan(x)-1/16/(1+x)-3/16*ln(1+x)-1/16/(x-1)+3/16*ln(x-1)

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Maxima [A]  time = 1.49676, size = 39, normalized size = 1.34 \begin{align*} -\frac{x^{3}}{4 \,{\left (x^{4} - 1\right )}} + \frac{3}{8} \, \arctan \left (x\right ) - \frac{3}{16} \, \log \left (x + 1\right ) + \frac{3}{16} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^8-2*x^4+1),x, algorithm="maxima")

[Out]

-1/4*x^3/(x^4 - 1) + 3/8*arctan(x) - 3/16*log(x + 1) + 3/16*log(x - 1)

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Fricas [B]  time = 1.45841, size = 134, normalized size = 4.62 \begin{align*} -\frac{4 \, x^{3} - 6 \,{\left (x^{4} - 1\right )} \arctan \left (x\right ) + 3 \,{\left (x^{4} - 1\right )} \log \left (x + 1\right ) - 3 \,{\left (x^{4} - 1\right )} \log \left (x - 1\right )}{16 \,{\left (x^{4} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^8-2*x^4+1),x, algorithm="fricas")

[Out]

-1/16*(4*x^3 - 6*(x^4 - 1)*arctan(x) + 3*(x^4 - 1)*log(x + 1) - 3*(x^4 - 1)*log(x - 1))/(x^4 - 1)

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Sympy [A]  time = 0.157046, size = 32, normalized size = 1.1 \begin{align*} - \frac{x^{3}}{4 x^{4} - 4} + \frac{3 \log{\left (x - 1 \right )}}{16} - \frac{3 \log{\left (x + 1 \right )}}{16} + \frac{3 \operatorname{atan}{\left (x \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(x**8-2*x**4+1),x)

[Out]

-x**3/(4*x**4 - 4) + 3*log(x - 1)/16 - 3*log(x + 1)/16 + 3*atan(x)/8

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Giac [A]  time = 1.10297, size = 42, normalized size = 1.45 \begin{align*} -\frac{x^{3}}{4 \,{\left (x^{4} - 1\right )}} + \frac{3}{8} \, \arctan \left (x\right ) - \frac{3}{16} \, \log \left ({\left | x + 1 \right |}\right ) + \frac{3}{16} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(x^8-2*x^4+1),x, algorithm="giac")

[Out]

-1/4*x^3/(x^4 - 1) + 3/8*arctan(x) - 3/16*log(abs(x + 1)) + 3/16*log(abs(x - 1))