3.4.3 \(\int \frac {x^2}{1-2 x^4+x^8} \, dx\) [303]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 29, normalized size of antiderivative = 1.00, 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, 296, 304, 209, 212} \begin {gather*} -\frac {\text {ArcTan}(x)}{8}+\frac {x^3}{4 \left (1-x^4\right )}+\frac {1}{8} \tanh ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(1 - 2*x^4 + x^8),x]

[Out]

x^3/(4*(1 - x^4)) - ArcTan[x]/8 + 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 296

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

Rule 304

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]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 33, normalized size = 1.14 \begin {gather*} \frac {1}{16} \left (-\frac {4 x^3}{-1+x^4}-2 \tan ^{-1}(x)-\log (1-x)+\log (1+x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(1 - 2*x^4 + x^8),x]

[Out]

((-4*x^3)/(-1 + x^4) - 2*ArcTan[x] - Log[1 - x] + Log[1 + x])/16

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Maple [A]
time = 0.03, size = 42, normalized size = 1.45

method	result	size
risch	\(-\frac {x^{3}}{4 \left (x^{4}-1\right )}+\frac {\ln \left (1+x \right )}{16}-\frac {\ln \left (-1+x \right )}{16}-\frac {\arctan \left (x \right )}{8}\)	\(30\)
default	\(-\frac {1}{16 \left (-1+x \right )}-\frac {\ln \left (-1+x \right )}{16}-\frac {x}{8 \left (x^{2}+1\right )}-\frac {\arctan \left (x \right )}{8}-\frac {1}{16 \left (1+x \right )}+\frac {\ln \left (1+x \right )}{16}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.55, size = 29, normalized size = 1.00 \begin {gather*} -\frac {x^{3}}{4 \, {\left (x^{4} - 1\right )}} - \frac {1}{8} \, \arctan \left (x\right ) + \frac {1}{16} \, \log \left (x + 1\right ) - \frac {1}{16} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 45 vs. \(2 (21) = 42\).
time = 0.36, size = 45, normalized size = 1.55 \begin {gather*} -\frac {4 \, x^{3} + 2 \, {\left (x^{4} - 1\right )} \arctan \left (x\right ) - {\left (x^{4} - 1\right )} \log \left (x + 1\right ) + {\left (x^{4} - 1\right )} \log \left (x - 1\right )}{16 \, {\left (x^{4} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(4*x^3 + 2*(x^4 - 1)*arctan(x) - (x^4 - 1)*log(x + 1) + (x^4 - 1)*log(x - 1))/(x^4 - 1)

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Sympy [A]
time = 0.06, size = 27, normalized size = 0.93 \begin {gather*} - \frac {x^{3}}{4 x^{4} - 4} - \frac {\log {\left (x - 1 \right )}}{16} + \frac {\log {\left (x + 1 \right )}}{16} - \frac {\operatorname {atan}{\left (x \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.18, size = 31, normalized size = 1.07 \begin {gather*} -\frac {x^{3}}{4 \, {\left (x^{4} - 1\right )}} - \frac {1}{8} \, \arctan \left (x\right ) + \frac {1}{16} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{16} \, \log \left ({\left | x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.03, size = 23, normalized size = 0.79 \begin {gather*} \frac {\mathrm {atanh}\left (x\right )}{8}-\frac {\mathrm {atan}\left (x\right )}{8}-\frac {x^3}{4\,\left (x^4-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^8 - 2*x^4 + 1),x)

[Out]

atanh(x)/8 - atan(x)/8 - x^3/(4*(x^4 - 1))

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