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

Optimal. Leaf size=50 \[ \frac {\log \left (x^4+\sqrt {3} x^2+1\right )}{4 \sqrt {3}}-\frac {\log \left (x^4-\sqrt {3} x^2+1\right )}{4 \sqrt {3}} \]

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Rubi [A]  time = 0.04, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1490, 1164, 628} \begin {gather*} \frac {\log \left (x^4+\sqrt {3} x^2+1\right )}{4 \sqrt {3}}-\frac {\log \left (x^4-\sqrt {3} x^2+1\right )}{4 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(1 - x^4))/(1 - x^4 + x^8),x]

[Out]

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1164

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e - b/c, 2]},
Dist[e/(2*c*q), Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x
- x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] &&  !GtQ[b^2
- 4*a*c, 0]

Rule 1490

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(d + e*x^(n/k))^q*(a + b*x^(n/k) + c*x^((2*n)/
k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, c, d, e, p, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] &&
IGtQ[n, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 44, normalized size = 0.88 \begin {gather*} \frac {\log \left (x^4+\sqrt {3} x^2+1\right )-\log \left (-x^4+\sqrt {3} x^2-1\right )}{4 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*(1 - x^4))/(1 - x^4 + x^8),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (1-x^4\right )}{1-x^4+x^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x*(1 - x^4))/(1 - x^4 + x^8),x]

[Out]

IntegrateAlgebraic[(x*(1 - x^4))/(1 - x^4 + x^8), x]

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fricas [A]  time = 0.87, size = 41, normalized size = 0.82 \begin {gather*} \frac {1}{12} \, \sqrt {3} \log \left (\frac {x^{8} + 5 \, x^{4} + 2 \, \sqrt {3} {\left (x^{6} + x^{2}\right )} + 1}{x^{8} - x^{4} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(3)*log((x^8 + 5*x^4 + 2*sqrt(3)*(x^6 + x^2) + 1)/(x^8 - x^4 + 1))

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giac [A]  time = 0.44, size = 31, normalized size = 0.62 \begin {gather*} -\frac {1}{12} \, \sqrt {3} \log \left (\frac {x^{2} - \sqrt {3} + \frac {1}{x^{2}}}{x^{2} + \sqrt {3} + \frac {1}{x^{2}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*sqrt(3)*log((x^2 - sqrt(3) + 1/x^2)/(x^2 + sqrt(3) + 1/x^2))

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maple [A]  time = 0.02, size = 39, normalized size = 0.78 \begin {gather*} -\frac {\sqrt {3}\, \ln \left (x^{4}-\sqrt {3}\, x^{2}+1\right )}{12}+\frac {\sqrt {3}\, \ln \left (x^{4}+\sqrt {3}\, x^{2}+1\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*3^(1/2)*ln(x^4-3^(1/2)*x^2+1)+1/12*3^(1/2)*ln(x^4+3^(1/2)*x^2+1)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (x^{4} - 1\right )} x}{x^{8} - x^{4} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.89, size = 20, normalized size = 0.40 \begin {gather*} \frac {\sqrt {3}\,\mathrm {atanh}\left (\frac {\sqrt {3}\,x^2}{x^4+1}\right )}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x*(x^4 - 1))/(x^8 - x^4 + 1),x)

[Out]

(3^(1/2)*atanh((3^(1/2)*x^2)/(x^4 + 1)))/6

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sympy [A]  time = 0.13, size = 42, normalized size = 0.84 \begin {gather*} - \frac {\sqrt {3} \log {\left (x^{4} - \sqrt {3} x^{2} + 1 \right )}}{12} + \frac {\sqrt {3} \log {\left (x^{4} + \sqrt {3} x^{2} + 1 \right )}}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(3)*log(x**4 - sqrt(3)*x**2 + 1)/12 + sqrt(3)*log(x**4 + sqrt(3)*x**2 + 1)/12

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