3.1.45 \(\int \frac {x^3 (1-x^4)}{1-x^4+x^8} \, dx\)

Optimal. Leaf size=39 \[ -\frac {\tan ^{-1}\left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{8} \log \left (x^8-x^4+1\right ) \]

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Rubi [A]  time = 0.04, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1468, 634, 618, 204, 628} \begin {gather*} -\frac {1}{8} \log \left (x^8-x^4+1\right )-\frac {\tan ^{-1}\left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(1 - x^4))/(1 - x^4 + x^8),x]

[Out]

-ArcTan[(1 - 2*x^4)/Sqrt[3]]/(4*Sqrt[3]) - Log[1 - x^4 + x^8]/8

Rule 204

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1468

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 39, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {2 x^4-1}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{8} \log \left (x^8-x^4+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(1 - x^4))/(1 - x^4 + x^8),x]

[Out]

ArcTan[(-1 + 2*x^4)/Sqrt[3]]/(4*Sqrt[3]) - Log[1 - x^4 + x^8]/8

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3 \left (1-x^4\right )}{1-x^4+x^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^3*(1 - x^4))/(1 - x^4 + x^8),x]

[Out]

IntegrateAlgebraic[(x^3*(1 - x^4))/(1 - x^4 + x^8), x]

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fricas [A]  time = 0.73, size = 32, normalized size = 0.82 \begin {gather*} \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{8} - x^{4} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-x^4+1)/(x^8-x^4+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.63, size = 32, normalized size = 0.82 \begin {gather*} \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{8} - x^{4} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-x^4+1)/(x^8-x^4+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 33, normalized size = 0.85 \begin {gather*} \frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{4}-1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (x^{8}-x^{4}+1\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(-x^4+1)/(x^8-x^4+1),x)

[Out]

-1/8*ln(x^8-x^4+1)+1/12*3^(1/2)*arctan(1/3*(2*x^4-1)*3^(1/2))

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maxima [A]  time = 1.51, size = 32, normalized size = 0.82 \begin {gather*} \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{8} - x^{4} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(-x^4+1)/(x^8-x^4+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.05, size = 34, normalized size = 0.87 \begin {gather*} -\frac {\ln \left (x^8-x^4+1\right )}{8}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}}{3}-\frac {2\,\sqrt {3}\,x^4}{3}\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^3*(x^4 - 1))/(x^8 - x^4 + 1),x)

[Out]

- log(x^8 - x^4 + 1)/8 - (3^(1/2)*atan(3^(1/2)/3 - (2*3^(1/2)*x^4)/3))/12

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sympy [A]  time = 0.15, size = 37, normalized size = 0.95 \begin {gather*} - \frac {\log {\left (x^{8} - x^{4} + 1 \right )}}{8} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{4}}{3} - \frac {\sqrt {3}}{3} \right )}}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(-x**4+1)/(x**8-x**4+1),x)

[Out]

-log(x**8 - x**4 + 1)/8 + sqrt(3)*atan(2*sqrt(3)*x**4/3 - sqrt(3)/3)/12

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