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

Optimal. Leaf size=89 \[ -\frac {1}{2 x^2}+\frac {1}{4} \tan ^{-1}\left (\sqrt {3}-2 x^2\right )-\frac {1}{4} \tan ^{-1}\left (2 x^2+\sqrt {3}\right )-\frac {\log \left (x^4-\sqrt {3} x^2+1\right )}{8 \sqrt {3}}+\frac {\log \left (x^4+\sqrt {3} x^2+1\right )}{8 \sqrt {3}} \]

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Rubi [A]  time = 0.09, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {1490, 1281, 1127, 1161, 618, 204, 1164, 628} \begin {gather*} -\frac {1}{2 x^2}-\frac {\log \left (x^4-\sqrt {3} x^2+1\right )}{8 \sqrt {3}}+\frac {\log \left (x^4+\sqrt {3} x^2+1\right )}{8 \sqrt {3}}+\frac {1}{4} \tan ^{-1}\left (\sqrt {3}-2 x^2\right )-\frac {1}{4} \tan ^{-1}\left (2 x^2+\sqrt {3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 1127

Int[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, Dist[1/2, Int[(q + x^2)/(
a + b*x^2 + c*x^4), x], x] - Dist[1/2, Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && Lt
Q[b^2 - 4*a*c, 0] && PosQ[a*c]

Rule 1161

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), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/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[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 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 1281

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(d*(
f*x)^(m + 1)*(a + b*x^2 + c*x^4)^(p + 1))/(a*f*(m + 1)), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + b
*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

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

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Mathematica [C]  time = 0.02, size = 49, normalized size = 0.55 \begin {gather*} -\frac {1}{4} \text {RootSum}\left [\text {$\#$1}^8-\text {$\#$1}^4+1\&,\frac {\text {$\#$1}^2 \log (x-\text {$\#$1})}{2 \text {$\#$1}^4-1}\&\right ]-\frac {1}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*1/x^2 - RootSum[1 - #1^4 + #1^8 & , (Log[x - #1]*#1^2)/(-1 + 2*#1^4) & ]/4

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 1.03, size = 188, normalized size = 2.11 \begin {gather*} \frac {4 \, \sqrt {6} \sqrt {3} \sqrt {2} x^{2} \arctan \left (-\frac {1}{3} \, \sqrt {6} \sqrt {3} \sqrt {2} x^{2} + \frac {1}{3} \, \sqrt {6} \sqrt {3} \sqrt {2 \, x^{4} + \sqrt {6} \sqrt {2} x^{2} + 2} - \sqrt {3}\right ) + 4 \, \sqrt {6} \sqrt {3} \sqrt {2} x^{2} \arctan \left (-\frac {1}{3} \, \sqrt {6} \sqrt {3} \sqrt {2} x^{2} + \frac {1}{3} \, \sqrt {6} \sqrt {3} \sqrt {2 \, x^{4} - \sqrt {6} \sqrt {2} x^{2} + 2} + \sqrt {3}\right ) + \sqrt {6} \sqrt {2} x^{2} \log \left (2 \, x^{4} + \sqrt {6} \sqrt {2} x^{2} + 2\right ) - \sqrt {6} \sqrt {2} x^{2} \log \left (2 \, x^{4} - \sqrt {6} \sqrt {2} x^{2} + 2\right ) - 24}{48 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(4*sqrt(6)*sqrt(3)*sqrt(2)*x^2*arctan(-1/3*sqrt(6)*sqrt(3)*sqrt(2)*x^2 + 1/3*sqrt(6)*sqrt(3)*sqrt(2*x^4 +
sqrt(6)*sqrt(2)*x^2 + 2) - sqrt(3)) + 4*sqrt(6)*sqrt(3)*sqrt(2)*x^2*arctan(-1/3*sqrt(6)*sqrt(3)*sqrt(2)*x^2 +
1/3*sqrt(6)*sqrt(3)*sqrt(2*x^4 - sqrt(6)*sqrt(2)*x^2 + 2) + sqrt(3)) + sqrt(6)*sqrt(2)*x^2*log(2*x^4 + sqrt(6
)*sqrt(2)*x^2 + 2) - sqrt(6)*sqrt(2)*x^2*log(2*x^4 - sqrt(6)*sqrt(2)*x^2 + 2) - 24)/x^2

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giac [A]  time = 0.53, size = 81, normalized size = 0.91 \begin {gather*} -\frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} + \sqrt {3} x^{2} + 1\right ) + \frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} - \sqrt {3} x^{2} + 1\right ) - \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} + \sqrt {3}\right ) - \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} - \sqrt {3}\right ) - \frac {1}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*sqrt(3)*x^4*log(x^4 + sqrt(3)*x^2 + 1) + 1/24*sqrt(3)*x^4*log(x^4 - sqrt(3)*x^2 + 1) - 1/4*x^4*arctan(2*
x^2 + sqrt(3)) - 1/4*x^4*arctan(2*x^2 - sqrt(3)) - 1/2/x^2

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maple [A]  time = 0.01, size = 70, normalized size = 0.79 \begin {gather*} -\frac {\arctan \left (2 x^{2}-\sqrt {3}\right )}{4}-\frac {\arctan \left (2 x^{2}+\sqrt {3}\right )}{4}-\frac {\sqrt {3}\, \ln \left (x^{4}-\sqrt {3}\, x^{2}+1\right )}{24}+\frac {\sqrt {3}\, \ln \left (x^{4}+\sqrt {3}\, x^{2}+1\right )}{24}-\frac {1}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x^2-1/4*arctan(2*x^2-3^(1/2))-1/4*arctan(2*x^2+3^(1/2))-1/24*3^(1/2)*ln(x^4-3^(1/2)*x^2+1)+1/24*3^(1/2)*l
n(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*} -\frac {1}{2 \, x^{2}} - \int \frac {x^{5}}{x^{8} - x^{4} + 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/x^2 - integrate(x^5/(x^8 - x^4 + 1), x)

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mupad [B]  time = 0.10, size = 56, normalized size = 0.63 \begin {gather*} \mathrm {atan}\left (\frac {2\,x^2}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atan}\left (\frac {2\,x^2}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\frac {1}{2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.23, size = 76, normalized size = 0.85 \begin {gather*} - \frac {\sqrt {3} \log {\left (x^{4} - \sqrt {3} x^{2} + 1 \right )}}{24} + \frac {\sqrt {3} \log {\left (x^{4} + \sqrt {3} x^{2} + 1 \right )}}{24} - \frac {\operatorname {atan}{\left (2 x^{2} - \sqrt {3} \right )}}{4} - \frac {\operatorname {atan}{\left (2 x^{2} + \sqrt {3} \right )}}{4} - \frac {1}{2 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(3)*log(x**4 - sqrt(3)*x**2 + 1)/24 + sqrt(3)*log(x**4 + sqrt(3)*x**2 + 1)/24 - atan(2*x**2 - sqrt(3))/4
- atan(2*x**2 + sqrt(3))/4 - 1/(2*x**2)

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