### 3.59 $$\int \frac{1-x^4}{x^3 (1-x^4+x^8)} \, dx$$

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

[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])

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Rubi [A]  time = 0.0903328, antiderivative size = 89, normalized size of antiderivative = 1., 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} $-\frac{1}{2 x^2}-\frac{\log \left (x^4-\sqrt{3} x^2+1\right )}{8 \sqrt{3}}+\frac{\log \left (x^4+\sqrt{3} x^2+1\right )}{8 \sqrt{3}}+\frac{1}{4} \tan ^{-1}\left (\sqrt{3}-2 x^2\right )-\frac{1}{4} \tan ^{-1}\left (2 x^2+\sqrt{3}\right )$

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 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]

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 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 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 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 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 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]

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*}

Mathematica [C]  time = 0.0158076, size = 49, normalized size = 0.55 $-\frac{1}{4} \text{RootSum}\left [\text{\#1}^8-\text{\#1}^4+1\& ,\frac{\text{\#1}^2 \log (x-\text{\#1})}{2 \text{\#1}^4-1}\& \right ]-\frac{1}{2 x^2}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 70, normalized size = 0.8 \begin{align*} -{\frac{1}{2\,{x}^{2}}}-{\frac{\arctan \left ( 2\,{x}^{2}-\sqrt{3} \right ) }{4}}-{\frac{\arctan \left ( 2\,{x}^{2}+\sqrt{3} \right ) }{4}}-{\frac{\ln \left ( 1+{x}^{4}-{x}^{2}\sqrt{3} \right ) \sqrt{3}}{24}}+{\frac{\ln \left ( 1+{x}^{4}+{x}^{2}\sqrt{3} \right ) \sqrt{3}}{24}} \end{align*}

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*ln(1+x^4-x^2*3^(1/2))*3^(1/2)+1/24*ln(1+x^4+
x^2*3^(1/2))*3^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2 \, x^{2}} - \int \frac{x^{5}}{x^{8} - x^{4} + 1}\,{d x} \end{align*}

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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Fricas [B]  time = 1.56864, size = 566, normalized size = 6.36 \begin{align*} \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{align*}

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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Sympy [A]  time = 0.262795, size = 76, normalized size = 0.85 \begin{align*} - \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{align*}

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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Giac [B]  time = 1.26255, size = 348, normalized size = 3.91 \begin{align*} -\frac{1}{48} \,{\left (\sqrt{6} - 3 \, \sqrt{2}\right )} \arctan \left (\frac{4 \, x + \sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}\right ) - \frac{1}{48} \,{\left (\sqrt{6} - 3 \, \sqrt{2}\right )} \arctan \left (\frac{4 \, x - \sqrt{6} + \sqrt{2}}{\sqrt{6} + \sqrt{2}}\right ) - \frac{1}{48} \,{\left (\sqrt{6} + 3 \, \sqrt{2}\right )} \arctan \left (\frac{4 \, x + \sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}}\right ) - \frac{1}{48} \,{\left (\sqrt{6} + 3 \, \sqrt{2}\right )} \arctan \left (\frac{4 \, x - \sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}}\right ) - \frac{1}{96} \,{\left (\sqrt{6} - 3 \, \sqrt{2}\right )} \log \left (x^{2} + \frac{1}{2} \, x{\left (\sqrt{6} + \sqrt{2}\right )} + 1\right ) + \frac{1}{96} \,{\left (\sqrt{6} - 3 \, \sqrt{2}\right )} \log \left (x^{2} - \frac{1}{2} \, x{\left (\sqrt{6} + \sqrt{2}\right )} + 1\right ) - \frac{1}{96} \,{\left (\sqrt{6} + 3 \, \sqrt{2}\right )} \log \left (x^{2} + \frac{1}{2} \, x{\left (\sqrt{6} - \sqrt{2}\right )} + 1\right ) + \frac{1}{96} \,{\left (\sqrt{6} + 3 \, \sqrt{2}\right )} \log \left (x^{2} - \frac{1}{2} \, x{\left (\sqrt{6} - \sqrt{2}\right )} + 1\right ) - \frac{1}{2 \, x^{2}} \end{align*}

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]

-1/48*(sqrt(6) - 3*sqrt(2))*arctan((4*x + sqrt(6) - sqrt(2))/(sqrt(6) + sqrt(2))) - 1/48*(sqrt(6) - 3*sqrt(2))
*arctan((4*x - sqrt(6) + sqrt(2))/(sqrt(6) + sqrt(2))) - 1/48*(sqrt(6) + 3*sqrt(2))*arctan((4*x + sqrt(6) + sq
rt(2))/(sqrt(6) - sqrt(2))) - 1/48*(sqrt(6) + 3*sqrt(2))*arctan((4*x - sqrt(6) - sqrt(2))/(sqrt(6) - sqrt(2)))
- 1/96*(sqrt(6) - 3*sqrt(2))*log(x^2 + 1/2*x*(sqrt(6) + sqrt(2)) + 1) + 1/96*(sqrt(6) - 3*sqrt(2))*log(x^2 -
1/2*x*(sqrt(6) + sqrt(2)) + 1) - 1/96*(sqrt(6) + 3*sqrt(2))*log(x^2 + 1/2*x*(sqrt(6) - sqrt(2)) + 1) + 1/96*(s
qrt(6) + 3*sqrt(2))*log(x^2 - 1/2*x*(sqrt(6) - sqrt(2)) + 1) - 1/2/x^2