\(\int \frac {x}{1-x^4+x^8} \, dx\) [353]

Optimal result

Integrand size = 14, antiderivative size = 82 \[ \int \frac {x}{1-x^4+x^8} \, dx=-\frac {1}{4} \arctan \left (\sqrt {3}-2 x^2\right )+\frac {1}{4} \arctan \left (\sqrt {3}+2 x^2\right )-\frac {\log \left (1-\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1373, 1108, 648, 632, 210, 642} \[ \int \frac {x}{1-x^4+x^8} \, dx=-\frac {1}{4} \arctan \left (\sqrt {3}-2 x^2\right )+\frac {1}{4} \arctan \left (2 x^2+\sqrt {3}\right )-\frac {\log \left (x^4-\sqrt {3} x^2+1\right )}{8 \sqrt {3}}+\frac {\log \left (x^4+\sqrt {3} x^2+1\right )}{8 \sqrt {3}} \]

[In]

Int[x/(1 - x^4 + x^8),x]

[Out]

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

Rule 210

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

Rule 632

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 642

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 648

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 1108

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

Rule 1373

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{1-x^2+x^4} \, dx,x,x^2\right ) \\ & = \frac {\text {Subst}\left (\int \frac {\sqrt {3}-x}{1-\sqrt {3} x+x^2} \, dx,x,x^2\right )}{4 \sqrt {3}}+\frac {\text {Subst}\left (\int \frac {\sqrt {3}+x}{1+\sqrt {3} x+x^2} \, dx,x,x^2\right )}{4 \sqrt {3}} \\ & = \frac {1}{8} \text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,x^2\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,x^2\right )-\frac {\text {Subst}\left (\int \frac {-\sqrt {3}+2 x}{1-\sqrt {3} x+x^2} \, dx,x,x^2\right )}{8 \sqrt {3}}+\frac {\text {Subst}\left (\int \frac {\sqrt {3}+2 x}{1+\sqrt {3} x+x^2} \, dx,x,x^2\right )}{8 \sqrt {3}} \\ & = -\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} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 x^2\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 x^2\right ) \\ & = -\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] (verified)

Result contains complex when optimal does not.

Time = 0.05 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.01 \[ \int \frac {x}{1-x^4+x^8} \, dx=\frac {i \left (\sqrt {-1-i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )-\sqrt {-1+i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )\right )}{2 \sqrt {6}} \]

[In]

Integrate[x/(1 - x^4 + x^8),x]

[Out]

((I/2)*(Sqrt[-1 - I*Sqrt[3]]*ArcTan[((1 - I*Sqrt[3])*x^2)/2] - Sqrt[-1 + I*Sqrt[3]]*ArcTan[((1 + I*Sqrt[3])*x^
2)/2]))/Sqrt[6]

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.39

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-3 \textit {\_R}^{3}+x^{2}+\textit {\_R} \right )\right )}{4}\)	\(32\)
default	\(\frac {\arctan \left (2 x^{2}-\sqrt {3}\right )}{4}+\frac {\arctan \left (2 x^{2}+\sqrt {3}\right )}{4}-\frac {\ln \left (1+x^{4}-x^{2} \sqrt {3}\right ) \sqrt {3}}{24}+\frac {\ln \left (1+x^{4}+x^{2} \sqrt {3}\right ) \sqrt {3}}{24}\)	\(65\)

[In]

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

[Out]

1/4*sum(_R*ln(-3*_R^3+x^2+_R),_R=RootOf(9*_Z^4+3*_Z^2+1))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.24 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.01 \[ \int \frac {x}{1-x^4+x^8} \, dx=-\frac {1}{24} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (12 \, x^{2} + \sqrt {6} \sqrt {i \, \sqrt {3} - 1} {\left (i \, \sqrt {3} - 3\right )}\right ) + \frac {1}{24} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (12 \, x^{2} + \sqrt {6} \sqrt {i \, \sqrt {3} - 1} {\left (-i \, \sqrt {3} + 3\right )}\right ) + \frac {1}{24} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (12 \, x^{2} + \sqrt {6} {\left (i \, \sqrt {3} + 3\right )} \sqrt {-i \, \sqrt {3} - 1}\right ) - \frac {1}{24} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (12 \, x^{2} + \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} {\left (-i \, \sqrt {3} - 3\right )}\right ) \]

[In]

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

[Out]

-1/24*sqrt(6)*sqrt(I*sqrt(3) - 1)*log(12*x^2 + sqrt(6)*sqrt(I*sqrt(3) - 1)*(I*sqrt(3) - 3)) + 1/24*sqrt(6)*sqr
t(I*sqrt(3) - 1)*log(12*x^2 + sqrt(6)*sqrt(I*sqrt(3) - 1)*(-I*sqrt(3) + 3)) + 1/24*sqrt(6)*sqrt(-I*sqrt(3) - 1
)*log(12*x^2 + sqrt(6)*(I*sqrt(3) + 3)*sqrt(-I*sqrt(3) - 1)) - 1/24*sqrt(6)*sqrt(-I*sqrt(3) - 1)*log(12*x^2 +
sqrt(6)*sqrt(-I*sqrt(3) - 1)*(-I*sqrt(3) - 3))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {x}{1-x^4+x^8} \, dx=- \frac {\sqrt {3} \log {\left (x^{4} - \sqrt {3} x^{2} + 1 \right )}}{24} + \frac {\sqrt {3} \log {\left (x^{4} + \sqrt {3} x^{2} + 1 \right )}}{24} + \frac {\operatorname {atan}{\left (2 x^{2} - \sqrt {3} \right )}}{4} + \frac {\operatorname {atan}{\left (2 x^{2} + \sqrt {3} \right )}}{4} \]

[In]

integrate(x/(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

Maxima [F]

\[ \int \frac {x}{1-x^4+x^8} \, dx=\int { \frac {x}{x^{8} - x^{4} + 1} \,d x } \]

[In]

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

[Out]

integrate(x/(x^8 - x^4 + 1), x)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.78 \[ \int \frac {x}{1-x^4+x^8} \, dx=\frac {1}{24} \, \sqrt {3} \log \left (x^{4} + \sqrt {3} x^{2} + 1\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{4} - \sqrt {3} x^{2} + 1\right ) + \frac {1}{4} \, \arctan \left (2 \, x^{2} + \sqrt {3}\right ) + \frac {1}{4} \, \arctan \left (2 \, x^{2} - \sqrt {3}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65 \[ \int \frac {x}{1-x^4+x^8} \, dx=-\mathrm {atan}\left (-\frac {x^2}{2}+\frac {\sqrt {3}\,x^2\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\mathrm {atan}\left (\frac {x^2}{2}+\frac {\sqrt {3}\,x^2\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right ) \]

[In]

int(x/(x^8 - x^4 + 1),x)

[Out]

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