\(\int \frac {1}{1+x^4+x^8} \, dx\) [342]

Optimal result

Integrand size = 10, antiderivative size = 88 \[ \int \frac {1}{1+x^4+x^8} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\log \left (1-\sqrt {3} x+x^2\right )}{4 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x+x^2\right )}{4 \sqrt {3}} \]

[Out]

-1/6*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+1/6*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)-1/12*ln(1+x^2-x*3^(1/2))*3^(1
/2)+1/12*ln(1+x^2+x*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {1360, 1178, 642, 1175, 632, 210} \[ \int \frac {1}{1+x^4+x^8} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\log \left (x^2-\sqrt {3} x+1\right )}{4 \sqrt {3}}+\frac {\log \left (x^2+\sqrt {3} x+1\right )}{4 \sqrt {3}} \]

[In]

Int[(1 + x^4 + x^8)^(-1),x]

[Out]

-1/2*ArcTan[(1 - 2*x)/Sqrt[3]]/Sqrt[3] + ArcTan[(1 + 2*x)/Sqrt[3]]/(2*Sqrt[3]) - Log[1 - Sqrt[3]*x + x^2]/(4*S
qrt[3]) + Log[1 + Sqrt[3]*x + x^2]/(4*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 1175

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 1178

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 1360

Int[((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(-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^(n/2))/(q - r*x^(n/2) + x^n), x], x] + Dist[1/(2*c*q*r), Int[(r + x^(n/2
))/(q + r*x^(n/2) + x^n), x], x]]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n/2,
0] && NegQ[b^2 - 4*a*c]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.77 \[ \int \frac {1}{1+x^4+x^8} \, dx=\frac {2 \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+2 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-\log \left (-1+\sqrt {3} x-x^2\right )+\log \left (1+\sqrt {3} x+x^2\right )}{4 \sqrt {3}} \]

[In]

Integrate[(1 + x^4 + x^8)^(-1),x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76

method	result	size
default	\(\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}-\frac {\ln \left (1+x^{2}-x \sqrt {3}\right ) \sqrt {3}}{12}+\frac {\ln \left (1+x^{2}+x \sqrt {3}\right ) \sqrt {3}}{12}\)	\(67\)
risch	\(-\frac {\ln \left (1+x^{2}-x \sqrt {3}\right ) \sqrt {3}}{12}+\frac {\ln \left (1+x^{2}+x \sqrt {3}\right ) \sqrt {3}}{12}+\frac {\sqrt {3}\, \arctan \left (\frac {x \sqrt {3}}{3}\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {x^{3} \sqrt {3}}{3}+\frac {2 x \sqrt {3}}{3}\right )}{6}\)	\(68\)

[In]

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

[Out]

1/6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))+1/6*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)-1/12*ln(1+x^2-x*3^(1/2))*3^(1/
2)+1/12*ln(1+x^2+x*3^(1/2))*3^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \frac {1}{1+x^4+x^8} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x^{3} + 2 \, x\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} x\right ) + \frac {1}{12} \, \sqrt {3} \log \left (\frac {x^{4} + 5 \, x^{2} + 2 \, \sqrt {3} {\left (x^{3} + x\right )} + 1}{x^{4} - x^{2} + 1}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.93 \[ \int \frac {1}{1+x^4+x^8} \, dx=\frac {\sqrt {3} \cdot \left (2 \operatorname {atan}{\left (\frac {\sqrt {3} x}{3} \right )} + 2 \operatorname {atan}{\left (\frac {\sqrt {3} x^{3}}{3} + \frac {2 \sqrt {3} x}{3} \right )}\right )}{12} - \frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{12} + \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{12} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/2*integrate((x^2 - 1
)/(x^4 - x^2 + 1), x)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.75 \[ \int \frac {1}{1+x^4+x^8} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) - \frac {1}{12} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) \]

[In]

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

[Out]

1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/12*sqrt(3)*log(x^2 +
sqrt(3)*x + 1) - 1/12*sqrt(3)*log(x^2 - sqrt(3)*x + 1)

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.45 \[ \int \frac {1}{1+x^4+x^8} \, dx=-\frac {\sqrt {3}\,\left (\mathrm {atan}\left (\frac {2\,\sqrt {3}\,x}{3\,\left (\frac {2\,x^2}{3}-\frac {2}{3}\right )}\right )-\mathrm {atanh}\left (\frac {2\,\sqrt {3}\,x}{3\,\left (\frac {2\,x^2}{3}+\frac {2}{3}\right )}\right )\right )}{6} \]

[In]

int(1/(x^4 + x^8 + 1),x)

[Out]

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