\(\int \frac {1}{2+x^6} \, dx\) [48]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [C] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 7, antiderivative size = 138 \[ \int \frac {1}{2+x^6} \, dx=\frac {\arctan \left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}-\frac {\arctan \left (\sqrt {3}-2^{5/6} x\right )}{6\ 2^{5/6}}+\frac {\arctan \left (\sqrt {3}+2^{5/6} x\right )}{6\ 2^{5/6}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [6]{2} \sqrt {3} x+x^2\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\log \left (\sqrt [3]{2}+\sqrt [6]{2} \sqrt {3} x+x^2\right )}{4\ 2^{5/6} \sqrt {3}} \]

[Out]

1/6*arctan(1/2*x*2^(5/6))*2^(1/6)+1/12*arctan(x*2^(5/6)-3^(1/2))*2^(1/6)+1/12*arctan(x*2^(5/6)+3^(1/2))*2^(1/6
)-1/24*ln(2^(1/3)+x^2-2^(1/6)*x*3^(1/2))*2^(1/6)*3^(1/2)+1/24*ln(2^(1/3)+x^2+2^(1/6)*x*3^(1/2))*2^(1/6)*3^(1/2
)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {215, 648, 632, 210, 642, 209} \[ \int \frac {1}{2+x^6} \, dx=\frac {\arctan \left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}-\frac {\arctan \left (\sqrt {3}-2^{5/6} x\right )}{6\ 2^{5/6}}+\frac {\arctan \left (2^{5/6} x+\sqrt {3}\right )}{6\ 2^{5/6}}-\frac {\log \left (x^2-\sqrt [6]{2} \sqrt {3} x+\sqrt [3]{2}\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\log \left (x^2+\sqrt [6]{2} \sqrt {3} x+\sqrt [3]{2}\right )}{4\ 2^{5/6} \sqrt {3}} \]

[In]

Int[(2 + x^6)^(-1),x]

[Out]

ArcTan[x/2^(1/6)]/(3*2^(5/6)) - ArcTan[Sqrt[3] - 2^(5/6)*x]/(6*2^(5/6)) + ArcTan[Sqrt[3] + 2^(5/6)*x]/(6*2^(5/
6)) - Log[2^(1/3) - 2^(1/6)*Sqrt[3]*x + x^2]/(4*2^(5/6)*Sqrt[3]) + Log[2^(1/3) + 2^(1/6)*Sqrt[3]*x + x^2]/(4*2
^(5/6)*Sqrt[3])

Rule 209

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

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 215

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/b, n]]
, k, u, v}, Simp[u = Int[(r - s*Cos[(2*k - 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] +
 Int[(r + s*Cos[(2*k - 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x]; 2*(r^2/(a*n))*Int[1/
(r^2 + s^2*x^2), x] + Dist[2*(r/(a*n)), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)
/4, 0] && PosQ[a/b]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {\sqrt [6]{2}-\frac {\sqrt {3} x}{2}}{\sqrt [3]{2}-\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{3\ 2^{5/6}}+\frac {\int \frac {\sqrt [6]{2}+\frac {\sqrt {3} x}{2}}{\sqrt [3]{2}+\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{3\ 2^{5/6}}+\frac {\int \frac {1}{\sqrt [3]{2}+x^2} \, dx}{3\ 2^{2/3}} \\ & = \frac {\arctan \left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}+\frac {\int \frac {1}{\sqrt [3]{2}-\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{12\ 2^{2/3}}+\frac {\int \frac {1}{\sqrt [3]{2}+\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{12\ 2^{2/3}}-\frac {\int \frac {-\sqrt [6]{2} \sqrt {3}+2 x}{\sqrt [3]{2}-\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{4\ 2^{5/6} \sqrt {3}}+\frac {\int \frac {\sqrt [6]{2} \sqrt {3}+2 x}{\sqrt [3]{2}+\sqrt [6]{2} \sqrt {3} x+x^2} \, dx}{4\ 2^{5/6} \sqrt {3}} \\ & = \frac {\arctan \left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [6]{2} \sqrt {3} x+x^2\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\log \left (\sqrt [3]{2}+\sqrt [6]{2} \sqrt {3} x+x^2\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\text {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1-\frac {2^{5/6} x}{\sqrt {3}}\right )}{6\ 2^{5/6} \sqrt {3}}-\frac {\text {Subst}\left (\int \frac {1}{-\frac {1}{3}-x^2} \, dx,x,1+\frac {2^{5/6} x}{\sqrt {3}}\right )}{6\ 2^{5/6} \sqrt {3}} \\ & = \frac {\arctan \left (\frac {x}{\sqrt [6]{2}}\right )}{3\ 2^{5/6}}-\frac {\arctan \left (\sqrt {3}-2^{5/6} x\right )}{6\ 2^{5/6}}+\frac {\arctan \left (\sqrt {3}+2^{5/6} x\right )}{6\ 2^{5/6}}-\frac {\log \left (\sqrt [3]{2}-\sqrt [6]{2} \sqrt {3} x+x^2\right )}{4\ 2^{5/6} \sqrt {3}}+\frac {\log \left (\sqrt [3]{2}+\sqrt [6]{2} \sqrt {3} x+x^2\right )}{4\ 2^{5/6} \sqrt {3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.83 \[ \int \frac {1}{2+x^6} \, dx=\frac {4 \arctan \left (\frac {x}{\sqrt [6]{2}}\right )-2 \arctan \left (\sqrt {3}-2^{5/6} x\right )+2 \arctan \left (\sqrt {3}+2^{5/6} x\right )-\sqrt {3} \log \left (2-2^{5/6} \sqrt {3} x+2^{2/3} x^2\right )+\sqrt {3} \log \left (2+2^{5/6} \sqrt {3} x+2^{2/3} x^2\right )}{12\ 2^{5/6}} \]

[In]

Integrate[(2 + x^6)^(-1),x]

[Out]

(4*ArcTan[x/2^(1/6)] - 2*ArcTan[Sqrt[3] - 2^(5/6)*x] + 2*ArcTan[Sqrt[3] + 2^(5/6)*x] - Sqrt[3]*Log[2 - 2^(5/6)
*Sqrt[3]*x + 2^(2/3)*x^2] + Sqrt[3]*Log[2 + 2^(5/6)*Sqrt[3]*x + 2^(2/3)*x^2])/(12*2^(5/6))

Maple [C] (verified)

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

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.16

method result size
risch \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}+2\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{5}}\right )}{6}\) \(22\)
default \(\frac {\arctan \left (\frac {x 2^{\frac {5}{6}}}{2}\right ) 2^{\frac {1}{6}}}{6}+\frac {\arctan \left (x 2^{\frac {5}{6}}-\sqrt {3}\right ) 2^{\frac {1}{6}}}{12}+\frac {\arctan \left (x 2^{\frac {5}{6}}+\sqrt {3}\right ) 2^{\frac {1}{6}}}{12}-\frac {\ln \left (2^{\frac {1}{3}}+x^{2}-2^{\frac {1}{6}} x \sqrt {3}\right ) 2^{\frac {1}{6}} \sqrt {3}}{24}+\frac {\ln \left (2^{\frac {1}{3}}+x^{2}+2^{\frac {1}{6}} x \sqrt {3}\right ) 2^{\frac {1}{6}} \sqrt {3}}{24}\) \(95\)
meijerg \(\frac {2^{\frac {1}{6}} \left (-\frac {x \sqrt {3}\, \ln \left (1-\frac {\sqrt {3}\, 2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}{2}+\frac {2^{\frac {2}{3}} \left (x^{6}\right )^{\frac {1}{3}}}{2}\right )}{2 \left (x^{6}\right )^{\frac {1}{6}}}+\frac {x \arctan \left (\frac {2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}{4-\sqrt {3}\, 2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}\right )}{\left (x^{6}\right )^{\frac {1}{6}}}+\frac {2 x \arctan \left (\frac {2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}{2}\right )}{\left (x^{6}\right )^{\frac {1}{6}}}+\frac {x \sqrt {3}\, \ln \left (1+\frac {\sqrt {3}\, 2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}{2}+\frac {2^{\frac {2}{3}} \left (x^{6}\right )^{\frac {1}{3}}}{2}\right )}{2 \left (x^{6}\right )^{\frac {1}{6}}}+\frac {x \arctan \left (\frac {2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}{4+\sqrt {3}\, 2^{\frac {5}{6}} \left (x^{6}\right )^{\frac {1}{6}}}\right )}{\left (x^{6}\right )^{\frac {1}{6}}}\right )}{12}\) \(170\)

[In]

int(1/(x^6+2),x,method=_RETURNVERBOSE)

[Out]

1/6*sum(1/_R^5*ln(x-_R),_R=RootOf(_Z^6+2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.19 \[ \int \frac {1}{2+x^6} \, dx=\frac {1}{384} \cdot 32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} {\left (\sqrt {-3} + 1\right )} \log \left (32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} {\left (\sqrt {-3} + 1\right )} + 32 \, x\right ) - \frac {1}{384} \cdot 32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} {\left (\sqrt {-3} + 1\right )} \log \left (-32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} {\left (\sqrt {-3} + 1\right )} + 32 \, x\right ) + \frac {1}{384} \cdot 32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} {\left (\sqrt {-3} - 1\right )} \log \left (32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} {\left (\sqrt {-3} - 1\right )} + 32 \, x\right ) - \frac {1}{384} \cdot 32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} {\left (\sqrt {-3} - 1\right )} \log \left (-32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} {\left (\sqrt {-3} - 1\right )} + 32 \, x\right ) + \frac {1}{192} \cdot 32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} \log \left (32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} + 16 \, x\right ) - \frac {1}{192} \cdot 32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} \log \left (-32^{\frac {5}{6}} \left (-1\right )^{\frac {1}{6}} + 16 \, x\right ) \]

[In]

integrate(1/(x^6+2),x, algorithm="fricas")

[Out]

1/384*32^(5/6)*(-1)^(1/6)*(sqrt(-3) + 1)*log(32^(5/6)*(-1)^(1/6)*(sqrt(-3) + 1) + 32*x) - 1/384*32^(5/6)*(-1)^
(1/6)*(sqrt(-3) + 1)*log(-32^(5/6)*(-1)^(1/6)*(sqrt(-3) + 1) + 32*x) + 1/384*32^(5/6)*(-1)^(1/6)*(sqrt(-3) - 1
)*log(32^(5/6)*(-1)^(1/6)*(sqrt(-3) - 1) + 32*x) - 1/384*32^(5/6)*(-1)^(1/6)*(sqrt(-3) - 1)*log(-32^(5/6)*(-1)
^(1/6)*(sqrt(-3) - 1) + 32*x) + 1/192*32^(5/6)*(-1)^(1/6)*log(32^(5/6)*(-1)^(1/6) + 16*x) - 1/192*32^(5/6)*(-1
)^(1/6)*log(-32^(5/6)*(-1)^(1/6) + 16*x)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.10 \[ \int \frac {1}{2+x^6} \, dx=\operatorname {RootSum} {\left (1492992 t^{6} + 1, \left ( t \mapsto t \log {\left (12 t + x \right )} \right )\right )} \]

[In]

integrate(1/(x**6+2),x)

[Out]

RootSum(1492992*_t**6 + 1, Lambda(_t, _t*log(12*_t + x)))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.78 \[ \int \frac {1}{2+x^6} \, dx=\frac {1}{24} \, \sqrt {3} 2^{\frac {1}{6}} \log \left (x^{2} + \sqrt {3} 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) - \frac {1}{24} \, \sqrt {3} 2^{\frac {1}{6}} \log \left (x^{2} - \sqrt {3} 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) + \frac {1}{12} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} {\left (2 \, x + \sqrt {3} 2^{\frac {1}{6}}\right )}\right ) + \frac {1}{12} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} {\left (2 \, x - \sqrt {3} 2^{\frac {1}{6}}\right )}\right ) + \frac {1}{6} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} x\right ) \]

[In]

integrate(1/(x^6+2),x, algorithm="maxima")

[Out]

1/24*sqrt(3)*2^(1/6)*log(x^2 + sqrt(3)*2^(1/6)*x + 2^(1/3)) - 1/24*sqrt(3)*2^(1/6)*log(x^2 - sqrt(3)*2^(1/6)*x
 + 2^(1/3)) + 1/12*2^(1/6)*arctan(1/2*2^(5/6)*(2*x + sqrt(3)*2^(1/6))) + 1/12*2^(1/6)*arctan(1/2*2^(5/6)*(2*x
- sqrt(3)*2^(1/6))) + 1/6*2^(1/6)*arctan(1/2*2^(5/6)*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.78 \[ \int \frac {1}{2+x^6} \, dx=\frac {1}{24} \, \sqrt {3} 2^{\frac {1}{6}} \log \left (x^{2} + \sqrt {3} 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) - \frac {1}{24} \, \sqrt {3} 2^{\frac {1}{6}} \log \left (x^{2} - \sqrt {3} 2^{\frac {1}{6}} x + 2^{\frac {1}{3}}\right ) + \frac {1}{12} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} {\left (2 \, x + \sqrt {3} 2^{\frac {1}{6}}\right )}\right ) + \frac {1}{12} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} {\left (2 \, x - \sqrt {3} 2^{\frac {1}{6}}\right )}\right ) + \frac {1}{6} \cdot 2^{\frac {1}{6}} \arctan \left (\frac {1}{2} \cdot 2^{\frac {5}{6}} x\right ) \]

[In]

integrate(1/(x^6+2),x, algorithm="giac")

[Out]

1/24*sqrt(3)*2^(1/6)*log(x^2 + sqrt(3)*2^(1/6)*x + 2^(1/3)) - 1/24*sqrt(3)*2^(1/6)*log(x^2 - sqrt(3)*2^(1/6)*x
 + 2^(1/3)) + 1/12*2^(1/6)*arctan(1/2*2^(5/6)*(2*x + sqrt(3)*2^(1/6))) + 1/12*2^(1/6)*arctan(1/2*2^(5/6)*(2*x
- sqrt(3)*2^(1/6))) + 1/6*2^(1/6)*arctan(1/2*2^(5/6)*x)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.98 \[ \int \frac {1}{2+x^6} \, dx=\frac {2^{1/6}\,\mathrm {atan}\left (\frac {2^{5/6}\,x}{2}\right )}{6}+\frac {2^{1/6}\,\mathrm {atan}\left (\frac {2^{1/6}\,x}{2\,\left (-\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}+\frac {2^{1/6}\,\sqrt {3}\,x\,1{}\mathrm {i}}{2\,\left (-\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}\right )\,\left (\sqrt {3}-\mathrm {i}\right )\,1{}\mathrm {i}}{12}+\frac {2^{1/6}\,\mathrm {atan}\left (\frac {2^{1/6}\,x}{2\,\left (\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}-\frac {2^{1/6}\,\sqrt {3}\,x\,1{}\mathrm {i}}{2\,\left (\frac {2^{1/3}}{2}+\frac {2^{1/3}\,\sqrt {3}\,1{}\mathrm {i}}{2}\right )}\right )\,\left (\sqrt {3}+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{12} \]

[In]

int(1/(x^6 + 2),x)

[Out]

(2^(1/6)*atan((2^(5/6)*x)/2))/6 + (2^(1/6)*atan((2^(1/6)*x)/(2*((2^(1/3)*3^(1/2)*1i)/2 - 2^(1/3)/2)) + (2^(1/6
)*3^(1/2)*x*1i)/(2*((2^(1/3)*3^(1/2)*1i)/2 - 2^(1/3)/2)))*(3^(1/2) - 1i)*1i)/12 + (2^(1/6)*atan((2^(1/6)*x)/(2
*((2^(1/3)*3^(1/2)*1i)/2 + 2^(1/3)/2)) - (2^(1/6)*3^(1/2)*x*1i)/(2*((2^(1/3)*3^(1/2)*1i)/2 + 2^(1/3)/2)))*(3^(
1/2) + 1i)*1i)/12