\(\int \frac {1+x^4}{1+x^6} \, dx\) [146]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int \frac {1+x^4}{1+x^6} \, dx=-\frac {1}{3} \arctan \left (\sqrt {3}-2 x\right )+\frac {2 \arctan (x)}{3}+\frac {1}{3} \arctan \left (\sqrt {3}+2 x\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 22, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {1890, 215, 648, 632, 210, 642, 209, 301} \[ \int \frac {1+x^4}{1+x^6} \, dx=-\frac {1}{3} \arctan \left (\sqrt {3}-2 x\right )+\frac {2 \arctan (x)}{3}+\frac {1}{3} \arctan \left (2 x+\sqrt {3}\right ) \]

[In]

Int[(1 + x^4)/(1 + x^6),x]

[Out]

-1/3*ArcTan[Sqrt[3] - 2*x] + (2*ArcTan[x])/3 + ArcTan[Sqrt[3] + 2*x]/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 301

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

Rule 1890

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = Sum[x^ii*((Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii
]*x^(n/2))/(a + b*x^n)), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ
[n/2, 0] && Expon[Pq, x] < n

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.60 \[ \int \frac {1+x^4}{1+x^6} \, dx=\frac {2 \arctan (x)}{3}-\frac {1}{3} \arctan \left (\frac {x}{-1+x^2}\right ) \]

[In]

Integrate[(1 + x^4)/(1 + x^6),x]

[Out]

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

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.29

method	result	size
risch	\(\arctan \left (x \right )+\frac {\arctan \left (x^{3}\right )}{3}\)	\(10\)
default	\(\frac {2 \arctan \left (x \right )}{3}+\frac {\arctan \left (2 x -\sqrt {3}\right )}{3}+\frac {\arctan \left (2 x +\sqrt {3}\right )}{3}\)	\(28\)
parallelrisch	\(\frac {i \ln \left (x +i\right )}{3}-\frac {i \ln \left (x -i\right )}{3}+\frac {i \ln \left (x^{2}+i x -1\right )}{6}-\frac {i \ln \left (x^{2}-i x -1\right )}{6}\)	\(44\)
meijerg	\(\frac {x^{5} \sqrt {3}\, \ln \left (1-\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{12 \left (x^{6}\right )^{\frac {5}{6}}}+\frac {x^{5} \arctan \left (\frac {\left (x^{6}\right )^{\frac {1}{6}}}{2-\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}\right )}{6 \left (x^{6}\right )^{\frac {5}{6}}}+\frac {x^{5} \arctan \left (\left (x^{6}\right )^{\frac {1}{6}}\right )}{3 \left (x^{6}\right )^{\frac {5}{6}}}-\frac {x^{5} \sqrt {3}\, \ln \left (1+\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{12 \left (x^{6}\right )^{\frac {5}{6}}}+\frac {x^{5} \arctan \left (\frac {\left (x^{6}\right )^{\frac {1}{6}}}{2+\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}\right )}{6 \left (x^{6}\right )^{\frac {5}{6}}}-\frac {x \sqrt {3}\, \ln \left (1-\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{12 \left (x^{6}\right )^{\frac {1}{6}}}+\frac {x \arctan \left (\frac {\left (x^{6}\right )^{\frac {1}{6}}}{2-\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}\right )}{6 \left (x^{6}\right )^{\frac {1}{6}}}+\frac {x \arctan \left (\left (x^{6}\right )^{\frac {1}{6}}\right )}{3 \left (x^{6}\right )^{\frac {1}{6}}}+\frac {x \sqrt {3}\, \ln \left (1+\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}+\left (x^{6}\right )^{\frac {1}{3}}\right )}{12 \left (x^{6}\right )^{\frac {1}{6}}}+\frac {x \arctan \left (\frac {\left (x^{6}\right )^{\frac {1}{6}}}{2+\sqrt {3}\, \left (x^{6}\right )^{\frac {1}{6}}}\right )}{6 \left (x^{6}\right )^{\frac {1}{6}}}\)	\(274\)

[In]

int((x^4+1)/(x^6+1),x,method=_RETURNVERBOSE)

[Out]

arctan(x)+1/3*arctan(x^3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.26 \[ \int \frac {1+x^4}{1+x^6} \, dx=\frac {1}{3} \, \arctan \left (x^{3}\right ) + \arctan \left (x\right ) \]

[In]

integrate((x^4+1)/(x^6+1),x, algorithm="fricas")

[Out]

1/3*arctan(x^3) + arctan(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.23 \[ \int \frac {1+x^4}{1+x^6} \, dx=\operatorname {atan}{\left (x \right )} + \frac {\operatorname {atan}{\left (x^{3} \right )}}{3} \]

[In]

integrate((x**4+1)/(x**6+1),x)

[Out]

atan(x) + atan(x**3)/3

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {1+x^4}{1+x^6} \, dx=\frac {1}{3} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{3} \, \arctan \left (2 \, x - \sqrt {3}\right ) + \frac {2}{3} \, \arctan \left (x\right ) \]

[In]

integrate((x^4+1)/(x^6+1),x, algorithm="maxima")

[Out]

1/3*arctan(2*x + sqrt(3)) + 1/3*arctan(2*x - sqrt(3)) + 2/3*arctan(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.77 \[ \int \frac {1+x^4}{1+x^6} \, dx=\frac {1}{3} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{3} \, \arctan \left (2 \, x - \sqrt {3}\right ) + \frac {2}{3} \, \arctan \left (x\right ) \]

[In]

integrate((x^4+1)/(x^6+1),x, algorithm="giac")

[Out]

1/3*arctan(2*x + sqrt(3)) + 1/3*arctan(2*x - sqrt(3)) + 2/3*arctan(x)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.26 \[ \int \frac {1+x^4}{1+x^6} \, dx=\frac {\mathrm {atan}\left (x^3\right )}{3}+\mathrm {atan}\left (x\right ) \]

[In]

int((x^4 + 1)/(x^6 + 1),x)

[Out]

atan(x^3)/3 + atan(x)