\(\int \frac {1+x^4}{1+x^4+x^8} \, dx\) [12]

Optimal result

Integrand size = 16, antiderivative size = 140 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4} \arctan \left (\sqrt {3}-2 x\right )+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{4} \arctan \left (\sqrt {3}+2 x\right )-\frac {1}{8} \log \left (1-x+x^2\right )+\frac {1}{8} \log \left (1+x+x^2\right )-\frac {\log \left (1-\sqrt {3} x+x^2\right )}{8 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x+x^2\right )}{8 \sqrt {3}} \]

[Out]

1/4*arctan(2*x-3^(1/2))+1/4*arctan(2*x+3^(1/2))-1/8*ln(x^2-x+1)+1/8*ln(x^2+x+1)-1/12*arctan(1/3*(1-2*x)*3^(1/2
))*3^(1/2)+1/12*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)-1/24*ln(1+x^2-x*3^(1/2))*3^(1/2)+1/24*ln(1+x^2+x*3^(1/2))*
3^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1433, 1108, 648, 632, 210, 642} \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{4} \arctan \left (\sqrt {3}-2 x\right )+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{4} \arctan \left (2 x+\sqrt {3}\right )-\frac {1}{8} \log \left (x^2-x+1\right )+\frac {1}{8} \log \left (x^2+x+1\right )-\frac {\log \left (x^2-\sqrt {3} x+1\right )}{8 \sqrt {3}}+\frac {\log \left (x^2+\sqrt {3} x+1\right )}{8 \sqrt {3}} \]

[In]

Int[(1 + x^4)/(1 + x^4 + x^8),x]

[Out]

-1/4*ArcTan[(1 - 2*x)/Sqrt[3]]/Sqrt[3] - ArcTan[Sqrt[3] - 2*x]/4 + ArcTan[(1 + 2*x)/Sqrt[3]]/(4*Sqrt[3]) + Arc
Tan[Sqrt[3] + 2*x]/4 - Log[1 - x + x^2]/8 + Log[1 + x + x^2]/8 - Log[1 - Sqrt[3]*x + x^2]/(8*Sqrt[3]) + Log[1
+ Sqrt[3]*x + x^2]/(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 1433

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[2*(d/e) -
b/c, 2]}, Dist[e/(2*c), Int[1/Simp[d/e + q*x^(n/2) + x^n, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x^(n/2
) + x^n, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2,
0] && IGtQ[n/2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d, e*Rt[a/c, 2]]))

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.96 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=\frac {1}{48} \left (4 i \sqrt {-6-6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )-4 i \sqrt {-6+6 i \sqrt {3}} \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )+4 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+4 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-6 \log \left (1-x+x^2\right )+6 \log \left (1+x+x^2\right )\right ) \]

[In]

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

[Out]

((4*I)*Sqrt[-6 - (6*I)*Sqrt[3]]*ArcTan[((1 - I*Sqrt[3])*x)/2] - (4*I)*Sqrt[-6 + (6*I)*Sqrt[3]]*ArcTan[((1 + I*
Sqrt[3])*x)/2] + 4*Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] + 4*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - 6*Log[1 - x + x^
2] + 6*Log[1 + x + x^2])/48

Maple [C] (verified)

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

Time = 0.14 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.64

method	result	size
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-3 \textit {\_R}^{3}+\textit {\_R} +x \right )\right )}{4}-\frac {\ln \left (4 x^{2}-4 x +4\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (4 x^{2}+4 x +4\right )}{8}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}\)	\(89\)
default	\(-\frac {\ln \left (x^{2}-x +1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (x^{2}+x +1\right )}{8}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}-\frac {\ln \left (1+x^{2}-x \sqrt {3}\right ) \sqrt {3}}{24}+\frac {\arctan \left (2 x -\sqrt {3}\right )}{4}+\frac {\ln \left (1+x^{2}+x \sqrt {3}\right ) \sqrt {3}}{24}+\frac {\arctan \left (2 x +\sqrt {3}\right )}{4}\)	\(109\)

[In]

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

[Out]

1/4*sum(_R*ln(-3*_R^3+_R+x),_R=RootOf(9*_Z^4+3*_Z^2+1))-1/8*ln(4*x^2-4*x+4)+1/12*3^(1/2)*arctan(1/3*(2*x-1)*3^
(1/2))+1/8*ln(4*x^2+4*x+4)+1/12*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.49 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=-\frac {1}{24} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (\sqrt {6} \sqrt {i \, \sqrt {3} - 1} {\left (i \, \sqrt {3} - 3\right )} + 12 \, x\right ) + \frac {1}{24} \, \sqrt {6} \sqrt {i \, \sqrt {3} - 1} \log \left (\sqrt {6} \sqrt {i \, \sqrt {3} - 1} {\left (-i \, \sqrt {3} + 3\right )} + 12 \, x\right ) + \frac {1}{24} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (\sqrt {6} {\left (i \, \sqrt {3} + 3\right )} \sqrt {-i \, \sqrt {3} - 1} + 12 \, x\right ) - \frac {1}{24} \, \sqrt {6} \sqrt {-i \, \sqrt {3} - 1} \log \left (\sqrt {6} \sqrt {-i \, \sqrt {3} - 1} {\left (-i \, \sqrt {3} - 3\right )} + 12 \, x\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \]

[In]

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

[Out]

-1/24*sqrt(6)*sqrt(I*sqrt(3) - 1)*log(sqrt(6)*sqrt(I*sqrt(3) - 1)*(I*sqrt(3) - 3) + 12*x) + 1/24*sqrt(6)*sqrt(
I*sqrt(3) - 1)*log(sqrt(6)*sqrt(I*sqrt(3) - 1)*(-I*sqrt(3) + 3) + 12*x) + 1/24*sqrt(6)*sqrt(-I*sqrt(3) - 1)*lo
g(sqrt(6)*(I*sqrt(3) + 3)*sqrt(-I*sqrt(3) - 1) + 12*x) - 1/24*sqrt(6)*sqrt(-I*sqrt(3) - 1)*log(sqrt(6)*sqrt(-I
*sqrt(3) - 1)*(-I*sqrt(3) - 3) + 12*x) + 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/12*sqrt(3)*arctan(1/3*
sqrt(3)*(2*x - 1)) + 1/8*log(x^2 + x + 1) - 1/8*log(x^2 - x + 1)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.34 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.36 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=\left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x - 1 - \frac {\sqrt {3} i}{3} + 9216 \left (- \frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{5} \right )} + \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x - 1 + 9216 \left (- \frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{5} + \frac {\sqrt {3} i}{3} \right )} + \left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right ) \log {\left (x + 1 - \frac {\sqrt {3} i}{3} + 9216 \left (\frac {1}{8} - \frac {\sqrt {3} i}{24}\right )^{5} \right )} + \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right ) \log {\left (x + 1 + 9216 \left (\frac {1}{8} + \frac {\sqrt {3} i}{24}\right )^{5} + \frac {\sqrt {3} i}{3} \right )} + \operatorname {RootSum} {\left (2304 t^{4} + 48 t^{2} + 1, \left ( t \mapsto t \log {\left (9216 t^{5} + 8 t + x \right )} \right )\right )} \]

[In]

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

[Out]

(-1/8 - sqrt(3)*I/24)*log(x - 1 - sqrt(3)*I/3 + 9216*(-1/8 - sqrt(3)*I/24)**5) + (-1/8 + sqrt(3)*I/24)*log(x -
1 + 9216*(-1/8 + sqrt(3)*I/24)**5 + sqrt(3)*I/3) + (1/8 - sqrt(3)*I/24)*log(x + 1 - sqrt(3)*I/3 + 9216*(1/8 -
sqrt(3)*I/24)**5) + (1/8 + sqrt(3)*I/24)*log(x + 1 + 9216*(1/8 + sqrt(3)*I/24)**5 + sqrt(3)*I/3) + RootSum(23
04*_t**4 + 48*_t**2 + 1, Lambda(_t, _t*log(9216*_t**5 + 8*_t + x)))

Maxima [F]

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

[In]

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

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/2*integrate(1/(x^4
- x^2 + 1), x) + 1/8*log(x^2 + x + 1) - 1/8*log(x^2 - x + 1)

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.77 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{24} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) - \frac {1}{24} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {1}{4} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{4} \, \arctan \left (2 \, x - \sqrt {3}\right ) + \frac {1}{8} \, \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, \log \left (x^{2} - x + 1\right ) \]

[In]

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

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/24*sqrt(3)*log(x^2
+ sqrt(3)*x + 1) - 1/24*sqrt(3)*log(x^2 - sqrt(3)*x + 1) + 1/4*arctan(2*x + sqrt(3)) + 1/4*arctan(2*x - sqrt(
3)) + 1/8*log(x^2 + x + 1) - 1/8*log(x^2 - x + 1)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.68 \[ \int \frac {1+x^4}{1+x^4+x^8} \, dx=\mathrm {atan}\left (\frac {2\,x}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atan}\left (\frac {2\,x}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}+\frac {1}{4}{}\mathrm {i}\right )+\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {\sqrt {3}}{12}-\frac {1}{4}{}\mathrm {i}\right ) \]

[In]

int((x^4 + 1)/(x^4 + x^8 + 1),x)

[Out]

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