3.13 \(\int \frac {1+x^4}{1+x^8} \, dx\)

Optimal. Leaf size=347 \[ -\frac {\log \left (x^2-\sqrt {2-\sqrt {2}} x+1\right )}{8 \sqrt {2-\sqrt {2}}}+\frac {\log \left (x^2+\sqrt {2-\sqrt {2}} x+1\right )}{8 \sqrt {2-\sqrt {2}}}-\frac {\log \left (x^2-\sqrt {2+\sqrt {2}} x+1\right )}{8 \sqrt {2+\sqrt {2}}}+\frac {\log \left (x^2+\sqrt {2+\sqrt {2}} x+1\right )}{8 \sqrt {2+\sqrt {2}}}-\frac {1}{4} \sqrt {\frac {1}{2} \left (2-\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{4} \sqrt {\frac {1}{2} \left (2+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (2-\sqrt {2}\right )} \tan ^{-1}\left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (2+\sqrt {2}\right )} \tan ^{-1}\left (\frac {2 x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 347, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {1413, 1094, 634, 618, 204, 628} \[ -\frac {\log \left (x^2-\sqrt {2-\sqrt {2}} x+1\right )}{8 \sqrt {2-\sqrt {2}}}+\frac {\log \left (x^2+\sqrt {2-\sqrt {2}} x+1\right )}{8 \sqrt {2-\sqrt {2}}}-\frac {\log \left (x^2-\sqrt {2+\sqrt {2}} x+1\right )}{8 \sqrt {2+\sqrt {2}}}+\frac {\log \left (x^2+\sqrt {2+\sqrt {2}} x+1\right )}{8 \sqrt {2+\sqrt {2}}}-\frac {1}{4} \sqrt {\frac {1}{2} \left (2-\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{4} \sqrt {\frac {1}{2} \left (2+\sqrt {2}\right )} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (2-\sqrt {2}\right )} \tan ^{-1}\left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (2+\sqrt {2}\right )} \tan ^{-1}\left (\frac {2 x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)/(1 + x^8),x]

[Out]

-(Sqrt[(2 - Sqrt[2])/2]*ArcTan[(Sqrt[2 - Sqrt[2]] - 2*x)/Sqrt[2 + Sqrt[2]]])/4 - (Sqrt[(2 + Sqrt[2])/2]*ArcTan
[(Sqrt[2 + Sqrt[2]] - 2*x)/Sqrt[2 - Sqrt[2]]])/4 + (Sqrt[(2 - Sqrt[2])/2]*ArcTan[(Sqrt[2 - Sqrt[2]] + 2*x)/Sqr
t[2 + Sqrt[2]]])/4 + (Sqrt[(2 + Sqrt[2])/2]*ArcTan[(Sqrt[2 + Sqrt[2]] + 2*x)/Sqrt[2 - Sqrt[2]]])/4 - Log[1 - S
qrt[2 - Sqrt[2]]*x + x^2]/(8*Sqrt[2 - Sqrt[2]]) + Log[1 + Sqrt[2 - Sqrt[2]]*x + x^2]/(8*Sqrt[2 - Sqrt[2]]) - L
og[1 - Sqrt[2 + Sqrt[2]]*x + x^2]/(8*Sqrt[2 + Sqrt[2]]) + Log[1 + Sqrt[2 + Sqrt[2]]*x + x^2]/(8*Sqrt[2 + Sqrt[
2]])

Rule 204

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

Rule 618

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 628

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 634

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 1094

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 1413

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.19, size = 258, normalized size = 0.74 \[ \frac {1}{8} \left (-\left (\sin \left (\frac {\pi }{8}\right )+\cos \left (\frac {\pi }{8}\right )\right ) \log \left (x^2-2 x \sin \left (\frac {\pi }{8}\right )+1\right )+\left (\sin \left (\frac {\pi }{8}\right )+\cos \left (\frac {\pi }{8}\right )\right ) \log \left (x^2+2 x \sin \left (\frac {\pi }{8}\right )+1\right )+\left (\sin \left (\frac {\pi }{8}\right )-\cos \left (\frac {\pi }{8}\right )\right ) \log \left (x^2-2 x \cos \left (\frac {\pi }{8}\right )+1\right )+\left (\cos \left (\frac {\pi }{8}\right )-\sin \left (\frac {\pi }{8}\right )\right ) \log \left (x^2+2 x \cos \left (\frac {\pi }{8}\right )+1\right )+2 \left (\sin \left (\frac {\pi }{8}\right )+\cos \left (\frac {\pi }{8}\right )\right ) \tan ^{-1}\left (\csc \left (\frac {\pi }{8}\right ) \left (x-\cos \left (\frac {\pi }{8}\right )\right )\right )+2 \left (\sin \left (\frac {\pi }{8}\right )+\cos \left (\frac {\pi }{8}\right )\right ) \tan ^{-1}\left (\csc \left (\frac {\pi }{8}\right ) \left (x+\cos \left (\frac {\pi }{8}\right )\right )\right )+2 \left (\cos \left (\frac {\pi }{8}\right )-\sin \left (\frac {\pi }{8}\right )\right ) \tan ^{-1}\left (\sec \left (\frac {\pi }{8}\right ) \left (x+\sin \left (\frac {\pi }{8}\right )\right )\right )+2 \left (\cos \left (\frac {\pi }{8}\right )-\sin \left (\frac {\pi }{8}\right )\right ) \tan ^{-1}\left (x \sec \left (\frac {\pi }{8}\right )-\tan \left (\frac {\pi }{8}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^4)/(1 + x^8),x]

[Out]

(2*ArcTan[Sec[Pi/8]*(x + Sin[Pi/8])]*(Cos[Pi/8] - Sin[Pi/8]) + 2*ArcTan[x*Sec[Pi/8] - Tan[Pi/8]]*(Cos[Pi/8] -
Sin[Pi/8]) + Log[1 + x^2 + 2*x*Cos[Pi/8]]*(Cos[Pi/8] - Sin[Pi/8]) + Log[1 + x^2 - 2*x*Cos[Pi/8]]*(-Cos[Pi/8] +
Sin[Pi/8]) + 2*ArcTan[(x - Cos[Pi/8])*Csc[Pi/8]]*(Cos[Pi/8] + Sin[Pi/8]) + 2*ArcTan[(x + Cos[Pi/8])*Csc[Pi/8]
]*(Cos[Pi/8] + Sin[Pi/8]) - Log[1 + x^2 - 2*x*Sin[Pi/8]]*(Cos[Pi/8] + Sin[Pi/8]) + Log[1 + x^2 + 2*x*Sin[Pi/8]
]*(Cos[Pi/8] + Sin[Pi/8]))/8

________________________________________________________________________________________

fricas [B]  time = 0.87, size = 991, normalized size = 2.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) + 2))*arctan(-(2*x - 2*sqrt(x^2 + x*sqrt(-sqrt(2) + 2) + 1) + sqrt(-sq
rt(2) + 2))/sqrt(sqrt(2) + 2)) - 1/8*(sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) + 2))*arctan(-(2*x - 2*sqrt(x^2 - x*sq
rt(-sqrt(2) + 2) + 1) - sqrt(-sqrt(2) + 2))/sqrt(sqrt(2) + 2)) - 1/8*(sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))*
arctan(-(2*x - 2*sqrt(x^2 + x*sqrt(sqrt(2) + 2) + 1) + sqrt(sqrt(2) + 2))/sqrt(-sqrt(2) + 2)) - 1/8*(sqrt(sqrt
(2) + 2) + sqrt(-sqrt(2) + 2))*arctan(-(2*x - 2*sqrt(x^2 - x*sqrt(sqrt(2) + 2) + 1) - sqrt(sqrt(2) + 2))/sqrt(
-sqrt(2) + 2)) - 1/8*sqrt(2)*sqrt(-sqrt(2) + 2)*arctan(-(2*sqrt(2)*x - 2*sqrt(2)*sqrt(x^2 + 1/2*sqrt(2)*x*sqrt
(sqrt(2) + 2) - 1/2*sqrt(2)*x*sqrt(-sqrt(2) + 2) + 1) + sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) + 2))/(sqrt(sqrt(2)
+ 2) + sqrt(-sqrt(2) + 2))) - 1/8*sqrt(2)*sqrt(-sqrt(2) + 2)*arctan(-(2*sqrt(2)*x - 2*sqrt(2)*sqrt(x^2 - 1/2*s
qrt(2)*x*sqrt(sqrt(2) + 2) + 1/2*sqrt(2)*x*sqrt(-sqrt(2) + 2) + 1) - sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))/(
sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))) + 1/8*sqrt(2)*sqrt(sqrt(2) + 2)*arctan((2*sqrt(2)*x - 2*sqrt(2)*sqrt(
x^2 + 1/2*sqrt(2)*x*sqrt(sqrt(2) + 2) + 1/2*sqrt(2)*x*sqrt(-sqrt(2) + 2) + 1) + sqrt(sqrt(2) + 2) + sqrt(-sqrt
(2) + 2))/(sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) + 2))) + 1/8*sqrt(2)*sqrt(sqrt(2) + 2)*arctan((2*sqrt(2)*x - 2*sq
rt(2)*sqrt(x^2 - 1/2*sqrt(2)*x*sqrt(sqrt(2) + 2) - 1/2*sqrt(2)*x*sqrt(-sqrt(2) + 2) + 1) - sqrt(sqrt(2) + 2) -
sqrt(-sqrt(2) + 2))/(sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) + 2))) + 1/32*sqrt(2)*sqrt(-sqrt(2) + 2)*log(x^2 + 1/2
*sqrt(2)*x*sqrt(sqrt(2) + 2) + 1/2*sqrt(2)*x*sqrt(-sqrt(2) + 2) + 1) + 1/32*sqrt(2)*sqrt(sqrt(2) + 2)*log(x^2
+ 1/2*sqrt(2)*x*sqrt(sqrt(2) + 2) - 1/2*sqrt(2)*x*sqrt(-sqrt(2) + 2) + 1) - 1/32*sqrt(2)*sqrt(sqrt(2) + 2)*log
(x^2 - 1/2*sqrt(2)*x*sqrt(sqrt(2) + 2) + 1/2*sqrt(2)*x*sqrt(-sqrt(2) + 2) + 1) - 1/32*sqrt(2)*sqrt(-sqrt(2) +
2)*log(x^2 - 1/2*sqrt(2)*x*sqrt(sqrt(2) + 2) - 1/2*sqrt(2)*x*sqrt(-sqrt(2) + 2) + 1) + 1/32*(sqrt(sqrt(2) + 2)
- sqrt(-sqrt(2) + 2))*log(x^2 + x*sqrt(sqrt(2) + 2) + 1) - 1/32*(sqrt(sqrt(2) + 2) - sqrt(-sqrt(2) + 2))*log(
x^2 - x*sqrt(sqrt(2) + 2) + 1) + 1/32*(sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))*log(x^2 + x*sqrt(-sqrt(2) + 2)
+ 1) - 1/32*(sqrt(sqrt(2) + 2) + sqrt(-sqrt(2) + 2))*log(x^2 - x*sqrt(-sqrt(2) + 2) + 1)

________________________________________________________________________________________

giac [A]  time = 0.88, size = 247, normalized size = 0.71 \[ \frac {1}{8} \, \sqrt {-2 \, \sqrt {2} + 4} \arctan \left (\frac {2 \, x + \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {-2 \, \sqrt {2} + 4} \arctan \left (\frac {2 \, x - \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {2 \, \sqrt {2} + 4} \arctan \left (\frac {2 \, x + \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {2 \, \sqrt {2} + 4} \arctan \left (\frac {2 \, x - \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{16} \, \sqrt {-2 \, \sqrt {2} + 4} \log \left (x^{2} + x \sqrt {\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {-2 \, \sqrt {2} + 4} \log \left (x^{2} - x \sqrt {\sqrt {2} + 2} + 1\right ) + \frac {1}{16} \, \sqrt {2 \, \sqrt {2} + 4} \log \left (x^{2} + x \sqrt {-\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {2 \, \sqrt {2} + 4} \log \left (x^{2} - x \sqrt {-\sqrt {2} + 2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(-2*sqrt(2) + 4)*arctan((2*x + sqrt(-sqrt(2) + 2))/sqrt(sqrt(2) + 2)) + 1/8*sqrt(-2*sqrt(2) + 4)*arcta
n((2*x - sqrt(-sqrt(2) + 2))/sqrt(sqrt(2) + 2)) + 1/8*sqrt(2*sqrt(2) + 4)*arctan((2*x + sqrt(sqrt(2) + 2))/sqr
t(-sqrt(2) + 2)) + 1/8*sqrt(2*sqrt(2) + 4)*arctan((2*x - sqrt(sqrt(2) + 2))/sqrt(-sqrt(2) + 2)) + 1/16*sqrt(-2
*sqrt(2) + 4)*log(x^2 + x*sqrt(sqrt(2) + 2) + 1) - 1/16*sqrt(-2*sqrt(2) + 4)*log(x^2 - x*sqrt(sqrt(2) + 2) + 1
) + 1/16*sqrt(2*sqrt(2) + 4)*log(x^2 + x*sqrt(-sqrt(2) + 2) + 1) - 1/16*sqrt(2*sqrt(2) + 4)*log(x^2 - x*sqrt(-
sqrt(2) + 2) + 1)

________________________________________________________________________________________

maple [C]  time = 0.01, size = 27, normalized size = 0.08 \[ \frac {\left (\RootOf \left (\textit {\_Z}^{8}+1\right )^{4}+1\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{8}+1\right )+x \right )}{8 \RootOf \left (\textit {\_Z}^{8}+1\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+1)/(x^8+1),x)

[Out]

1/8*sum((_R^4+1)/_R^7*ln(-_R+x),_R=RootOf(_Z^8+1))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4} + 1}{x^{8} + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((x^4 + 1)/(x^8 + 1), x)

________________________________________________________________________________________

mupad [B]  time = 2.28, size = 311, normalized size = 0.90 \[ -\ln \left ({\left (\frac {\sqrt {-2\,\sqrt {2}-4}}{16}-\frac {\sqrt {4-2\,\sqrt {2}}}{16}\right )}^3\,\left (65536\,x-16384\,\sqrt {-2\,\sqrt {2}-4}+16384\,\sqrt {4-2\,\sqrt {2}}\right )+256\right )\,\left (\frac {\sqrt {-2\,\sqrt {2}-4}}{16}-\frac {\sqrt {4-2\,\sqrt {2}}}{16}\right )+\mathrm {atan}\left (\frac {x\,\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{2}+\frac {x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{2}+\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{2}-\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {2}\,\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{8}+\frac {\sqrt {2}\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )-\frac {\mathrm {atan}\left (x\,{\left (\sqrt {2}+2\right )}^{3/2}\,\left (1-\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,x\,{\left (\sqrt {2}+2\right )}^{3/2}\,\left (-\frac {3}{4}+\frac {1}{4}{}\mathrm {i}\right )\right )\,\left (-2+\sqrt {2}\,\left (1-\mathrm {i}\right )\right )\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}+\frac {\mathrm {atan}\left (x\,{\left (\sqrt {2}+2\right )}^{3/2}\,\left (\frac {1}{2}+1{}\mathrm {i}\right )+\sqrt {2}\,x\,{\left (\sqrt {2}+2\right )}^{3/2}\,\left (-\frac {1}{4}-\frac {3}{4}{}\mathrm {i}\right )\right )\,\left (\sqrt {2}\,\left (1+1{}\mathrm {i}\right )-2{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}+\sqrt {2}\,\ln \left (x+{\left (\sqrt {2}+2\right )}^{3/2}\,\left (-\frac {1}{2}-\mathrm {i}\right )+\sqrt {2}\,{\left (\sqrt {2}+2\right )}^{3/2}\,\left (\frac {1}{4}+\frac {3}{4}{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}-2}}{16}+\frac {\sqrt {\sqrt {2}+2}}{16}\right )\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4 + 1)/(x^8 + 1),x)

[Out]

atan((x*(2^(1/2) - 2)^(1/2)*1i)/2 + (x*(2^(1/2) + 2)^(1/2)*1i)/2 + (2^(1/2)*x*(2^(1/2) - 2)^(1/2)*1i)/2 - (2^(
1/2)*x*(2^(1/2) + 2)^(1/2)*1i)/2)*((2^(1/2)*(2^(1/2) - 2)^(1/2)*1i)/8 + (2^(1/2)*(2^(1/2) + 2)^(1/2)*1i)/8) -
log(((- 2*2^(1/2) - 4)^(1/2)/16 - (4 - 2*2^(1/2))^(1/2)/16)^3*(65536*x - 16384*(- 2*2^(1/2) - 4)^(1/2) + 16384
*(4 - 2*2^(1/2))^(1/2)) + 256)*((- 2*2^(1/2) - 4)^(1/2)/16 - (4 - 2*2^(1/2))^(1/2)/16) - (atan(x*(2^(1/2) + 2)
^(3/2)*(1 - 1i/2) - 2^(1/2)*x*(2^(1/2) + 2)^(3/2)*(3/4 - 1i/4))*(2^(1/2)*(1 - 1i) - 2)*(2^(1/2) + 2)^(1/2)*1i)
/8 + (atan(x*(2^(1/2) + 2)^(3/2)*(1/2 + 1i) - 2^(1/2)*x*(2^(1/2) + 2)^(3/2)*(1/4 + 3i/4))*(2^(1/2)*(1 + 1i) -
2i)*(2^(1/2) + 2)^(1/2)*1i)/8 + 2^(1/2)*log(x - (2^(1/2) + 2)^(3/2)*(1/2 + 1i) + 2^(1/2)*(2^(1/2) + 2)^(3/2)*(
1/4 + 3i/4))*((2^(1/2) - 2)^(1/2)/16 + (2^(1/2) + 2)^(1/2)/16)*1i

________________________________________________________________________________________

sympy [A]  time = 2.78, size = 19, normalized size = 0.05 \[ \operatorname {RootSum} {\left (1048576 t^{8} + 1, \left (t \mapsto t \log {\left (4096 t^{5} + 4 t + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(1048576*_t**8 + 1, Lambda(_t, _t*log(4096*_t**5 + 4*_t + x)))

________________________________________________________________________________________