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

Optimal. Leaf size=339 \[ -\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (x^2-\sqrt {2-\sqrt {2}} x+1\right )+\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (x^2+\sqrt {2-\sqrt {2}} x+1\right )-\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (x^2-\sqrt {2+\sqrt {2}} x+1\right )+\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (x^2+\sqrt {2+\sqrt {2}} x+1\right )-\frac {\tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}+\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}} \]

[Out]

-1/16*ln(1+x^2-x*(2-2^(1/2))^(1/2))*(2-2^(1/2))^(1/2)+1/16*ln(1+x^2+x*(2-2^(1/2))^(1/2))*(2-2^(1/2))^(1/2)-1/4
*arctan((-2*x+(2-2^(1/2))^(1/2))/(2+2^(1/2))^(1/2))/(4-2*2^(1/2))^(1/2)+1/4*arctan((2*x+(2-2^(1/2))^(1/2))/(2+
2^(1/2))^(1/2))/(4-2*2^(1/2))^(1/2)-1/16*ln(1+x^2-x*(2+2^(1/2))^(1/2))*(2+2^(1/2))^(1/2)+1/16*ln(1+x^2+x*(2+2^
(1/2))^(1/2))*(2+2^(1/2))^(1/2)-1/4*arctan((-2*x+(2+2^(1/2))^(1/2))/(2-2^(1/2))^(1/2))/(4+2*2^(1/2))^(1/2)+1/4
*arctan((2*x+(2+2^(1/2))^(1/2))/(2-2^(1/2))^(1/2))/(4+2*2^(1/2))^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 339, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 6, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {213, 1169, 634, 618, 204, 628} \[ -\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (x^2-\sqrt {2-\sqrt {2}} x+1\right )+\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (x^2+\sqrt {2-\sqrt {2}} x+1\right )-\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (x^2-\sqrt {2+\sqrt {2}} x+1\right )+\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (x^2+\sqrt {2+\sqrt {2}} x+1\right )-\frac {\tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}+\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^8)^(-1),x]

[Out]

-ArcTan[(Sqrt[2 - Sqrt[2]] - 2*x)/Sqrt[2 + Sqrt[2]]]/(4*Sqrt[2*(2 - Sqrt[2])]) - ArcTan[(Sqrt[2 + Sqrt[2]] - 2
*x)/Sqrt[2 - Sqrt[2]]]/(4*Sqrt[2*(2 + Sqrt[2])]) + ArcTan[(Sqrt[2 - Sqrt[2]] + 2*x)/Sqrt[2 + Sqrt[2]]]/(4*Sqrt
[2*(2 - Sqrt[2])]) + ArcTan[(Sqrt[2 + Sqrt[2]] + 2*x)/Sqrt[2 - Sqrt[2]]]/(4*Sqrt[2*(2 + Sqrt[2])]) - (Sqrt[2 -
Sqrt[2]]*Log[1 - Sqrt[2 - Sqrt[2]]*x + x^2])/16 + (Sqrt[2 - Sqrt[2]]*Log[1 + Sqrt[2 - Sqrt[2]]*x + x^2])/16 -
(Sqrt[2 + Sqrt[2]]*Log[1 - Sqrt[2 + Sqrt[2]]*x + x^2])/16 + (Sqrt[2 + Sqrt[2]]*Log[1 + Sqrt[2 + Sqrt[2]]*x +
x^2])/16

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 213

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 4]], s = Denominator[Rt[a/b, 4]]},
Dist[r/(2*Sqrt[2]*a), Int[(Sqrt[2]*r - s*x^(n/4))/(r^2 - Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x] + Dist[r/
(2*Sqrt[2]*a), Int[(Sqrt[2]*r + s*x^(n/4))/(r^2 + Sqrt[2]*r*s*x^(n/4) + s^2*x^(n/2)), x], x]] /; FreeQ[{a, b},
x] && IGtQ[n/4, 1] && GtQ[a/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 1169

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r =
Rt[2*q - b/c, 2]}, Dist[1/(2*c*q*r), Int[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(
d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2
- b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {1}{1+x^8} \, dx &=\frac {\int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2}+x^2}{1+\sqrt {2} x^2+x^4} \, dx}{2 \sqrt {2}}\\ &=\frac {\int \frac {\sqrt {2 \left (2-\sqrt {2}\right )}-\left (-1+\sqrt {2}\right ) x}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\int \frac {\sqrt {2 \left (2-\sqrt {2}\right )}+\left (-1+\sqrt {2}\right ) x}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\int \frac {\sqrt {2 \left (2+\sqrt {2}\right )}-\left (1+\sqrt {2}\right ) x}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}+\frac {\int \frac {\sqrt {2 \left (2+\sqrt {2}\right )}+\left (1+\sqrt {2}\right ) x}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}\\ &=\frac {1}{8} \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )} \int \frac {1}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx+\frac {1}{8} \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )} \int \frac {1}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx-\frac {1}{16} \sqrt {2-\sqrt {2}} \int \frac {-\sqrt {2-\sqrt {2}}+2 x}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx+\frac {1}{16} \sqrt {2-\sqrt {2}} \int \frac {\sqrt {2-\sqrt {2}}+2 x}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx-\frac {1}{16} \sqrt {2+\sqrt {2}} \int \frac {-\sqrt {2+\sqrt {2}}+2 x}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx+\frac {1}{16} \sqrt {2+\sqrt {2}} \int \frac {\sqrt {2+\sqrt {2}}+2 x}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx+\frac {1}{8} \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )} \int \frac {1}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx+\frac {1}{8} \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )} \int \frac {1}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx\\ &=-\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (1-\sqrt {2-\sqrt {2}} x+x^2\right )+\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (1+\sqrt {2-\sqrt {2}} x+x^2\right )-\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (1-\sqrt {2+\sqrt {2}} x+x^2\right )+\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (1+\sqrt {2+\sqrt {2}} x+x^2\right )-\frac {1}{4} \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )} \operatorname {Subst}\left (\int \frac {1}{-2+\sqrt {2}-x^2} \, dx,x,-\sqrt {2+\sqrt {2}}+2 x\right )-\frac {1}{4} \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )} \operatorname {Subst}\left (\int \frac {1}{-2+\sqrt {2}-x^2} \, dx,x,\sqrt {2+\sqrt {2}}+2 x\right )-\frac {1}{4} \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )} \operatorname {Subst}\left (\int \frac {1}{-2-\sqrt {2}-x^2} \, dx,x,-\sqrt {2-\sqrt {2}}+2 x\right )-\frac {1}{4} \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )} \operatorname {Subst}\left (\int \frac {1}{-2-\sqrt {2}-x^2} \, dx,x,\sqrt {2-\sqrt {2}}+2 x\right )\\ &=-\frac {1}{8} \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{8} \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{8} \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}+2 x}{\sqrt {2+\sqrt {2}}}\right )+\frac {1}{8} \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}+2 x}{\sqrt {2-\sqrt {2}}}\right )-\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (1-\sqrt {2-\sqrt {2}} x+x^2\right )+\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (1+\sqrt {2-\sqrt {2}} x+x^2\right )-\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (1-\sqrt {2+\sqrt {2}} x+x^2\right )+\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (1+\sqrt {2+\sqrt {2}} x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 209, normalized size = 0.62 \[ -\frac {1}{8} \sin \left (\frac {\pi }{8}\right ) \log \left (x^2-2 x \sin \left (\frac {\pi }{8}\right )+1\right )+\frac {1}{8} \sin \left (\frac {\pi }{8}\right ) \log \left (x^2+2 x \sin \left (\frac {\pi }{8}\right )+1\right )-\frac {1}{8} \cos \left (\frac {\pi }{8}\right ) \log \left (x^2-2 x \cos \left (\frac {\pi }{8}\right )+1\right )+\frac {1}{8} \cos \left (\frac {\pi }{8}\right ) \log \left (x^2+2 x \cos \left (\frac {\pi }{8}\right )+1\right )+\frac {1}{4} \sin \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\csc \left (\frac {\pi }{8}\right ) \left (x-\cos \left (\frac {\pi }{8}\right )\right )\right )+\frac {1}{4} \sin \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\csc \left (\frac {\pi }{8}\right ) \left (x+\cos \left (\frac {\pi }{8}\right )\right )\right )+\frac {1}{4} \cos \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\sec \left (\frac {\pi }{8}\right ) \left (x-\sin \left (\frac {\pi }{8}\right )\right )\right )+\frac {1}{4} \cos \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\sec \left (\frac {\pi }{8}\right ) \left (x+\sin \left (\frac {\pi }{8}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^8)^(-1),x]

[Out]

(ArcTan[Sec[Pi/8]*(x - Sin[Pi/8])]*Cos[Pi/8])/4 + (ArcTan[Sec[Pi/8]*(x + Sin[Pi/8])]*Cos[Pi/8])/4 - (Cos[Pi/8]
*Log[1 + x^2 - 2*x*Cos[Pi/8]])/8 + (Cos[Pi/8]*Log[1 + x^2 + 2*x*Cos[Pi/8]])/8 + (ArcTan[(x - Cos[Pi/8])*Csc[Pi
/8]]*Sin[Pi/8])/4 + (ArcTan[(x + Cos[Pi/8])*Csc[Pi/8]]*Sin[Pi/8])/4 - (Log[1 + x^2 - 2*x*Sin[Pi/8]]*Sin[Pi/8])
/8 + (Log[1 + x^2 + 2*x*Sin[Pi/8]]*Sin[Pi/8])/8

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fricas [B]  time = 0.43, size = 1027, normalized size = 3.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(sqrt(2)*sqrt(sqrt(2) + 2) + 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))) - 1/16*(sqrt(2)*sqrt(sqrt(2) + 2) + sqrt(2)*sqrt(-sqrt(2) + 2))*arc
tan(-(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))) + 1/16*(sqrt(2)*sqrt(sq
rt(2) + 2) - 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) - s
qrt(-sqrt(2) + 2))) + 1/16*(sqrt(2)*sqrt(sqrt(2) + 2) - 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))) + 1/64*(sqrt(2)*sqrt(sqrt(2) + 2) + sqrt(2)*sqr
t(-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) + 1/64*(sqr
t(2)*sqrt(sqrt(2) + 2) - 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) - 1/64*(sqrt(2)*sqrt(sqrt(2) + 2) - 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) - 1/64*(sqrt(2)*sqrt(sqrt(2) + 2) + 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) - 1/8*sqrt(sq
rt(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)*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)*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)*arctan(-(2*x - 2*sqrt(x^2 - x*sqrt(sqrt(2) + 2) + 1) - sqrt(sqrt(2) + 2
))/sqrt(-sqrt(2) + 2)) + 1/32*sqrt(sqrt(2) + 2)*log(x^2 + x*sqrt(sqrt(2) + 2) + 1) - 1/32*sqrt(sqrt(2) + 2)*lo
g(x^2 - x*sqrt(sqrt(2) + 2) + 1) + 1/32*sqrt(-sqrt(2) + 2)*log(x^2 + x*sqrt(-sqrt(2) + 2) + 1) - 1/32*sqrt(-sq
rt(2) + 2)*log(x^2 - x*sqrt(-sqrt(2) + 2) + 1)

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giac [A]  time = 1.03, size = 239, normalized size = 0.71 \[ \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {2 \, x + \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {2 \, x - \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {2 \, x + \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {2 \, x - \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \log \left (x^{2} + x \sqrt {\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \log \left (x^{2} - x \sqrt {\sqrt {2} + 2} + 1\right ) + \frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \log \left (x^{2} + x \sqrt {-\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \log \left (x^{2} - x \sqrt {-\sqrt {2} + 2} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(sqrt(2) + 2)*arctan((2*x + sqrt(-sqrt(2) + 2))/sqrt(sqrt(2) + 2)) + 1/8*sqrt(sqrt(2) + 2)*arctan((2*x
- sqrt(-sqrt(2) + 2))/sqrt(sqrt(2) + 2)) + 1/8*sqrt(-sqrt(2) + 2)*arctan((2*x + sqrt(sqrt(2) + 2))/sqrt(-sqrt
(2) + 2)) + 1/8*sqrt(-sqrt(2) + 2)*arctan((2*x - sqrt(sqrt(2) + 2))/sqrt(-sqrt(2) + 2)) + 1/16*sqrt(sqrt(2) +
2)*log(x^2 + x*sqrt(sqrt(2) + 2) + 1) - 1/16*sqrt(sqrt(2) + 2)*log(x^2 - x*sqrt(sqrt(2) + 2) + 1) + 1/16*sqrt(
-sqrt(2) + 2)*log(x^2 + x*sqrt(-sqrt(2) + 2) + 1) - 1/16*sqrt(-sqrt(2) + 2)*log(x^2 - x*sqrt(-sqrt(2) + 2) + 1
)

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maple [C]  time = 0.00, size = 22, normalized size = 0.06 \[ \frac {\ln \left (-\RootOf \left (\textit {\_Z}^{8}+1\right )+x \right )}{8 \RootOf \left (\textit {\_Z}^{8}+1\right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(x^8 + 1), x)

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mupad [B]  time = 0.30, size = 288, normalized size = 0.85 \[ \mathrm {atan}\left (\frac {x\,\sqrt {-\sqrt {2}-2}\,1{}\mathrm {i}}{\sqrt {2-\sqrt {2}}\,\sqrt {-\sqrt {2}-2}+\sqrt {2}}-\frac {x\,\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{\sqrt {2-\sqrt {2}}\,\sqrt {-\sqrt {2}-2}+\sqrt {2}}\right )\,\left (\frac {\sqrt {-\sqrt {2}-2}\,1{}\mathrm {i}}{8}-\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )-\mathrm {atan}\left (\frac {x\,\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{\sqrt {2}+\sqrt {\sqrt {2}-2}\,\sqrt {\sqrt {2}+2}}+\frac {x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{\sqrt {2}+\sqrt {\sqrt {2}-2}\,\sqrt {\sqrt {2}+2}}\right )\,\left (\frac {\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{8}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )+\mathrm {atan}\left (-\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}+2}}{2}+x\,\sqrt {\sqrt {2}+2}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {2}\,1{}\mathrm {i}}{16}-\frac {1}{16}-\frac {1}{16}{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,2{}\mathrm {i}-\mathrm {atan}\left (x\,\sqrt {\sqrt {2}+2}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {2}}{16}-\frac {1}{16}+\frac {1}{16}{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,2{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^8 + 1),x)

[Out]

atan((x*(- 2^(1/2) - 2)^(1/2)*1i)/((2 - 2^(1/2))^(1/2)*(- 2^(1/2) - 2)^(1/2) + 2^(1/2)) - (x*(2 - 2^(1/2))^(1/
2)*1i)/((2 - 2^(1/2))^(1/2)*(- 2^(1/2) - 2)^(1/2) + 2^(1/2)))*(((- 2^(1/2) - 2)^(1/2)*1i)/8 - ((2 - 2^(1/2))^(
1/2)*1i)/8) - atan((x*(2^(1/2) - 2)^(1/2)*1i)/(2^(1/2) + (2^(1/2) - 2)^(1/2)*(2^(1/2) + 2)^(1/2)) + (x*(2^(1/2
) + 2)^(1/2)*1i)/(2^(1/2) + (2^(1/2) - 2)^(1/2)*(2^(1/2) + 2)^(1/2)))*(((2^(1/2) - 2)^(1/2)*1i)/8 + ((2^(1/2)
+ 2)^(1/2)*1i)/8) + atan(x*(2^(1/2) + 2)^(1/2)*(1/2 + 1i/2) - (2^(1/2)*x*(2^(1/2) + 2)^(1/2))/2)*((2^(1/2)*1i)
/16 - (1/16 + 1i/16))*(2^(1/2) + 2)^(1/2)*2i - atan(x*(2^(1/2) + 2)^(1/2)*(1/2 - 1i/2) + (2^(1/2)*x*(2^(1/2) +
2)^(1/2)*1i)/2)*(2^(1/2)/16 - (1/16 - 1i/16))*(2^(1/2) + 2)^(1/2)*2i

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sympy [A]  time = 2.82, size = 14, normalized size = 0.04 \[ \operatorname {RootSum} {\left (16777216 t^{8} + 1, \left (t \mapsto t \log {\left (8 t + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(16777216*_t**8 + 1, Lambda(_t, _t*log(8*_t + x)))

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