3.27.37 \(\int \frac {1+x^4}{(-1-x^2+x^4) \sqrt [4]{-x^2+x^4}} \, dx\)

Optimal. Leaf size=233 \[ \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^2}}\right )-\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}} x}{\sqrt [4]{x^4-x^2}}\right )-\sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^4-x^2}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4-x^2}}\right )-\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {\sqrt {5}}{2}-\frac {1}{2}} x}{\sqrt [4]{x^4-x^2}}\right )-\sqrt {\frac {1}{2} \left (\sqrt {5}-1\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{x^4-x^2}}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 1.28, antiderivative size = 393, normalized size of antiderivative = 1.69, number of steps used = 18, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {2056, 6715, 6728, 240, 212, 206, 203, 377} \begin {gather*} \frac {\sqrt {x} \sqrt [4]{x^2-1} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt {x} \sqrt [4]{x^2-1} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}}-\frac {\sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \sqrt {x} \sqrt [4]{x^2-1} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}}+\frac {\sqrt {x} \sqrt [4]{x^2-1} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt {x} \sqrt [4]{x^2-1} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}}-\frac {\sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \sqrt {x} \sqrt [4]{x^2-1} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )}{\sqrt [4]{x^4-x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^4)/((-1 - x^2 + x^4)*(-x^2 + x^4)^(1/4)),x]

[Out]

(Sqrt[x]*(-1 + x^2)^(1/4)*ArcTan[Sqrt[x]/(-1 + x^2)^(1/4)])/(-x^2 + x^4)^(1/4) - (((3 + Sqrt[5])/2)^(1/4)*Sqrt
[x]*(-1 + x^2)^(1/4)*ArcTan[((2/(3 + Sqrt[5]))^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)])/(-x^2 + x^4)^(1/4) - (((3 - S
qrt[5])/2)^(1/4)*Sqrt[x]*(-1 + x^2)^(1/4)*ArcTan[(((3 + Sqrt[5])/2)^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)])/(-x^2 +
x^4)^(1/4) + (Sqrt[x]*(-1 + x^2)^(1/4)*ArcTanh[Sqrt[x]/(-1 + x^2)^(1/4)])/(-x^2 + x^4)^(1/4) - (((3 + Sqrt[5])
/2)^(1/4)*Sqrt[x]*(-1 + x^2)^(1/4)*ArcTanh[((2/(3 + Sqrt[5]))^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)])/(-x^2 + x^4)^(
1/4) - (((3 - Sqrt[5])/2)^(1/4)*Sqrt[x]*(-1 + x^2)^(1/4)*ArcTanh[(((3 + Sqrt[5])/2)^(1/4)*Sqrt[x])/(-1 + x^2)^
(1/4)])/(-x^2 + x^4)^(1/4)

Rule 203

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

Rule 206

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

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 240

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1+x^4}{\left (-1-x^2+x^4\right ) \sqrt [4]{-x^2+x^4}} \, dx &=\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \int \frac {1+x^4}{\sqrt {x} \sqrt [4]{-1+x^2} \left (-1-x^2+x^4\right )} \, dx}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1+x^8}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1}{\sqrt [4]{-1+x^4}}+\frac {2+x^4}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {2+x^4}{\sqrt [4]{-1+x^4} \left (-1-x^4+x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (2 \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {1+\sqrt {5}}{\sqrt [4]{-1+x^4} \left (-1-\sqrt {5}+2 x^4\right )}+\frac {1-\sqrt {5}}{\sqrt [4]{-1+x^4} \left (-1+\sqrt {5}+2 x^4\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (\sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (2 \left (1-\sqrt {5}\right ) \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^4} \left (-1+\sqrt {5}+2 x^4\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (2 \left (1+\sqrt {5}\right ) \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{-1+x^4} \left (-1-\sqrt {5}+2 x^4\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (2 \left (1-\sqrt {5}\right ) \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1+\sqrt {5}-\left (1+\sqrt {5}\right ) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (2 \left (1+\sqrt {5}\right ) \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-\sqrt {5}-\left (1-\sqrt {5}\right ) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}\\ &=\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\left (\left (1-\sqrt {5}\right ) \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-\sqrt {5}}-\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt {2} \sqrt [4]{-x^2+x^4}}+\frac {\left (\left (1-\sqrt {5}\right ) \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {3-\sqrt {5}}+\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt {2} \sqrt [4]{-x^2+x^4}}-\frac {\left (\left (1+\sqrt {5}\right ) \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {3+\sqrt {5}}-\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt {2} \sqrt [4]{-x^2+x^4}}-\frac {\left (\left (1+\sqrt {5}\right ) \sqrt {x} \sqrt [4]{-1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {3+\sqrt {5}}+\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt {2} \sqrt [4]{-x^2+x^4}}\\ &=\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \sqrt {x} \sqrt [4]{-1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}+\frac {\sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}-\frac {\sqrt [4]{\frac {1}{2} \left (3-\sqrt {5}\right )} \sqrt {x} \sqrt [4]{-1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt {x}}{\sqrt [4]{-1+x^2}}\right )}{\sqrt [4]{-x^2+x^4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.29, size = 268, normalized size = 1.15 \begin {gather*} \frac {\sqrt {x} \sqrt [4]{x^2-1} \left (2 \tan ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )-2^{3/4} \sqrt [4]{3+\sqrt {5}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )-2^{3/4} \sqrt [4]{3-\sqrt {5}} \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )+2 \tanh ^{-1}\left (\frac {\sqrt {x}}{\sqrt [4]{x^2-1}}\right )-2^{3/4} \sqrt [4]{3+\sqrt {5}} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {2}{3+\sqrt {5}}} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )-2^{3/4} \sqrt [4]{3-\sqrt {5}} \tanh ^{-1}\left (\frac {\sqrt [4]{\frac {1}{2} \left (3+\sqrt {5}\right )} \sqrt {x}}{\sqrt [4]{x^2-1}}\right )\right )}{2 \sqrt [4]{x^2 \left (x^2-1\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^4)/((-1 - x^2 + x^4)*(-x^2 + x^4)^(1/4)),x]

[Out]

(Sqrt[x]*(-1 + x^2)^(1/4)*(2*ArcTan[Sqrt[x]/(-1 + x^2)^(1/4)] - 2^(3/4)*(3 + Sqrt[5])^(1/4)*ArcTan[((2/(3 + Sq
rt[5]))^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)] - 2^(3/4)*(3 - Sqrt[5])^(1/4)*ArcTan[(((3 + Sqrt[5])/2)^(1/4)*Sqrt[x]
)/(-1 + x^2)^(1/4)] + 2*ArcTanh[Sqrt[x]/(-1 + x^2)^(1/4)] - 2^(3/4)*(3 + Sqrt[5])^(1/4)*ArcTanh[((2/(3 + Sqrt[
5]))^(1/4)*Sqrt[x])/(-1 + x^2)^(1/4)] - 2^(3/4)*(3 - Sqrt[5])^(1/4)*ArcTanh[(((3 + Sqrt[5])/2)^(1/4)*Sqrt[x])/
(-1 + x^2)^(1/4)]))/(2*(x^2*(-1 + x^2))^(1/4))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.74, size = 233, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right )-\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-x^2+x^4}}\right )-\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-x^2+x^4}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-x^2+x^4}}\right )-\sqrt {\frac {1}{2} \left (1+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {-\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-x^2+x^4}}\right )-\sqrt {\frac {1}{2} \left (-1+\sqrt {5}\right )} \tanh ^{-1}\left (\frac {\sqrt {\frac {1}{2}+\frac {\sqrt {5}}{2}} x}{\sqrt [4]{-x^2+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^4)/((-1 - x^2 + x^4)*(-x^2 + x^4)^(1/4)),x]

[Out]

ArcTan[x/(-x^2 + x^4)^(1/4)] - Sqrt[(1 + Sqrt[5])/2]*ArcTan[(Sqrt[-1/2 + Sqrt[5]/2]*x)/(-x^2 + x^4)^(1/4)] - S
qrt[(-1 + Sqrt[5])/2]*ArcTan[(Sqrt[1/2 + Sqrt[5]/2]*x)/(-x^2 + x^4)^(1/4)] + ArcTanh[x/(-x^2 + x^4)^(1/4)] - S
qrt[(1 + Sqrt[5])/2]*ArcTanh[(Sqrt[-1/2 + Sqrt[5]/2]*x)/(-x^2 + x^4)^(1/4)] - Sqrt[(-1 + Sqrt[5])/2]*ArcTanh[(
Sqrt[1/2 + Sqrt[5]/2]*x)/(-x^2 + x^4)^(1/4)]

________________________________________________________________________________________

fricas [B]  time = 74.55, size = 1300, normalized size = 5.58

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)/(x^4-x^2-1)/(x^4-x^2)^(1/4),x, algorithm="fricas")

[Out]

-1/2*sqrt(2)*sqrt(sqrt(5) + 1)*arctan(-1/163724*(sqrt(2)*(sqrt(5)*sqrt(2)*(224*x^5 - 310*x^3 + 43*x) - sqrt(2)
*(215*x^5 - 663*x^3 + 224*x) - sqrt(x^4 - x^2)*(sqrt(5)*sqrt(2)*(86*x^3 - 267*x) - sqrt(2)*(448*x^3 - 439*x)))
*sqrt(40157*sqrt(5) + 36899)*sqrt(sqrt(5) + 1) - 81862*((x^4 - x^2)^(3/4)*(sqrt(2)*(2*x^2 - 1) + sqrt(5)*sqrt(
2)) + (x^4 - x^2)^(1/4)*(sqrt(5)*sqrt(2)*(x^4 - x^2) - sqrt(2)*(x^4 - 3*x^2)))*sqrt(sqrt(5) + 1))/(x^5 - x^3 -
x)) + 1/2*sqrt(2)*sqrt(sqrt(5) - 1)*arctan(1/163724*(sqrt(2)*(sqrt(5)*sqrt(2)*(224*x^5 - 310*x^3 + 43*x) + sq
rt(2)*(215*x^5 - 663*x^3 + 224*x) + sqrt(x^4 - x^2)*(sqrt(5)*sqrt(2)*(86*x^3 - 267*x) + sqrt(2)*(448*x^3 - 439
*x)))*sqrt(40157*sqrt(5) - 36899)*sqrt(sqrt(5) - 1) + 81862*((x^4 - x^2)^(3/4)*(sqrt(2)*(2*x^2 - 1) - sqrt(5)*
sqrt(2)) + (x^4 - x^2)^(1/4)*(sqrt(5)*sqrt(2)*(x^4 - x^2) + sqrt(2)*(x^4 - 3*x^2)))*sqrt(sqrt(5) - 1))/(x^5 -
x^3 - x)) + 1/8*sqrt(2)*sqrt(sqrt(5) + 1)*log((4*(x^4 - x^2)^(3/4)*(448*x^2 + sqrt(5)*(86*x^2 + 181) - 9) + (s
qrt(5)*sqrt(2)*(9*x^5 - 371*x^3 + 181*x) - sqrt(2)*(905*x^5 - 923*x^3 + 9*x) - 2*sqrt(x^4 - x^2)*(sqrt(5)*sqrt
(2)*(181*x^3 - 95*x) - sqrt(2)*(9*x^3 - 457*x)))*sqrt(sqrt(5) + 1) - 4*(9*x^4 - 457*x^2 - sqrt(5)*(181*x^4 - 9
5*x^2))*(x^4 - x^2)^(1/4))/(x^5 - x^3 - x)) - 1/8*sqrt(2)*sqrt(sqrt(5) + 1)*log((4*(x^4 - x^2)^(3/4)*(448*x^2
+ sqrt(5)*(86*x^2 + 181) - 9) - (sqrt(5)*sqrt(2)*(9*x^5 - 371*x^3 + 181*x) - sqrt(2)*(905*x^5 - 923*x^3 + 9*x)
- 2*sqrt(x^4 - x^2)*(sqrt(5)*sqrt(2)*(181*x^3 - 95*x) - sqrt(2)*(9*x^3 - 457*x)))*sqrt(sqrt(5) + 1) - 4*(9*x^
4 - 457*x^2 - sqrt(5)*(181*x^4 - 95*x^2))*(x^4 - x^2)^(1/4))/(x^5 - x^3 - x)) - 1/8*sqrt(2)*sqrt(sqrt(5) - 1)*
log((4*(x^4 - x^2)^(3/4)*(448*x^2 - sqrt(5)*(86*x^2 + 181) - 9) + (sqrt(5)*sqrt(2)*(9*x^5 - 371*x^3 + 181*x) +
sqrt(2)*(905*x^5 - 923*x^3 + 9*x) + 2*sqrt(x^4 - x^2)*(sqrt(5)*sqrt(2)*(181*x^3 - 95*x) + sqrt(2)*(9*x^3 - 45
7*x)))*sqrt(sqrt(5) - 1) + 4*(9*x^4 - 457*x^2 + sqrt(5)*(181*x^4 - 95*x^2))*(x^4 - x^2)^(1/4))/(x^5 - x^3 - x)
) + 1/8*sqrt(2)*sqrt(sqrt(5) - 1)*log((4*(x^4 - x^2)^(3/4)*(448*x^2 - sqrt(5)*(86*x^2 + 181) - 9) - (sqrt(5)*s
qrt(2)*(9*x^5 - 371*x^3 + 181*x) + sqrt(2)*(905*x^5 - 923*x^3 + 9*x) + 2*sqrt(x^4 - x^2)*(sqrt(5)*sqrt(2)*(181
*x^3 - 95*x) + sqrt(2)*(9*x^3 - 457*x)))*sqrt(sqrt(5) - 1) + 4*(9*x^4 - 457*x^2 + sqrt(5)*(181*x^4 - 95*x^2))*
(x^4 - x^2)^(1/4))/(x^5 - x^3 - x)) - 1/2*arctan(2*((x^4 - x^2)^(1/4)*x^2 + (x^4 - x^2)^(3/4))/x) + 1/2*log((2
*x^3 + 2*(x^4 - x^2)^(1/4)*x^2 + 2*sqrt(x^4 - x^2)*x - x + 2*(x^4 - x^2)^(3/4))/x)

________________________________________________________________________________________

giac [A]  time = 0.48, size = 232, normalized size = 1.00 \begin {gather*} -\frac {1}{2} \, \sqrt {2 \, \sqrt {5} - 2} \arctan \left (\frac {{\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}}{\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}}}\right ) - \frac {1}{2} \, \sqrt {2 \, \sqrt {5} + 2} \arctan \left (\frac {{\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}}{\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}}}\right ) + \frac {1}{4} \, \sqrt {2 \, \sqrt {5} - 2} \log \left (\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} \, \sqrt {2 \, \sqrt {5} - 2} \log \left (\sqrt {\frac {1}{2} \, \sqrt {5} + \frac {1}{2}} - {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{4} \, \sqrt {2 \, \sqrt {5} + 2} \log \left (\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{4} \, \sqrt {2 \, \sqrt {5} + 2} \log \left ({\left | -\sqrt {\frac {1}{2} \, \sqrt {5} - \frac {1}{2}} + {\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} \right |}\right ) + \arctan \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{2} \, \log \left ({\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{2} \, \log \left (-{\left (-\frac {1}{x^{2}} + 1\right )}^{\frac {1}{4}} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)/(x^4-x^2-1)/(x^4-x^2)^(1/4),x, algorithm="giac")

[Out]

-1/2*sqrt(2*sqrt(5) - 2)*arctan((-1/x^2 + 1)^(1/4)/sqrt(1/2*sqrt(5) + 1/2)) - 1/2*sqrt(2*sqrt(5) + 2)*arctan((
-1/x^2 + 1)^(1/4)/sqrt(1/2*sqrt(5) - 1/2)) + 1/4*sqrt(2*sqrt(5) - 2)*log(sqrt(1/2*sqrt(5) + 1/2) + (-1/x^2 + 1
)^(1/4)) - 1/4*sqrt(2*sqrt(5) - 2)*log(sqrt(1/2*sqrt(5) + 1/2) - (-1/x^2 + 1)^(1/4)) + 1/4*sqrt(2*sqrt(5) + 2)
*log(sqrt(1/2*sqrt(5) - 1/2) + (-1/x^2 + 1)^(1/4)) - 1/4*sqrt(2*sqrt(5) + 2)*log(abs(-sqrt(1/2*sqrt(5) - 1/2)
+ (-1/x^2 + 1)^(1/4))) + arctan((-1/x^2 + 1)^(1/4)) - 1/2*log((-1/x^2 + 1)^(1/4) + 1) + 1/2*log(-(-1/x^2 + 1)^
(1/4) + 1)

________________________________________________________________________________________

maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {x^{4}+1}{\left (x^{4}-x^{2}-1\right ) \left (x^{4}-x^{2}\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+1)/(x^4-x^2-1)/(x^4-x^2)^(1/4),x)

[Out]

int((x^4+1)/(x^4-x^2-1)/(x^4-x^2)^(1/4),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} + 1}{{\left (x^{4} - x^{2}\right )}^{\frac {1}{4}} {\left (x^{4} - x^{2} - 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+1)/(x^4-x^2-1)/(x^4-x^2)^(1/4),x, algorithm="maxima")

[Out]

integrate((x^4 + 1)/((x^4 - x^2)^(1/4)*(x^4 - x^2 - 1)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {x^4+1}{{\left (x^4-x^2\right )}^{1/4}\,\left (-x^4+x^2+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^4 + 1)/((x^4 - x^2)^(1/4)*(x^2 - x^4 + 1)),x)

[Out]

int(-(x^4 + 1)/((x^4 - x^2)^(1/4)*(x^2 - x^4 + 1)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} + 1}{\sqrt [4]{x^{2} \left (x - 1\right ) \left (x + 1\right )} \left (x^{4} - x^{2} - 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+1)/(x**4-x**2-1)/(x**4-x**2)**(1/4),x)

[Out]

Integral((x**4 + 1)/((x**2*(x - 1)*(x + 1))**(1/4)*(x**4 - x**2 - 1)), x)

________________________________________________________________________________________