3.24.1 \(\int \frac {1-x^4+2 x^8}{\sqrt [4]{1+x^4} (-1-2 x^4+x^8)} \, dx\)

Optimal. Leaf size=176 \[ \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {3 \tan ^{-1}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [8]{2}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {3 \tanh ^{-1}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [8]{2}}+\frac {3 \tan ^{-1}\left (\frac {2^{5/8} x \sqrt [4]{x^4+1}}{\sqrt [4]{2} x^2-\sqrt {x^4+1}}\right )}{4\ 2^{5/8}}-\frac {3 \tanh ^{-1}\left (\frac {2\ 2^{3/8} x \sqrt [4]{x^4+1}}{2^{3/4} \sqrt {x^4+1}+2 x^2}\right )}{4\ 2^{5/8}} \]

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Rubi [A]  time = 0.49, antiderivative size = 306, normalized size of antiderivative = 1.74, number of steps used = 31, number of rules used = 14, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {6728, 240, 212, 206, 203, 1428, 408, 377, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {3 \tan ^{-1}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [8]{2}}-\frac {3 \left (1-\sqrt {2}\right ) \tan ^{-1}\left (1-\frac {2^{5/8} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [8]{2} \left (2-\sqrt {2}\right )}+\frac {3 \left (1-\sqrt {2}\right ) \tan ^{-1}\left (\frac {2^{5/8} x}{\sqrt [4]{x^4+1}}+1\right )}{4 \sqrt [8]{2} \left (2-\sqrt {2}\right )}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {3 \tanh ^{-1}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{x^4+1}}\right )}{4 \sqrt [8]{2}}-\frac {3 \left (1-\sqrt {2}\right ) \log \left (-\frac {2 x}{\sqrt [4]{x^4+1}}+\frac {2^{5/8} x^2}{\sqrt {x^4+1}}+2^{3/8}\right )}{8 \sqrt [8]{2} \left (2-\sqrt {2}\right )}+\frac {3 \left (1-\sqrt {2}\right ) \log \left (\frac {2^{5/8} x}{\sqrt [4]{x^4+1}}+\frac {\sqrt [4]{2} x^2}{\sqrt {x^4+1}}+1\right )}{8 \sqrt [8]{2} \left (2-\sqrt {2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x^4 + 2*x^8)/((1 + x^4)^(1/4)*(-1 - 2*x^4 + x^8)),x]

[Out]

ArcTan[x/(1 + x^4)^(1/4)] - (3*ArcTan[(2^(1/8)*x)/(1 + x^4)^(1/4)])/(4*2^(1/8)) - (3*(1 - Sqrt[2])*ArcTan[1 -
(2^(5/8)*x)/(1 + x^4)^(1/4)])/(4*2^(1/8)*(2 - Sqrt[2])) + (3*(1 - Sqrt[2])*ArcTan[1 + (2^(5/8)*x)/(1 + x^4)^(1
/4)])/(4*2^(1/8)*(2 - Sqrt[2])) + ArcTanh[x/(1 + x^4)^(1/4)] - (3*ArcTanh[(2^(1/8)*x)/(1 + x^4)^(1/4)])/(4*2^(
1/8)) - (3*(1 - Sqrt[2])*Log[2^(3/8) + (2^(5/8)*x^2)/Sqrt[1 + x^4] - (2*x)/(1 + x^4)^(1/4)])/(8*2^(1/8)*(2 - S
qrt[2])) + (3*(1 - Sqrt[2])*Log[1 + (2^(1/4)*x^2)/Sqrt[1 + x^4] + (2^(5/8)*x)/(1 + x^4)^(1/4)])/(8*2^(1/8)*(2
- Sqrt[2]))

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 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 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 211

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

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 408

Int[((a_) + (b_.)*(x_)^4)^(p_)/((c_) + (d_.)*(x_)^4), x_Symbol] :> Dist[b/d, Int[(a + b*x^4)^(p - 1), x], x] -
Dist[(b*c - a*d)/d, Int[(a + b*x^4)^(p - 1)/(c + d*x^4), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0
] && (EqQ[p, 3/4] || EqQ[p, 5/4])

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
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 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

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

Rule 1428

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

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

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Mathematica [F]  time = 0.16, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1-x^4+2 x^8}{\sqrt [4]{1+x^4} \left (-1-2 x^4+x^8\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(1 - x^4 + 2*x^8)/((1 + x^4)^(1/4)*(-1 - 2*x^4 + x^8)),x]

[Out]

Integrate[(1 - x^4 + 2*x^8)/((1 + x^4)^(1/4)*(-1 - 2*x^4 + x^8)), x]

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IntegrateAlgebraic [A]  time = 0.65, size = 176, normalized size = 1.00 \begin {gather*} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3 \tan ^{-1}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [8]{2}}+\frac {3 \tan ^{-1}\left (\frac {2^{5/8} x \sqrt [4]{1+x^4}}{\sqrt [4]{2} x^2-\sqrt {1+x^4}}\right )}{4\ 2^{5/8}}+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {3 \tanh ^{-1}\left (\frac {\sqrt [8]{2} x}{\sqrt [4]{1+x^4}}\right )}{4 \sqrt [8]{2}}-\frac {3 \tanh ^{-1}\left (\frac {2\ 2^{3/8} x \sqrt [4]{1+x^4}}{2 x^2+2^{3/4} \sqrt {1+x^4}}\right )}{4\ 2^{5/8}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 - x^4 + 2*x^8)/((1 + x^4)^(1/4)*(-1 - 2*x^4 + x^8)),x]

[Out]

ArcTan[x/(1 + x^4)^(1/4)] - (3*ArcTan[(2^(1/8)*x)/(1 + x^4)^(1/4)])/(4*2^(1/8)) + (3*ArcTan[(2^(5/8)*x*(1 + x^
4)^(1/4))/(2^(1/4)*x^2 - Sqrt[1 + x^4])])/(4*2^(5/8)) + ArcTanh[x/(1 + x^4)^(1/4)] - (3*ArcTanh[(2^(1/8)*x)/(1
+ x^4)^(1/4)])/(4*2^(1/8)) - (3*ArcTanh[(2*2^(3/8)*x*(1 + x^4)^(1/4))/(2*x^2 + 2^(3/4)*Sqrt[1 + x^4])])/(4*2^
(5/8))

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fricas [B]  time = 0.49, size = 353, normalized size = 2.01 \begin {gather*} -\frac {3}{4} \cdot 2^{\frac {7}{8}} \arctan \left (\frac {2^{\frac {7}{8}} x \sqrt {\frac {2^{\frac {1}{4}} x^{2} + \sqrt {x^{4} + 1}}{x^{2}}} - 2^{\frac {7}{8}} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{2 \, x}\right ) - \frac {3}{16} \cdot 2^{\frac {7}{8}} \log \left (\frac {2^{\frac {1}{8}} x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {3}{16} \cdot 2^{\frac {7}{8}} \log \left (-\frac {2^{\frac {1}{8}} x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{4} \cdot 2^{\frac {3}{8}} \arctan \left (\frac {2^{\frac {3}{8}} x \sqrt {\frac {2^{\frac {1}{4}} x^{2} + 2^{\frac {5}{8}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x + \sqrt {x^{4} + 1}}{x^{2}}} - x - 2^{\frac {3}{8}} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{4} \cdot 2^{\frac {3}{8}} \arctan \left (\frac {2^{\frac {3}{8}} x \sqrt {\frac {2^{\frac {1}{4}} x^{2} - 2^{\frac {5}{8}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x + \sqrt {x^{4} + 1}}{x^{2}}} + x - 2^{\frac {3}{8}} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {3}{16} \cdot 2^{\frac {3}{8}} \log \left (\frac {4 \, {\left (2^{\frac {1}{4}} x^{2} + 2^{\frac {5}{8}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x + \sqrt {x^{4} + 1}\right )}}{x^{2}}\right ) + \frac {3}{16} \cdot 2^{\frac {3}{8}} \log \left (\frac {4 \, {\left (2^{\frac {1}{4}} x^{2} - 2^{\frac {5}{8}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x + \sqrt {x^{4} + 1}\right )}}{x^{2}}\right ) - \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{2} \, \log \left (\frac {x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{2} \, \log \left (-\frac {x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8-x^4+1)/(x^4+1)^(1/4)/(x^8-2*x^4-1),x, algorithm="fricas")

[Out]

-3/4*2^(7/8)*arctan(1/2*(2^(7/8)*x*sqrt((2^(1/4)*x^2 + sqrt(x^4 + 1))/x^2) - 2^(7/8)*(x^4 + 1)^(1/4))/x) - 3/1
6*2^(7/8)*log((2^(1/8)*x + (x^4 + 1)^(1/4))/x) + 3/16*2^(7/8)*log(-(2^(1/8)*x - (x^4 + 1)^(1/4))/x) - 3/4*2^(3
/8)*arctan((2^(3/8)*x*sqrt((2^(1/4)*x^2 + 2^(5/8)*(x^4 + 1)^(1/4)*x + sqrt(x^4 + 1))/x^2) - x - 2^(3/8)*(x^4 +
1)^(1/4))/x) - 3/4*2^(3/8)*arctan((2^(3/8)*x*sqrt((2^(1/4)*x^2 - 2^(5/8)*(x^4 + 1)^(1/4)*x + sqrt(x^4 + 1))/x
^2) + x - 2^(3/8)*(x^4 + 1)^(1/4))/x) - 3/16*2^(3/8)*log(4*(2^(1/4)*x^2 + 2^(5/8)*(x^4 + 1)^(1/4)*x + sqrt(x^4
+ 1))/x^2) + 3/16*2^(3/8)*log(4*(2^(1/4)*x^2 - 2^(5/8)*(x^4 + 1)^(1/4)*x + sqrt(x^4 + 1))/x^2) - arctan((x^4
+ 1)^(1/4)/x) + 1/2*log((x + (x^4 + 1)^(1/4))/x) - 1/2*log(-(x - (x^4 + 1)^(1/4))/x)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8-x^4+1)/(x^4+1)^(1/4)/(x^8-2*x^4-1),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to convert to real 1/4 Error: Bad Argument ValueUnable to convert to real 1/4 Error: Bad Argument Value

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {2 x^{8}-x^{4}+1}{\left (x^{4}+1\right )^{\frac {1}{4}} \left (x^{8}-2 x^{4}-1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^8-x^4+1)/(x^4+1)^(1/4)/(x^8-2*x^4-1),x)

[Out]

int((2*x^8-x^4+1)/(x^4+1)^(1/4)/(x^8-2*x^4-1),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, x^{8} - x^{4} + 1}{{\left (x^{8} - 2 \, x^{4} - 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^8-x^4+1)/(x^4+1)^(1/4)/(x^8-2*x^4-1),x, algorithm="maxima")

[Out]

integrate((2*x^8 - x^4 + 1)/((x^8 - 2*x^4 - 1)*(x^4 + 1)^(1/4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {2\,x^8-x^4+1}{{\left (x^4+1\right )}^{1/4}\,\left (-x^8+2\,x^4+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x^8 - x^4 + 1)/((x^4 + 1)^(1/4)*(2*x^4 - x^8 + 1)),x)

[Out]

int(-(2*x^8 - x^4 + 1)/((x^4 + 1)^(1/4)*(2*x^4 - x^8 + 1)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**8-x**4+1)/(x**4+1)**(1/4)/(x**8-2*x**4-1),x)

[Out]

Timed out

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