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

Optimal. Leaf size=137 \[ \frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{6 \sqrt [4]{2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{6 \sqrt [4]{2}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{x^4+1}}{\sqrt {x^4+1}-x^2}\right )}{6 \sqrt {2}}+\frac {5 \tanh ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{x^4+1}}{\sqrt {x^4+1}+x^2}\right )}{6 \sqrt {2}} \]

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Rubi [A]  time = 0.27, antiderivative size = 193, normalized size of antiderivative = 1.41, number of steps used = 16, number of rules used = 11, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {6728, 377, 212, 206, 203, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{6 \sqrt [4]{2}}-\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}\right )}{6 \sqrt {2}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}+1\right )}{6 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )}{6 \sqrt [4]{2}}-\frac {5 \log \left (-\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}+\frac {x^2}{\sqrt {x^4+1}}+1\right )}{12 \sqrt {2}}+\frac {5 \log \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}+\frac {x^2}{\sqrt {x^4+1}}+1\right )}{12 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(2^(1/4)*x)/(1 + x^4)^(1/4)]/(6*2^(1/4)) - (5*ArcTan[1 - (Sqrt[2]*x)/(1 + x^4)^(1/4)])/(6*Sqrt[2]) + (5
*ArcTan[1 + (Sqrt[2]*x)/(1 + x^4)^(1/4)])/(6*Sqrt[2]) + ArcTanh[(2^(1/4)*x)/(1 + x^4)^(1/4)]/(6*2^(1/4)) - (5*
Log[1 + x^2/Sqrt[1 + x^4] - (Sqrt[2]*x)/(1 + x^4)^(1/4)])/(12*Sqrt[2]) + (5*Log[1 + x^2/Sqrt[1 + x^4] + (Sqrt[
2]*x)/(1 + x^4)^(1/4)])/(12*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 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 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 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 {-2+x^4}{\sqrt [4]{1+x^4} \left (-1-x^4+2 x^8\right )} \, dx &=\int \left (-\frac {4}{3 \sqrt [4]{1+x^4} \left (-4+4 x^4\right )}+\frac {10}{3 \sqrt [4]{1+x^4} \left (2+4 x^4\right )}\right ) \, dx\\ &=-\left (\frac {4}{3} \int \frac {1}{\sqrt [4]{1+x^4} \left (-4+4 x^4\right )} \, dx\right )+\frac {10}{3} \int \frac {1}{\sqrt [4]{1+x^4} \left (2+4 x^4\right )} \, dx\\ &=-\left (\frac {4}{3} \operatorname {Subst}\left (\int \frac {1}{-4+8 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\right )+\frac {10}{3} \operatorname {Subst}\left (\int \frac {1}{2+2 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {5}{3} \operatorname {Subst}\left (\int \frac {1-x^2}{2+2 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {5}{3} \operatorname {Subst}\left (\int \frac {1+x^2}{2+2 x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}+\frac {5}{12} \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {5}{12} \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {5 \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}}-\frac {5 \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}-\frac {5 \log \left (1+\frac {x^2}{\sqrt {1+x^4}}-\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}}+\frac {5 \log \left (1+\frac {x^2}{\sqrt {1+x^4}}+\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}-\frac {5 \tan ^{-1}\left (1-\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}+\frac {5 \tan ^{-1}\left (1+\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}-\frac {5 \log \left (1+\frac {x^2}{\sqrt {1+x^4}}-\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}}+\frac {5 \log \left (1+\frac {x^2}{\sqrt {1+x^4}}+\frac {\sqrt {2} x}{\sqrt [4]{1+x^4}}\right )}{12 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 171, normalized size = 1.25 \begin {gather*} \frac {2 \sqrt [4]{2} \tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )+2 \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+1}}\right )-5 \left (2 \tan ^{-1}\left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}\right )-2 \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}+1\right )+\log \left (-\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}+\frac {x^2}{\sqrt {x^4+1}}+1\right )-\log \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+1}}+\frac {x^2}{\sqrt {x^4+1}}+1\right )\right )}{12 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*2^(1/4)*ArcTan[(2^(1/4)*x)/(1 + x^4)^(1/4)] + 2*2^(1/4)*ArcTanh[(2^(1/4)*x)/(1 + x^4)^(1/4)] - 5*(2*ArcTan[
1 - (Sqrt[2]*x)/(1 + x^4)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*x)/(1 + x^4)^(1/4)] + Log[1 + x^2/Sqrt[1 + x^4] - (Sq
rt[2]*x)/(1 + x^4)^(1/4)] - Log[1 + x^2/Sqrt[1 + x^4] + (Sqrt[2]*x)/(1 + x^4)^(1/4)]))/(12*Sqrt[2])

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IntegrateAlgebraic [A]  time = 0.40, size = 137, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}+\frac {5 \tan ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{1+x^4}}{-x^2+\sqrt {1+x^4}}\right )}{6 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{1+x^4}}\right )}{6 \sqrt [4]{2}}+\frac {5 \tanh ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{1+x^4}}{x^2+\sqrt {1+x^4}}\right )}{6 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(2^(1/4)*x)/(1 + x^4)^(1/4)]/(6*2^(1/4)) + (5*ArcTan[(Sqrt[2]*x*(1 + x^4)^(1/4))/(-x^2 + Sqrt[1 + x^4])
])/(6*Sqrt[2]) + ArcTanh[(2^(1/4)*x)/(1 + x^4)^(1/4)]/(6*2^(1/4)) + (5*ArcTanh[(Sqrt[2]*x*(1 + x^4)^(1/4))/(x^
2 + Sqrt[1 + x^4])])/(6*Sqrt[2])

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fricas [B]  time = 10.38, size = 610, normalized size = 4.45 \begin {gather*} -\frac {1}{12} \cdot 2^{\frac {3}{4}} \arctan \left (\frac {4 \cdot 2^{\frac {3}{4}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 2^{\frac {3}{4}} {\left (2 \cdot 2^{\frac {3}{4}} \sqrt {x^{4} + 1} x^{2} + 2^{\frac {1}{4}} {\left (3 \, x^{4} + 1\right )}\right )}}{2 \, {\left (x^{4} - 1\right )}}\right ) + \frac {1}{48} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) - \frac {1}{48} \cdot 2^{\frac {3}{4}} \log \left (\frac {4 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 2^{\frac {3}{4}} {\left (3 \, x^{4} + 1\right )} + 4 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} - 1}\right ) + \frac {5}{12} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x^{2} - \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {5}{4}} - {\left (2 \, x^{5} - \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x^{2} - \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {5}{4}} + 2 \, x\right )} \sqrt {\frac {2 \, x^{4} + 2 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{4} + 1} x^{2} + 2 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 1}{2 \, x^{4} + 1}}}{2 \, {\left (x^{5} + x\right )}}\right ) + \frac {5}{12} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x^{2} - \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {5}{4}} + {\left (2 \, x^{5} + \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x^{2} + \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {5}{4}} + 2 \, x\right )} \sqrt {\frac {2 \, x^{4} - 2 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{4} + 1} x^{2} - 2 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 1}{2 \, x^{4} + 1}}}{2 \, {\left (x^{5} + x\right )}}\right ) + \frac {5}{48} \, \sqrt {2} \log \left (\frac {2 \, x^{4} + 2 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{4} + 1} x^{2} + 2 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 1}{2 \, x^{4} + 1}\right ) - \frac {5}{48} \, \sqrt {2} \log \left (\frac {2 \, x^{4} - 2 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \, \sqrt {x^{4} + 1} x^{2} - 2 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x + 1}{2 \, x^{4} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*2^(3/4)*arctan(1/2*(4*2^(3/4)*(x^4 + 1)^(1/4)*x^3 + 4*2^(1/4)*(x^4 + 1)^(3/4)*x + 2^(3/4)*(2*2^(3/4)*sqr
t(x^4 + 1)*x^2 + 2^(1/4)*(3*x^4 + 1)))/(x^4 - 1)) + 1/48*2^(3/4)*log((4*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 4*2^(1/4
)*sqrt(x^4 + 1)*x^2 + 2^(3/4)*(3*x^4 + 1) + 4*(x^4 + 1)^(3/4)*x)/(x^4 - 1)) - 1/48*2^(3/4)*log((4*sqrt(2)*(x^4
+ 1)^(1/4)*x^3 - 4*2^(1/4)*sqrt(x^4 + 1)*x^2 - 2^(3/4)*(3*x^4 + 1) + 4*(x^4 + 1)^(3/4)*x)/(x^4 - 1)) + 5/12*s
qrt(2)*arctan(1/2*(sqrt(2)*(x^4 + 1)^(3/4)*x^2 - sqrt(2)*(x^4 + 1)^(5/4) - (2*x^5 - sqrt(2)*(x^4 + 1)^(3/4)*x^
2 - sqrt(2)*(x^4 + 1)^(5/4) + 2*x)*sqrt((2*x^4 + 2*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 4*sqrt(x^4 + 1)*x^2 + 2*sqrt(
2)*(x^4 + 1)^(3/4)*x + 1)/(2*x^4 + 1)))/(x^5 + x)) + 5/12*sqrt(2)*arctan(1/2*(sqrt(2)*(x^4 + 1)^(3/4)*x^2 - sq
rt(2)*(x^4 + 1)^(5/4) + (2*x^5 + sqrt(2)*(x^4 + 1)^(3/4)*x^2 + sqrt(2)*(x^4 + 1)^(5/4) + 2*x)*sqrt((2*x^4 - 2*
sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 4*sqrt(x^4 + 1)*x^2 - 2*sqrt(2)*(x^4 + 1)^(3/4)*x + 1)/(2*x^4 + 1)))/(x^5 + x))
+ 5/48*sqrt(2)*log((2*x^4 + 2*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 4*sqrt(x^4 + 1)*x^2 + 2*sqrt(2)*(x^4 + 1)^(3/4)*x
+ 1)/(2*x^4 + 1)) - 5/48*sqrt(2)*log((2*x^4 - 2*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 4*sqrt(x^4 + 1)*x^2 - 2*sqrt(2)*
(x^4 + 1)^(3/4)*x + 1)/(2*x^4 + 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} - 2}{{\left (2 \, x^{8} - 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((x^4-2)/(x^4+1)^(1/4)/(2*x^8-x^4-1),x, algorithm="giac")

[Out]

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

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maple [C]  time = 5.32, size = 646, normalized size = 4.72

method	result	size
trager	\(\frac {5 \RootOf \left (\textit {\_Z}^{4}+1\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{3}-2 \left (x^{4}+1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{4}+1\right )}{2 x^{4}+1}\right )}{12}-\frac {5 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (-\frac {-2 \left (x^{4}+1\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{3}-2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{4}+1\right ) x^{2}-2 \left (x^{4}+1\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{4}+1\right )^{3}}{2 x^{4}+1}\right )}{12}-\frac {\RootOf \left (\textit {\_Z}^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )\right ) \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} x^{2}-4 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \left (x^{4}+1\right )^{\frac {1}{4}} x^{3}-2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )\right ) \RootOf \left (\textit {\_Z}^{4}+1\right ) x^{2}-3 \RootOf \left (\textit {\_Z}^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )\right ) x^{4}+4 \RootOf \left (\textit {\_Z}^{4}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}} x^{3}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{24}-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{2} \RootOf \left (\textit {\_Z}^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )\right ) \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} x^{2}-3 \RootOf \left (\textit {\_Z}^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )\right ) \RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{4}+4 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \left (x^{4}+1\right )^{\frac {1}{4}} x^{3}+2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )\right ) \RootOf \left (\textit {\_Z}^{4}+1\right ) x^{2}-4 \RootOf \left (\textit {\_Z}^{4}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}} x^{3}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{4}+1\right )^{2} \RootOf \left (\textit {\_Z}^{2}+2 \RootOf \left (\textit {\_Z}^{4}+1\right )^{3}-2 \RootOf \left (\textit {\_Z}^{4}+1\right )\right )}{\left (-1+x \right ) \left (1+x \right ) \left (x^{2}+1\right )}\right )}{24}\)	\(646\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/12*RootOf(_Z^4+1)*ln(-(2*(x^4+1)^(1/2)*RootOf(_Z^4+1)^3*x^2+2*(x^4+1)^(1/4)*RootOf(_Z^4+1)^2*x^3-2*(x^4+1)^(
3/4)*x-RootOf(_Z^4+1))/(2*x^4+1))-5/12*RootOf(_Z^4+1)^3*ln(-(-2*(x^4+1)^(1/4)*RootOf(_Z^4+1)^2*x^3-2*(x^4+1)^(
1/2)*RootOf(_Z^4+1)*x^2-2*(x^4+1)^(3/4)*x+RootOf(_Z^4+1)^3)/(2*x^4+1))-1/24*RootOf(_Z^2+2*RootOf(_Z^4+1)^3-2*R
ootOf(_Z^4+1))*ln(-(2*(x^4+1)^(1/2)*RootOf(_Z^2+2*RootOf(_Z^4+1)^3-2*RootOf(_Z^4+1))*RootOf(_Z^4+1)^3*x^2-4*Ro
otOf(_Z^4+1)^3*(x^4+1)^(1/4)*x^3-2*(x^4+1)^(1/2)*RootOf(_Z^2+2*RootOf(_Z^4+1)^3-2*RootOf(_Z^4+1))*RootOf(_Z^4+
1)*x^2-3*RootOf(_Z^2+2*RootOf(_Z^4+1)^3-2*RootOf(_Z^4+1))*x^4+4*RootOf(_Z^4+1)*(x^4+1)^(1/4)*x^3+4*(x^4+1)^(3/
4)*x-RootOf(_Z^2+2*RootOf(_Z^4+1)^3-2*RootOf(_Z^4+1)))/(-1+x)/(1+x)/(x^2+1))-1/24*RootOf(_Z^4+1)^2*RootOf(_Z^2
+2*RootOf(_Z^4+1)^3-2*RootOf(_Z^4+1))*ln(-(2*(x^4+1)^(1/2)*RootOf(_Z^2+2*RootOf(_Z^4+1)^3-2*RootOf(_Z^4+1))*Ro
otOf(_Z^4+1)^3*x^2-3*RootOf(_Z^2+2*RootOf(_Z^4+1)^3-2*RootOf(_Z^4+1))*RootOf(_Z^4+1)^2*x^4+4*RootOf(_Z^4+1)^3*
(x^4+1)^(1/4)*x^3+2*(x^4+1)^(1/2)*RootOf(_Z^2+2*RootOf(_Z^4+1)^3-2*RootOf(_Z^4+1))*RootOf(_Z^4+1)*x^2-4*RootOf
(_Z^4+1)*(x^4+1)^(1/4)*x^3+4*(x^4+1)^(3/4)*x-RootOf(_Z^4+1)^2*RootOf(_Z^2+2*RootOf(_Z^4+1)^3-2*RootOf(_Z^4+1))
)/(-1+x)/(1+x)/(x^2+1))

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

[Out]

integrate((x^4 - 2)/((2*x^8 - 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 {x^4-2}{{\left (x^4+1\right )}^{1/4}\,\left (-2\,x^8+x^4+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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