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

Optimal. Leaf size=139 \[ \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {\tan ^{-1}\left (\frac {\left (\sqrt {2} x-\sqrt {2}\right ) \sqrt [4]{x^4+1}}{\sqrt {x^4+1}-x^2+2 x-1}\right )}{\sqrt {2}}-\frac {\tanh ^{-1}\left (\frac {\left (\sqrt {2} x-\sqrt {2}\right ) \sqrt [4]{x^4+1}}{\sqrt {x^4+1}+x^2-2 x+1}\right )}{\sqrt {2}} \]

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Rubi [C]  time = 1.85, antiderivative size = 851, normalized size of antiderivative = 6.12, number of steps used = 53, number of rules used = 21, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.778, Rules used = {1593, 6728, 240, 212, 206, 203, 2153, 1240, 377, 208, 205, 510, 1248, 746, 399, 490, 1213, 537, 444, 63, 298} \begin {gather*} -\frac {1}{6} \left (1+i \sqrt {3}\right ) F_1\left (\frac {3}{4};\frac {1}{4},1;\frac {7}{4};-x^4,-\frac {2 x^4}{1-i \sqrt {3}}\right ) x^3-\frac {1}{6} \left (1-i \sqrt {3}\right ) F_1\left (\frac {3}{4};\frac {1}{4},1;\frac {7}{4};-x^4,-\frac {2 x^4}{1+i \sqrt {3}}\right ) x^3+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {1}{4} \left (1+i \sqrt {3}\right ) \sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \tan ^{-1}\left (\frac {x}{\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \sqrt [4]{x^4+1}}\right )-\frac {1}{2} \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} x}{\sqrt [4]{x^4+1}}\right )-\frac {1}{2} \left (\frac {1}{2} \left (1-i \sqrt {3}\right )\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x^4+1}}{\sqrt [4]{1-i \sqrt {3}}}\right )-\frac {1}{2} \left (\frac {1}{2} \left (1+i \sqrt {3}\right )\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x^4+1}}{\sqrt [4]{1+i \sqrt {3}}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {1}{4} \left (1+i \sqrt {3}\right ) \sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \sqrt [4]{x^4+1}}\right )-\frac {1}{2} \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} x}{\sqrt [4]{x^4+1}}\right )+\frac {1}{2} \left (\frac {1}{2} \left (1-i \sqrt {3}\right )\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x^4+1}}{\sqrt [4]{1-i \sqrt {3}}}\right )+\frac {1}{2} \left (\frac {1}{2} \left (1+i \sqrt {3}\right )\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{x^4+1}}{\sqrt [4]{1+i \sqrt {3}}}\right )-\frac {i \sqrt {-x^4} \Pi \left (\frac {1}{2} \left (-i-\sqrt {3}\right );\left .\sin ^{-1}\left (\sqrt [4]{x^4+1}\right )\right |-1\right )}{2 x^2}+\frac {i \sqrt {-x^4} \Pi \left (\frac {1}{2} \left (i-\sqrt {3}\right );\left .\sin ^{-1}\left (\sqrt [4]{x^4+1}\right )\right |-1\right )}{2 x^2}+\frac {i \sqrt {-x^4} \Pi \left (\frac {1}{\sqrt {\frac {1}{2} \left (1-i \sqrt {3}\right )}};\left .\sin ^{-1}\left (\sqrt [4]{x^4+1}\right )\right |-1\right )}{2 x^2}-\frac {i \sqrt {-x^4} \Pi \left (\frac {1}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}};\left .\sin ^{-1}\left (\sqrt [4]{x^4+1}\right )\right |-1\right )}{2 x^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/6*((1 + I*Sqrt[3])*x^3*AppellF1[3/4, 1/4, 1, 7/4, -x^4, (-2*x^4)/(1 - I*Sqrt[3])]) - ((1 - I*Sqrt[3])*x^3*A
ppellF1[3/4, 1/4, 1, 7/4, -x^4, (-2*x^4)/(1 + I*Sqrt[3])])/6 + ArcTan[x/(1 + x^4)^(1/4)]/2 - ((1 + I*Sqrt[3])*
(-((I - Sqrt[3])/(I + Sqrt[3])))^(1/4)*ArcTan[x/((-((I - Sqrt[3])/(I + Sqrt[3])))^(1/4)*(1 + x^4)^(1/4))])/4 -
((-((I - Sqrt[3])/(I + Sqrt[3])))^(3/4)*ArcTan[((-((I - Sqrt[3])/(I + Sqrt[3])))^(1/4)*x)/(1 + x^4)^(1/4)])/2
- (((1 - I*Sqrt[3])/2)^(3/4)*ArcTan[(2^(1/4)*(1 + x^4)^(1/4))/(1 - I*Sqrt[3])^(1/4)])/2 - (((1 + I*Sqrt[3])/2
)^(3/4)*ArcTan[(2^(1/4)*(1 + x^4)^(1/4))/(1 + I*Sqrt[3])^(1/4)])/2 + ArcTanh[x/(1 + x^4)^(1/4)]/2 - ((1 + I*Sq
rt[3])*(-((I - Sqrt[3])/(I + Sqrt[3])))^(1/4)*ArcTanh[x/((-((I - Sqrt[3])/(I + Sqrt[3])))^(1/4)*(1 + x^4)^(1/4
))])/4 - ((-((I - Sqrt[3])/(I + Sqrt[3])))^(3/4)*ArcTanh[((-((I - Sqrt[3])/(I + Sqrt[3])))^(1/4)*x)/(1 + x^4)^
(1/4)])/2 + (((1 - I*Sqrt[3])/2)^(3/4)*ArcTanh[(2^(1/4)*(1 + x^4)^(1/4))/(1 - I*Sqrt[3])^(1/4)])/2 + (((1 + I*
Sqrt[3])/2)^(3/4)*ArcTanh[(2^(1/4)*(1 + x^4)^(1/4))/(1 + I*Sqrt[3])^(1/4)])/2 - ((I/2)*Sqrt[-x^4]*EllipticPi[(
-I - Sqrt[3])/2, ArcSin[(1 + x^4)^(1/4)], -1])/x^2 + ((I/2)*Sqrt[-x^4]*EllipticPi[(I - Sqrt[3])/2, ArcSin[(1 +
x^4)^(1/4)], -1])/x^2 + ((I/2)*Sqrt[-x^4]*EllipticPi[1/Sqrt[(1 - I*Sqrt[3])/2], ArcSin[(1 + x^4)^(1/4)], -1])
/x^2 - ((I/2)*Sqrt[-x^4]*EllipticPi[1/Sqrt[(1 + I*Sqrt[3])/2], ArcSin[(1 + x^4)^(1/4)], -1])/x^2

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 205

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

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/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 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), 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 399

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/x, Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 444

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

Rule 490

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

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b,
c, d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-(f/e), -(d/c)
])

Rule 746

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

Rule 1213

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

Rule 1240

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^4)^p, (d/
(d^2 - e^2*x^4) - (e*x^2)/(d^2 - e^2*x^4))^(-q), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&&  !IntegerQ[p] && ILtQ[q, 0]

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2153

Int[((c_) + (d_.)*(x_)^(n_.))^(q_)*((a_) + (b_.)*(x_)^(nn_.))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^
nn)^p, (c/(c^2 - d^2*x^(2*n)) - (d*x^n)/(c^2 - d^2*x^(2*n)))^(-q), x], x] /; FreeQ[{a, b, c, d, n, nn, p}, x]
&&  !IntegerQ[p] && ILtQ[q, 0] && IGtQ[Log[2, nn/n], 0]

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+x^2}{\left (1-x+x^2\right ) \sqrt [4]{1+x^4}} \, dx &=\int \frac {(-2+x) x}{\left (1-x+x^2\right ) \sqrt [4]{1+x^4}} \, dx\\ &=\int \left (\frac {1}{\sqrt [4]{1+x^4}}-\frac {1+x}{\left (1-x+x^2\right ) \sqrt [4]{1+x^4}}\right ) \, dx\\ &=\int \frac {1}{\sqrt [4]{1+x^4}} \, dx-\int \frac {1+x}{\left (1-x+x^2\right ) \sqrt [4]{1+x^4}} \, dx\\ &=-\int \left (\frac {1-i \sqrt {3}}{\left (-1-i \sqrt {3}+2 x\right ) \sqrt [4]{1+x^4}}+\frac {1+i \sqrt {3}}{\left (-1+i \sqrt {3}+2 x\right ) \sqrt [4]{1+x^4}}\right ) \, dx+\operatorname {Subst}\left (\int \frac {1}{1-x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\left (1-i \sqrt {3}\right ) \int \frac {1}{\left (-1-i \sqrt {3}+2 x\right ) \sqrt [4]{1+x^4}} \, dx-\left (1+i \sqrt {3}\right ) \int \frac {1}{\left (-1+i \sqrt {3}+2 x\right ) \sqrt [4]{1+x^4}} \, dx\\ &=\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\left (1-i \sqrt {3}\right ) \int \left (\frac {i-\sqrt {3}}{2 \left (i+\sqrt {3}+2 i x^2\right ) \sqrt [4]{1+x^4}}+\frac {x}{\left (1-i \sqrt {3}+2 x^2\right ) \sqrt [4]{1+x^4}}\right ) \, dx-\left (1+i \sqrt {3}\right ) \int \left (\frac {-i-\sqrt {3}}{2 \left (-i+\sqrt {3}-2 i x^2\right ) \sqrt [4]{1+x^4}}+\frac {x}{\left (1+i \sqrt {3}+2 x^2\right ) \sqrt [4]{1+x^4}}\right ) \, dx\\ &=\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+2 i \int \frac {1}{\left (-i+\sqrt {3}-2 i x^2\right ) \sqrt [4]{1+x^4}} \, dx-2 i \int \frac {1}{\left (i+\sqrt {3}+2 i x^2\right ) \sqrt [4]{1+x^4}} \, dx-\left (1-i \sqrt {3}\right ) \int \frac {x}{\left (1-i \sqrt {3}+2 x^2\right ) \sqrt [4]{1+x^4}} \, dx-\left (1+i \sqrt {3}\right ) \int \frac {x}{\left (1+i \sqrt {3}+2 x^2\right ) \sqrt [4]{1+x^4}} \, dx\\ &=\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+2 i \int \left (\frac {1+i \sqrt {3}}{2 \left (i+\sqrt {3}+2 i x^4\right ) \sqrt [4]{1+x^4}}+\frac {i x^2}{\sqrt [4]{1+x^4} \left (1-i \sqrt {3}+2 x^4\right )}\right ) \, dx-2 i \int \left (\frac {1-i \sqrt {3}}{2 \left (-i+\sqrt {3}-2 i x^4\right ) \sqrt [4]{1+x^4}}-\frac {i x^2}{\sqrt [4]{1+x^4} \left (1+i \sqrt {3}+2 x^4\right )}\right ) \, dx-\frac {1}{2} \left (1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-i \sqrt {3}+2 x\right ) \sqrt [4]{1+x^2}} \, dx,x,x^2\right )-\frac {1}{2} \left (1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+i \sqrt {3}+2 x\right ) \sqrt [4]{1+x^2}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-2 \int \frac {x^2}{\sqrt [4]{1+x^4} \left (1-i \sqrt {3}+2 x^4\right )} \, dx-2 \int \frac {x^2}{\sqrt [4]{1+x^4} \left (1+i \sqrt {3}+2 x^4\right )} \, dx+\left (i-\sqrt {3}\right ) \int \frac {1}{\left (i+\sqrt {3}+2 i x^4\right ) \sqrt [4]{1+x^4}} \, dx-\left (-1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (1-i \sqrt {3}\right )^2-4 x^2\right ) \sqrt [4]{1+x^2}} \, dx,x,x^2\right )-\left (-1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (\left (1+i \sqrt {3}\right )^2-4 x^2\right ) \sqrt [4]{1+x^2}} \, dx,x,x^2\right )-\left (-1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (\left (1-i \sqrt {3}\right )^2-4 x^2\right ) \sqrt [4]{1+x^2}} \, dx,x,x^2\right )-\frac {1}{2} \left (1+i \sqrt {3}\right )^2 \operatorname {Subst}\left (\int \frac {1}{\left (\left (1+i \sqrt {3}\right )^2-4 x^2\right ) \sqrt [4]{1+x^2}} \, dx,x,x^2\right )-\left (i+\sqrt {3}\right ) \int \frac {1}{\left (-i+\sqrt {3}-2 i x^4\right ) \sqrt [4]{1+x^4}} \, dx\\ &=-\frac {2 x^3 F_1\left (\frac {3}{4};\frac {1}{4},1;\frac {7}{4};-x^4,-\frac {2 x^4}{1-i \sqrt {3}}\right )}{3 \left (1-i \sqrt {3}\right )}-\frac {2 x^3 F_1\left (\frac {3}{4};\frac {1}{4},1;\frac {7}{4};-x^4,-\frac {2 x^4}{1+i \sqrt {3}}\right )}{3 \left (1+i \sqrt {3}\right )}+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\left (i-\sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{i+\sqrt {3}-\left (-i+\sqrt {3}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \left (-1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (1+i \sqrt {3}\right )^2-4 x\right ) \sqrt [4]{1+x}} \, dx,x,x^4\right )-\frac {1}{2} \left (-1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\left (1-i \sqrt {3}\right )^2-4 x\right ) \sqrt [4]{1+x}} \, dx,x,x^4\right )-\left (i+\sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{-i+\sqrt {3}-\left (i+\sqrt {3}\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {\left (2 \left (-1-i \sqrt {3}\right ) \sqrt {-x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (4+\left (1-i \sqrt {3}\right )^2-4 x^4\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{1+x^4}\right )}{x^2}-\frac {\left (\left (1+i \sqrt {3}\right )^2 \sqrt {-x^4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (4+\left (1+i \sqrt {3}\right )^2-4 x^4\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{1+x^4}\right )}{x^2}\\ &=-\frac {2 x^3 F_1\left (\frac {3}{4};\frac {1}{4},1;\frac {7}{4};-x^4,-\frac {2 x^4}{1-i \sqrt {3}}\right )}{3 \left (1-i \sqrt {3}\right )}-\frac {2 x^3 F_1\left (\frac {3}{4};\frac {1}{4},1;\frac {7}{4};-x^4,-\frac {2 x^4}{1+i \sqrt {3}}\right )}{3 \left (1+i \sqrt {3}\right )}+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\left (2 \left (1-i \sqrt {3}\right )\right ) \operatorname {Subst}\left (\int \frac {x^2}{4+\left (1-i \sqrt {3}\right )^2-4 x^4} \, dx,x,\sqrt [4]{1+x^4}\right )+\left (2 \left (1+i \sqrt {3}\right )\right ) \operatorname {Subst}\left (\int \frac {x^2}{4+\left (1+i \sqrt {3}\right )^2-4 x^4} \, dx,x,\sqrt [4]{1+x^4}\right )+\frac {\left (i-\sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {i+\sqrt {3}}-\sqrt {-i+\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {i+\sqrt {3}}}+\frac {\left (i-\sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {i+\sqrt {3}}+\sqrt {-i+\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {i+\sqrt {3}}}-\frac {\left (i+\sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-i+\sqrt {3}}-\sqrt {i+\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {-i+\sqrt {3}}}-\frac {\left (i+\sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-i+\sqrt {3}}+\sqrt {i+\sqrt {3}} x^2} \, dx,x,\frac {x}{\sqrt [4]{1+x^4}}\right )}{2 \sqrt {-i+\sqrt {3}}}-\frac {\left (\left (-1-i \sqrt {3}\right ) \sqrt {-x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {1-i \sqrt {3}}-\sqrt {2} x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{1+x^4}\right )}{2 \sqrt {2} x^2}+\frac {\left (\left (-1-i \sqrt {3}\right ) \sqrt {-x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {1-i \sqrt {3}}+\sqrt {2} x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{1+x^4}\right )}{2 \sqrt {2} x^2}-\frac {\left (\left (1+i \sqrt {3}\right )^2 \sqrt {-x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {1+i \sqrt {3}}-\sqrt {2} x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{1+x^4}\right )}{4 \sqrt {2} x^2}+\frac {\left (\left (1+i \sqrt {3}\right )^2 \sqrt {-x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (\sqrt {1+i \sqrt {3}}+\sqrt {2} x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{1+x^4}\right )}{4 \sqrt {2} x^2}\\ &=-\frac {2 x^3 F_1\left (\frac {3}{4};\frac {1}{4},1;\frac {7}{4};-x^4,-\frac {2 x^4}{1-i \sqrt {3}}\right )}{3 \left (1-i \sqrt {3}\right )}-\frac {2 x^3 F_1\left (\frac {3}{4};\frac {1}{4},1;\frac {7}{4};-x^4,-\frac {2 x^4}{1+i \sqrt {3}}\right )}{3 \left (1+i \sqrt {3}\right )}+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {\tan ^{-1}\left (\frac {x}{\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \sqrt [4]{1+x^4}}\right )}{2 \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4}}-\frac {1}{2} \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \sqrt [4]{1+x^4}}\right )}{2 \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4}}-\frac {1}{2} \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} x}{\sqrt [4]{1+x^4}}\right )+\frac {\left (1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-i \sqrt {3}}-\sqrt {2} x^2} \, dx,x,\sqrt [4]{1+x^4}\right )}{2 \sqrt {2}}-\frac {\left (1-i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-i \sqrt {3}}+\sqrt {2} x^2} \, dx,x,\sqrt [4]{1+x^4}\right )}{2 \sqrt {2}}+\frac {\left (1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+i \sqrt {3}}-\sqrt {2} x^2} \, dx,x,\sqrt [4]{1+x^4}\right )}{2 \sqrt {2}}-\frac {\left (1+i \sqrt {3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+i \sqrt {3}}+\sqrt {2} x^2} \, dx,x,\sqrt [4]{1+x^4}\right )}{2 \sqrt {2}}-\frac {\left (\left (-1-i \sqrt {3}\right ) \sqrt {-x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {1-i \sqrt {3}}-\sqrt {2} x^2\right )} \, dx,x,\sqrt [4]{1+x^4}\right )}{2 \sqrt {2} x^2}+\frac {\left (\left (-1-i \sqrt {3}\right ) \sqrt {-x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {1-i \sqrt {3}}+\sqrt {2} x^2\right )} \, dx,x,\sqrt [4]{1+x^4}\right )}{2 \sqrt {2} x^2}-\frac {\left (\left (1+i \sqrt {3}\right )^2 \sqrt {-x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {1+i \sqrt {3}}-\sqrt {2} x^2\right )} \, dx,x,\sqrt [4]{1+x^4}\right )}{4 \sqrt {2} x^2}+\frac {\left (\left (1+i \sqrt {3}\right )^2 \sqrt {-x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {1+i \sqrt {3}}+\sqrt {2} x^2\right )} \, dx,x,\sqrt [4]{1+x^4}\right )}{4 \sqrt {2} x^2}\\ &=-\frac {2 x^3 F_1\left (\frac {3}{4};\frac {1}{4},1;\frac {7}{4};-x^4,-\frac {2 x^4}{1-i \sqrt {3}}\right )}{3 \left (1-i \sqrt {3}\right )}-\frac {2 x^3 F_1\left (\frac {3}{4};\frac {1}{4},1;\frac {7}{4};-x^4,-\frac {2 x^4}{1+i \sqrt {3}}\right )}{3 \left (1+i \sqrt {3}\right )}+\frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {\tan ^{-1}\left (\frac {x}{\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \sqrt [4]{1+x^4}}\right )}{2 \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4}}-\frac {1}{2} \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} x}{\sqrt [4]{1+x^4}}\right )-\frac {1}{2} \left (\frac {1}{2} \left (1-i \sqrt {3}\right )\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{1+x^4}}{\sqrt [4]{1-i \sqrt {3}}}\right )-\frac {1}{2} \left (\frac {1}{2} \left (1+i \sqrt {3}\right )\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{1+x^4}}{\sqrt [4]{1+i \sqrt {3}}}\right )+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} \sqrt [4]{1+x^4}}\right )}{2 \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4}}-\frac {1}{2} \left (-\frac {i-\sqrt {3}}{i+\sqrt {3}}\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{-\frac {i-\sqrt {3}}{i+\sqrt {3}}} x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \left (\frac {1}{2} \left (1-i \sqrt {3}\right )\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{1+x^4}}{\sqrt [4]{1-i \sqrt {3}}}\right )+\frac {1}{2} \left (\frac {1}{2} \left (1+i \sqrt {3}\right )\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt [4]{1+x^4}}{\sqrt [4]{1+i \sqrt {3}}}\right )-\frac {\left (1+i \sqrt {3}\right ) \sqrt {-x^4} \Pi \left (\frac {1}{2} \left (-i-\sqrt {3}\right );\left .\sin ^{-1}\left (\sqrt [4]{1+x^4}\right )\right |-1\right )}{2 \sqrt {2} \sqrt {1-i \sqrt {3}} x^2}+\frac {\left (1+i \sqrt {3}\right )^{3/2} \sqrt {-x^4} \Pi \left (\frac {1}{2} \left (i-\sqrt {3}\right );\left .\sin ^{-1}\left (\sqrt [4]{1+x^4}\right )\right |-1\right )}{4 \sqrt {2} x^2}+\frac {\left (1+i \sqrt {3}\right ) \sqrt {-x^4} \Pi \left (\frac {1}{\sqrt {\frac {1}{2} \left (1-i \sqrt {3}\right )}};\left .\sin ^{-1}\left (\sqrt [4]{1+x^4}\right )\right |-1\right )}{2 \sqrt {2} \sqrt {1-i \sqrt {3}} x^2}-\frac {\left (1+i \sqrt {3}\right )^{3/2} \sqrt {-x^4} \Pi \left (\frac {1}{\sqrt {\frac {1}{2} \left (1+i \sqrt {3}\right )}};\left .\sin ^{-1}\left (\sqrt [4]{1+x^4}\right )\right |-1\right )}{4 \sqrt {2} x^2}\\ \end {align*}

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 12.01, size = 139, normalized size = 1.00 \begin {gather*} \frac {1}{2} \tan ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {\tan ^{-1}\left (\frac {\left (-\sqrt {2}+\sqrt {2} x\right ) \sqrt [4]{1+x^4}}{-1+2 x-x^2+\sqrt {1+x^4}}\right )}{\sqrt {2}}+\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )-\frac {\tanh ^{-1}\left (\frac {\left (-\sqrt {2}+\sqrt {2} x\right ) \sqrt [4]{1+x^4}}{1-2 x+x^2+\sqrt {1+x^4}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 12.14, size = 1172, normalized size = 8.43

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*arctan(-(x^8 - 4*x^7 + 10*x^6 - 16*x^5 + 19*x^4 - 16*x^3 + sqrt(2)*(x^5 - 7*x^4 + 15*x^3 - 15*x^2
+ 7*x - 1)*(x^4 + 1)^(3/4) + 10*x^2 - sqrt(2)*(x^7 - x^6 - 6*x^5 + 16*x^4 - 16*x^3 + 6*x^2 + x - 1)*(x^4 + 1)^
(1/4) + 2*(x^6 - 4*x^5 + 8*x^4 - 10*x^3 + 8*x^2 - 4*x + 1)*sqrt(x^4 + 1) - (sqrt(2)*(x^6 - 8*x^5 + 22*x^4 - 30
*x^3 + 22*x^2 - 8*x + 1)*sqrt(x^4 + 1) + 4*(x^5 - 5*x^4 + 10*x^3 - 10*x^2 + 5*x - 1)*(x^4 + 1)^(3/4) + sqrt(2)
*(2*x^8 - 10*x^7 + 19*x^6 - 22*x^5 + 21*x^4 - 22*x^3 + 19*x^2 - 10*x + 2) + 2*(x^7 - 5*x^6 + 12*x^5 - 18*x^4 +
18*x^3 - 12*x^2 + 5*x - 1)*(x^4 + 1)^(1/4))*sqrt((x^4 - 2*x^3 - sqrt(2)*(x^4 + 1)^(3/4)*(x - 1) + 3*x^2 - sqr
t(2)*(x^4 + 1)^(1/4)*(x^3 - 3*x^2 + 3*x - 1) + 2*sqrt(x^4 + 1)*(x^2 - 2*x + 1) - 2*x + 1)/(x^4 - 2*x^3 + 3*x^2
- 2*x + 1)) - 4*x + 1)/(3*x^8 - 12*x^7 + 14*x^6 - 11*x^4 + 14*x^2 - 12*x + 3)) - 1/2*sqrt(2)*arctan(-(x^8 - 4
*x^7 + 10*x^6 - 16*x^5 + 19*x^4 - 16*x^3 - sqrt(2)*(x^5 - 7*x^4 + 15*x^3 - 15*x^2 + 7*x - 1)*(x^4 + 1)^(3/4) +
10*x^2 + sqrt(2)*(x^7 - x^6 - 6*x^5 + 16*x^4 - 16*x^3 + 6*x^2 + x - 1)*(x^4 + 1)^(1/4) + 2*(x^6 - 4*x^5 + 8*x
^4 - 10*x^3 + 8*x^2 - 4*x + 1)*sqrt(x^4 + 1) + (sqrt(2)*(x^6 - 8*x^5 + 22*x^4 - 30*x^3 + 22*x^2 - 8*x + 1)*sqr
t(x^4 + 1) - 4*(x^5 - 5*x^4 + 10*x^3 - 10*x^2 + 5*x - 1)*(x^4 + 1)^(3/4) + sqrt(2)*(2*x^8 - 10*x^7 + 19*x^6 -
22*x^5 + 21*x^4 - 22*x^3 + 19*x^2 - 10*x + 2) - 2*(x^7 - 5*x^6 + 12*x^5 - 18*x^4 + 18*x^3 - 12*x^2 + 5*x - 1)*
(x^4 + 1)^(1/4))*sqrt((x^4 - 2*x^3 + sqrt(2)*(x^4 + 1)^(3/4)*(x - 1) + 3*x^2 + sqrt(2)*(x^4 + 1)^(1/4)*(x^3 -
3*x^2 + 3*x - 1) + 2*sqrt(x^4 + 1)*(x^2 - 2*x + 1) - 2*x + 1)/(x^4 - 2*x^3 + 3*x^2 - 2*x + 1)) - 4*x + 1)/(3*x
^8 - 12*x^7 + 14*x^6 - 11*x^4 + 14*x^2 - 12*x + 3)) - 1/8*sqrt(2)*log(4*(x^4 - 2*x^3 + sqrt(2)*(x^4 + 1)^(3/4)
*(x - 1) + 3*x^2 + sqrt(2)*(x^4 + 1)^(1/4)*(x^3 - 3*x^2 + 3*x - 1) + 2*sqrt(x^4 + 1)*(x^2 - 2*x + 1) - 2*x + 1
)/(x^4 - 2*x^3 + 3*x^2 - 2*x + 1)) + 1/8*sqrt(2)*log(4*(x^4 - 2*x^3 - sqrt(2)*(x^4 + 1)^(3/4)*(x - 1) + 3*x^2
- sqrt(2)*(x^4 + 1)^(1/4)*(x^3 - 3*x^2 + 3*x - 1) + 2*sqrt(x^4 + 1)*(x^2 - 2*x + 1) - 2*x + 1)/(x^4 - 2*x^3 +
3*x^2 - 2*x + 1)) + 1/4*arctan(2*(x^4 + 1)^(1/4)*x^3 + 2*(x^4 + 1)^(3/4)*x) + 1/4*log(2*x^4 + 2*(x^4 + 1)^(1/4
)*x^3 + 2*sqrt(x^4 + 1)*x^2 + 2*(x^4 + 1)^(3/4)*x + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 9.94, size = 621, normalized size = 4.47

method	result	size
trager	\(\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{4}+1}\, x^{2}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{4}+2 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}+\RootOf \left (\textit {\_Z}^{2}+1\right )\right )}{4}-\frac {\ln \left (2 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 x^{2} \sqrt {x^{4}+1}+2 x^{3} \left (x^{4}+1\right )^{\frac {1}{4}}-2 x^{4}-1\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \ln \left (\frac {\sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \RootOf \left (\textit {\_Z}^{2}+1\right ) x -\RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}} x^{3}+\left (x^{4}+1\right )^{\frac {3}{4}} x +\sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \RootOf \left (\textit {\_Z}^{2}+1\right )+3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}} x^{2}-2 \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) x^{3}-\left (x^{4}+1\right )^{\frac {3}{4}}-3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}} x +3 \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) x^{2}+\RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}}-2 \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) x}{\left (x^{2}-x +1\right )^{2}}\right )}{2}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \ln \left (\frac {-\sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) x^{2}+\RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}} x^{3}+2 \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+\left (x^{4}+1\right )^{\frac {3}{4}} x +2 \sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) x -3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}} x^{2}-3 \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \RootOf \left (\textit {\_Z}^{2}+1\right ) x^{2}-\left (x^{4}+1\right )^{\frac {3}{4}}-\sqrt {x^{4}+1}\, \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right )+3 \RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}} x +2 \RootOf \left (\textit {\_Z}^{2}-\RootOf \left (\textit {\_Z}^{2}+1\right )\right ) \RootOf \left (\textit {\_Z}^{2}+1\right ) x -\RootOf \left (\textit {\_Z}^{2}+1\right ) \left (x^{4}+1\right )^{\frac {1}{4}}}{\left (x^{2}-x +1\right )^{2}}\right )}{2}\)	\(621\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*RootOf(_Z^2+1)*ln(-2*RootOf(_Z^2+1)*(x^4+1)^(1/2)*x^2+2*RootOf(_Z^2+1)*x^4+2*(x^4+1)^(3/4)*x-2*x^3*(x^4+1)
^(1/4)+RootOf(_Z^2+1))-1/4*ln(2*(x^4+1)^(3/4)*x-2*x^2*(x^4+1)^(1/2)+2*x^3*(x^4+1)^(1/4)-2*x^4-1)+1/2*RootOf(_Z
^2-RootOf(_Z^2+1))*ln(((x^4+1)^(1/2)*RootOf(_Z^2-RootOf(_Z^2+1))*RootOf(_Z^2+1)*x^2-2*(x^4+1)^(1/2)*RootOf(_Z^
2-RootOf(_Z^2+1))*RootOf(_Z^2+1)*x-RootOf(_Z^2+1)*(x^4+1)^(1/4)*x^3+(x^4+1)^(3/4)*x+(x^4+1)^(1/2)*RootOf(_Z^2-
RootOf(_Z^2+1))*RootOf(_Z^2+1)+3*RootOf(_Z^2+1)*(x^4+1)^(1/4)*x^2-2*RootOf(_Z^2-RootOf(_Z^2+1))*x^3-(x^4+1)^(3
/4)-3*RootOf(_Z^2+1)*(x^4+1)^(1/4)*x+3*RootOf(_Z^2-RootOf(_Z^2+1))*x^2+RootOf(_Z^2+1)*(x^4+1)^(1/4)-2*RootOf(_
Z^2-RootOf(_Z^2+1))*x)/(x^2-x+1)^2)-1/2*RootOf(_Z^2+1)*RootOf(_Z^2-RootOf(_Z^2+1))*ln((-(x^4+1)^(1/2)*RootOf(_
Z^2-RootOf(_Z^2+1))*x^2+RootOf(_Z^2+1)*(x^4+1)^(1/4)*x^3+2*RootOf(_Z^2-RootOf(_Z^2+1))*RootOf(_Z^2+1)*x^3+(x^4
+1)^(3/4)*x+2*(x^4+1)^(1/2)*RootOf(_Z^2-RootOf(_Z^2+1))*x-3*RootOf(_Z^2+1)*(x^4+1)^(1/4)*x^2-3*RootOf(_Z^2-Roo
tOf(_Z^2+1))*RootOf(_Z^2+1)*x^2-(x^4+1)^(3/4)-(x^4+1)^(1/2)*RootOf(_Z^2-RootOf(_Z^2+1))+3*RootOf(_Z^2+1)*(x^4+
1)^(1/4)*x+2*RootOf(_Z^2-RootOf(_Z^2+1))*RootOf(_Z^2+1)*x-RootOf(_Z^2+1)*(x^4+1)^(1/4))/(x^2-x+1)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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