3.24.72 \(\int \frac {x^4 \sqrt [4]{-x^2+x^4}}{1+x^4+x^8} \, dx\) [2372]

Optimal. Leaf size=189 \[ \frac {1}{4} \text {RootSum}\left [3-3 \text {$\#$1}^4+\text {$\#$1}^8\& ,\frac {-3 \log (x)+3 \log \left (\sqrt [4]{-x^2+x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{-x^2+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-3 \text {$\#$1}^3+2 \text {$\#$1}^7}\& \right ]-\frac {1}{4} \text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\& ,\frac {-\log (x)+\log \left (\sqrt [4]{-x^2+x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{-x^2+x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-\text {$\#$1}^3+2 \text {$\#$1}^7}\& \right ] \]

[Out]

Unintegrable

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Rubi [C] Result contains complex when optimal does not.
time = 0.75, antiderivative size = 273, normalized size of antiderivative = 1.44, number of steps used = 29, number of rules used = 12, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2081, 6860, 1284, 1531, 285, 338, 304, 209, 212, 1543, 525, 524} \begin {gather*} \frac {i x \sqrt [4]{x^4-x^2} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,-\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )}{3 \sqrt {3} \sqrt [4]{1-x^2}}+\frac {i x \sqrt [4]{x^4-x^2} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )}{3 \sqrt {3} \sqrt [4]{1-x^2}}-\frac {i x \sqrt [4]{x^4-x^2} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,-\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )}{3 \sqrt {3} \sqrt [4]{1-x^2}}-\frac {i x \sqrt [4]{x^4-x^2} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )}{3 \sqrt {3} \sqrt [4]{1-x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*x*(-x^2 + x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, x^2, -1/2*((1 - I*Sqrt[3])*x^2)])/(Sqrt[3]*(1 - x^2)^(
1/4)) + ((I/3)*x*(-x^2 + x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, x^2, ((1 - I*Sqrt[3])*x^2)/2])/(Sqrt[3]*(1 - x
^2)^(1/4)) - ((I/3)*x*(-x^2 + x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, x^2, -1/2*((1 + I*Sqrt[3])*x^2)])/(Sqrt[3
]*(1 - x^2)^(1/4)) - ((I/3)*x*(-x^2 + x^4)^(1/4)*AppellF1[3/4, -1/4, 1, 7/4, x^2, ((1 + I*Sqrt[3])*x^2)/2])/(S
qrt[3]*(1 - x^2)^(1/4))

Rule 209

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

Rule 212

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

Rule 285

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + n
*p + 1))), x] + Dist[a*n*(p/(m + n*p + 1)), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 304

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 338

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 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)] && IntegersQ[m, p + (m + 1)/n]

Rule 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{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])

Rule 1284

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{k = Denominat
or[m]}, Dist[k/f, Subst[Int[x^(k*(m + 1) - 1)*(d + e*(x^(2*k)/f))^q*(a + c*(x^(4*k)/f))^p, x], x, (f*x)^(1/k)]
, x]] /; FreeQ[{a, c, d, e, f, p, q}, x] && FractionQ[m] && IntegerQ[p]

Rule 1531

Int[(((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.)), x_Symbol] :> Dist[f^(2*n)/
c, Int[(f*x)^(m - 2*n)*(d + e*x^n)^q, x], x] - Dist[a*(f^(2*n)/c), Int[(f*x)^(m - 2*n)*((d + e*x^n)^q/(a + c*x
^(2*n))), x], x] /; FreeQ[{a, c, d, e, f, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] &&  !IntegerQ[q] && GtQ[m, 2*n
- 1]

Rule 1543

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.)), x_Symbol] :> Int[ExpandInte
grand[(d + e*x^n)^q, (f*x)^m/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, f, q, n}, x] && EqQ[n2, 2*n] && IGt
Q[n, 0] &&  !IntegerQ[q] && IntegerQ[m]

Rule 2081

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 6860

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 {x^4 \sqrt [4]{-x^2+x^4}}{1+x^4+x^8} \, dx &=\frac {\sqrt [4]{-x^2+x^4} \int \frac {x^{9/2} \sqrt [4]{-1+x^2}}{1+x^4+x^8} \, dx}{\sqrt {x} \sqrt [4]{-1+x^2}}\\ &=\frac {\sqrt [4]{-x^2+x^4} \int \left (\frac {2 i x^{9/2} \sqrt [4]{-1+x^2}}{\sqrt {3} \left (-1+i \sqrt {3}-2 x^4\right )}+\frac {2 i x^{9/2} \sqrt [4]{-1+x^2}}{\sqrt {3} \left (1+i \sqrt {3}+2 x^4\right )}\right ) \, dx}{\sqrt {x} \sqrt [4]{-1+x^2}}\\ &=\frac {\left (2 i \sqrt [4]{-x^2+x^4}\right ) \int \frac {x^{9/2} \sqrt [4]{-1+x^2}}{-1+i \sqrt {3}-2 x^4} \, dx}{\sqrt {3} \sqrt {x} \sqrt [4]{-1+x^2}}+\frac {\left (2 i \sqrt [4]{-x^2+x^4}\right ) \int \frac {x^{9/2} \sqrt [4]{-1+x^2}}{1+i \sqrt {3}+2 x^4} \, dx}{\sqrt {3} \sqrt {x} \sqrt [4]{-1+x^2}}\\ &=\frac {\left (4 i \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^{10} \sqrt [4]{-1+x^4}}{-1+i \sqrt {3}-2 x^8} \, dx,x,\sqrt {x}\right )}{\sqrt {3} \sqrt {x} \sqrt [4]{-1+x^2}}+\frac {\left (4 i \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^{10} \sqrt [4]{-1+x^4}}{1+i \sqrt {3}+2 x^8} \, dx,x,\sqrt {x}\right )}{\sqrt {3} \sqrt {x} \sqrt [4]{-1+x^2}}\\ &=-\frac {\left (2 i \left (1-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{-1+i \sqrt {3}-2 x^8} \, dx,x,\sqrt {x}\right )}{\sqrt {3} \sqrt {x} \sqrt [4]{-1+x^2}}-\frac {\left (2 i \left (1+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{1+i \sqrt {3}+2 x^8} \, dx,x,\sqrt {x}\right )}{\sqrt {3} \sqrt {x} \sqrt [4]{-1+x^2}}\\ &=-\frac {\left (2 i \left (1-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \left (-\frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1+i \sqrt {3}\right )} \left (-\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4\right )}+\frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1+i \sqrt {3}\right )} \left (\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {3} \sqrt {x} \sqrt [4]{-1+x^2}}-\frac {\left (2 i \left (1+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \left (\frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1-i \sqrt {3}\right )} \left (-\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4\right )}-\frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1-i \sqrt {3}\right )} \left (\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt {3} \sqrt {x} \sqrt [4]{-1+x^2}}\\ &=\frac {\left (i \left (1-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{-\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {\frac {3}{2} \left (-1+i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{-1+x^2}}-\frac {\left (i \left (1-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {\frac {3}{2} \left (-1+i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{-1+x^2}}-\frac {\left (i \left (1+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{-\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {\frac {3}{2} \left (-1-i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{-1+x^2}}+\frac {\left (i \left (1+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{-1+x^4}}{\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {\frac {3}{2} \left (-1-i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{-1+x^2}}\\ &=\frac {\left (i \left (1-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{1-x^4}}{-\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {\frac {3}{2} \left (-1+i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{1-x^2}}-\frac {\left (i \left (1-i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{1-x^4}}{\sqrt {2 \left (-1+i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {\frac {3}{2} \left (-1+i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{1-x^2}}-\frac {\left (i \left (1+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{1-x^4}}{-\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {\frac {3}{2} \left (-1-i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{1-x^2}}+\frac {\left (i \left (1+i \sqrt {3}\right ) \sqrt [4]{-x^2+x^4}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt [4]{1-x^4}}{\sqrt {2 \left (-1-i \sqrt {3}\right )}+2 x^4} \, dx,x,\sqrt {x}\right )}{\sqrt {\frac {3}{2} \left (-1-i \sqrt {3}\right )} \sqrt {x} \sqrt [4]{1-x^2}}\\ &=\frac {i x \sqrt [4]{-x^2+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,-\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )}{3 \sqrt {3} \sqrt [4]{1-x^2}}+\frac {i x \sqrt [4]{-x^2+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,\frac {1}{2} \left (1-i \sqrt {3}\right ) x^2\right )}{3 \sqrt {3} \sqrt [4]{1-x^2}}-\frac {i x \sqrt [4]{-x^2+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,-\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )}{3 \sqrt {3} \sqrt [4]{1-x^2}}-\frac {i x \sqrt [4]{-x^2+x^4} F_1\left (\frac {3}{4};-\frac {1}{4},1;\frac {7}{4};x^2,\frac {1}{2} \left (1+i \sqrt {3}\right ) x^2\right )}{3 \sqrt {3} \sqrt [4]{1-x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 216, normalized size = 1.14 \begin {gather*} \frac {x^{3/2} \left (-1+x^2\right )^{3/4} \left (\text {RootSum}\left [3-3 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-3 \log (x)+6 \log \left (\sqrt [4]{-1+x^2}-\sqrt {x} \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-2 \log \left (\sqrt [4]{-1+x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{-3 \text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ]-\text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+2 \log \left (\sqrt [4]{-1+x^2}-\sqrt {x} \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-2 \log \left (\sqrt [4]{-1+x^2}-\sqrt {x} \text {$\#$1}\right ) \text {$\#$1}^4}{-\text {$\#$1}^3+2 \text {$\#$1}^7}\&\right ]\right )}{8 \left (x^2 \left (-1+x^2\right )\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(3/2)*(-1 + x^2)^(3/4)*(RootSum[3 - 3*#1^4 + #1^8 & , (-3*Log[x] + 6*Log[(-1 + x^2)^(1/4) - Sqrt[x]*#1] + L
og[x]*#1^4 - 2*Log[(-1 + x^2)^(1/4) - Sqrt[x]*#1]*#1^4)/(-3*#1^3 + 2*#1^7) & ] - RootSum[1 - #1^4 + #1^8 & , (
-Log[x] + 2*Log[(-1 + x^2)^(1/4) - Sqrt[x]*#1] + Log[x]*#1^4 - 2*Log[(-1 + x^2)^(1/4) - Sqrt[x]*#1]*#1^4)/(-#1
^3 + 2*#1^7) & ]))/(8*(x^2*(-1 + x^2))^(3/4))

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 1.
time = 167.54, size = 11367, normalized size = 60.14

method	result	size
trager	\(\text {Expression too large to display}\)	\(11367\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)] Timed out
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(x^4*(x^4-x^2)^(1/4)/(x^8+x^4+1),x, algorithm="fricas")

[Out]

Timed out

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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