3.26.39 \(\int \frac {\sqrt [3]{x^2+x^4}}{x (-1+x^2)} \, dx\)

Optimal. Leaf size=213 \[ \frac {\log \left (2^{2/3} \sqrt [3]{x^4+x^2}-2 x\right )}{2\ 2^{2/3}}+\frac {\log \left (2^{2/3} \sqrt [3]{x^4+x^2}+2 x\right )}{2\ 2^{2/3}}-\frac {\log \left (-2 x^2+2^{2/3} \sqrt [3]{x^4+x^2} x-\sqrt [3]{2} \left (x^4+x^2\right )^{2/3}\right )}{4\ 2^{2/3}}-\frac {\log \left (2 x^2+2^{2/3} \sqrt [3]{x^4+x^2} x+\sqrt [3]{2} \left (x^4+x^2\right )^{2/3}\right )}{4\ 2^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x^2}{x^2+\sqrt [3]{2} \left (x^4+x^2\right )^{2/3}}\right )}{2\ 2^{2/3}} \]

________________________________________________________________________________________

Rubi [C]  time = 0.68, antiderivative size = 382, normalized size of antiderivative = 1.79, number of steps used = 9, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {1311, 2013, 622, 619, 236, 219, 2034, 758, 133} \begin {gather*} \frac {3^{3/4} \sqrt {2-\sqrt {3}} \left (-x^4-x^2\right )^{2/3} \left (1-2^{2/3} \sqrt [3]{-x^2 \left (x^2+1\right )}\right ) \sqrt {\frac {2 \sqrt [3]{2} \left (-x^2 \left (x^2+1\right )\right )^{2/3}+2^{2/3} \sqrt [3]{-x^2 \left (x^2+1\right )}+1}{\left (-2^{2/3} \sqrt [3]{-x^2 \left (x^2+1\right )}-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-2^{2/3} \sqrt [3]{-x^2 \left (x^2+1\right )}+\sqrt {3}+1}{-2^{2/3} \sqrt [3]{-x^2 \left (x^2+1\right )}-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{2^{2/3} \left (2 x^2+1\right ) \left (x^4+x^2\right )^{2/3} \sqrt {-\frac {1-2^{2/3} \sqrt [3]{-x^2 \left (x^2+1\right )}}{\left (-2^{2/3} \sqrt [3]{-x^2 \left (x^2+1\right )}-\sqrt {3}+1\right )^2}}}-\frac {3 \left (-\frac {x^2}{1-x^2}\right )^{2/3} \left (-\frac {x^2+1}{1-x^2}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},\frac {2}{3};\frac {7}{3};\frac {1}{1-x^2},\frac {2}{1-x^2}\right )}{4 \left (x^4+x^2\right )^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Int[(x^2 + x^4)^(1/3)/(x*(-1 + x^2)),x]

[Out]

(-3*(-(x^2/(1 - x^2)))^(2/3)*(-((1 + x^2)/(1 - x^2)))^(2/3)*AppellF1[4/3, 2/3, 2/3, 7/3, (1 - x^2)^(-1), 2/(1
- x^2)])/(4*(x^2 + x^4)^(2/3)) + (3^(3/4)*Sqrt[2 - Sqrt[3]]*(-x^2 - x^4)^(2/3)*(1 - 2^(2/3)*(-(x^2*(1 + x^2)))
^(1/3))*Sqrt[(1 + 2^(2/3)*(-(x^2*(1 + x^2)))^(1/3) + 2*2^(1/3)*(-(x^2*(1 + x^2)))^(2/3))/(1 - Sqrt[3] - 2^(2/3
)*(-(x^2*(1 + x^2)))^(1/3))^2]*EllipticF[ArcSin[(1 + Sqrt[3] - 2^(2/3)*(-(x^2*(1 + x^2)))^(1/3))/(1 - Sqrt[3]
- 2^(2/3)*(-(x^2*(1 + x^2)))^(1/3))], -7 + 4*Sqrt[3]])/(2^(2/3)*(1 + 2*x^2)*(x^2 + x^4)^(2/3)*Sqrt[-((1 - 2^(2
/3)*(-(x^2*(1 + x^2)))^(1/3))/(1 - Sqrt[3] - 2^(2/3)*(-(x^2*(1 + x^2)))^(1/3))^2)])

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 219

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 - Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 + Sqrt[3
])*s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[-((s*(s + r*x))/((1 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 236

Int[((a_) + (b_.)*(x_)^2)^(-2/3), x_Symbol] :> Dist[(3*Sqrt[b*x^2])/(2*b*x), Subst[Int[1/Sqrt[-a + x^3], x], x
, (a + b*x^2)^(1/3)], x] /; FreeQ[{a, b}, x]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 622

Int[((b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(b*x + c*x^2)^p/(-((c*(b*x + c*x^2))/b^2))^p, Int[(-((
c*x)/b) - (c^2*x^2)/b^2)^p, x], x] /; FreeQ[{b, c}, x] && RationalQ[p] && 3 <= Denominator[p] <= 4

Rule 758

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, -Dist[((1/(d + e*x))^(2*p)*(a + b*x + c*x^2)^p)/(e*((e*(b - q + 2*c*x))/(2*c*(d + e*x)))^p*((e*(b + q +
2*c*x))/(2*c*(d + e*x)))^p), Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - (e*(b - q))/(2*c))*x, x]^p*Simp[1 - (d
- (e*(b + q))/(2*c))*x, x]^p, x], x, 1/(d + e*x)], x]] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b^2 - 4*a*c, 0]
&& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p] && ILtQ[m, 0]

Rule 1311

Int[(((f_.)*(x_))^(m_)*((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.))/((d_.) + (e_.)*(x_)^2), x_Symbol] :> Dist[
1/(d*e), Int[(f*x)^m*(a*e + c*d*x^2)*(a + b*x^2 + c*x^4)^(p - 1), x], x] - Dist[(c*d^2 - b*d*e + a*e^2)/(d*e*f
^2), Int[((f*x)^(m + 2)*(a + b*x^2 + c*x^4)^(p - 1))/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && Ne
Q[b^2 - 4*a*c, 0] && GtQ[p, 0] && LtQ[m, 0]

Rule 2013

Int[(x_)^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[(a*x^Simplify[j/n]
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && IntegerQ[Simplify[j
/n]] && EqQ[Simplify[m - n + 1], 0]

Rule 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rubi steps

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

________________________________________________________________________________________

Mathematica [C]  time = 0.06, size = 46, normalized size = 0.22 \begin {gather*} -\frac {3 \sqrt [3]{x^4+x^2} F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};-\frac {1}{x^2},\frac {1}{x^2}\right )}{2 \sqrt [3]{\frac {1}{x^2}+1} x^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^2 + x^4)^(1/3)/(x*(-1 + x^2)),x]

[Out]

(-3*(x^2 + x^4)^(1/3)*AppellF1[1/3, -1/3, 1, 4/3, -x^(-2), x^(-2)])/(2*(1 + x^(-2))^(1/3)*x^2)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.44, size = 213, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x^2}{x^2+\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}}\right )}{2\ 2^{2/3}}+\frac {\log \left (-2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{2\ 2^{2/3}}+\frac {\log \left (2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{2\ 2^{2/3}}-\frac {\log \left (-2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}-\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{4\ 2^{2/3}}-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}+\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{4\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^2 + x^4)^(1/3)/(x*(-1 + x^2)),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 2.82, size = 326, normalized size = 1.53 \begin {gather*} \frac {1}{12} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (-\frac {4^{\frac {1}{6}} \sqrt {3} {\left (6 \cdot 4^{\frac {2}{3}} {\left (x^{10} + 33 \, x^{8} + 110 \, x^{6} + 110 \, x^{4} + 33 \, x^{2} + 1\right )} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} - 48 \, {\left (x^{8} + 2 \, x^{6} - 6 \, x^{4} + 2 \, x^{2} + 1\right )} {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}} - 4^{\frac {1}{3}} {\left (x^{12} - 42 \, x^{10} - 417 \, x^{8} - 812 \, x^{6} - 417 \, x^{4} - 42 \, x^{2} + 1\right )}\right )}}{6 \, {\left (x^{12} + 102 \, x^{10} + 447 \, x^{8} + 628 \, x^{6} + 447 \, x^{4} + 102 \, x^{2} + 1\right )}}\right ) - \frac {1}{48} \cdot 4^{\frac {2}{3}} \log \left (\frac {24 \cdot 4^{\frac {1}{3}} {\left (x^{4} + 4 \, x^{2} + 1\right )} {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}} + 4^{\frac {2}{3}} {\left (x^{8} + 32 \, x^{6} + 78 \, x^{4} + 32 \, x^{2} + 1\right )} + 12 \, {\left (x^{6} + 11 \, x^{4} + 11 \, x^{2} + 1\right )} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}}}{x^{8} - 4 \, x^{6} + 6 \, x^{4} - 4 \, x^{2} + 1}\right ) + \frac {1}{24} \cdot 4^{\frac {2}{3}} \log \left (-\frac {3 \cdot 4^{\frac {2}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )} - 4^{\frac {1}{3}} {\left (x^{4} - 2 \, x^{2} + 1\right )} - 12 \, {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}}}{x^{4} - 2 \, x^{2} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+x^2)^(1/3)/x/(x^2-1),x, algorithm="fricas")

[Out]

1/12*4^(1/6)*sqrt(3)*arctan(-1/6*4^(1/6)*sqrt(3)*(6*4^(2/3)*(x^10 + 33*x^8 + 110*x^6 + 110*x^4 + 33*x^2 + 1)*(
x^4 + x^2)^(1/3) - 48*(x^8 + 2*x^6 - 6*x^4 + 2*x^2 + 1)*(x^4 + x^2)^(2/3) - 4^(1/3)*(x^12 - 42*x^10 - 417*x^8
- 812*x^6 - 417*x^4 - 42*x^2 + 1))/(x^12 + 102*x^10 + 447*x^8 + 628*x^6 + 447*x^4 + 102*x^2 + 1)) - 1/48*4^(2/
3)*log((24*4^(1/3)*(x^4 + 4*x^2 + 1)*(x^4 + x^2)^(2/3) + 4^(2/3)*(x^8 + 32*x^6 + 78*x^4 + 32*x^2 + 1) + 12*(x^
6 + 11*x^4 + 11*x^2 + 1)*(x^4 + x^2)^(1/3))/(x^8 - 4*x^6 + 6*x^4 - 4*x^2 + 1)) + 1/24*4^(2/3)*log(-(3*4^(2/3)*
(x^4 + x^2)^(1/3)*(x^2 + 1) - 4^(1/3)*(x^4 - 2*x^2 + 1) - 12*(x^4 + x^2)^(2/3))/(x^4 - 2*x^2 + 1))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}}}{{\left (x^{2} - 1\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+x^2)^(1/3)/x/(x^2-1),x, algorithm="giac")

[Out]

integrate((x^4 + x^2)^(1/3)/((x^2 - 1)*x), x)

________________________________________________________________________________________

maple [C]  time = 24.31, size = 1110, normalized size = 5.21

method	result	size
trager	\(\text {Expression too large to display}\)	\(1110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+x^2)^(1/3)/x/(x^2-1),x,method=_RETURNVERBOSE)

[Out]

1/4*RootOf(_Z^3-2)*ln((2247248*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^4*x^4+210350
4*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^3*x^4-9550804*RootOf(RootOf(_Z^3-2)^2+2
*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^4*x^2-8939892*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*
RootOf(_Z^3-2)^3*x^2-3511325*RootOf(_Z^3-2)^2*x^4-3286725*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*
RootOf(_Z^3-2)*x^4+2247248*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^4+2103504*RootOf
(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^3+34713150*(x^4+x^2)^(2/3)*RootOf(RootOf(_Z^3-2
)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2-30056942*RootOf(_Z^3-2)^2*x^2-28134366*RootOf(_Z^3-2)*RootOf(
RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x^2+10021887*RootOf(_Z^3-2)*(x^4+x^2)^(1/3)*x^2-14669376*RootOf(R
ootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*(x^4+x^2)^(1/3)*x^2-3511325*RootOf(_Z^3-2)^2-3286725*RootOf(RootOf
(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)[phone]*(x^4+x^2)^(1/3)*RootOf(_Z^3-2)-14669376*(x^4+x^
2)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)[phone]*(x^4+x^2)^(2/3))/(-1+x)^2/(1+x)^2)+1/2*R
ootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*ln((2247248*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_
Z^2)*RootOf(_Z^3-2)^4*x^4+2390992*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^3*x^4-9
550804*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^4*x^2-10161716*RootOf(RootOf(_Z^3-2)
^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^3*x^2+5758573*RootOf(_Z^3-2)^2*x^4+6126917*RootOf(RootOf(_Z^3-
2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)*x^4+2247248*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2
)*RootOf(_Z^3-2)^4+2390992*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)^2*RootOf(_Z^3-2)^3-34713150*(x^
4+x^2)^(2/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)^2+20506138*RootOf(_Z^3-2)^2*x^
2+21817802*RootOf(_Z^3-2)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*x^2-7334688*RootOf(_Z^3-2)*(x^4+
x^2)^(1/3)*x^2+20043774*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*(x^4+x^2)^(1/3)*x^2+5758573*RootOf
(_Z^3-2)^2+6126917*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)*RootOf(_Z^3-2)-7334688*(x^4+x^2)^(1/3)*
RootOf(_Z^3-2)[phone]*(x^4+x^2)^(1/3)*RootOf(RootOf(_Z^3-2)^2+2*_Z*RootOf(_Z^3-2)+4*_Z^2)-20043774*(x^4+x^2)
^(2/3))/(-1+x)^2/(1+x)^2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}}}{{\left (x^{2} - 1\right )} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4+x^2)^(1/3)/x/(x^2-1),x, algorithm="maxima")

[Out]

integrate((x^4 + x^2)^(1/3)/((x^2 - 1)*x), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} -\int \frac {{\left (x^4+x^2\right )}^{1/3}}{x-x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + x^4)^(1/3)/(x*(x^2 - 1)),x)

[Out]

-int((x^2 + x^4)^(1/3)/(x - x^3), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [3]{x^{2} \left (x^{2} + 1\right )}}{x \left (x - 1\right ) \left (x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4+x**2)**(1/3)/x/(x**2-1),x)

[Out]

Integral((x**2*(x**2 + 1))**(1/3)/(x*(x - 1)*(x + 1)), x)

________________________________________________________________________________________