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

Optimal. Leaf size=208 \[ \frac {\log \left (-2 x^2+2^{2/3} \sqrt {3} \sqrt [3]{x^4-x^2} x-\sqrt [3]{2} \left (x^4-x^2\right )^{2/3}\right )}{4\ 2^{2/3}}-\frac {\log \left (2 x^2+\sqrt [3]{2} \left (x^4-x^2\right )^{2/3}\right )}{2\ 2^{2/3}}+\frac {\log \left (2 x^2+2^{2/3} \sqrt {3} \sqrt [3]{x^4-x^2} x+\sqrt [3]{2} \left (x^4-x^2\right )^{2/3}\right )}{4\ 2^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x^2}{\sqrt [3]{2} \left (x^4-x^2\right )^{2/3}-x^2}\right )}{2\ 2^{2/3}} \]

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Rubi [C]  time = 0.60, antiderivative size = 370, normalized size of antiderivative = 1.78, number of steps used = 9, number of rules used = 9, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1311, 2013, 622, 619, 236, 219, 2034, 758, 133} \begin {gather*} \frac {3 \left (\frac {x^2}{x^2+1}\right )^{2/3} \left (-\frac {1-x^2}{x^2+1}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},\frac {2}{3};\frac {7}{3};\frac {2}{x^2+1},\frac {1}{x^2+1}\right )}{4 \left (x^4-x^2\right )^{2/3}}-\frac {3^{3/4} \sqrt {2-\sqrt {3}} \left (x^2-x^4\right )^{2/3} \left (1-2^{2/3} \sqrt [3]{x^2-x^4}\right ) \sqrt {\frac {2 \sqrt [3]{2} \left (x^2-x^4\right )^{2/3}+2^{2/3} \sqrt [3]{x^2-x^4}+1}{\left (-2^{2/3} \sqrt [3]{x^2-x^4}-\sqrt {3}+1\right )^2}} F\left (\sin ^{-1}\left (\frac {-2^{2/3} \sqrt [3]{x^2-x^4}+\sqrt {3}+1}{-2^{2/3} \sqrt [3]{x^2-x^4}-\sqrt {3}+1}\right )|-7+4 \sqrt {3}\right )}{2^{2/3} \left (1-2 x^2\right ) \left (x^4-x^2\right )^{2/3} \sqrt {-\frac {1-2^{2/3} \sqrt [3]{x^2-x^4}}{\left (-2^{2/3} \sqrt [3]{x^2-x^4}-\sqrt {3}+1\right )^2}}} \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, 2/(1 + x^2), (1 + x^2)^(-
1)])/(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 - x^4)^(1/3))*Sqr
t[(1 + 2^(2/3)*(x^2 - x^4)^(1/3) + 2*2^(1/3)*(x^2 - x^4)^(2/3))/(1 - Sqrt[3] - 2^(2/3)*(x^2 - x^4)^(1/3))^2]*E
llipticF[ArcSin[(1 + Sqrt[3] - 2^(2/3)*(x^2 - x^4)^(1/3))/(1 - Sqrt[3] - 2^(2/3)*(x^2 - x^4)^(1/3))], -7 + 4*S
qrt[3]])/(2^(2/3)*(1 - 2*x^2)*(-x^2 + x^4)^(2/3)*Sqrt[-((1 - 2^(2/3)*(x^2 - x^4)^(1/3))/(1 - Sqrt[3] - 2^(2/3)
*(x^2 - x^4)^(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 &=-\left (2 \int \frac {x}{\left (1+x^2\right ) \left (-x^2+x^4\right )^{2/3}} \, dx\right )+\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}}{(1-2 x)^{2/3} (1-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 {2}{1+x^2},\frac {1}{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 {2}{1+x^2},\frac {1}{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 {2}{1+x^2},\frac {1}{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*}

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Mathematica [C]  time = 0.03, size = 47, normalized size = 0.23 \begin {gather*} \frac {3 \sqrt [3]{x^2 \left (x^2-1\right )} F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};x^2,-x^2\right )}{2 \sqrt [3]{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*(-1 + x^2))^(1/3)*AppellF1[1/3, -1/3, 1, 4/3, x^2, -x^2])/(2*(1 - x^2)^(1/3))

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IntegrateAlgebraic [A]  time = 0.39, size = 208, 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^{2/3} \sqrt {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+\sqrt [3]{2} \left (-x^2+x^4\right )^{2/3}\right )}{2\ 2^{2/3}}+\frac {\log \left (2 x^2+2^{2/3} \sqrt {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^(2/3)*Sqrt[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^(1/3)*(-x^2 + x^4)^(2/3)]/(2*
2^(2/3)) + Log[2*x^2 + 2^(2/3)*Sqrt[3]*x*(-x^2 + x^4)^(1/3) + 2^(1/3)*(-x^2 + x^4)^(2/3)]/(4*2^(2/3))

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fricas [B]  time = 2.18, size = 365, normalized size = 1.75 \begin {gather*} -\frac {1}{12} \cdot 4^{\frac {1}{6}} \sqrt {3} \left (-1\right )^{\frac {1}{3}} \arctan \left (-\frac {4^{\frac {1}{6}} \sqrt {3} {\left (6 \cdot 4^{\frac {2}{3}} \left (-1\right )^{\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 (-1\right )^{\frac {1}{3}} {\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}} \left (-1\right )^{\frac {1}{3}} \log \left (\frac {24 \cdot 4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{4} - x^{2}\right )}^{\frac {2}{3}} {\left (x^{4} - 4 \, x^{2} + 1\right )} - 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{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}} \left (-1\right )^{\frac {1}{3}} \log \left (-\frac {3 \cdot 4^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{4} - x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} - 1\right )} - 4^{\frac {1}{3}} \left (-1\right )^{\frac {2}{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)*(-1)^(1/3)*arctan(-1/6*4^(1/6)*sqrt(3)*(6*4^(2/3)*(-1)^(2/3)*(x^10 - 33*x^8 + 110*x^6 -
110*x^4 + 33*x^2 - 1)*(x^4 - x^2)^(1/3) - 48*(-1)^(1/3)*(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 + 44
7*x^4 - 102*x^2 + 1)) - 1/48*4^(2/3)*(-1)^(1/3)*log((24*4^(1/3)*(-1)^(2/3)*(x^4 - x^2)^(2/3)*(x^4 - 4*x^2 + 1)
- 4^(2/3)*(-1)^(1/3)*(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)*(-1)^(1/3)*log(-(3*4^(2/3)*(-1)^(1/3)*(x^4 - x^2)^(1/3)*(x^
2 - 1) - 4^(1/3)*(-1)^(2/3)*(x^4 + 2*x^2 + 1) - 12*(x^4 - x^2)^(2/3))/(x^4 + 2*x^2 + 1))

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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)

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maple [C]  time = 25.62, size = 1702, normalized size = 8.18

method	result	size
trager	\(\text {Expression too large to display}\)	\(1702\)

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*ln(-(5842000*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^4*x^4+11606848*RootOf(Roo
tOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^3*x^4-24828500*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(
_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^4*x^2-49329104*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^
3+2)^3*x^2-2555875*x^4*RootOf(_Z^3+2)^2-5077996*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^
3+2)*x^4+33610788*(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+5842000
*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^4+11606848*RootOf(RootOf(_Z^3+2)^2+2*_Z*Ro
otOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^3+1339950*(x^4-x^2)^(1/3)*RootOf(_Z^3+2)*x^2-33610788*(x^4-x^2)^(1/3)*Ro
otOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*x^2+28479750*x^2*RootOf(_Z^3+2)^2+56583384*x^2*RootOf(RootOf
(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)+2679900*(x^4-x^2)^(2/3)-1339950*(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)-2555875*RootOf(_Z^3+2)^2-5
077996*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2))/(x^2+1)^2)*RootOf(_Z^3+2)-1/2*ln(-(
5842000*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^4*x^4+11606848*RootOf(RootOf(_Z^3+2
)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^3*x^4-24828500*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*
_Z^2)*RootOf(_Z^3+2)^4*x^2-49329104*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^3*x^2
-2555875*x^4*RootOf(_Z^3+2)^2-5077996*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)*x^4+3
3610788*(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+5842000*RootOf(Ro
otOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^4+11606848*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+
2)+4*_Z^2)^2*RootOf(_Z^3+2)^3+1339950*(x^4-x^2)^(1/3)*RootOf(_Z^3+2)*x^2-33610788*(x^4-x^2)^(1/3)*RootOf(RootO
f(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*x^2+28479750*x^2*RootOf(_Z^3+2)^2+56583384*x^2*RootOf(RootOf(_Z^3+2)^2
+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)+2679900*(x^4-x^2)^(2/3)-1339950*(x^4-x^2)^(1/3)*RootOf(_Z^3+2)+336
10788*(x^4-x^2)^(1/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)-2555875*RootOf(_Z^3+2)^2-5077996*Roo
tOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2))/(x^2+1)^2)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootO
f(_Z^3+2)+4*_Z^2)+1/2*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*ln((38576*RootOf(RootOf(_Z^3+2)^2+2*
_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^4*x^4-11606848*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*
RootOf(_Z^3+2)^3*x^4-163948*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)^4*x^2+49329104*
RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^3*x^2-21699*x^4*RootOf(_Z^3+2)^2+6528852*
RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)*x^4+33610788*(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+38576*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^
2)*RootOf(_Z^3+2)^4-11606848*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)^2*RootOf(_Z^3+2)^3-18145344*(
x^4-x^2)^(1/3)*RootOf(_Z^3+2)*x^2-33610788*(x^4-x^2)^(1/3)*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)
*x^2-24110*x^2*RootOf(_Z^3+2)^2+7254280*x^2*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*_Z^2)*RootOf(_Z^3+2)
-36290688*(x^4-x^2)^(2/3)[phone]*(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)-21699*RootOf(_Z^3+2)^2+6528852*RootOf(RootOf(_Z^3+2)^2+2*_Z*RootOf(_Z^3+2)+4*
_Z^2)*RootOf(_Z^3+2))/(x^2+1)^2)

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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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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