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

Optimal. Leaf size=140 \[ -\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{x^3+x}-x\right )}{4 \sqrt [3]{2}}-\frac {3 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{2} \sqrt [3]{x^3+x}+x}\right )}{4 \sqrt [3]{2}}-\frac {3 \sqrt [3]{x^3+x} \left (2 x^2-1\right )}{4 x^3}+\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{x^3+x} x+2^{2/3} \left (x^3+x\right )^{2/3}+x^2\right )}{8 \sqrt [3]{2}} \]

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Rubi [A]  time = 0.38, antiderivative size = 228, normalized size of antiderivative = 1.63, number of steps used = 13, number of rules used = 13, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {2056, 580, 583, 12, 466, 465, 494, 292, 31, 634, 617, 204, 628} \begin {gather*} -\frac {3 \sqrt [3]{x^3+x}}{2 x}+\frac {3 \sqrt [3]{x^3+x}}{4 x^3}-\frac {3 \sqrt [3]{x^3+x} \log \left (\sqrt [3]{2}-\frac {x^{2/3}}{\sqrt [3]{x^2+1}}\right )}{4 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{x^2+1}}+\frac {3 \sqrt [3]{x^3+x} \log \left (\frac {x^{4/3}}{\left (x^2+1\right )^{2/3}}+\frac {\sqrt [3]{2} x^{2/3}}{\sqrt [3]{x^2+1}}+2^{2/3}\right )}{8 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{x^2+1}}-\frac {3 \sqrt {3} \sqrt [3]{x^3+x} \tan ^{-1}\left (\frac {\frac {2^{2/3} x^{2/3}}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{4 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(x + x^3)^(1/3))/(4*x^3) - (3*(x + x^3)^(1/3))/(2*x) - (3*Sqrt[3]*(x + x^3)^(1/3)*ArcTan[(1 + (2^(2/3)*x^(2
/3))/(1 + x^2)^(1/3))/Sqrt[3]])/(4*2^(1/3)*x^(1/3)*(1 + x^2)^(1/3)) - (3*(x + x^3)^(1/3)*Log[2^(1/3) - x^(2/3)
/(1 + x^2)^(1/3)])/(4*2^(1/3)*x^(1/3)*(1 + x^2)^(1/3)) + (3*(x + x^3)^(1/3)*Log[2^(2/3) + x^(4/3)/(1 + x^2)^(2
/3) + (2^(1/3)*x^(2/3))/(1 + x^2)^(1/3)])/(8*2^(1/3)*x^(1/3)*(1 + x^2)^(1/3))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 292

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

Rule 465

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

Rule 466

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> With[{k = Deno
minator[m]}, Dist[k/e, Subst[Int[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/e^n)^p*(c + (d*x^(k*n))/e^n)^q, x], x, (e*
x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && FractionQ[m] && Intege
rQ[p]

Rule 494

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[(k*a^(p + (m + 1)/n))/n, Subst[Int[(x^((k*(m + 1))/n - 1)*(c - (b*c - a*d)*x^k)^q)/(1 - b*x^k)^(p
+ q + (m + 1)/n + 1), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 0]

Rule 580

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*g*(m + 1)), x] - Dist[1/(a*g^n*(m + 1
)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c*(p + 1) + a*d*q)
+ d*((b*e - a*f)*(m + 1) + b*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && IGtQ[n
, 0] && GtQ[q, 0] && LtQ[m, -1] &&  !(EqQ[q, 1] && SimplerQ[e + f*x^n, c + d*x^n])

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

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 634

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

Rule 2056

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]

Rubi steps

\begin {align*} \int \frac {\left (-4+x^2\right ) \sqrt [3]{x+x^3}}{x^4 \left (2+x^2\right )} \, dx &=\frac {\sqrt [3]{x+x^3} \int \frac {\left (-4+x^2\right ) \sqrt [3]{1+x^2}}{x^{11/3} \left (2+x^2\right )} \, dx}{\sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {3 \sqrt [3]{x+x^3}}{4 x^3}+\frac {\left (3 \sqrt [3]{x+x^3}\right ) \int \frac {\frac {32}{3}+\frac {40 x^2}{3}}{x^{5/3} \left (1+x^2\right )^{2/3} \left (2+x^2\right )} \, dx}{16 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {3 \sqrt [3]{x+x^3}}{4 x^3}-\frac {3 \sqrt [3]{x+x^3}}{2 x}-\frac {\left (9 \sqrt [3]{x+x^3}\right ) \int -\frac {32 \sqrt [3]{x}}{3 \left (1+x^2\right )^{2/3} \left (2+x^2\right )} \, dx}{64 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {3 \sqrt [3]{x+x^3}}{4 x^3}-\frac {3 \sqrt [3]{x+x^3}}{2 x}+\frac {\left (3 \sqrt [3]{x+x^3}\right ) \int \frac {\sqrt [3]{x}}{\left (1+x^2\right )^{2/3} \left (2+x^2\right )} \, dx}{2 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {3 \sqrt [3]{x+x^3}}{4 x^3}-\frac {3 \sqrt [3]{x+x^3}}{2 x}+\frac {\left (9 \sqrt [3]{x+x^3}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (1+x^6\right )^{2/3} \left (2+x^6\right )} \, dx,x,\sqrt [3]{x}\right )}{2 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {3 \sqrt [3]{x+x^3}}{4 x^3}-\frac {3 \sqrt [3]{x+x^3}}{2 x}+\frac {\left (9 \sqrt [3]{x+x^3}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+x^3\right )^{2/3} \left (2+x^3\right )} \, dx,x,x^{2/3}\right )}{4 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {3 \sqrt [3]{x+x^3}}{4 x^3}-\frac {3 \sqrt [3]{x+x^3}}{2 x}+\frac {\left (9 \sqrt [3]{x+x^3}\right ) \operatorname {Subst}\left (\int \frac {x}{2-x^3} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {3 \sqrt [3]{x+x^3}}{4 x^3}-\frac {3 \sqrt [3]{x+x^3}}{2 x}+\frac {\left (3 \sqrt [3]{x+x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{2}-x} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{1+x^2}}-\frac {\left (3 \sqrt [3]{x+x^3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{2}-x}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {3 \sqrt [3]{x+x^3}}{4 x^3}-\frac {3 \sqrt [3]{x+x^3}}{2 x}-\frac {3 \sqrt [3]{x+x^3} \log \left (\sqrt [3]{2}-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{1+x^2}}-\frac {\left (9 \sqrt [3]{x+x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{8 \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left (3 \sqrt [3]{x+x^3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{2}+2 x}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{8 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {3 \sqrt [3]{x+x^3}}{4 x^3}-\frac {3 \sqrt [3]{x+x^3}}{2 x}-\frac {3 \sqrt [3]{x+x^3} \log \left (\sqrt [3]{2}-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {3 \sqrt [3]{x+x^3} \log \left (2^{2/3}+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {\sqrt [3]{2} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{8 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {\left (9 \sqrt [3]{x+x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2^{2/3} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ &=\frac {3 \sqrt [3]{x+x^3}}{4 x^3}-\frac {3 \sqrt [3]{x+x^3}}{2 x}-\frac {3 \sqrt {3} \sqrt [3]{x+x^3} \tan ^{-1}\left (\frac {1+\frac {2^{2/3} x^{2/3}}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{4 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{1+x^2}}-\frac {3 \sqrt [3]{x+x^3} \log \left (\sqrt [3]{2}-\frac {x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{4 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{1+x^2}}+\frac {3 \sqrt [3]{x+x^3} \log \left (2^{2/3}+\frac {x^{4/3}}{\left (1+x^2\right )^{2/3}}+\frac {\sqrt [3]{2} x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{8 \sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{1+x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.09, size = 102, normalized size = 0.73 \begin {gather*} \frac {3 \sqrt [3]{x^3+x} \left (9 \left (x^2+2\right ) x^2 \, _2F_1\left (\frac {2}{3},2;\frac {5}{3};\frac {x^2}{2 x^2+2}\right )+\left (9 x^4-6 x^2\right ) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {x^2}{2 x^2+2}\right )-4 \left (11 x^4+7 x^2-4\right )\right )}{64 x^3 \left (x^2+1\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*(x + x^3)^(1/3)*(-4*(-4 + 7*x^2 + 11*x^4) + (-6*x^2 + 9*x^4)*Hypergeometric2F1[2/3, 1, 5/3, x^2/(2 + 2*x^2)
] + 9*x^2*(2 + x^2)*Hypergeometric2F1[2/3, 2, 5/3, x^2/(2 + 2*x^2)]))/(64*x^3*(1 + x^2))

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IntegrateAlgebraic [A]  time = 0.41, size = 140, normalized size = 1.00 \begin {gather*} -\frac {3 \left (-1+2 x^2\right ) \sqrt [3]{x+x^3}}{4 x^3}-\frac {3 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{2} \sqrt [3]{x+x^3}}\right )}{4 \sqrt [3]{2}}-\frac {3 \log \left (-x+\sqrt [3]{2} \sqrt [3]{x+x^3}\right )}{4 \sqrt [3]{2}}+\frac {3 \log \left (x^2+\sqrt [3]{2} x \sqrt [3]{x+x^3}+2^{2/3} \left (x+x^3\right )^{2/3}\right )}{8 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(-1 + 2*x^2)*(x + x^3)^(1/3))/(4*x^3) - (3*Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x + 2*2^(1/3)*(x + x^3)^(1/3))])/(4
*2^(1/3)) - (3*Log[-x + 2^(1/3)*(x + x^3)^(1/3)])/(4*2^(1/3)) + (3*Log[x^2 + 2^(1/3)*x*(x + x^3)^(1/3) + 2^(2/
3)*(x + x^3)^(2/3)])/(8*2^(1/3))

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fricas [B]  time = 2.52, size = 288, normalized size = 2.06 \begin {gather*} \frac {2 \, \sqrt {3} 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} x^{3} \arctan \left (\frac {\sqrt {3} 2^{\frac {1}{6}} {\left (24 \, \sqrt {2} \left (-1\right )^{\frac {1}{3}} {\left (2 \, x^{4} + 5 \, x^{2} + 2\right )} {\left (x^{3} + x\right )}^{\frac {2}{3}} - 12 \cdot 2^{\frac {1}{6}} \left (-1\right )^{\frac {2}{3}} {\left (19 \, x^{5} + 22 \, x^{3} + 4 \, x\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}} - 2^{\frac {5}{6}} {\left (91 \, x^{6} + 168 \, x^{4} + 84 \, x^{2} + 8\right )}\right )}}{6 \, {\left (53 \, x^{6} + 48 \, x^{4} - 12 \, x^{2} - 8\right )}}\right ) - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} x^{3} \log \left (\frac {12 \cdot 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{3} + x\right )}^{\frac {2}{3}} {\left (2 \, x^{2} + 1\right )} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (19 \, x^{4} + 22 \, x^{2} + 4\right )} + 6 \, {\left (5 \, x^{3} + 4 \, x\right )} {\left (x^{3} + x\right )}^{\frac {1}{3}}}{x^{4} + 4 \, x^{2} + 4}\right ) + 2 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} x^{3} \log \left (\frac {3 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{3} + x\right )}^{\frac {1}{3}} x - 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{2} + 2\right )} + 6 \, {\left (x^{3} + x\right )}^{\frac {2}{3}}}{x^{2} + 2}\right ) - 12 \, {\left (x^{3} + x\right )}^{\frac {1}{3}} {\left (2 \, x^{2} - 1\right )}}{16 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*sqrt(3)*2^(2/3)*(-1)^(1/3)*x^3*arctan(1/6*sqrt(3)*2^(1/6)*(24*sqrt(2)*(-1)^(1/3)*(2*x^4 + 5*x^2 + 2)*(
x^3 + x)^(2/3) - 12*2^(1/6)*(-1)^(2/3)*(19*x^5 + 22*x^3 + 4*x)*(x^3 + x)^(1/3) - 2^(5/6)*(91*x^6 + 168*x^4 + 8
4*x^2 + 8))/(53*x^6 + 48*x^4 - 12*x^2 - 8)) - 2^(2/3)*(-1)^(1/3)*x^3*log((12*2^(1/3)*(-1)^(2/3)*(x^3 + x)^(2/3
)*(2*x^2 + 1) - 2^(2/3)*(-1)^(1/3)*(19*x^4 + 22*x^2 + 4) + 6*(5*x^3 + 4*x)*(x^3 + x)^(1/3))/(x^4 + 4*x^2 + 4))
+ 2*2^(2/3)*(-1)^(1/3)*x^3*log((3*2^(2/3)*(-1)^(1/3)*(x^3 + x)^(1/3)*x - 2^(1/3)*(-1)^(2/3)*(x^2 + 2) + 6*(x^
3 + x)^(2/3))/(x^2 + 2)) - 12*(x^3 + x)^(1/3)*(2*x^2 - 1))/x^3

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giac [A]  time = 0.18, size = 97, normalized size = 0.69 \begin {gather*} \frac {3}{4} \, \sqrt {3} \left (\frac {1}{2}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} \left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (\left (\frac {1}{2}\right )^{\frac {1}{3}} + 2 \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}}\right )}\right ) + \frac {3}{4} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {4}{3}} + \frac {3}{16} \cdot 4^{\frac {1}{3}} \log \left (\left (\frac {1}{2}\right )^{\frac {2}{3}} + \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {2}{3}}\right ) - \frac {3}{4} \, \left (\frac {1}{2}\right )^{\frac {1}{3}} \log \left ({\left | -\left (\frac {1}{2}\right )^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \right |}\right ) - \frac {9}{4} \, {\left (\frac {1}{x^{2}} + 1\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*sqrt(3)*(1/2)^(1/3)*arctan(2/3*sqrt(3)*(1/2)^(2/3)*((1/2)^(1/3) + 2*(1/x^2 + 1)^(1/3))) + 3/4*(1/x^2 + 1)^
(4/3) + 3/16*4^(1/3)*log((1/2)^(2/3) + (1/2)^(1/3)*(1/x^2 + 1)^(1/3) + (1/x^2 + 1)^(2/3)) - 3/4*(1/2)^(1/3)*lo
g(abs(-(1/2)^(1/3) + (1/x^2 + 1)^(1/3))) - 9/4*(1/x^2 + 1)^(1/3)

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maple [C]  time = 19.53, size = 1101, normalized size = 7.86

method	result	size
trager	\(\text {Expression too large to display}\)	\(1101\)
risch	\(\text {Expression too large to display}\)	\(1570\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/4*(2*x^2-1)*(x^3+x)^(1/3)/x^3-3/8*ln((36*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)^2*RootOf(_Z
^3+4)^3*x^2-6*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)^4*x^2-36*RootOf(RootOf(_Z^
3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)^2*RootOf(_Z^3+4)^3+6*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z
^2)*RootOf(_Z^3+4)^4+144*(x^3+x)^(2/3)*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)^2
+66*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)*x^2-11*RootOf(_Z^3+4)^2*x^2-108*Root
Of(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*(x^3+x)^(1/3)*x-24*(x^3+x)^(1/3)*RootOf(_Z^3+4)*x+60*RootOf
(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)-10*RootOf(_Z^3+4)^2-30*(x^3+x)^(2/3))/(x^2+2))
*RootOf(_Z^3+4)-9/2*ln((36*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)^2*RootOf(_Z^3+4)^3*x^2-6*Roo
tOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)^4*x^2-36*RootOf(RootOf(_Z^3+4)^2+12*_Z*Root
Of(_Z^3+4)+144*_Z^2)^2*RootOf(_Z^3+4)^3+6*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4
)^4+144*(x^3+x)^(2/3)*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)^2+66*RootOf(RootOf
(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)*x^2-11*RootOf(_Z^3+4)^2*x^2-108*RootOf(RootOf(_Z^3+4)
^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*(x^3+x)^(1/3)*x-24*(x^3+x)^(1/3)*RootOf(_Z^3+4)*x+60*RootOf(RootOf(_Z^3+4)^2
+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)-10*RootOf(_Z^3+4)^2-30*(x^3+x)^(2/3))/(x^2+2))*RootOf(RootOf(_Z
^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)+3/8*RootOf(_Z^3+4)*ln((144*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)
+144*_Z^2)^2*RootOf(_Z^3+4)^3*x^2-6*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)^4*x^
2-144*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)^2*RootOf(_Z^3+4)^3+6*RootOf(RootOf(_Z^3+4)^2+12*_
Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)^4-288*(x^3+x)^(2/3)*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144
*_Z^2)*RootOf(_Z^3+4)^2-312*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)*x^2+13*RootO
f(_Z^3+4)^2*x^2+360*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*(x^3+x)^(1/3)*x+48*(x^3+x)^(1/3)*Ro
otOf(_Z^3+4)*x-192*RootOf(RootOf(_Z^3+4)^2+12*_Z*RootOf(_Z^3+4)+144*_Z^2)*RootOf(_Z^3+4)+8*RootOf(_Z^3+4)^2+36
*(x^3+x)^(2/3))/(x^2+2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {3 \, {\left (18 \, x^{5} + 7 \, {\left (x^{3} + x\right )} x^{2} + 8 \, x^{3} - 10 \, x\right )} {\left (x^{2} + 1\right )}^{\frac {1}{3}}}{56 \, {\left (x^{\frac {17}{3}} + 2 \, x^{\frac {11}{3}}\right )}} + \int \frac {9 \, {\left (3 \, x^{4} - x^{2} - 4\right )} {\left (x^{2} + 1\right )}^{\frac {1}{3}}}{7 \, {\left (x^{\frac {23}{3}} + 4 \, x^{\frac {17}{3}} + 4 \, x^{\frac {11}{3}}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/56*(18*x^5 + 7*(x^3 + x)*x^2 + 8*x^3 - 10*x)*(x^2 + 1)^(1/3)/(x^(17/3) + 2*x^(11/3)) + integrate(9/7*(3*x^4
- x^2 - 4)*(x^2 + 1)^(1/3)/(x^(23/3) + 4*x^(17/3) + 4*x^(11/3)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (x^2-4\right )\,{\left (x^3+x\right )}^{1/3}}{x^4\,\left (x^2+2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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