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

Optimal. Leaf size=142 \[ \sqrt [3]{3} \log \left (3^{2/3} \sqrt [3]{2 x^3+x}-3 x\right )+3^{5/6} \tan ^{-1}\left (\frac {3^{5/6} x}{2 \sqrt [3]{2 x^3+x}+\sqrt [3]{3} x}\right )+\frac {3 \sqrt [3]{2 x^3+x} \left (10 x^2+1\right )}{8 x^3}-\frac {1}{2} \sqrt [3]{3} \log \left (3^{2/3} \sqrt [3]{2 x^3+x} x+\sqrt [3]{3} \left (2 x^3+x\right )^{2/3}+3 x^2\right ) \]

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Rubi [A]  time = 0.42, antiderivative size = 242, normalized size of antiderivative = 1.70, number of steps used = 13, number of rules used = 13, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.482, Rules used = {2056, 580, 583, 12, 466, 465, 494, 292, 31, 634, 617, 204, 628} \begin {gather*} \frac {15 \sqrt [3]{2 x^3+x}}{4 x}+\frac {3 \sqrt [3]{2 x^3+x}}{8 x^3}+\frac {\sqrt [3]{3} \sqrt [3]{2 x^3+x} \log \left (1-\frac {\sqrt [3]{3} x^{2/3}}{\sqrt [3]{2 x^2+1}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}-\frac {\sqrt [3]{3} \sqrt [3]{2 x^3+x} \log \left (\frac {3^{2/3} x^{4/3}}{\left (2 x^2+1\right )^{2/3}}+\frac {\sqrt [3]{3} x^{2/3}}{\sqrt [3]{2 x^2+1}}+1\right )}{2 \sqrt [3]{x} \sqrt [3]{2 x^2+1}}+\frac {3^{5/6} \sqrt [3]{2 x^3+x} \tan ^{-1}\left (\frac {2 x^{2/3}}{\sqrt [6]{3} \sqrt [3]{2 x^2+1}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(x + 2*x^3)^(1/3))/(8*x^3) + (15*(x + 2*x^3)^(1/3))/(4*x) + (3^(5/6)*(x + 2*x^3)^(1/3)*ArcTan[1/Sqrt[3] + (
2*x^(2/3))/(3^(1/6)*(1 + 2*x^2)^(1/3))])/(x^(1/3)*(1 + 2*x^2)^(1/3)) + (3^(1/3)*(x + 2*x^3)^(1/3)*Log[1 - (3^(
1/3)*x^(2/3))/(1 + 2*x^2)^(1/3)])/(x^(1/3)*(1 + 2*x^2)^(1/3)) - (3^(1/3)*(x + 2*x^3)^(1/3)*Log[1 + (3^(2/3)*x^
(4/3))/(1 + 2*x^2)^(2/3) + (3^(1/3)*x^(2/3))/(1 + 2*x^2)^(1/3)])/(2*x^(1/3)*(1 + 2*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 (1+x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (-1+x^2\right )} \, dx &=\frac {\sqrt [3]{x+2 x^3} \int \frac {\left (1+x^2\right ) \sqrt [3]{1+2 x^2}}{x^{11/3} \left (-1+x^2\right )} \, dx}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {3 \sqrt [3]{x+2 x^3}}{8 x^3}-\frac {\left (3 \sqrt [3]{x+2 x^3}\right ) \int \frac {-\frac {20}{3}-\frac {28 x^2}{3}}{x^{5/3} \left (-1+x^2\right ) \left (1+2 x^2\right )^{2/3}} \, dx}{8 \sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {3 \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {15 \sqrt [3]{x+2 x^3}}{4 x}-\frac {\left (9 \sqrt [3]{x+2 x^3}\right ) \int -\frac {32 \sqrt [3]{x}}{3 \left (-1+x^2\right ) \left (1+2 x^2\right )^{2/3}} \, dx}{16 \sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {3 \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {15 \sqrt [3]{x+2 x^3}}{4 x}+\frac {\left (6 \sqrt [3]{x+2 x^3}\right ) \int \frac {\sqrt [3]{x}}{\left (-1+x^2\right ) \left (1+2 x^2\right )^{2/3}} \, dx}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {3 \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {15 \sqrt [3]{x+2 x^3}}{4 x}+\frac {\left (18 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (-1+x^6\right ) \left (1+2 x^6\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {3 \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {15 \sqrt [3]{x+2 x^3}}{4 x}+\frac {\left (9 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (-1+x^3\right ) \left (1+2 x^3\right )^{2/3}} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {3 \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {15 \sqrt [3]{x+2 x^3}}{4 x}+\frac {\left (9 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {x}{-1+3 x^3} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {3 \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {15 \sqrt [3]{x+2 x^3}}{4 x}+\frac {\left (3^{2/3} \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{-1+\sqrt [3]{3} x} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}-\frac {\left (3^{2/3} \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {-1+\sqrt [3]{3} x}{1+\sqrt [3]{3} x+3^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {3 \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {15 \sqrt [3]{x+2 x^3}}{4 x}+\frac {\sqrt [3]{3} \sqrt [3]{x+2 x^3} \log \left (1-\frac {\sqrt [3]{3} x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}-\frac {\left (\sqrt [3]{3} \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{3}+2\ 3^{2/3} x}{1+\sqrt [3]{3} x+3^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{2 \sqrt [3]{x} \sqrt [3]{1+2 x^2}}+\frac {\left (3\ 3^{2/3} \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt [3]{3} x+3^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{2 \sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {3 \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {15 \sqrt [3]{x+2 x^3}}{4 x}+\frac {\sqrt [3]{3} \sqrt [3]{x+2 x^3} \log \left (1-\frac {\sqrt [3]{3} x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}-\frac {\sqrt [3]{3} \sqrt [3]{x+2 x^3} \log \left (1+\frac {3^{2/3} x^{4/3}}{\left (1+2 x^2\right )^{2/3}}+\frac {\sqrt [3]{3} x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{2 \sqrt [3]{x} \sqrt [3]{1+2 x^2}}-\frac {\left (3 \sqrt [3]{3} \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{3} x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {3 \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {15 \sqrt [3]{x+2 x^3}}{4 x}+\frac {3^{5/6} \sqrt [3]{x+2 x^3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{3} x^{2/3}}{\sqrt [3]{1+2 x^2}}}{\sqrt {3}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}+\frac {\sqrt [3]{3} \sqrt [3]{x+2 x^3} \log \left (1-\frac {\sqrt [3]{3} x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}-\frac {\sqrt [3]{3} \sqrt [3]{x+2 x^3} \log \left (1+\frac {3^{2/3} x^{4/3}}{\left (1+2 x^2\right )^{2/3}}+\frac {\sqrt [3]{3} x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{2 \sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ \end {align*}

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Mathematica [C]  time = 4.23, size = 103, normalized size = 0.73 \begin {gather*} \frac {3 \left (2 x^2+1\right ) \sqrt [3]{2 x^3+x} \left (2 \left (6 x^4-7 x^2+1\right ) \, _2F_1\left (1,1;\frac {2}{3};\frac {3 x^2}{x^2-1}\right )+27 \left (2 x^4+x^2\right ) \, _2F_1\left (2,2;\frac {5}{3};\frac {3 x^2}{x^2-1}\right )-\left (x^2-1\right )^2\right )}{8 x^3 \left (x^2-1\right )^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*(1 + 2*x^2)*(x + 2*x^3)^(1/3)*(-(-1 + x^2)^2 + 2*(1 - 7*x^2 + 6*x^4)*Hypergeometric2F1[1, 1, 2/3, (3*x^2)/(
-1 + x^2)] + 27*(x^2 + 2*x^4)*Hypergeometric2F1[2, 2, 5/3, (3*x^2)/(-1 + x^2)]))/(8*x^3*(-1 + x^2)^2)

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IntegrateAlgebraic [A]  time = 0.39, size = 142, normalized size = 1.00 \begin {gather*} \frac {3 \left (1+10 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}+3^{5/6} \tan ^{-1}\left (\frac {3^{5/6} x}{\sqrt [3]{3} x+2 \sqrt [3]{x+2 x^3}}\right )+\sqrt [3]{3} \log \left (-3 x+3^{2/3} \sqrt [3]{x+2 x^3}\right )-\frac {1}{2} \sqrt [3]{3} \log \left (3 x^2+3^{2/3} x \sqrt [3]{x+2 x^3}+\sqrt [3]{3} \left (x+2 x^3\right )^{2/3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(1 + 10*x^2)*(x + 2*x^3)^(1/3))/(8*x^3) + 3^(5/6)*ArcTan[(3^(5/6)*x)/(3^(1/3)*x + 2*(x + 2*x^3)^(1/3))] + 3
^(1/3)*Log[-3*x + 3^(2/3)*(x + 2*x^3)^(1/3)] - (3^(1/3)*Log[3*x^2 + 3^(2/3)*x*(x + 2*x^3)^(1/3) + 3^(1/3)*(x +
2*x^3)^(2/3)])/2

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fricas [B]  time = 1.79, size = 263, normalized size = 1.85 \begin {gather*} \frac {8 \cdot 3^{\frac {5}{6}} x^{3} \arctan \left (\frac {6 \cdot 3^{\frac {5}{6}} {\left (8 \, x^{4} - 7 \, x^{2} - 1\right )} {\left (2 \, x^{3} + x\right )}^{\frac {2}{3}} - \sqrt {3} {\left (377 \, x^{6} + 300 \, x^{4} + 51 \, x^{2} + 1\right )} - 18 \cdot 3^{\frac {1}{6}} {\left (55 \, x^{5} + 25 \, x^{3} + x\right )} {\left (2 \, x^{3} + x\right )}^{\frac {1}{3}}}{3 \, {\left (487 \, x^{6} + 240 \, x^{4} + 3 \, x^{2} - 1\right )}}\right ) - 4 \cdot 3^{\frac {1}{3}} x^{3} \log \left (\frac {3 \cdot 3^{\frac {2}{3}} {\left (2 \, x^{3} + x\right )}^{\frac {2}{3}} {\left (8 \, x^{2} + 1\right )} + 3^{\frac {1}{3}} {\left (55 \, x^{4} + 25 \, x^{2} + 1\right )} + 9 \, {\left (7 \, x^{3} + 2 \, x\right )} {\left (2 \, x^{3} + x\right )}^{\frac {1}{3}}}{x^{4} - 2 \, x^{2} + 1}\right ) + 8 \cdot 3^{\frac {1}{3}} x^{3} \log \left (-\frac {3^{\frac {2}{3}} {\left (x^{2} - 1\right )} - 9 \cdot 3^{\frac {1}{3}} {\left (2 \, x^{3} + x\right )}^{\frac {1}{3}} x + 9 \, {\left (2 \, x^{3} + x\right )}^{\frac {2}{3}}}{x^{2} - 1}\right ) + 9 \, {\left (2 \, x^{3} + x\right )}^{\frac {1}{3}} {\left (10 \, x^{2} + 1\right )}}{24 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(8*3^(5/6)*x^3*arctan(1/3*(6*3^(5/6)*(8*x^4 - 7*x^2 - 1)*(2*x^3 + x)^(2/3) - sqrt(3)*(377*x^6 + 300*x^4 +
51*x^2 + 1) - 18*3^(1/6)*(55*x^5 + 25*x^3 + x)*(2*x^3 + x)^(1/3))/(487*x^6 + 240*x^4 + 3*x^2 - 1)) - 4*3^(1/3
)*x^3*log((3*3^(2/3)*(2*x^3 + x)^(2/3)*(8*x^2 + 1) + 3^(1/3)*(55*x^4 + 25*x^2 + 1) + 9*(7*x^3 + 2*x)*(2*x^3 +
x)^(1/3))/(x^4 - 2*x^2 + 1)) + 8*3^(1/3)*x^3*log(-(3^(2/3)*(x^2 - 1) - 9*3^(1/3)*(2*x^3 + x)^(1/3)*x + 9*(2*x^
3 + x)^(2/3))/(x^2 - 1)) + 9*(2*x^3 + x)^(1/3)*(10*x^2 + 1))/x^3

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giac [A]  time = 0.23, size = 90, normalized size = 0.63 \begin {gather*} -3^{\frac {5}{6}} \arctan \left (\frac {1}{3} \cdot 3^{\frac {1}{6}} {\left (3^{\frac {1}{3}} + 2 \, {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {1}{3}}\right )}\right ) + \frac {3}{8} \, {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {4}{3}} - \frac {1}{2} \cdot 3^{\frac {1}{3}} \log \left (3^{\frac {2}{3}} + 3^{\frac {1}{3}} {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {2}{3}}\right ) + 3^{\frac {1}{3}} \log \left ({\left | -3^{\frac {1}{3}} + {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {1}{3}} \right |}\right ) + 3 \, {\left (\frac {1}{x^{2}} + 2\right )}^{\frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3^(5/6)*arctan(1/3*3^(1/6)*(3^(1/3) + 2*(1/x^2 + 2)^(1/3))) + 3/8*(1/x^2 + 2)^(4/3) - 1/2*3^(1/3)*log(3^(2/3)
+ 3^(1/3)*(1/x^2 + 2)^(1/3) + (1/x^2 + 2)^(2/3)) + 3^(1/3)*log(abs(-3^(1/3) + (1/x^2 + 2)^(1/3))) + 3*(1/x^2
+ 2)^(1/3)

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maple [C]  time = 20.50, size = 1155, normalized size = 8.13

method	result	size
trager	\(\text {Expression too large to display}\)	\(1155\)
risch	\(\text {Expression too large to display}\)	\(1860\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(10*x^2+1)*(2*x^3+x)^(1/3)/x^3-ln((102880*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*RootOf
(_Z^3-3)^4*x^2+6755328*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)^2*RootOf(_Z^3-3)^3*x^2+465408
*(2*x^3+x)^(2/3)*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*RootOf(_Z^3-3)^2-411520*RootOf(Root
Of(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*RootOf(_Z^3-3)^4-27021312*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf
(_Z^3-3)+82944*_Z^2)^2*RootOf(_Z^3-3)^3+28935*RootOf(_Z^3-3)^2*x^2+1899936*RootOf(RootOf(_Z^3-3)^2+288*_Z*Root
Of(_Z^3-3)+82944*_Z^2)*RootOf(_Z^3-3)*x^2+32883*(2*x^3+x)^(1/3)*RootOf(_Z^3-3)*x+1396224*(2*x^3+x)^(1/3)*RootO
f(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*x+32883*(2*x^3+x)^(2/3)+3215*RootOf(_Z^3-3)^2+211104*Root
Of(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*RootOf(_Z^3-3))/(-1+x)/(1+x))*RootOf(_Z^3-3)-288*ln((102
880*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*RootOf(_Z^3-3)^4*x^2+6755328*RootOf(RootOf(_Z^3-
3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)^2*RootOf(_Z^3-3)^3*x^2+465408*(2*x^3+x)^(2/3)*RootOf(RootOf(_Z^3-3)^2+2
88*_Z*RootOf(_Z^3-3)+82944*_Z^2)*RootOf(_Z^3-3)^2-411520*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_
Z^2)*RootOf(_Z^3-3)^4-27021312*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)^2*RootOf(_Z^3-3)^3+28
935*RootOf(_Z^3-3)^2*x^2+1899936*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*RootOf(_Z^3-3)*x^2+
32883*(2*x^3+x)^(1/3)*RootOf(_Z^3-3)*x+1396224*(2*x^3+x)^(1/3)*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+8
2944*_Z^2)*x+32883*(2*x^3+x)^(2/3)+3215*RootOf(_Z^3-3)^2+211104*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+
82944*_Z^2)*RootOf(_Z^3-3))/(-1+x)/(1+x))*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)+288*RootOf
(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*ln(-(238272*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+
82944*_Z^2)*RootOf(_Z^3-3)^4*x^2-20265984*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)^2*RootOf(_
Z^3-3)^3*x^2+1396224*(2*x^3+x)^(2/3)*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*RootOf(_Z^3-3)^
2-953088*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*RootOf(_Z^3-3)^4+81063936*RootOf(RootOf(_Z^
3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)^2*RootOf(_Z^3-3)^3-64532*RootOf(_Z^3-3)^2*x^2+5488704*RootOf(RootOf(_
Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*RootOf(_Z^3-3)*x^2-84105*(2*x^3+x)^(1/3)*RootOf(_Z^3-3)*x+4188672*(
2*x^3+x)^(1/3)*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*x-84105*(2*x^3+x)^(2/3)-17374*RootOf(
_Z^3-3)^2+1477728*RootOf(RootOf(_Z^3-3)^2+288*_Z*RootOf(_Z^3-3)+82944*_Z^2)*RootOf(_Z^3-3))/(-1+x)/(1+x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(((x + 2*x^3)^(1/3)*(x^2 + 1))/(x^4*(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 \left (2 x^{2} + 1\right )} \left (x^{2} + 1\right )}{x^{4} \left (x - 1\right ) \left (x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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