3.28.38 \(\int \frac {1+x+x^2}{(-1+x^2) \sqrt [3]{x^2+x^4}} \, dx\) [2738]

Optimal. Leaf size=254 \[ -\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} x}{-x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{4 \sqrt [3]{2}}-\frac {3 \sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{4 \sqrt [3]{2}}+\frac {3 \log \left (-2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{4 \sqrt [3]{2}}-\frac {\log \left (2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{4 \sqrt [3]{2}}+\frac {\log \left (-2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}-\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{8 \sqrt [3]{2}}-\frac {3 \log \left (2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}+\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{8 \sqrt [3]{2}} \]

[Out]

-1/8*3^(1/2)*arctan(3^(1/2)*x/(-x+2^(2/3)*(x^4+x^2)^(1/3)))*2^(2/3)-3/8*3^(1/2)*arctan(3^(1/2)*x/(x+2^(2/3)*(x
^4+x^2)^(1/3)))*2^(2/3)+3/8*ln(-2*x+2^(2/3)*(x^4+x^2)^(1/3))*2^(2/3)-1/8*ln(2*x+2^(2/3)*(x^4+x^2)^(1/3))*2^(2/
3)+1/16*ln(-2*x^2+2^(2/3)*x*(x^4+x^2)^(1/3)-2^(1/3)*(x^4+x^2)^(2/3))*2^(2/3)-3/16*ln(2*x^2+2^(2/3)*x*(x^4+x^2)
^(1/3)+2^(1/3)*(x^4+x^2)^(2/3))*2^(2/3)

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Rubi [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 0.86, antiderivative size = 477, normalized size of antiderivative = 1.88, number of steps used = 32, number of rules used = 15, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {2081, 6857, 371, 973, 477, 440, 476, 502, 2174, 206, 31, 648, 631, 210, 642} \begin {gather*} -\frac {6 x \sqrt [3]{x^2+1} F_1\left (\frac {1}{6};1,\frac {1}{3};\frac {7}{6};x^2,-x^2\right )}{\sqrt [3]{x^4+x^2}}-\frac {\sqrt {3} x^{2/3} \sqrt [3]{x^2+1} \text {ArcTan}\left (\frac {1-\frac {2 \sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt [3]{x^4+x^2}}-\frac {\sqrt {3} x^{2/3} \sqrt [3]{x^2+1} \text {ArcTan}\left (\frac {\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{4 \sqrt [3]{2} \sqrt [3]{x^4+x^2}}+\frac {3 x \sqrt [3]{x^2+1} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^4+x^2}}-\frac {x^{2/3} \sqrt [3]{x^2+1} \log \left (\left (1-x^{2/3}\right )^2 \left (x^{2/3}+1\right )\right )}{8 \sqrt [3]{2} \sqrt [3]{x^4+x^2}}-\frac {x^{2/3} \sqrt [3]{x^2+1} \log \left (\frac {2^{2/3} \left (x^{2/3}+1\right )^2}{\left (x^2+1\right )^{2/3}}-\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1\right )}{4 \sqrt [3]{2} \sqrt [3]{x^4+x^2}}+\frac {x^{2/3} \sqrt [3]{x^2+1} \log \left (\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1\right )}{2 \sqrt [3]{2} \sqrt [3]{x^4+x^2}}+\frac {3 x^{2/3} \sqrt [3]{x^2+1} \log \left (x^{2/3}-2^{2/3} \sqrt [3]{x^2+1}+1\right )}{8 \sqrt [3]{2} \sqrt [3]{x^4+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*x*(1 + x^2)^(1/3)*AppellF1[1/6, 1, 1/3, 7/6, x^2, -x^2])/(x^2 + x^4)^(1/3) - (Sqrt[3]*x^(2/3)*(1 + x^2)^(1
/3)*ArcTan[(1 - (2*2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3))/Sqrt[3]])/(2*2^(1/3)*(x^2 + x^4)^(1/3)) - (Sqrt[3]*
x^(2/3)*(1 + x^2)^(1/3)*ArcTan[(1 + (2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3))/Sqrt[3]])/(4*2^(1/3)*(x^2 + x^4)^
(1/3)) + (3*x*(1 + x^2)^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -x^2])/(x^2 + x^4)^(1/3) - (x^(2/3)*(1 + x^2)^(
1/3)*Log[(1 - x^(2/3))^2*(1 + x^(2/3))])/(8*2^(1/3)*(x^2 + x^4)^(1/3)) - (x^(2/3)*(1 + x^2)^(1/3)*Log[1 + (2^(
2/3)*(1 + x^(2/3))^2)/(1 + x^2)^(2/3) - (2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3)])/(4*2^(1/3)*(x^2 + x^4)^(1/3)
) + (x^(2/3)*(1 + x^2)^(1/3)*Log[1 + (2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3)])/(2*2^(1/3)*(x^2 + x^4)^(1/3)) +
(3*x^(2/3)*(1 + x^2)^(1/3)*Log[1 + x^(2/3) - 2^(2/3)*(1 + x^2)^(1/3)])/(8*2^(1/3)*(x^2 + x^4)^(1/3))

Rule 31

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

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 210

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

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 476

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 477

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 502

Int[(x_)/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[b/a, 3]}, Dist[-q^2/(3
*d), Int[1/((1 - q*x)*(a + b*x^3)^(1/3)), x], x] + Dist[q/d, Subst[Int[1/(1 + 2*a*x^3), x], x, (1 + q*x)/(a +
b*x^3)^(1/3)], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[b*c + a*d, 0]

Rule 631

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 642

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 648

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 973

Int[(((g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[d*((g*x)^n/x^n), In
t[(x^n*(a + c*x^2)^p)/(d^2 - e^2*x^2), x], x] - Dist[e*((g*x)^n/x^n), Int[(x^(n + 1)*(a + c*x^2)^p)/(d^2 - e^2
*x^2), x], x] /; FreeQ[{a, c, d, e, g, n, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] &&  !IntegersQ[n, 2
*p]

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 2174

Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Simp[Sqrt[3]*(ArcTan[(1 - 2^(1/3)*Rt[b,
3]*((c - d*x)/(d*(a + b*x^3)^(1/3))))/Sqrt[3]]/(2^(4/3)*Rt[b, 3]*c)), x] + (Simp[Log[(c + d*x)^2*(c - d*x)]/(2
^(7/3)*Rt[b, 3]*c), x] - Simp[(3*Log[Rt[b, 3]*(c - d*x) + 2^(2/3)*d*(a + b*x^3)^(1/3)])/(2^(7/3)*Rt[b, 3]*c),
x]) /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 + a*d^3, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1+x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx &=\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1+x+x^2}{x^{2/3} \left (-1+x^2\right ) \sqrt [3]{1+x^2}} \, dx}{\sqrt [3]{x^2+x^4}}\\ &=\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \left (\frac {1}{x^{2/3} \sqrt [3]{1+x^2}}+\frac {2+x}{x^{2/3} \left (-1+x^2\right ) \sqrt [3]{1+x^2}}\right ) \, dx}{\sqrt [3]{x^2+x^4}}\\ &=\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1}{x^{2/3} \sqrt [3]{1+x^2}} \, dx}{\sqrt [3]{x^2+x^4}}+\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {2+x}{x^{2/3} \left (-1+x^2\right ) \sqrt [3]{1+x^2}} \, dx}{\sqrt [3]{x^2+x^4}}\\ &=\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}+\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \left (-\frac {3}{2 (1-x) x^{2/3} \sqrt [3]{1+x^2}}-\frac {1}{2 x^{2/3} (1+x) \sqrt [3]{1+x^2}}\right ) \, dx}{\sqrt [3]{x^2+x^4}}\\ &=\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1}{x^{2/3} (1+x) \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}-\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1}{(1-x) x^{2/3} \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}\\ &=\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1}{x^{2/3} \left (1-x^2\right ) \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}+\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {\sqrt [3]{x}}{\left (1-x^2\right ) \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}-\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1}{x^{2/3} \left (1-x^2\right ) \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}-\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {\sqrt [3]{x}}{\left (1-x^2\right ) \sqrt [3]{1+x^2}} \, dx}{2 \sqrt [3]{x^2+x^4}}\\ &=\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^6\right ) \sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{2 \sqrt [3]{x^2+x^4}}+\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {x^3}{\left (1-x^6\right ) \sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{2 \sqrt [3]{x^2+x^4}}-\frac {\left (9 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^6\right ) \sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{2 \sqrt [3]{x^2+x^4}}-\frac {\left (9 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {x^3}{\left (1-x^6\right ) \sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{2 \sqrt [3]{x^2+x^4}}\\ &=-\frac {6 x \sqrt [3]{1+x^2} F_1\left (\frac {1}{6};1,\frac {1}{3};\frac {7}{6};x^2,-x^2\right )}{\sqrt [3]{x^2+x^4}}+\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}+\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {x}{\left (1-x^3\right ) \sqrt [3]{1+x^3}} \, dx,x,x^{2/3}\right )}{4 \sqrt [3]{x^2+x^4}}-\frac {\left (9 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {x}{\left (1-x^3\right ) \sqrt [3]{1+x^3}} \, dx,x,x^{2/3}\right )}{4 \sqrt [3]{x^2+x^4}}\\ &=-\frac {6 x \sqrt [3]{1+x^2} F_1\left (\frac {1}{6};1,\frac {1}{3};\frac {7}{6};x^2,-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {3 x^2 \sqrt [3]{1+x^2} F_1\left (\frac {2}{3};1,\frac {1}{3};\frac {5}{3};x^2,-x^2\right )}{4 \sqrt [3]{x^2+x^4}}+\frac {3 x \sqrt [3]{1+x^2} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^2+x^4}}\\ \end {align*}

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Mathematica [A]
time = 0.74, size = 263, normalized size = 1.04 \begin {gather*} \frac {x^{2/3} \sqrt [3]{1+x^2} \left (2 \sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}-2^{2/3} \sqrt [3]{1+x^2}}\right )-6 \sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}}\right )+6 \log \left (-2 \sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}\right )-2 \log \left (2 \sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}\right )+\log \left (-2 x^{2/3}+2^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}-\sqrt [3]{2} \left (1+x^2\right )^{2/3}\right )-3 \log \left (2 x^{2/3}+2^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}+\sqrt [3]{2} \left (1+x^2\right )^{2/3}\right )\right )}{8 \sqrt [3]{2} \sqrt [3]{x^2+x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(2/3)*(1 + x^2)^(1/3)*(2*Sqrt[3]*ArcTan[(Sqrt[3]*x^(1/3))/(x^(1/3) - 2^(2/3)*(1 + x^2)^(1/3))] - 6*Sqrt[3]*
ArcTan[(Sqrt[3]*x^(1/3))/(x^(1/3) + 2^(2/3)*(1 + x^2)^(1/3))] + 6*Log[-2*x^(1/3) + 2^(2/3)*(1 + x^2)^(1/3)] -
2*Log[2*x^(1/3) + 2^(2/3)*(1 + x^2)^(1/3)] + Log[-2*x^(2/3) + 2^(2/3)*x^(1/3)*(1 + x^2)^(1/3) - 2^(1/3)*(1 + x
^2)^(2/3)] - 3*Log[2*x^(2/3) + 2^(2/3)*x^(1/3)*(1 + x^2)^(1/3) + 2^(1/3)*(1 + x^2)^(2/3)]))/(8*2^(1/3)*(x^2 +
x^4)^(1/3))

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {x^{2}+x +1}{\left (x^{2}-1\right ) \left (x^{4}+x^{2}\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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

[Out]

integral((x^4 + x^2)^(2/3)*(x^2 + x + 1)/(x^6 - x^2), x)

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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