\(\int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx\) [1790]

Optimal result

Integrand size = 20, antiderivative size = 121 \[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{\sqrt [3]{2}}+\frac {\log \left (-2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{\sqrt [3]{2}}-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}+\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{2 \sqrt [3]{2}} \]

[Out]

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

Rubi [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 0.55 (sec) , antiderivative size = 472, normalized size of antiderivative = 3.90, number of steps used = 17, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {2081, 6865, 6857, 251, 1452, 440, 476, 502, 2174, 206, 31, 648, 631, 210, 642} \[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=-\frac {6 x \sqrt [3]{x^2+1} \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{3},\frac {7}{6},x^2,-x^2\right )}{\sqrt [3]{x^4+x^2}}-\frac {\sqrt {3} x^{2/3} \sqrt [3]{x^2+1} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{x^4+x^2}}-\frac {\sqrt {3} x^{2/3} \sqrt [3]{x^2+1} \arctan \left (\frac {\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt [3]{x^4+x^2}}+\frac {3 x \sqrt [3]{x^2+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-x^2\right )}{\sqrt [3]{x^4+x^2}}-\frac {x^{2/3} \sqrt [3]{x^2+1} \log \left (\left (1-x^{2/3}\right )^2 \left (x^{2/3}+1\right )\right )}{4 \sqrt [3]{2} \sqrt [3]{x^4+x^2}}-\frac {x^{2/3} \sqrt [3]{x^2+1} \log \left (\frac {2^{2/3} \left (x^{2/3}+1\right )^2}{\left (x^2+1\right )^{2/3}}-\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1\right )}{2 \sqrt [3]{2} \sqrt [3]{x^4+x^2}}+\frac {x^{2/3} \sqrt [3]{x^2+1} \log \left (\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1\right )}{\sqrt [3]{2} \sqrt [3]{x^4+x^2}}+\frac {3 x^{2/3} \sqrt [3]{x^2+1} \log \left (x^{2/3}-2^{2/3} \sqrt [3]{x^2+1}+1\right )}{4 \sqrt [3]{2} \sqrt [3]{x^4+x^2}} \]

[In]

Int[(1 + x)/((-1 + x)*(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^(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]])/(2*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))])/(4*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)])/(2*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^(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)])/(4*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 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || 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 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 1452

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^(2
*n))^p, (d/(d^2 - e^2*x^(2*n)) - e*(x^n/(d^2 - e^2*x^(2*n))))^(-q), x], x] /; FreeQ[{a, c, d, e, n, p}, x] &&
EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[q, 0]

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]

Rule 6865

Int[(u_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k, Subst[Int[x^(k*(m + 1) - 1)*(u /. x -> x^k
), x], x, x^(1/k)], x]] /; FractionQ[m]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \int \frac {1+x}{(-1+x) x^{2/3} \sqrt [3]{1+x^2}} \, dx}{\sqrt [3]{x^2+x^4}} \\ & = \frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {1+x^3}{\left (-1+x^3\right ) \sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2+x^4}} \\ & = \frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \left (\frac {1}{\sqrt [3]{1+x^6}}+\frac {2}{\left (-1+x^3\right ) \sqrt [3]{1+x^6}}\right ) \, dx,x,\sqrt [3]{x}\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}{\sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2+x^4}}+\frac {\left (6 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{\left (-1+x^3\right ) \sqrt [3]{1+x^6}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2+x^4}} \\ & = \frac {3 x \sqrt [3]{1+x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-x^2\right )}{\sqrt [3]{x^2+x^4}}+\frac {\left (6 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \left (\frac {1}{\left (-1+x^6\right ) \sqrt [3]{1+x^6}}+\frac {x^3}{\left (-1+x^6\right ) \sqrt [3]{1+x^6}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x^2+x^4}} \\ & = \frac {3 x \sqrt [3]{1+x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-x^2\right )}{\sqrt [3]{x^2+x^4}}+\frac {\left (6 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 )}{\sqrt [3]{x^2+x^4}}+\frac {\left (6 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 )}{\sqrt [3]{x^2+x^4}} \\ & = -\frac {6 x \sqrt [3]{1+x^2} \operatorname {AppellF1}\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} \operatorname {Hypergeometric2F1}\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 )}{\sqrt [3]{x^2+x^4}} \\ & = -\frac {6 x \sqrt [3]{1+x^2} \operatorname {AppellF1}\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} \operatorname {Hypergeometric2F1}\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 ) \text {Subst}\left (\int \frac {1}{(1-x) \sqrt [3]{1+x^3}} \, dx,x,x^{2/3}\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}{1+2 x^3} \, dx,x,\frac {1+x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{\sqrt [3]{x^2+x^4}} \\ & = -\frac {6 x \sqrt [3]{1+x^2} \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{3},\frac {7}{6},x^2,-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {\sqrt {3} x^{2/3} \sqrt [3]{1+x^2} \arctan \left (\frac {1+\frac {\sqrt [3]{2} \left (1+x^{2/3}\right )}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {3 x \sqrt [3]{1+x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {x^{2/3} \sqrt [3]{1+x^2} \log \left (\left (1-x^{2/3}\right )^2 \left (1+x^{2/3}\right )\right )}{4 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {3 x^{2/3} \sqrt [3]{1+x^2} \log \left (1+x^{2/3}-2^{2/3} \sqrt [3]{1+x^2}\right )}{4 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt [3]{2} x} \, dx,x,\frac {1+x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{\sqrt [3]{x^2+x^4}}+\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {2-\sqrt [3]{2} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx,x,\frac {1+x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{\sqrt [3]{x^2+x^4}} \\ & = -\frac {6 x \sqrt [3]{1+x^2} \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{3},\frac {7}{6},x^2,-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {\sqrt {3} x^{2/3} \sqrt [3]{1+x^2} \arctan \left (\frac {1+\frac {\sqrt [3]{2} \left (1+x^{2/3}\right )}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {3 x \sqrt [3]{1+x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {x^{2/3} \sqrt [3]{1+x^2} \log \left (\left (1-x^{2/3}\right )^2 \left (1+x^{2/3}\right )\right )}{4 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {x^{2/3} \sqrt [3]{1+x^2} \log \left (1+\frac {\sqrt [3]{2} \left (1+x^{2/3}\right )}{\sqrt [3]{1+x^2}}\right )}{\sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {3 x^{2/3} \sqrt [3]{1+x^2} \log \left (1+x^{2/3}-2^{2/3} \sqrt [3]{1+x^2}\right )}{4 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx,x,\frac {1+x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 \sqrt [3]{x^2+x^4}}-\frac {\left (x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {-\sqrt [3]{2}+2\ 2^{2/3} x}{1-\sqrt [3]{2} x+2^{2/3} x^2} \, dx,x,\frac {1+x^{2/3}}{\sqrt [3]{1+x^2}}\right )}{2 \sqrt [3]{2} \sqrt [3]{x^2+x^4}} \\ & = -\frac {6 x \sqrt [3]{1+x^2} \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{3},\frac {7}{6},x^2,-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {\sqrt {3} x^{2/3} \sqrt [3]{1+x^2} \arctan \left (\frac {1+\frac {\sqrt [3]{2} \left (1+x^{2/3}\right )}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {3 x \sqrt [3]{1+x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {x^{2/3} \sqrt [3]{1+x^2} \log \left (\left (1-x^{2/3}\right )^2 \left (1+x^{2/3}\right )\right )}{4 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}-\frac {x^{2/3} \sqrt [3]{1+x^2} \log \left (1+\frac {2^{2/3} \left (1+x^{2/3}\right )^2}{\left (1+x^2\right )^{2/3}}-\frac {\sqrt [3]{2} \left (1+x^{2/3}\right )}{\sqrt [3]{1+x^2}}\right )}{2 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {x^{2/3} \sqrt [3]{1+x^2} \log \left (1+\frac {\sqrt [3]{2} \left (1+x^{2/3}\right )}{\sqrt [3]{1+x^2}}\right )}{\sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {3 x^{2/3} \sqrt [3]{1+x^2} \log \left (1+x^{2/3}-2^{2/3} \sqrt [3]{1+x^2}\right )}{4 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {\left (3 x^{2/3} \sqrt [3]{1+x^2}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{2} \left (1+x^{2/3}\right )}{\sqrt [3]{1+x^2}}\right )}{\sqrt [3]{2} \sqrt [3]{x^2+x^4}} \\ & = -\frac {6 x \sqrt [3]{1+x^2} \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{3},\frac {7}{6},x^2,-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {\sqrt {3} x^{2/3} \sqrt [3]{1+x^2} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \left (1+x^{2/3}\right )}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{x^2+x^4}}-\frac {\sqrt {3} x^{2/3} \sqrt [3]{1+x^2} \arctan \left (\frac {1+\frac {\sqrt [3]{2} \left (1+x^{2/3}\right )}{\sqrt [3]{1+x^2}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {3 x \sqrt [3]{1+x^2} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-x^2\right )}{\sqrt [3]{x^2+x^4}}-\frac {x^{2/3} \sqrt [3]{1+x^2} \log \left (\left (1-x^{2/3}\right )^2 \left (1+x^{2/3}\right )\right )}{4 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}-\frac {x^{2/3} \sqrt [3]{1+x^2} \log \left (1+\frac {2^{2/3} \left (1+x^{2/3}\right )^2}{\left (1+x^2\right )^{2/3}}-\frac {\sqrt [3]{2} \left (1+x^{2/3}\right )}{\sqrt [3]{1+x^2}}\right )}{2 \sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {x^{2/3} \sqrt [3]{1+x^2} \log \left (1+\frac {\sqrt [3]{2} \left (1+x^{2/3}\right )}{\sqrt [3]{1+x^2}}\right )}{\sqrt [3]{2} \sqrt [3]{x^2+x^4}}+\frac {3 x^{2/3} \sqrt [3]{1+x^2} \log \left (1+x^{2/3}-2^{2/3} \sqrt [3]{1+x^2}\right )}{4 \sqrt [3]{2} \sqrt [3]{x^2+x^4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.76 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.21 \[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=-\frac {x^{2/3} \sqrt [3]{1+x^2} \left (2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}}\right )-2 \log \left (-2 \sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}\right )+\log \left (2 x^{2/3}+2^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}+\sqrt [3]{2} \left (1+x^2\right )^{2/3}\right )\right )}{2 \sqrt [3]{2} \sqrt [3]{x^2+x^4}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 7.22 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.87

method	result	size
pseudoelliptic	\(\frac {2^{\frac {2}{3}} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+x \right )}{3 x}\right )+2 \ln \left (\frac {-2^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {2^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )\right )}{4}\)	\(105\)
trager	\(\text {Expression too large to display}\)	\(1463\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 307 vs. \(2 (94) = 188\).

Time = 2.44 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.54 \[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=-\frac {1}{6} \cdot 4^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {3 \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (x^{4} + 2 \, x^{3} - 6 \, x^{2} + 2 \, x + 1\right )} {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}} + 6 \cdot 4^{\frac {1}{3}} \sqrt {3} {\left (x^{5} + 14 \, x^{4} + 6 \, x^{3} + 14 \, x^{2} + x\right )} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} + \sqrt {3} {\left (x^{7} + 30 \, x^{6} + 51 \, x^{5} + 52 \, x^{4} + 51 \, x^{3} + 30 \, x^{2} + x\right )}}{3 \, {\left (x^{7} - 6 \, x^{6} - 93 \, x^{5} - 20 \, x^{4} - 93 \, x^{3} - 6 \, x^{2} + x\right )}}\right ) - \frac {1}{12} \cdot 4^{\frac {1}{3}} \log \left (\frac {6 \cdot 4^{\frac {1}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}} {\left (x^{2} + 4 \, x + 1\right )} + 4^{\frac {2}{3}} {\left (x^{5} + 14 \, x^{4} + 6 \, x^{3} + 14 \, x^{2} + x\right )} + 24 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{3} + x^{2} + x\right )}}{x^{5} - 4 \, x^{4} + 6 \, x^{3} - 4 \, x^{2} + x}\right ) + \frac {1}{6} \cdot 4^{\frac {1}{3}} \log \left (-\frac {3 \cdot 4^{\frac {2}{3}} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x + 4^{\frac {1}{3}} {\left (x^{3} - 2 \, x^{2} + x\right )} - 6 \, {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}}}{x^{3} - 2 \, x^{2} + x}\right ) \]

[In]

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

[Out]

-1/6*4^(1/3)*sqrt(3)*arctan(1/3*(3*4^(2/3)*sqrt(3)*(x^4 + 2*x^3 - 6*x^2 + 2*x + 1)*(x^4 + x^2)^(2/3) + 6*4^(1/
3)*sqrt(3)*(x^5 + 14*x^4 + 6*x^3 + 14*x^2 + x)*(x^4 + x^2)^(1/3) + sqrt(3)*(x^7 + 30*x^6 + 51*x^5 + 52*x^4 + 5
1*x^3 + 30*x^2 + x))/(x^7 - 6*x^6 - 93*x^5 - 20*x^4 - 93*x^3 - 6*x^2 + x)) - 1/12*4^(1/3)*log((6*4^(1/3)*(x^4
+ x^2)^(2/3)*(x^2 + 4*x + 1) + 4^(2/3)*(x^5 + 14*x^4 + 6*x^3 + 14*x^2 + x) + 24*(x^4 + x^2)^(1/3)*(x^3 + x^2 +
x))/(x^5 - 4*x^4 + 6*x^3 - 4*x^2 + x)) + 1/6*4^(1/3)*log(-(3*4^(2/3)*(x^4 + x^2)^(1/3)*x + 4^(1/3)*(x^3 - 2*x
^2 + x) - 6*(x^4 + x^2)^(2/3))/(x^3 - 2*x^2 + x))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1+x}{(-1+x) \sqrt [3]{x^2+x^4}} \, dx=\int \frac {x+1}{{\left (x^4+x^2\right )}^{1/3}\,\left (x-1\right )} \,d x \]

[In]

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

[Out]

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