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

Optimal. Leaf size=115 \[ \frac {1}{2} \text {RootSum}\left [\text {$\#$1}^6-4 \text {$\#$1}^3+6\& ,\frac {-\text {$\#$1}^3 \log \left (\sqrt [3]{2 x^3+x}-\text {$\#$1} x\right )+\text {$\#$1}^3 \log (x)+6 \log \left (\sqrt [3]{2 x^3+x}-\text {$\#$1} x\right )-6 \log (x)}{\text {$\#$1}^5-2 \text {$\#$1}^2}\& \right ]-\frac {3 \left (10 x^2+1\right ) \sqrt [3]{2 x^3+x}}{8 x^3} \]

________________________________________________________________________________________

Rubi [C]  time = 0.87, antiderivative size = 363, normalized size of antiderivative = 3.16, number of steps used = 20, number of rules used = 17, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.548, Rules used = {2056, 1270, 1491, 1521, 264, 6725, 277, 331, 292, 31, 634, 617, 204, 628, 21, 1529, 510} \begin {gather*} -\frac {3 x \sqrt [3]{2 x^3+x} F_1\left (\frac {2}{3};-\frac {4}{3},1;\frac {5}{3};-2 x^2,-i \sqrt {2} x^2\right )}{4 \sqrt [3]{2 x^2+1}}-\frac {3 x \sqrt [3]{2 x^3+x} F_1\left (\frac {2}{3};-\frac {4}{3},1;\frac {5}{3};-2 x^2,i \sqrt {2} x^2\right )}{4 \sqrt [3]{2 x^2+1}}-\frac {3 \sqrt [3]{2 x^3+x}}{x}-\frac {3 \sqrt [3]{2 x^3+x} \left (2 x^2+1\right )}{8 x^3}-\frac {\sqrt [3]{2} \sqrt [3]{2 x^3+x} \log \left (1-\frac {\sqrt [3]{2} x^{2/3}}{\sqrt [3]{2 x^2+1}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}}+\frac {\sqrt [3]{2 x^3+x} \log \left (\frac {2^{2/3} x^{4/3}}{\left (2 x^2+1\right )^{2/3}}+\frac {\sqrt [3]{2} x^{2/3}}{\sqrt [3]{2 x^2+1}}+1\right )}{2^{2/3} \sqrt [3]{x} \sqrt [3]{2 x^2+1}}-\frac {\sqrt [3]{2} \sqrt {3} \sqrt [3]{2 x^3+x} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{2} x^{2/3}}{\sqrt [3]{2 x^2+1}}+1}{\sqrt {3}}\right )}{\sqrt [3]{x} \sqrt [3]{2 x^2+1}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*(x + 2*x^3)^(1/3))/x - (3*(1 + 2*x^2)*(x + 2*x^3)^(1/3))/(8*x^3) - (3*x*(x + 2*x^3)^(1/3)*AppellF1[2/3, -4
/3, 1, 5/3, -2*x^2, (-I)*Sqrt[2]*x^2])/(4*(1 + 2*x^2)^(1/3)) - (3*x*(x + 2*x^3)^(1/3)*AppellF1[2/3, -4/3, 1, 5
/3, -2*x^2, I*Sqrt[2]*x^2])/(4*(1 + 2*x^2)^(1/3)) - (2^(1/3)*Sqrt[3]*(x + 2*x^3)^(1/3)*ArcTan[(1 + (2*2^(1/3)*
x^(2/3))/(1 + 2*x^2)^(1/3))/Sqrt[3]])/(x^(1/3)*(1 + 2*x^2)^(1/3)) - (2^(1/3)*(x + 2*x^3)^(1/3)*Log[1 - (2^(1/3
)*x^(2/3))/(1 + 2*x^2)^(1/3)])/(x^(1/3)*(1 + 2*x^2)^(1/3)) + ((x + 2*x^3)^(1/3)*Log[1 + (2^(2/3)*x^(4/3))/(1 +
2*x^2)^(2/3) + (2^(1/3)*x^(2/3))/(1 + 2*x^2)^(1/3)])/(2^(2/3)*x^(1/3)*(1 + 2*x^2)^(1/3))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + 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 264

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

Rule 277

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

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 510

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

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 1270

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{k = Denominat
or[m]}, Dist[k/f, Subst[Int[x^(k*(m + 1) - 1)*(d + (e*x^(2*k))/f)^q*(a + (c*x^(4*k))/f)^p, x], x, (f*x)^(1/k)]
, x]] /; FreeQ[{a, c, d, e, f, p, q}, x] && FractionQ[m] && IntegerQ[p]

Rule 1491

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = GCD[m +
1, n]}, Dist[1/k, Subst[Int[x^((m + 1)/k - 1)*(d + e*x^(n/k))^q*(a + c*x^((2*n)/k))^p, x], x, x^k], x] /; k !=
1] /; FreeQ[{a, c, d, e, p, q}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] && IntegerQ[m]

Rule 1521

Int[(((f_.)*(x_))^(m_)*((d_.) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.)), x_Symbol] :> Dist[d/a, Int[
(f*x)^m*(d + e*x^n)^(q - 1), x], x] + Dist[1/(a*f^n), Int[((f*x)^(m + n)*(d + e*x^n)^(q - 1)*Simp[a*e - c*d*x^
n, x])/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, f}, x] && EqQ[n2, 2*n] && IGtQ[n, 0] &&  !IntegerQ[q] &&
GtQ[q, 0] && LtQ[m, 0]

Rule 1529

Int[(((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))^(q_))/((a_) + (c_.)*(x_)^(n2_.)), x_Symbol] :> Int[ExpandInte
grand[(d + e*x^n)^q, (f*x)^m/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, f, q, n}, x] && EqQ[n2, 2*n] && IGt
Q[n, 0] &&  !IntegerQ[q] && IntegerQ[m]

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]

Rule 6725

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 {\left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (1+2 x^4\right )} \, dx &=\frac {\sqrt [3]{x+2 x^3} \int \frac {\left (1+2 x^2\right )^{4/3}}{x^{11/3} \left (1+2 x^4\right )} \, dx}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {\left (3 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {\left (1+2 x^6\right )^{4/3}}{x^9 \left (1+2 x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {\left (3 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {\left (1+2 x^3\right )^{4/3}}{x^5 \left (1+2 x^6\right )} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=\frac {\left (3 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1+2 x^3}}{x^5} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x} \sqrt [3]{1+2 x^2}}+\frac {\left (3 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {\left (2-2 x^3\right ) \sqrt [3]{1+2 x^3}}{x^2 \left (1+2 x^6\right )} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=-\frac {3 \left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {\left (3 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \left (\frac {2 \sqrt [3]{1+2 x^3}}{x^2}+\frac {2 x \left (-1-2 x^3\right ) \sqrt [3]{1+2 x^3}}{1+2 x^6}\right ) \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=-\frac {3 \left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {\left (3 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {\sqrt [3]{1+2 x^3}}{x^2} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}+\frac {\left (3 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {x \left (-1-2 x^3\right ) \sqrt [3]{1+2 x^3}}{1+2 x^6} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=-\frac {3 \sqrt [3]{x+2 x^3}}{x}-\frac {3 \left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}-\frac {\left (3 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {x \left (1+2 x^3\right )^{4/3}}{1+2 x^6} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}+\frac {\left (6 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {x}{\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}}{x}-\frac {3 \left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}-\frac {\left (3 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \left (-\frac {i x \left (1+2 x^3\right )^{4/3}}{\sqrt {2} \left (-i \sqrt {2}+2 x^3\right )}+\frac {i x \left (1+2 x^3\right )^{4/3}}{\sqrt {2} \left (i \sqrt {2}+2 x^3\right )}\right ) \, dx,x,x^{2/3}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}+\frac {\left (6 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {x}{1-2 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}}{x}-\frac {3 \left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {\left (3 i \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {x \left (1+2 x^3\right )^{4/3}}{-i \sqrt {2}+2 x^3} \, dx,x,x^{2/3}\right )}{\sqrt {2} \sqrt [3]{x} \sqrt [3]{1+2 x^2}}-\frac {\left (3 i \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {x \left (1+2 x^3\right )^{4/3}}{i \sqrt {2}+2 x^3} \, dx,x,x^{2/3}\right )}{\sqrt {2} \sqrt [3]{x} \sqrt [3]{1+2 x^2}}+\frac {\left (2^{2/3} \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt [3]{2} 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 (2^{2/3} \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {1-\sqrt [3]{2} x}{1+\sqrt [3]{2} x+2^{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}}{x}-\frac {3 \left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}-\frac {3 x \sqrt [3]{x+2 x^3} F_1\left (\frac {2}{3};-\frac {4}{3},1;\frac {5}{3};-2 x^2,-i \sqrt {2} x^2\right )}{4 \sqrt [3]{1+2 x^2}}-\frac {3 x \sqrt [3]{x+2 x^3} F_1\left (\frac {2}{3};-\frac {4}{3},1;\frac {5}{3};-2 x^2,i \sqrt {2} x^2\right )}{4 \sqrt [3]{1+2 x^2}}-\frac {\sqrt [3]{2} \sqrt [3]{x+2 x^3} \log \left (1-\frac {\sqrt [3]{2} x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}+\frac {\sqrt [3]{x+2 x^3} \operatorname {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 {x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{2^{2/3} \sqrt [3]{x} \sqrt [3]{1+2 x^2}}-\frac {\left (3 \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt [3]{2} x+2^{2/3} x^2} \, dx,x,\frac {x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{\sqrt [3]{2} \sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=-\frac {3 \sqrt [3]{x+2 x^3}}{x}-\frac {3 \left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}-\frac {3 x \sqrt [3]{x+2 x^3} F_1\left (\frac {2}{3};-\frac {4}{3},1;\frac {5}{3};-2 x^2,-i \sqrt {2} x^2\right )}{4 \sqrt [3]{1+2 x^2}}-\frac {3 x \sqrt [3]{x+2 x^3} F_1\left (\frac {2}{3};-\frac {4}{3},1;\frac {5}{3};-2 x^2,i \sqrt {2} x^2\right )}{4 \sqrt [3]{1+2 x^2}}-\frac {\sqrt [3]{2} \sqrt [3]{x+2 x^3} \log \left (1-\frac {\sqrt [3]{2} x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}+\frac {\sqrt [3]{x+2 x^3} \log \left (1+\frac {2^{2/3} x^{4/3}}{\left (1+2 x^2\right )^{2/3}}+\frac {\sqrt [3]{2} x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{2^{2/3} \sqrt [3]{x} \sqrt [3]{1+2 x^2}}+\frac {\left (3 \sqrt [3]{2} \sqrt [3]{x+2 x^3}\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 x^{2/3}}{\sqrt [3]{\frac {1}{2}+x^2}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ &=-\frac {3 \sqrt [3]{x+2 x^3}}{x}-\frac {3 \left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}-\frac {3 x \sqrt [3]{x+2 x^3} F_1\left (\frac {2}{3};-\frac {4}{3},1;\frac {5}{3};-2 x^2,-i \sqrt {2} x^2\right )}{4 \sqrt [3]{1+2 x^2}}-\frac {3 x \sqrt [3]{x+2 x^3} F_1\left (\frac {2}{3};-\frac {4}{3},1;\frac {5}{3};-2 x^2,i \sqrt {2} x^2\right )}{4 \sqrt [3]{1+2 x^2}}-\frac {\sqrt [3]{2} \sqrt {3} \sqrt [3]{x+2 x^3} \tan ^{-1}\left (\frac {1+\frac {2 \sqrt [3]{2} x^{2/3}}{\sqrt [3]{1+2 x^2}}}{\sqrt {3}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}-\frac {\sqrt [3]{2} \sqrt [3]{x+2 x^3} \log \left (1-\frac {\sqrt [3]{2} x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{\sqrt [3]{x} \sqrt [3]{1+2 x^2}}+\frac {\sqrt [3]{x+2 x^3} \log \left (1+\frac {2^{2/3} x^{4/3}}{\left (1+2 x^2\right )^{2/3}}+\frac {\sqrt [3]{2} x^{2/3}}{\sqrt [3]{1+2 x^2}}\right )}{2^{2/3} \sqrt [3]{x} \sqrt [3]{1+2 x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 3.74, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1+2 x^2\right ) \sqrt [3]{x+2 x^3}}{x^4 \left (1+2 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.30, size = 115, normalized size = 1.00 \begin {gather*} -\frac {3 \left (1+10 x^2\right ) \sqrt [3]{x+2 x^3}}{8 x^3}+\frac {1}{2} \text {RootSum}\left [6-4 \text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-6 \log (x)+6 \log \left (\sqrt [3]{x+2 x^3}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^3-\log \left (\sqrt [3]{x+2 x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-2 \text {$\#$1}^2+\text {$\#$1}^5}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(1 + 10*x^2)*(x + 2*x^3)^(1/3))/(8*x^3) + RootSum[6 - 4*#1^3 + #1^6 & , (-6*Log[x] + 6*Log[(x + 2*x^3)^(1/
3) - x*#1] + Log[x]*#1^3 - Log[(x + 2*x^3)^(1/3) - x*#1]*#1^3)/(-2*#1^2 + #1^5) & ]/2

________________________________________________________________________________________

fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 229.81, size = 9342, normalized size = 81.23

method	result	size
trager	\(\text {Expression too large to display}\)	\(9342\)
risch	\(\text {Expression too large to display}\)	\(14917\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

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 (2 \, x^{2} + 1\right )}}{{\left (2 \, x^{4} + 1\right )} x^{4}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________