3.24.5 \(\int \frac {(1+x^2) \sqrt [3]{-1-x^2+x^4+x^6}}{x} \, dx\)

Optimal. Leaf size=330 \[ \frac {\left (x^2-1\right )^{2/3} \left (x^2+1\right )^{4/3} \left (\frac {1}{12} \sqrt [3]{x^2-1} \left (3 \left (x^2+1\right )^{5/3}-2 \left (x^2+1\right )^{2/3}\right )+\frac {1}{2} \sqrt [3]{x^2-1} \left (x^2+1\right )^{2/3}+\frac {1}{18} \log \left (\sqrt [3]{x^2-1}-\sqrt [3]{x^2+1}\right )-\frac {1}{2} \log \left (\sqrt [3]{x^2-1}+\sqrt [3]{x^2+1}\right )+\frac {1}{4} \log \left (\left (x^2-1\right )^{2/3}-\sqrt [3]{x^2+1} \sqrt [3]{x^2-1}+\left (x^2+1\right )^{2/3}\right )-\frac {1}{36} \log \left (\left (x^2-1\right )^{2/3}+\sqrt [3]{x^2+1} \sqrt [3]{x^2-1}+\left (x^2+1\right )^{2/3}\right )+\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x^2+1}}{2 \sqrt [3]{x^2-1}-\sqrt [3]{x^2+1}}\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x^2+1}}{2 \sqrt [3]{x^2-1}+\sqrt [3]{x^2+1}}\right )}{6 \sqrt {3}}\right )}{\left (\left (x^2-1\right ) \left (x^2+1\right )^2\right )^{2/3}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.55, antiderivative size = 488, normalized size of antiderivative = 1.48, number of steps used = 9, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {6719, 101, 157, 50, 59, 105, 91} \begin {gather*} \frac {1}{4} \sqrt [3]{x^6+x^4-x^2-1} \left (x^2+1\right )+\frac {1}{3} \sqrt [3]{x^6+x^4-x^2-1}+\frac {\sqrt [3]{x^6+x^4-x^2-1} \log \left (x^2\right )}{4 \sqrt [3]{x^2-1} \left (x^2+1\right )^{2/3}}+\frac {\sqrt [3]{x^6+x^4-x^2-1} \log \left (x^2-1\right )}{36 \sqrt [3]{x^2-1} \left (x^2+1\right )^{2/3}}-\frac {3 \sqrt [3]{x^6+x^4-x^2-1} \log \left (-\sqrt [3]{x^2-1}-\sqrt [3]{x^2+1}\right )}{4 \sqrt [3]{x^2-1} \left (x^2+1\right )^{2/3}}+\frac {\sqrt [3]{x^6+x^4-x^2-1} \log \left (\frac {\sqrt [3]{x^2+1}}{\sqrt [3]{x^2-1}}-1\right )}{12 \sqrt [3]{x^2-1} \left (x^2+1\right )^{2/3}}-\frac {\sqrt {3} \sqrt [3]{x^6+x^4-x^2-1} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x^2+1}}{\sqrt {3} \sqrt [3]{x^2-1}}\right )}{2 \sqrt [3]{x^2-1} \left (x^2+1\right )^{2/3}}+\frac {\sqrt {3} \sqrt [3]{x^6+x^4-x^2-1} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^2+1}}{\sqrt {3} \sqrt [3]{x^2-1}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt [3]{x^2-1} \left (x^2+1\right )^{2/3}}-\frac {4 \sqrt [3]{x^6+x^4-x^2-1} \tan ^{-1}\left (\frac {2 \sqrt [3]{x^2+1}}{\sqrt {3} \sqrt [3]{x^2-1}}+\frac {1}{\sqrt {3}}\right )}{3 \sqrt {3} \sqrt [3]{x^2-1} \left (x^2+1\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + x^2)*(-1 - x^2 + x^4 + x^6)^(1/3))/x,x]

[Out]

(-1 - x^2 + x^4 + x^6)^(1/3)/3 + ((1 + x^2)*(-1 - x^2 + x^4 + x^6)^(1/3))/4 - (Sqrt[3]*(-1 - x^2 + x^4 + x^6)^
(1/3)*ArcTan[1/Sqrt[3] - (2*(1 + x^2)^(1/3))/(Sqrt[3]*(-1 + x^2)^(1/3))])/(2*(-1 + x^2)^(1/3)*(1 + x^2)^(2/3))
- (4*(-1 - x^2 + x^4 + x^6)^(1/3)*ArcTan[1/Sqrt[3] + (2*(1 + x^2)^(1/3))/(Sqrt[3]*(-1 + x^2)^(1/3))])/(3*Sqrt
[3]*(-1 + x^2)^(1/3)*(1 + x^2)^(2/3)) + (Sqrt[3]*(-1 - x^2 + x^4 + x^6)^(1/3)*ArcTan[1/Sqrt[3] + (2*(1 + x^2)^
(1/3))/(Sqrt[3]*(-1 + x^2)^(1/3))])/(2*(-1 + x^2)^(1/3)*(1 + x^2)^(2/3)) + ((-1 - x^2 + x^4 + x^6)^(1/3)*Log[x
^2])/(4*(-1 + x^2)^(1/3)*(1 + x^2)^(2/3)) + ((-1 - x^2 + x^4 + x^6)^(1/3)*Log[-1 + x^2])/(36*(-1 + x^2)^(1/3)*
(1 + x^2)^(2/3)) - (3*(-1 - x^2 + x^4 + x^6)^(1/3)*Log[-(-1 + x^2)^(1/3) - (1 + x^2)^(1/3)])/(4*(-1 + x^2)^(1/
3)*(1 + x^2)^(2/3)) + ((-1 - x^2 + x^4 + x^6)^(1/3)*Log[-1 + (1 + x^2)^(1/3)/(-1 + x^2)^(1/3)])/(12*(-1 + x^2)
^(1/3)*(1 + x^2)^(2/3))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 59

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

Rule 91

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.)*(x_))), x_Symbol] :> With[{q = Rt[
(d*e - c*f)/(b*e - a*f), 3]}, -Simp[(Sqrt[3]*q*ArcTan[1/Sqrt[3] + (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/
3))])/(d*e - c*f), x] + (Simp[(q*Log[e + f*x])/(2*(d*e - c*f)), x] - Simp[(3*q*Log[q*(a + b*x)^(1/3) - (c + d*
x)^(1/3)])/(2*(d*e - c*f)), x])] /; FreeQ[{a, b, c, d, e, f}, x]

Rule 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

\begin {align*} \int \frac {\left (1+x^2\right ) \sqrt [3]{-1-x^2+x^4+x^6}}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(1+x) \sqrt [3]{(-1+x) (1+x)^2}}{x} \, dx,x,x^2\right )\\ &=\frac {\sqrt [3]{\left (-1+x^2\right ) \left (1+x^2\right )^2} \operatorname {Subst}\left (\int \frac {\sqrt [3]{-1+x} (1+x)^{5/3}}{x} \, dx,x,x^2\right )}{2 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}\\ &=\frac {1}{4} \left (1+x^2\right ) \sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )}-\frac {\sqrt [3]{\left (-1+x^2\right ) \left (1+x^2\right )^2} \operatorname {Subst}\left (\int \frac {\left (2-\frac {4 x}{3}\right ) (1+x)^{2/3}}{(-1+x)^{2/3} x} \, dx,x,x^2\right )}{4 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}\\ &=\frac {1}{4} \left (1+x^2\right ) \sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )}+\frac {\sqrt [3]{\left (-1+x^2\right ) \left (1+x^2\right )^2} \operatorname {Subst}\left (\int \frac {(1+x)^{2/3}}{(-1+x)^{2/3}} \, dx,x,x^2\right )}{3 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}-\frac {\sqrt [3]{\left (-1+x^2\right ) \left (1+x^2\right )^2} \operatorname {Subst}\left (\int \frac {(1+x)^{2/3}}{(-1+x)^{2/3} x} \, dx,x,x^2\right )}{2 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}\\ &=\frac {1}{3} \sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )}+\frac {1}{4} \left (1+x^2\right ) \sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )}+\frac {\left (4 \sqrt [3]{\left (-1+x^2\right ) \left (1+x^2\right )^2}\right ) \operatorname {Subst}\left (\int \frac {1}{(-1+x)^{2/3} \sqrt [3]{1+x}} \, dx,x,x^2\right )}{9 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}-\frac {\sqrt [3]{\left (-1+x^2\right ) \left (1+x^2\right )^2} \operatorname {Subst}\left (\int \frac {1}{(-1+x)^{2/3} \sqrt [3]{1+x}} \, dx,x,x^2\right )}{2 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}-\frac {\sqrt [3]{\left (-1+x^2\right ) \left (1+x^2\right )^2} \operatorname {Subst}\left (\int \frac {1}{(-1+x)^{2/3} x \sqrt [3]{1+x}} \, dx,x,x^2\right )}{2 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}\\ &=\frac {1}{3} \sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )}+\frac {1}{4} \left (1+x^2\right ) \sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )}-\frac {\sqrt {3} \sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1+x^2}}{\sqrt {3} \sqrt [3]{-1+x^2}}\right )}{2 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}-\frac {4 \sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^2}}{\sqrt {3} \sqrt [3]{-1+x^2}}\right )}{3 \sqrt {3} \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}+\frac {\sqrt {3} \sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^2}}{\sqrt {3} \sqrt [3]{-1+x^2}}\right )}{2 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}+\frac {\sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )} \log (x)}{2 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}+\frac {\sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )} \log \left (1-x^2\right )}{36 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}-\frac {3 \sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )} \log \left (\sqrt [3]{-1+x^2}+\sqrt [3]{1+x^2}\right )}{4 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}+\frac {\sqrt [3]{-\left (\left (1-x^2\right ) \left (1+x^2\right )^2\right )} \log \left (1-\frac {\sqrt [3]{1+x^2}}{\sqrt [3]{-1+x^2}}\right )}{12 \sqrt [3]{-1+x^2} \left (1+x^2\right )^{2/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.11, size = 188, normalized size = 0.57 \begin {gather*} -\frac {3 \sqrt [3]{\left (x^2-1\right ) \left (x^2+1\right )^2} \left (2^{2/3} x^2 \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};\frac {1}{2} \left (1-x^2\right )\right )-4\ 2^{2/3} \left (x^2+1\right ) \, _2F_1\left (-\frac {5}{3},\frac {1}{3};\frac {4}{3};\frac {1}{2} \left (1-x^2\right )\right )+2\ 2^{2/3} \left (x^2+1\right ) \, _2F_1\left (-\frac {2}{3},\frac {1}{3};\frac {4}{3};\frac {1}{2} \left (1-x^2\right )\right )+2^{2/3} \, _2F_1\left (\frac {1}{3},\frac {1}{3};\frac {4}{3};\frac {1}{2} \left (1-x^2\right )\right )+2 \left (x^2+1\right )^{2/3} \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {1-x^2}{x^2+1}\right )\right )}{4 \left (x^2+1\right )^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + x^2)*(-1 - x^2 + x^4 + x^6)^(1/3))/x,x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 8.00, size = 308, normalized size = 0.93 \begin {gather*} \frac {\left (x^2-1\right )^{2/3} \left (x^2+1\right )^{4/3} \left (\frac {1}{12} \sqrt [3]{x^2-1} \left (3 \left (x^2+1\right )^{5/3}+4 \left (x^2+1\right )^{2/3}\right )+\frac {1}{18} \log \left (\sqrt [3]{x^2-1}-\sqrt [3]{x^2+1}\right )-\frac {1}{2} \log \left (\sqrt [3]{x^2-1}+\sqrt [3]{x^2+1}\right )+\frac {1}{4} \log \left (\left (x^2-1\right )^{2/3}-\sqrt [3]{x^2+1} \sqrt [3]{x^2-1}+\left (x^2+1\right )^{2/3}\right )-\frac {1}{36} \log \left (\left (x^2-1\right )^{2/3}+\sqrt [3]{x^2+1} \sqrt [3]{x^2-1}+\left (x^2+1\right )^{2/3}\right )+\frac {1}{2} \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x^2+1}}{2 \sqrt [3]{x^2-1}-\sqrt [3]{x^2+1}}\right )+\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x^2+1}}{2 \sqrt [3]{x^2-1}+\sqrt [3]{x^2+1}}\right )}{6 \sqrt {3}}\right )}{\left (\left (x^2-1\right ) \left (x^2+1\right )^2\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((1 + x^2)*(-1 - x^2 + x^4 + x^6)^(1/3))/x,x]

[Out]

((-1 + x^2)^(2/3)*(1 + x^2)^(4/3)*(((-1 + x^2)^(1/3)*(4*(1 + x^2)^(2/3) + 3*(1 + x^2)^(5/3)))/12 + (Sqrt[3]*Ar
cTan[(Sqrt[3]*(1 + x^2)^(1/3))/(2*(-1 + x^2)^(1/3) - (1 + x^2)^(1/3))])/2 + ArcTan[(Sqrt[3]*(1 + x^2)^(1/3))/(
2*(-1 + x^2)^(1/3) + (1 + x^2)^(1/3))]/(6*Sqrt[3]) + Log[(-1 + x^2)^(1/3) - (1 + x^2)^(1/3)]/18 - Log[(-1 + x^
2)^(1/3) + (1 + x^2)^(1/3)]/2 + Log[(-1 + x^2)^(2/3) - (-1 + x^2)^(1/3)*(1 + x^2)^(1/3) + (1 + x^2)^(2/3)]/4 -
Log[(-1 + x^2)^(2/3) + (-1 + x^2)^(1/3)*(1 + x^2)^(1/3) + (1 + x^2)^(2/3)]/36))/((-1 + x^2)*(1 + x^2)^2)^(2/3
)

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 305, normalized size = 0.92 \begin {gather*} -\frac {1}{18} \, \sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x^{2} + 1\right )} + 2 \, \sqrt {3} {\left (x^{6} + x^{4} - x^{2} - 1\right )}^{\frac {1}{3}}}{3 \, {\left (x^{2} + 1\right )}}\right ) - \frac {1}{2} \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} {\left (x^{2} + 1\right )} - 2 \, \sqrt {3} {\left (x^{6} + x^{4} - x^{2} - 1\right )}^{\frac {1}{3}}}{3 \, {\left (x^{2} + 1\right )}}\right ) + \frac {1}{12} \, {\left (x^{6} + x^{4} - x^{2} - 1\right )}^{\frac {1}{3}} {\left (3 \, x^{2} + 7\right )} - \frac {1}{36} \, \log \left (\frac {x^{4} + 2 \, x^{2} + {\left (x^{6} + x^{4} - x^{2} - 1\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )} + {\left (x^{6} + x^{4} - x^{2} - 1\right )}^{\frac {2}{3}} + 1}{x^{4} + 2 \, x^{2} + 1}\right ) + \frac {1}{4} \, \log \left (\frac {x^{4} + 2 \, x^{2} - {\left (x^{6} + x^{4} - x^{2} - 1\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )} + {\left (x^{6} + x^{4} - x^{2} - 1\right )}^{\frac {2}{3}} + 1}{x^{4} + 2 \, x^{2} + 1}\right ) - \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{6} + x^{4} - x^{2} - 1\right )}^{\frac {1}{3}} + 1}{x^{2} + 1}\right ) + \frac {1}{18} \, \log \left (-\frac {x^{2} - {\left (x^{6} + x^{4} - x^{2} - 1\right )}^{\frac {1}{3}} + 1}{x^{2} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^6+x^4-x^2-1)^(1/3)/x,x, algorithm="fricas")

[Out]

-1/18*sqrt(3)*arctan(1/3*(sqrt(3)*(x^2 + 1) + 2*sqrt(3)*(x^6 + x^4 - x^2 - 1)^(1/3))/(x^2 + 1)) - 1/2*sqrt(3)*
arctan(-1/3*(sqrt(3)*(x^2 + 1) - 2*sqrt(3)*(x^6 + x^4 - x^2 - 1)^(1/3))/(x^2 + 1)) + 1/12*(x^6 + x^4 - x^2 - 1
)^(1/3)*(3*x^2 + 7) - 1/36*log((x^4 + 2*x^2 + (x^6 + x^4 - x^2 - 1)^(1/3)*(x^2 + 1) + (x^6 + x^4 - x^2 - 1)^(2
/3) + 1)/(x^4 + 2*x^2 + 1)) + 1/4*log((x^4 + 2*x^2 - (x^6 + x^4 - x^2 - 1)^(1/3)*(x^2 + 1) + (x^6 + x^4 - x^2
- 1)^(2/3) + 1)/(x^4 + 2*x^2 + 1)) - 1/2*log((x^2 + (x^6 + x^4 - x^2 - 1)^(1/3) + 1)/(x^2 + 1)) + 1/18*log(-(x
^2 - (x^6 + x^4 - x^2 - 1)^(1/3) + 1)/(x^2 + 1))

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} + x^{4} - x^{2} - 1\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^6+x^4-x^2-1)^(1/3)/x,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 49.13, size = 6304, normalized size = 19.10 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{6} + x^{4} - x^{2} - 1\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+1)*(x^6+x^4-x^2-1)^(1/3)/x,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (x^2+1\right )\,{\left (x^6+x^4-x^2-1\right )}^{1/3}}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________