3.12.39 \(\int \frac {-1+x^2}{(1+x+x^2) \sqrt [3]{x^2+x^4}} \, dx\) [1139]

Optimal. Leaf size=85 \[ -\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} x}{-x+2 \sqrt [3]{x^2+x^4}}\right )-\log \left (x+\sqrt [3]{x^2+x^4}\right )+\frac {1}{2} \log \left (x^2-x \sqrt [3]{x^2+x^4}+\left (x^2+x^4\right )^{2/3}\right ) \]

[Out]

-3^(1/2)*arctan(3^(1/2)*x/(-x+2*(x^4+x^2)^(1/3)))-ln(x+(x^4+x^2)^(1/3))+1/2*ln(x^2-x*(x^4+x^2)^(1/3)+(x^4+x^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 = 0.70, antiderivative size = 297, normalized size of antiderivative = 3.49, number of steps used = 18, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2081, 6860, 371, 973, 477, 440, 476, 524} \begin {gather*} \frac {3 \sqrt [3]{x^2+1} x^2 F_1\left (\frac {2}{3};1,\frac {1}{3};\frac {5}{3};-\frac {4 x^2}{\left (i-\sqrt {3}\right )^2},-x^2\right )}{2 \left (1+i \sqrt {3}\right ) \sqrt [3]{x^4+x^2}}+\frac {3 \sqrt [3]{x^2+1} x^2 F_1\left (\frac {2}{3};1,\frac {1}{3};\frac {5}{3};-\frac {4 x^2}{\left (i+\sqrt {3}\right )^2},-x^2\right )}{2 \left (1-i \sqrt {3}\right ) \sqrt [3]{x^4+x^2}}-\frac {3 \sqrt [3]{x^2+1} x F_1\left (\frac {1}{6};1,\frac {1}{3};\frac {7}{6};-\frac {4 x^2}{\left (i-\sqrt {3}\right )^2},-x^2\right )}{\sqrt [3]{x^4+x^2}}-\frac {3 \sqrt [3]{x^2+1} x F_1\left (\frac {1}{6};1,\frac {1}{3};\frac {7}{6};-\frac {4 x^2}{\left (i+\sqrt {3}\right )^2},-x^2\right )}{\sqrt [3]{x^4+x^2}}+\frac {3 \sqrt [3]{x^2+1} x \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^2\right )}{\sqrt [3]{x^4+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x*(1 + x^2)^(1/3)*AppellF1[1/6, 1, 1/3, 7/6, (-4*x^2)/(I - Sqrt[3])^2, -x^2])/(x^2 + x^4)^(1/3) - (3*x*(1
+ x^2)^(1/3)*AppellF1[1/6, 1, 1/3, 7/6, (-4*x^2)/(I + Sqrt[3])^2, -x^2])/(x^2 + x^4)^(1/3) + (3*x^2*(1 + x^2)^
(1/3)*AppellF1[2/3, 1, 1/3, 5/3, (-4*x^2)/(I - Sqrt[3])^2, -x^2])/(2*(1 + I*Sqrt[3])*(x^2 + x^4)^(1/3)) + (3*x
^2*(1 + x^2)^(1/3)*AppellF1[2/3, 1, 1/3, 5/3, (-4*x^2)/(I + Sqrt[3])^2, -x^2])/(2*(1 - I*Sqrt[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)

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 524

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)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], 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 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 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]
time = 0.75, size = 118, normalized size = 1.39 \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 \sqrt [3]{1+x^2}}\right )-2 \log \left (\sqrt [3]{x}+\sqrt [3]{1+x^2}\right )+\log \left (x^{2/3}-\sqrt [3]{x} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )\right )}{2 \sqrt [3]{x^2+x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.85, size = 371, normalized size = 4.36

method	result	size
trager	\(\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (-\frac {-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}+3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{4}+x^{2}\right )^{\frac {2}{3}}-3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{4}+x^{2}\right )^{\frac {1}{3}} x +2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+2 x^{3}-3 \left (x^{4}+x^{2}\right )^{\frac {2}{3}}+3 x \left (x^{4}+x^{2}\right )^{\frac {1}{3}}-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -x^{2}+2 x}{\left (x^{2}+x +1\right ) x}\right )-\ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}+3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{4}+x^{2}\right )^{\frac {2}{3}}-3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{4}+x^{2}\right )^{\frac {1}{3}} x +2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-x^{3}-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -x^{2}-x}{\left (x^{2}+x +1\right ) x}\right ) \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+\ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}+3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{4}+x^{2}\right )^{\frac {2}{3}}-3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{4}+x^{2}\right )^{\frac {1}{3}} x +2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-x^{3}-\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -x^{2}-x}{\left (x^{2}+x +1\right ) x}\right )\)	\(371\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootOf(_Z^2-_Z+1)*ln(-(-RootOf(_Z^2-_Z+1)*x^3+3*RootOf(_Z^2-_Z+1)*(x^4+x^2)^(2/3)-3*RootOf(_Z^2-_Z+1)*(x^4+x^2
)^(1/3)*x+2*RootOf(_Z^2-_Z+1)*x^2+2*x^3-3*(x^4+x^2)^(2/3)+3*x*(x^4+x^2)^(1/3)-RootOf(_Z^2-_Z+1)*x-x^2+2*x)/(x^
2+x+1)/x)-ln((-RootOf(_Z^2-_Z+1)*x^3+3*RootOf(_Z^2-_Z+1)*(x^4+x^2)^(2/3)-3*RootOf(_Z^2-_Z+1)*(x^4+x^2)^(1/3)*x
+2*RootOf(_Z^2-_Z+1)*x^2-x^3-RootOf(_Z^2-_Z+1)*x-x^2-x)/(x^2+x+1)/x)*RootOf(_Z^2-_Z+1)+ln((-RootOf(_Z^2-_Z+1)*
x^3+3*RootOf(_Z^2-_Z+1)*(x^4+x^2)^(2/3)-3*RootOf(_Z^2-_Z+1)*(x^4+x^2)^(1/3)*x+2*RootOf(_Z^2-_Z+1)*x^2-x^3-Root
Of(_Z^2-_Z+1)*x-x^2-x)/(x^2+x+1)/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^2-1)/(x^2+x+1)/(x^4+x^2)^(1/3),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.05, size = 106, normalized size = 1.25 \begin {gather*} -\sqrt {3} \arctan \left (\frac {\sqrt {3} x^{2} + 2 \, \sqrt {3} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x + 4 \, \sqrt {3} {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}}}{8 \, x^{3} - x^{2} + 8 \, x}\right ) - \frac {1}{2} \, \log \left (\frac {x^{3} + x^{2} + 3 \, {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} x + x + 3 \, {\left (x^{4} + x^{2}\right )}^{\frac {2}{3}}}{x^{3} + x^{2} + x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(3)*arctan((sqrt(3)*x^2 + 2*sqrt(3)*(x^4 + x^2)^(1/3)*x + 4*sqrt(3)*(x^4 + x^2)^(2/3))/(8*x^3 - x^2 + 8*x
)) - 1/2*log((x^3 + x^2 + 3*(x^4 + x^2)^(1/3)*x + x + 3*(x^4 + x^2)^(2/3))/(x^3 + x^2 + x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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