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

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

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Rubi [C]  time = 1.15, 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 = {2056, 6728, 364, 959, 466, 429, 465, 510} \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*}

Warning: Unable to verify antiderivative.

[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 364

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

Rule 429

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 465

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 466

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 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 959

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 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 6728

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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 [F]  time = 0.76, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+x^2}{\left (1+x+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.32, size = 85, normalized size = 1.00 \begin {gather*} -\sqrt {3} \tan ^{-1}\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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.46, 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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} - 1}{{\left (x^{4} + x^{2}\right )}^{\frac {1}{3}} {\left (x^{2} + x + 1\right )}}\,{d x} \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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maple [C]  time = 1.70, size = 307, normalized size = 3.61

method	result	size
trager	\(-\ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}-2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}-3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}+\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{4}+x^{2}\right )^{\frac {2}{3}}+3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+2 x^{3}-3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +3 \left (x^{4}+x^{2}\right )^{\frac {2}{3}}+3 x \left (x^{4}+x^{2}\right )^{\frac {1}{3}}-x^{2}+2 x}{x \left (x^{2}+x +1\right )}\right )+\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (-\frac {\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}-2 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+\RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{4}+x^{2}\right )^{\frac {2}{3}}-3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{4}+x^{2}\right )^{\frac {1}{3}} x +3 \RootOf \left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-x^{3}+3 x \left (x^{4}+x^{2}\right )^{\frac {1}{3}}-x^{2}-x}{x \left (x^{2}+x +1\right )}\right )\)	\(307\)

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]

-ln((RootOf(_Z^2-_Z+1)^2*x^3-2*RootOf(_Z^2-_Z+1)^2*x^2-3*RootOf(_Z^2-_Z+1)*x^3+RootOf(_Z^2-_Z+1)^2*x-3*RootOf(
_Z^2-_Z+1)*(x^4+x^2)^(2/3)+3*RootOf(_Z^2-_Z+1)*x^2+2*x^3-3*RootOf(_Z^2-_Z+1)*x+3*(x^4+x^2)^(2/3)+3*x*(x^4+x^2)
^(1/3)-x^2+2*x)/x/(x^2+x+1))+RootOf(_Z^2-_Z+1)*ln(-(RootOf(_Z^2-_Z+1)^2*x^3-2*RootOf(_Z^2-_Z+1)^2*x^2+RootOf(_
Z^2-_Z+1)^2*x-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+3*RootOf(_Z^2-_Z+1)*x^
2-x^3+3*x*(x^4+x^2)^(1/3)-x^2-x)/x/(x^2+x+1))

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