3.18.92 \(\int \frac {-1+x^2}{(1+x^2) \sqrt [3]{x+x^5}} \, dx\)

Optimal. Leaf size=121 \[ -\frac {\log \left (2^{2/3} \sqrt [3]{x^5+x}+2 x\right )}{2 \sqrt [3]{2}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2^{2/3} \sqrt [3]{x^5+x}-x}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (2^{2/3} \sqrt [3]{x^5+x} x-\sqrt [3]{2} \left (x^5+x\right )^{2/3}-2 x^2\right )}{4 \sqrt [3]{2}} \]

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Rubi [C]  time = 0.62, antiderivative size = 123, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2056, 6715, 6725, 245, 1438, 429, 465, 510} \begin {gather*} -\frac {3 \sqrt [3]{x^4+1} x F_1\left (\frac {1}{6};1,\frac {1}{3};\frac {7}{6};x^4,-x^4\right )}{\sqrt [3]{x^5+x}}+\frac {3 \sqrt [3]{x^4+1} x^3 F_1\left (\frac {2}{3};1,\frac {1}{3};\frac {5}{3};x^4,-x^4\right )}{4 \sqrt [3]{x^5+x}}+\frac {3 \sqrt [3]{x^4+1} x \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^4\right )}{2 \sqrt [3]{x^5+x}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*x*(1 + x^4)^(1/3)*AppellF1[1/6, 1, 1/3, 7/6, x^4, -x^4])/(x + x^5)^(1/3) + (3*x^3*(1 + x^4)^(1/3)*AppellF1
[2/3, 1, 1/3, 5/3, x^4, -x^4])/(4*(x + x^5)^(1/3)) + (3*x*(1 + x^4)^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -x^
4])/(2*(x + x^5)^(1/3))

Rule 245

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 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 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 1438

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 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 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

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 {-1+x^2}{\left (1+x^2\right ) \sqrt [3]{x+x^5}} \, dx &=\frac {\left (\sqrt [3]{x} \sqrt [3]{1+x^4}\right ) \int \frac {-1+x^2}{\sqrt [3]{x} \left (1+x^2\right ) \sqrt [3]{1+x^4}} \, dx}{\sqrt [3]{x+x^5}}\\ &=\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^4}\right ) \operatorname {Subst}\left (\int \frac {-1+x^3}{\left (1+x^3\right ) \sqrt [3]{1+x^6}} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x+x^5}}\\ &=\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^4}\right ) \operatorname {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,x^{2/3}\right )}{2 \sqrt [3]{x+x^5}}\\ &=\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1+x^6}} \, dx,x,x^{2/3}\right )}{2 \sqrt [3]{x+x^5}}-\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^3\right ) \sqrt [3]{1+x^6}} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x+x^5}}\\ &=\frac {3 x \sqrt [3]{1+x^4} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^4\right )}{2 \sqrt [3]{x+x^5}}-\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^4}\right ) \operatorname {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,x^{2/3}\right )}{\sqrt [3]{x+x^5}}\\ &=\frac {3 x \sqrt [3]{1+x^4} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^4\right )}{2 \sqrt [3]{x+x^5}}-\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^4}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^6\right ) \sqrt [3]{1+x^6}} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x+x^5}}-\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^4}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (-1+x^6\right ) \sqrt [3]{1+x^6}} \, dx,x,x^{2/3}\right )}{\sqrt [3]{x+x^5}}\\ &=-\frac {3 x \sqrt [3]{1+x^4} F_1\left (\frac {1}{6};1,\frac {1}{3};\frac {7}{6};x^4,-x^4\right )}{\sqrt [3]{x+x^5}}+\frac {3 x \sqrt [3]{1+x^4} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^4\right )}{2 \sqrt [3]{x+x^5}}-\frac {\left (3 \sqrt [3]{x} \sqrt [3]{1+x^4}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (-1+x^3\right ) \sqrt [3]{1+x^3}} \, dx,x,x^{4/3}\right )}{2 \sqrt [3]{x+x^5}}\\ &=-\frac {3 x \sqrt [3]{1+x^4} F_1\left (\frac {1}{6};1,\frac {1}{3};\frac {7}{6};x^4,-x^4\right )}{\sqrt [3]{x+x^5}}+\frac {3 x^3 \sqrt [3]{1+x^4} F_1\left (\frac {2}{3};1,\frac {1}{3};\frac {5}{3};x^4,-x^4\right )}{4 \sqrt [3]{x+x^5}}+\frac {3 x \sqrt [3]{1+x^4} \, _2F_1\left (\frac {1}{6},\frac {1}{3};\frac {7}{6};-x^4\right )}{2 \sqrt [3]{x+x^5}}\\ \end {align*}

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Mathematica [F]  time = 0.42, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [3]{x+x^5}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.42, size = 121, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{-x+2^{2/3} \sqrt [3]{x+x^5}}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (2 x+2^{2/3} \sqrt [3]{x+x^5}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-2 x^2+2^{2/3} x \sqrt [3]{x+x^5}-\sqrt [3]{2} \left (x+x^5\right )^{2/3}\right )}{4 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 3.84, size = 320, normalized size = 2.64 \begin {gather*} \frac {1}{12} \, \sqrt {3} 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {1}{6}} {\left (12 \cdot 2^{\frac {1}{6}} \left (-1\right )^{\frac {2}{3}} {\left (x^{8} - 14 \, x^{6} + 6 \, x^{4} - 14 \, x^{2} + 1\right )} {\left (x^{5} + x\right )}^{\frac {2}{3}} - 24 \, \sqrt {2} \left (-1\right )^{\frac {1}{3}} {\left (x^{9} + x^{7} + x^{3} + x\right )} {\left (x^{5} + x\right )}^{\frac {1}{3}} + 2^{\frac {5}{6}} {\left (x^{12} + 24 \, x^{10} - 57 \, x^{8} + 56 \, x^{6} - 57 \, x^{4} + 24 \, x^{2} + 1\right )}\right )}}{6 \, {\left (x^{12} - 48 \, x^{10} + 15 \, x^{8} - 88 \, x^{6} + 15 \, x^{4} - 48 \, x^{2} + 1\right )}}\right ) - \frac {1}{24} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (\frac {12 \cdot 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{5} - x^{3} + x\right )} {\left (x^{5} + x\right )}^{\frac {1}{3}} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{8} - 14 \, x^{6} + 6 \, x^{4} - 14 \, x^{2} + 1\right )} - 6 \, {\left (x^{5} + x\right )}^{\frac {2}{3}} {\left (x^{4} - 4 \, x^{2} + 1\right )}}{x^{8} + 4 \, x^{6} + 6 \, x^{4} + 4 \, x^{2} + 1}\right ) + \frac {1}{12} \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} \log \left (-\frac {2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (x^{4} + 2 \, x^{2} + 1\right )} - 3 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{5} + x\right )}^{\frac {2}{3}} + 6 \, {\left (x^{5} + x\right )}^{\frac {1}{3}} x}{x^{4} + 2 \, x^{2} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)/(x^2+1)/(x^5+x)^(1/3),x, algorithm="fricas")

[Out]

1/12*sqrt(3)*2^(2/3)*(-1)^(1/3)*arctan(1/6*sqrt(3)*2^(1/6)*(12*2^(1/6)*(-1)^(2/3)*(x^8 - 14*x^6 + 6*x^4 - 14*x
^2 + 1)*(x^5 + x)^(2/3) - 24*sqrt(2)*(-1)^(1/3)*(x^9 + x^7 + x^3 + x)*(x^5 + x)^(1/3) + 2^(5/6)*(x^12 + 24*x^1
0 - 57*x^8 + 56*x^6 - 57*x^4 + 24*x^2 + 1))/(x^12 - 48*x^10 + 15*x^8 - 88*x^6 + 15*x^4 - 48*x^2 + 1)) - 1/24*2
^(2/3)*(-1)^(1/3)*log((12*2^(1/3)*(-1)^(2/3)*(x^5 - x^3 + x)*(x^5 + x)^(1/3) - 2^(2/3)*(-1)^(1/3)*(x^8 - 14*x^
6 + 6*x^4 - 14*x^2 + 1) - 6*(x^5 + x)^(2/3)*(x^4 - 4*x^2 + 1))/(x^8 + 4*x^6 + 6*x^4 + 4*x^2 + 1)) + 1/12*2^(2/
3)*(-1)^(1/3)*log(-(2^(1/3)*(-1)^(2/3)*(x^4 + 2*x^2 + 1) - 3*2^(2/3)*(-1)^(1/3)*(x^5 + x)^(2/3) + 6*(x^5 + x)^
(1/3)*x)/(x^4 + 2*x^2 + 1))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} - 1}{{\left (x^{5} + x\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)/(x^2+1)/(x^5+x)^(1/3),x, algorithm="giac")

[Out]

integrate((x^2 - 1)/((x^5 + x)^(1/3)*(x^2 + 1)), x)

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maple [C]  time = 18.25, size = 1133, normalized size = 9.36

method	result	size
trager	\(\text {Expression too large to display}\)	\(1133\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-1)/(x^2+1)/(x^5+x)^(1/3),x,method=_RETURNVERBOSE)

[Out]

1/4*RootOf(_Z^3+4)*ln(-(RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3*x^4+4*RootOf(Root
Of(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*RootOf(_Z^3+4)^2*x^4-2*RootOf(_Z^3+4)^3*RootOf(RootOf(_Z^3+4)^2+2*_
Z*RootOf(_Z^3+4)+4*_Z^2)*x^2-8*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x^2+6*(x
^5+x)^(2/3)*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)+RootOf(RootOf(_Z^3+4)^2+2*_Z*
RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3+4*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*RootOf(_Z^3+4)
^2-12*(x^5+x)^(1/3)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)*x+4*RootOf(_Z^3+4)*x^2+
16*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^2-12*(x^5+x)^(2/3))/(x^2+1)^2)-1/4*ln(-(RootOf(RootOf
(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3*x^4-2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z
^2)^2*RootOf(_Z^3+4)^2*x^4-2*RootOf(_Z^3+4)^3*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^2+4*RootOf
(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x^2-6*(x^5+x)^(2/3)*RootOf(_Z^3+4)^2*RootOf(R
ootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)-2*RootOf(_Z^3+4)*x^4+4*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)
+4*_Z^2)*x^4+RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3-2*RootOf(RootOf(_Z^3+4)^2+2*
_Z*RootOf(_Z^3+4)+4*_Z^2)^2*RootOf(_Z^3+4)^2-6*(x^5+x)^(1/3)*RootOf(_Z^3+4)^2*x-2*RootOf(_Z^3+4)+4*RootOf(Root
Of(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2))/(x^2+1)^2)*RootOf(_Z^3+4)-1/2*ln(-(RootOf(RootOf(_Z^3+4)^2+2*_Z*Root
Of(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3*x^4-2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*RootOf(_Z^3+4)
^2*x^4-2*RootOf(_Z^3+4)^3*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^2+4*RootOf(_Z^3+4)^2*RootOf(Ro
otOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)^2*x^2-6*(x^5+x)^(2/3)*RootOf(_Z^3+4)^2*RootOf(RootOf(_Z^3+4)^2+2*_Z
*RootOf(_Z^3+4)+4*_Z^2)-2*RootOf(_Z^3+4)*x^4+4*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*x^4+RootOf(
RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)*RootOf(_Z^3+4)^3-2*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*
_Z^2)^2*RootOf(_Z^3+4)^2-6*(x^5+x)^(1/3)*RootOf(_Z^3+4)^2*x-2*RootOf(_Z^3+4)+4*RootOf(RootOf(_Z^3+4)^2+2*_Z*Ro
otOf(_Z^3+4)+4*_Z^2))/(x^2+1)^2)*RootOf(RootOf(_Z^3+4)^2+2*_Z*RootOf(_Z^3+4)+4*_Z^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} - 1}{{\left (x^{5} + x\right )}^{\frac {1}{3}} {\left (x^{2} + 1\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)/(x^2+1)/(x^5+x)^(1/3),x, algorithm="maxima")

[Out]

integrate((x^2 - 1)/((x^5 + x)^(1/3)*(x^2 + 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^2+1\right )\,{\left (x^5+x\right )}^{1/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - 1)/((x^2 + 1)*(x + x^5)^(1/3)),x)

[Out]

int((x^2 - 1)/((x^2 + 1)*(x + x^5)^(1/3)), 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 \left (x^{4} + 1\right )} \left (x^{2} + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-1)/(x**2+1)/(x**5+x)**(1/3),x)

[Out]

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

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