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

Optimal. Leaf size=131 \[ \frac {\log \left (\sqrt [3]{3 x^2-1}-\sqrt [3]{2} x\right )}{3 \sqrt [3]{2}}-\frac {\log \left (2^{2/3} x^2+\sqrt [3]{2} \sqrt [3]{3 x^2-1} x+\left (3 x^2-1\right )^{2/3}\right )}{6 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{3 x^2-1}}{\sqrt [3]{3 x^2-1}+2 \sqrt [3]{2} x}\right )}{\sqrt [3]{2} \sqrt {3}} \]

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Rubi [C]  time = 0.61, antiderivative size = 265, normalized size of antiderivative = 2.02, number of steps used = 20, number of rules used = 10, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {6742, 757, 430, 429, 444, 55, 617, 204, 31, 56} \begin {gather*} -\frac {2 x \sqrt [3]{1-3 x^2} F_1\left (\frac {1}{2};\frac {1}{3},1;\frac {3}{2};3 x^2,x^2\right )}{3 \sqrt [3]{3 x^2-1}}-\frac {x \sqrt [3]{1-3 x^2} F_1\left (\frac {1}{2};1,\frac {1}{3};\frac {3}{2};4 x^2,3 x^2\right )}{3 \sqrt [3]{3 x^2-1}}-\frac {\log \left (1-4 x^2\right )}{12 \sqrt [3]{2}}-\frac {\log \left (1-x^2\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{3 x^2-1}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (2^{2/3} \sqrt [3]{3 x^2-1}+1\right )}{4 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {1-2\ 2^{2/3} \sqrt [3]{3 x^2-1}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{3 x^2-1}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*x*(1 - 3*x^2)^(1/3)*AppellF1[1/2, 1/3, 1, 3/2, 3*x^2, x^2])/(3*(-1 + 3*x^2)^(1/3)) - (x*(1 - 3*x^2)^(1/3)*
AppellF1[1/2, 1, 1/3, 3/2, 4*x^2, 3*x^2])/(3*(-1 + 3*x^2)^(1/3)) + ArcTan[(1 - 2*2^(2/3)*(-1 + 3*x^2)^(1/3))/S
qrt[3]]/(2*2^(1/3)*Sqrt[3]) + ArcTan[(1 + 2^(2/3)*(-1 + 3*x^2)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3]) - Log[1 - 4*x
^2]/(12*2^(1/3)) - Log[1 - x^2]/(6*2^(1/3)) + Log[2^(1/3) - (-1 + 3*x^2)^(1/3)]/(2*2^(1/3)) + Log[1 + 2^(2/3)*
(-1 + 3*x^2)^(1/3)]/(4*2^(1/3))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 56

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[-((b*c - a*d)/b), 3]}, Simp
[Log[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 - q*x + x^2), x], x, (c + d*x)^(
1/3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q + x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && Ne
gQ[(b*c - a*d)/b]

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 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 430

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPart[p]*(a + b*x^n)^F
racPart[p])/(1 + (b*x^n)/a)^FracPart[p], Int[(1 + (b*x^n)/a)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 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 757

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + c*x^2)^p, (d/(d
^2 - e^2*x^2) - (e*x)/(d^2 - e^2*x^2))^(-m), x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&
!IntegerQ[p] && ILtQ[m, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [F]  time = 1.12, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1+x}{(-1+x) (1+2 x) \sqrt [3]{-1+3 x^2}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.20, size = 131, normalized size = 1.00 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{-1+3 x^2}}{2 \sqrt [3]{2} x+\sqrt [3]{-1+3 x^2}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {\log \left (-\sqrt [3]{2} x+\sqrt [3]{-1+3 x^2}\right )}{3 \sqrt [3]{2}}-\frac {\log \left (2^{2/3} x^2+\sqrt [3]{2} x \sqrt [3]{-1+3 x^2}+\left (-1+3 x^2\right )^{2/3}\right )}{6 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (re
sidue poly has multiple non-linear factors)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 29.40, size = 651, normalized size = 4.97

method	result	size
trager	\(\frac {\RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right ) \ln \left (\frac {3 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-4\right )^{3} x^{3}+9 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} x^{3}-3 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} \left (3 x^{2}-1\right )^{\frac {2}{3}} x +2 \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} \left (3 x^{2}-1\right )^{\frac {1}{3}} x^{2}+12 \RootOf \left (\textit {\_Z}^{3}-4\right ) \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right ) \left (3 x^{2}-1\right )^{\frac {1}{3}} x^{2}-6 \RootOf \left (\textit {\_Z}^{3}-4\right ) x^{2}-18 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right ) x^{2}+4 \left (3 x^{2}-1\right )^{\frac {2}{3}} x +2 \RootOf \left (\textit {\_Z}^{3}-4\right )+6 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right )}{\left (-1+x \right )^{2} \left (1+2 x \right )}\right )}{2}+\frac {\RootOf \left (\textit {\_Z}^{3}-4\right ) \ln \left (-\frac {3 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-4\right )^{3} x^{3}+9 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right )^{2} \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} x^{3}-3 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right ) \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} \left (3 x^{2}-1\right )^{\frac {2}{3}} x +2 \RootOf \left (\textit {\_Z}^{3}-4\right )^{2} \left (3 x^{2}-1\right )^{\frac {1}{3}} x^{2}-6 \RootOf \left (\textit {\_Z}^{3}-4\right ) \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right ) \left (3 x^{2}-1\right )^{\frac {1}{3}} x^{2}+4 \RootOf \left (\textit {\_Z}^{3}-4\right ) x^{3}+12 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right ) x^{3}+6 \RootOf \left (\textit {\_Z}^{3}-4\right ) x^{2}+18 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right ) x^{2}-8 \left (3 x^{2}-1\right )^{\frac {2}{3}} x -2 \RootOf \left (\textit {\_Z}^{3}-4\right )-6 \RootOf \left (\RootOf \left (\textit {\_Z}^{3}-4\right )^{2}+3 \textit {\_Z} \RootOf \left (\textit {\_Z}^{3}-4\right )+9 \textit {\_Z}^{2}\right )}{\left (-1+x \right )^{2} \left (1+2 x \right )}\right )}{6}\)	\(651\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*ln((3*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z
^2)*RootOf(_Z^3-4)^3*x^3+9*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)^2*RootOf(_Z^3-4)^2*x^3-3*RootOf
(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*RootOf(_Z^3-4)^2*(3*x^2-1)^(2/3)*x+2*RootOf(_Z^3-4)^2*(3*x^2-1)^
(1/3)*x^2+12*RootOf(_Z^3-4)*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*(3*x^2-1)^(1/3)*x^2-6*RootOf(_
Z^3-4)*x^2-18*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*x^2+4*(3*x^2-1)^(2/3)*x+2*RootOf(_Z^3-4)+6*R
ootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2))/(-1+x)^2/(1+2*x))+1/6*RootOf(_Z^3-4)*ln(-(3*RootOf(RootOf(
_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*RootOf(_Z^3-4)^3*x^3+9*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^
2)^2*RootOf(_Z^3-4)^2*x^3-3*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*RootOf(_Z^3-4)^2*(3*x^2-1)^(2/
3)*x+2*RootOf(_Z^3-4)^2*(3*x^2-1)^(1/3)*x^2-6*RootOf(_Z^3-4)*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^
2)*(3*x^2-1)^(1/3)*x^2+4*RootOf(_Z^3-4)*x^3+12*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*x^3+6*RootO
f(_Z^3-4)*x^2+18*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2)*x^2-8*(3*x^2-1)^(2/3)*x-2*RootOf(_Z^3-4)-
6*RootOf(RootOf(_Z^3-4)^2+3*_Z*RootOf(_Z^3-4)+9*_Z^2))/(-1+x)^2/(1+2*x))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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