3.1.46 \(\int \frac {a+b x+c x^2}{(1-x+x^2) \sqrt [3]{1-x^3}} \, dx\) [46]

Optimal. Leaf size=493 \[ \frac {(a+b) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {(a+b) \tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}-\frac {c \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {(a-c) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {(b+c) \tan ^{-1}\left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {(a+b) \log \left ((1-x) (1+x)^2\right )}{12 \sqrt [3]{2}}-\frac {(a-c) \log \left (1+x^3\right )}{6 \sqrt [3]{2}}-\frac {(b+c) \log \left (1+x^3\right )}{6 \sqrt [3]{2}}+\frac {(a+b) \log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{6 \sqrt [3]{2}}-\frac {(a+b) \log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{3 \sqrt [3]{2}}+\frac {(b+c) \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}+\frac {(a-c) \log \left (-\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}+\frac {1}{2} c \log \left (x+\sqrt [3]{1-x^3}\right )-\frac {(a+b) \log \left (-1+x+2^{2/3} \sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2}} \]

[Out]

1/24*(a+b)*ln((1-x)*(1+x)^2)*2^(2/3)-1/12*(a-c)*ln(x^3+1)*2^(2/3)-1/12*(b+c)*ln(x^3+1)*2^(2/3)+1/12*(a+b)*ln(1
+2^(2/3)*(1-x)^2/(-x^3+1)^(2/3)-2^(1/3)*(1-x)/(-x^3+1)^(1/3))*2^(2/3)-1/6*(a+b)*ln(1+2^(1/3)*(1-x)/(-x^3+1)^(1
/3))*2^(2/3)+1/4*(b+c)*ln(2^(1/3)-(-x^3+1)^(1/3))*2^(2/3)+1/4*(a-c)*ln(-2^(1/3)*x-(-x^3+1)^(1/3))*2^(2/3)+1/2*
c*ln(x+(-x^3+1)^(1/3))-1/8*(a+b)*ln(-1+x+2^(2/3)*(-x^3+1)^(1/3))*2^(2/3)+1/6*(a+b)*arctan(1/3*(1-2*2^(1/3)*(1-
x)/(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)+1/12*(a+b)*arctan(1/3*(1+2^(1/3)*(1-x)/(-x^3+1)^(1/3))*3^(1/2))*2^
(2/3)*3^(1/2)-1/3*c*arctan(1/3*(1-2*x/(-x^3+1)^(1/3))*3^(1/2))*3^(1/2)-1/6*(a-c)*arctan(1/3*(1-2*2^(1/3)*x/(-x
^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)+1/6*(b+c)*arctan(1/3*(1+2^(2/3)*(-x^3+1)^(1/3))*3^(1/2))*2^(2/3)*3^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.29, antiderivative size = 570, normalized size of antiderivative = 1.16, number of steps used = 19, number of rules used = 14, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {2183, 384, 502, 2174, 206, 31, 648, 631, 210, 642, 455, 57, 494, 245} \begin {gather*} \frac {(a+b) \log \left (\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{6 \sqrt [3]{2}}-\frac {(a+b) \log \left (\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{3 \sqrt [3]{2}}-\frac {(a+b) \log \left (2^{2/3} \sqrt [3]{1-x^3}+x-1\right )}{4 \sqrt [3]{2}}+\frac {(a+b) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {(a+b) \tan ^{-1}\left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{2} \sqrt {3}}+\frac {(a+b) \log \left ((1-x) (x+1)^2\right )}{12 \sqrt [3]{2}}-\frac {a \log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {a \log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}-\frac {a \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {(b+c) \log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {(b+c) \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}+\frac {(b+c) \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}+\frac {c \log \left (x^3+1\right )}{6 \sqrt [3]{2}}-\frac {c \log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}+\frac {1}{2} c \log \left (\sqrt [3]{1-x^3}+x\right )-\frac {c \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {c \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)*ArcTan[(1 - (2*2^(1/3)*(1 - x))/(1 - x^3)^(1/3))/Sqrt[3]])/(2^(1/3)*Sqrt[3]) + ((a + b)*ArcTan[(1 + (
2^(1/3)*(1 - x))/(1 - x^3)^(1/3))/Sqrt[3]])/(2*2^(1/3)*Sqrt[3]) - (c*ArcTan[(1 - (2*x)/(1 - x^3)^(1/3))/Sqrt[3
]])/Sqrt[3] - (a*ArcTan[(1 - (2*2^(1/3)*x)/(1 - x^3)^(1/3))/Sqrt[3]])/(2^(1/3)*Sqrt[3]) + (c*ArcTan[(1 - (2*2^
(1/3)*x)/(1 - x^3)^(1/3))/Sqrt[3]])/(2^(1/3)*Sqrt[3]) + ((b + c)*ArcTan[(1 + 2^(2/3)*(1 - x^3)^(1/3))/Sqrt[3]]
)/(2^(1/3)*Sqrt[3]) + ((a + b)*Log[(1 - x)*(1 + x)^2])/(12*2^(1/3)) - (a*Log[1 + x^3])/(6*2^(1/3)) + (c*Log[1
+ x^3])/(6*2^(1/3)) - ((b + c)*Log[1 + x^3])/(6*2^(1/3)) + ((a + b)*Log[1 + (2^(2/3)*(1 - x)^2)/(1 - x^3)^(2/3
) - (2^(1/3)*(1 - x))/(1 - x^3)^(1/3)])/(6*2^(1/3)) - ((a + b)*Log[1 + (2^(1/3)*(1 - x))/(1 - x^3)^(1/3)])/(3*
2^(1/3)) + ((b + c)*Log[2^(1/3) - (1 - x^3)^(1/3)])/(2*2^(1/3)) + (a*Log[-(2^(1/3)*x) - (1 - x^3)^(1/3)])/(2*2
^(1/3)) - (c*Log[-(2^(1/3)*x) - (1 - x^3)^(1/3)])/(2*2^(1/3)) + (c*Log[x + (1 - x^3)^(1/3)])/2 - ((a + b)*Log[
-1 + x + 2^(2/3)*(1 - x^3)^(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 57

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 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 455

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 494

Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_)^(n_))^(q_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Dist[e^n/b, Int[
(e*x)^(m - n)*(c + d*x^n)^q, x], x] - Dist[a*(e^n/b), Int[(e*x)^(m - n)*((c + d*x^n)^q/(a + b*x^n)), x], x] /;
FreeQ[{a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1] && IntBinomialQ[a, b
, c, d, e, m, n, -1, q, x]

Rule 502

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

Rule 631

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 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 2174

Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Simp[Sqrt[3]*(ArcTan[(1 - 2^(1/3)*Rt[b,
3]*((c - d*x)/(d*(a + b*x^3)^(1/3))))/Sqrt[3]]/(2^(4/3)*Rt[b, 3]*c)), x] + (Simp[Log[(c + d*x)^2*(c - d*x)]/(2
^(7/3)*Rt[b, 3]*c), x] - Simp[(3*Log[Rt[b, 3]*(c - d*x) + 2^(2/3)*d*(a + b*x^3)^(1/3)])/(2^(7/3)*Rt[b, 3]*c),
x]) /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 + a*d^3, 0]

Rule 2183

Int[(Px_.)*((c_) + (d_.)*(x_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^3)^(p_.), x_Symbol] :> Dist[1/c^q, Int[E
xpandIntegrand[(c^3 - d^3*x^3)^q*(a + b*x^3)^p, Px/(c - d*x)^q, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] &&
PolyQ[Px, x] && EqQ[d^2 - c*e, 0] && ILtQ[q, 0] && RationalQ[p] && EqQ[Denominator[p], 3]

Rubi steps

\begin {align*} \int \frac {a+b x+c x^2}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx &=\int \left (\frac {c}{\sqrt [3]{1-x^3}}+\frac {a-c+(b+c) x}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}}\right ) \, dx\\ &=c \int \frac {1}{\sqrt [3]{1-x^3}} \, dx+\int \frac {a-c+(b+c) x}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx\\ &=-\frac {c \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} c \log \left (x+\sqrt [3]{1-x^3}\right )+\int \left (\frac {b-\frac {i (2 a+b-c)}{\sqrt {3}}+c}{\left (-1-i \sqrt {3}+2 x\right ) \sqrt [3]{1-x^3}}+\frac {b+\frac {i (2 a+b-c)}{\sqrt {3}}+c}{\left (-1+i \sqrt {3}+2 x\right ) \sqrt [3]{1-x^3}}\right ) \, dx\\ &=-\frac {c \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} c \log \left (x+\sqrt [3]{1-x^3}\right )+\frac {1}{3} \left (3 b-i \sqrt {3} (2 a+b-c)+3 c\right ) \int \frac {1}{\left (-1-i \sqrt {3}+2 x\right ) \sqrt [3]{1-x^3}} \, dx+\frac {1}{3} \left (3 b+i \sqrt {3} (2 a+b-c)+3 c\right ) \int \frac {1}{\left (-1+i \sqrt {3}+2 x\right ) \sqrt [3]{1-x^3}} \, dx\\ &=-\frac {c \tan ^{-1}\left (\frac {1-\frac {2 x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {\left (2 a+b-i \sqrt {3} b-c-i \sqrt {3} c\right ) \tan ^{-1}\left (\frac {2-\frac {\sqrt [3]{2} \left (1-i \sqrt {3}+2 x\right )}{\sqrt [3]{1-x^3}}}{2 \sqrt {3}}\right )}{2 \sqrt [3]{2} \left (i+\sqrt {3}\right )}+\frac {\left (2 a+b+i \sqrt {3} b-c+i \sqrt {3} c\right ) \tan ^{-1}\left (\frac {2-\frac {\sqrt [3]{2} \left (1+i \sqrt {3}+2 x\right )}{\sqrt [3]{1-x^3}}}{2 \sqrt {3}}\right )}{2 \sqrt [3]{2} \left (i-\sqrt {3}\right )}+\frac {\left (3 i b-\sqrt {3} \left (2 a+b-c-i \sqrt {3} c\right )\right ) \log \left (-\left (1-i \sqrt {3}-2 x\right )^2 \left (1-i \sqrt {3}+2 x\right )\right )}{12 \sqrt [3]{2} \left (i+\sqrt {3}\right )}+\frac {\left (3 i b+\sqrt {3} \left (2 a+b-c+i \sqrt {3} c\right )\right ) \log \left (-\left (1+i \sqrt {3}-2 x\right )^2 \left (1+i \sqrt {3}+2 x\right )\right )}{12 \sqrt [3]{2} \left (i-\sqrt {3}\right )}+\frac {1}{2} c \log \left (x+\sqrt [3]{1-x^3}\right )-\frac {\left (3 i b-\sqrt {3} \left (2 a+b-c-i \sqrt {3} c\right )\right ) \log \left (1-i \sqrt {3}+2 x+2\ 2^{2/3} \sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2} \left (i+\sqrt {3}\right )}-\frac {\left (3 i b+\sqrt {3} \left (2 a+b-c+i \sqrt {3} c\right )\right ) \log \left (1+i \sqrt {3}+2 x+2\ 2^{2/3} \sqrt [3]{1-x^3}\right )}{4 \sqrt [3]{2} \left (i-\sqrt {3}\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]
time = 10.24, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a+b x+c x^2}{\left (1-x+x^2\right ) \sqrt [3]{1-x^3}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded while calling a Python object} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[(a + b*x + c*x^2)/((1 - x + x^2)*(1 - x^3)^(1/3)),x]')

[Out]

cought exception: maximum recursion depth exceeded while calling a Python object

________________________________________________________________________________________

Maple [F]
time = 0.09, size = 0, normalized size = 0.00 \[\int \frac {c \,x^{2}+b x +a}{\left (x^{2}-x +1\right ) \left (-x^{3}+1\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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((c*x^2+b*x+a)/(x^2-x+1)/(-x^3+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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F] N/A
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((c*x^2+b*x+a)/(x^2-x+1)/(-x^3+1)^(1/3),x)

[Out]

Could not integrate

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {c\,x^2+b\,x+a}{{\left (1-x^3\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((a + b*x + c*x^2)/((1 - x^3)^(1/3)*(x^2 - x + 1)),x)

[Out]

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

________________________________________________________________________________________