3.49 \(\int \frac {a+b x+c x^2}{1-x^3} \, dx\)

Optimal. Leaf size=55 \[ \frac {1}{6} \log \left (x^2+x+1\right ) (a+b-2 c)-\frac {1}{3} \log (1-x) (a+b+c)+\frac {(a-b) \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1875, 31, 634, 618, 204, 628} \[ \frac {1}{6} \log \left (x^2+x+1\right ) (a+b-2 c)-\frac {1}{3} \log (1-x) (a+b+c)+\frac {(a-b) \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)/(1 - x^3),x]

[Out]

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

Rule 31

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

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 618

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

Rule 628

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 634

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 1875

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2], q = (-(a/b))^(1/3)}, Dist[(q*(A + B*q + C*q^2))/(3*a), Int[1/(q - x), x], x] + Dist[q/(3*a), Int[(q*(2*A
- B*q - C*q^2) + (A + B*q - 2*C*q^2)*x)/(q^2 + q*x + x^2), x], x] /; NeQ[a*B^3 - b*A^3, 0] && NeQ[A + B*q + C*
q^2, 0]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2] && LtQ[a/b, 0]

Rubi steps

\begin {align*} \int \frac {a+b x+c x^2}{1-x^3} \, dx &=\frac {1}{3} \int \frac {2 a-b-c+(a+b-2 c) x}{1+x+x^2} \, dx+\frac {1}{3} (a+b+c) \int \frac {1}{1-x} \, dx\\ &=-\frac {1}{3} (a+b+c) \log (1-x)+\frac {1}{2} (a-b) \int \frac {1}{1+x+x^2} \, dx+\frac {1}{6} (a+b-2 c) \int \frac {1+2 x}{1+x+x^2} \, dx\\ &=-\frac {1}{3} (a+b+c) \log (1-x)+\frac {1}{6} (a+b-2 c) \log \left (1+x+x^2\right )+(-a+b) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=\frac {(a-b) \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} (a+b+c) \log (1-x)+\frac {1}{6} (a+b-2 c) \log \left (1+x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 62, normalized size = 1.13 \[ \frac {1}{6} \left ((a+b) \log \left (x^2+x+1\right )-2 (a+b) \log (1-x)+2 \sqrt {3} (a-b) \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )-2 c \log \left (1-x^3\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.88, size = 47, normalized size = 0.85 \[ \frac {1}{3} \, \sqrt {3} {\left (a - b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, {\left (a + b - 2 \, c\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{3} \, {\left (a + b + c\right )} \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*(a - b)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*(a + b - 2*c)*log(x^2 + x + 1) - 1/3*(a + b + c)*log(x
- 1)

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giac [A]  time = 0.17, size = 52, normalized size = 0.95 \[ \frac {1}{3} \, {\left (\sqrt {3} a - \sqrt {3} b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, {\left (a + b - 2 \, c\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{3} \, {\left (a + b + c\right )} \log \left ({\left | x - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(sqrt(3)*a - sqrt(3)*b)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*(a + b - 2*c)*log(x^2 + x + 1) - 1/3*(a + b +
c)*log(abs(x - 1))

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maple [A]  time = 0.05, size = 87, normalized size = 1.58 \[ \frac {\sqrt {3}\, a \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}-\frac {a \ln \left (x -1\right )}{3}+\frac {a \ln \left (x^{2}+x +1\right )}{6}-\frac {\sqrt {3}\, b \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}-\frac {b \ln \left (x -1\right )}{3}+\frac {b \ln \left (x^{2}+x +1\right )}{6}-\frac {c \ln \left (x -1\right )}{3}-\frac {c \ln \left (x^{2}+x +1\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*c*ln(x-1)-1/3*ln(x-1)*b-1/3*a*ln(x-1)+1/6*a*ln(x^2+x+1)+1/6*ln(x^2+x+1)*b-1/3*c*ln(x^2+x+1)+1/3*3^(1/2)*a
rctan(1/3*(2*x+1)*3^(1/2))*a-1/3*3^(1/2)*arctan(1/3*(2*x+1)*3^(1/2))*b

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maxima [A]  time = 2.99, size = 47, normalized size = 0.85 \[ \frac {1}{3} \, \sqrt {3} {\left (a - b\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, {\left (a + b - 2 \, c\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{3} \, {\left (a + b + c\right )} \log \left (x - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(3)*(a - b)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*(a + b - 2*c)*log(x^2 + x + 1) - 1/3*(a + b + c)*log(x
- 1)

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mupad [B]  time = 4.95, size = 87, normalized size = 1.58 \[ \ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {a}{6}+\frac {b}{6}-\frac {c}{3}-\frac {\sqrt {3}\,a\,1{}\mathrm {i}}{6}+\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{6}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {a}{6}+\frac {b}{6}-\frac {c}{3}+\frac {\sqrt {3}\,a\,1{}\mathrm {i}}{6}-\frac {\sqrt {3}\,b\,1{}\mathrm {i}}{6}\right )-\ln \left (x-1\right )\,\left (\frac {a}{3}+\frac {b}{3}+\frac {c}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a + b*x + c*x^2)/(x^3 - 1),x)

[Out]

log(x - (3^(1/2)*1i)/2 + 1/2)*(a/6 + b/6 - c/3 - (3^(1/2)*a*1i)/6 + (3^(1/2)*b*1i)/6) + log(x + (3^(1/2)*1i)/2
+ 1/2)*(a/6 + b/6 - c/3 + (3^(1/2)*a*1i)/6 - (3^(1/2)*b*1i)/6) - log(x - 1)*(a/3 + b/3 + c/3)

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sympy [C]  time = 1.89, size = 323, normalized size = 5.87 \[ - \frac {\left (a + b + c\right ) \log {\left (x + \frac {a^{2} c - a^{2} \left (a + b + c\right ) - 2 a b^{2} + b c^{2} - 2 b c \left (a + b + c\right ) + b \left (a + b + c\right )^{2}}{a^{3} - b^{3}} \right )}}{3} - \left (- \frac {a}{6} - \frac {b}{6} + \frac {c}{3} - \frac {\sqrt {3} i \left (a - b\right )}{6}\right ) \log {\left (x + \frac {a^{2} c - 3 a^{2} \left (- \frac {a}{6} - \frac {b}{6} + \frac {c}{3} - \frac {\sqrt {3} i \left (a - b\right )}{6}\right ) - 2 a b^{2} + b c^{2} - 6 b c \left (- \frac {a}{6} - \frac {b}{6} + \frac {c}{3} - \frac {\sqrt {3} i \left (a - b\right )}{6}\right ) + 9 b \left (- \frac {a}{6} - \frac {b}{6} + \frac {c}{3} - \frac {\sqrt {3} i \left (a - b\right )}{6}\right )^{2}}{a^{3} - b^{3}} \right )} - \left (- \frac {a}{6} - \frac {b}{6} + \frac {c}{3} + \frac {\sqrt {3} i \left (a - b\right )}{6}\right ) \log {\left (x + \frac {a^{2} c - 3 a^{2} \left (- \frac {a}{6} - \frac {b}{6} + \frac {c}{3} + \frac {\sqrt {3} i \left (a - b\right )}{6}\right ) - 2 a b^{2} + b c^{2} - 6 b c \left (- \frac {a}{6} - \frac {b}{6} + \frac {c}{3} + \frac {\sqrt {3} i \left (a - b\right )}{6}\right ) + 9 b \left (- \frac {a}{6} - \frac {b}{6} + \frac {c}{3} + \frac {\sqrt {3} i \left (a - b\right )}{6}\right )^{2}}{a^{3} - b^{3}} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(a + b + c)*log(x + (a**2*c - a**2*(a + b + c) - 2*a*b**2 + b*c**2 - 2*b*c*(a + b + c) + b*(a + b + c)**2)/(a
**3 - b**3))/3 - (-a/6 - b/6 + c/3 - sqrt(3)*I*(a - b)/6)*log(x + (a**2*c - 3*a**2*(-a/6 - b/6 + c/3 - sqrt(3)
*I*(a - b)/6) - 2*a*b**2 + b*c**2 - 6*b*c*(-a/6 - b/6 + c/3 - sqrt(3)*I*(a - b)/6) + 9*b*(-a/6 - b/6 + c/3 - s
qrt(3)*I*(a - b)/6)**2)/(a**3 - b**3)) - (-a/6 - b/6 + c/3 + sqrt(3)*I*(a - b)/6)*log(x + (a**2*c - 3*a**2*(-a
/6 - b/6 + c/3 + sqrt(3)*I*(a - b)/6) - 2*a*b**2 + b*c**2 - 6*b*c*(-a/6 - b/6 + c/3 + sqrt(3)*I*(a - b)/6) + 9
*b*(-a/6 - b/6 + c/3 + sqrt(3)*I*(a - b)/6)**2)/(a**3 - b**3))

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