3.418 \(\int \frac {-8+2 x+3 x^2}{8+x^3} \, dx\)

Optimal. Leaf size=32 \[ \frac {3}{2} \log \left (x^2-2 x+4\right )+\frac {\tan ^{-1}\left (\frac {1-x}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

3/2*ln(x^2-2*x+4)+1/3*arctan(1/3*(1-x)*3^(1/2))*3^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1872, 634, 618, 204, 628} \[ \frac {3}{2} \log \left (x^2-2 x+4\right )+\frac {\tan ^{-1}\left (\frac {1-x}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[(-8 + 2*x + 3*x^2)/(8 + x^3),x]

[Out]

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

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 1872

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 31, normalized size = 0.97 \[ \frac {3}{2} \log \left (x^2-2 x+4\right )-\frac {\tan ^{-1}\left (\frac {x-1}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(-8 + 2*x + 3*x^2)/(8 + x^3),x]

[Out]

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

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fricas [A]  time = 0.88, size = 26, normalized size = 0.81 \[ -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x - 1\right )}\right ) + \frac {3}{2} \, \log \left (x^{2} - 2 \, x + 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2*x-8)/(x^3+8),x, algorithm="fricas")

[Out]

-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(x - 1)) + 3/2*log(x^2 - 2*x + 4)

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giac [A]  time = 0.33, size = 26, normalized size = 0.81 \[ -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x - 1\right )}\right ) + \frac {3}{2} \, \log \left (x^{2} - 2 \, x + 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2*x-8)/(x^3+8),x, algorithm="giac")

[Out]

-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(x - 1)) + 3/2*log(x^2 - 2*x + 4)

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maple [A]  time = 0.00, size = 29, normalized size = 0.91 \[ -\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -2\right ) \sqrt {3}}{6}\right )}{3}+\frac {3 \ln \left (x^{2}-2 x +4\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2+2*x-8)/(x^3+8),x)

[Out]

-1/3*3^(1/2)*arctan(1/6*(2*x-2)*3^(1/2))+3/2*ln(x^2-2*x+4)

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maxima [A]  time = 2.33, size = 26, normalized size = 0.81 \[ -\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x - 1\right )}\right ) + \frac {3}{2} \, \log \left (x^{2} - 2 \, x + 4\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2+2*x-8)/(x^3+8),x, algorithm="maxima")

[Out]

-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(x - 1)) + 3/2*log(x^2 - 2*x + 4)

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mupad [B]  time = 0.03, size = 30, normalized size = 0.94 \[ \frac {3\,\ln \left (x^2-2\,x+4\right )}{2}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}\,x}{3}-\frac {\sqrt {3}}{3}\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + 3*x^2 - 8)/(x^3 + 8),x)

[Out]

(3*log(x^2 - 2*x + 4))/2 - (3^(1/2)*atan((3^(1/2)*x)/3 - 3^(1/2)/3))/3

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sympy [A]  time = 0.12, size = 36, normalized size = 1.12 \[ \frac {3 \log {\left (x^{2} - 2 x + 4 \right )}}{2} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {\sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2+2*x-8)/(x**3+8),x)

[Out]

3*log(x**2 - 2*x + 4)/2 - sqrt(3)*atan(sqrt(3)*x/3 - sqrt(3)/3)/3

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