3.78 \(\int \frac{1-x+4 x^3}{1+x^3} \, dx\)

Optimal. Leaf size=44 \[ \frac{1}{3} \log \left (x^2-x+1\right )+4 x-\frac{2}{3} \log (x+1)+\frac{4 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

[Out]

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

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Rubi [A]  time = 0.0426502, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {1887, 1860, 31, 634, 618, 204, 628} \[ \frac{1}{3} \log \left (x^2-x+1\right )+4 x-\frac{2}{3} \log (x+1)+\frac{4 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 - x + 4*x^3)/(1 + x^3),x]

[Out]

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

Rule 1887

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x
] && PolyQ[Pq, x] && IntegerQ[n]

Rule 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s
*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/
b]

Rule 31

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

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1-x+4 x^3}{1+x^3} \, dx &=\int \left (4-\frac{3+x}{1+x^3}\right ) \, dx\\ &=4 x-\int \frac{3+x}{1+x^3} \, dx\\ &=4 x-\frac{1}{3} \int \frac{7-2 x}{1-x+x^2} \, dx-\frac{2}{3} \int \frac{1}{1+x} \, dx\\ &=4 x-\frac{2}{3} \log (1+x)+\frac{1}{3} \int \frac{-1+2 x}{1-x+x^2} \, dx-2 \int \frac{1}{1-x+x^2} \, dx\\ &=4 x-\frac{2}{3} \log (1+x)+\frac{1}{3} \log \left (1-x+x^2\right )+4 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=4 x+\frac{4 \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{2}{3} \log (1+x)+\frac{1}{3} \log \left (1-x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0078517, size = 44, normalized size = 1. \[ \frac{1}{3} \log \left (x^2-x+1\right )+4 x-\frac{2}{3} \log (x+1)-\frac{4 \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )}{\sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - x + 4*x^3)/(1 + x^3),x]

[Out]

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

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Maple [A]  time = 0.004, size = 38, normalized size = 0.9 \begin{align*} 4\,x+{\frac{\ln \left ({x}^{2}-x+1 \right ) }{3}}-{\frac{4\,\sqrt{3}}{3}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }-{\frac{2\,\ln \left ( 1+x \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^3-x+1)/(x^3+1),x)

[Out]

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

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Maxima [A]  time = 1.42637, size = 50, normalized size = 1.14 \begin{align*} -\frac{4}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 4 \, x + \frac{1}{3} \, \log \left (x^{2} - x + 1\right ) - \frac{2}{3} \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-x+1)/(x^3+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.54927, size = 122, normalized size = 2.77 \begin{align*} -\frac{4}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 4 \, x + \frac{1}{3} \, \log \left (x^{2} - x + 1\right ) - \frac{2}{3} \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-x+1)/(x^3+1),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.119879, size = 48, normalized size = 1.09 \begin{align*} 4 x - \frac{2 \log{\left (x + 1 \right )}}{3} + \frac{\log{\left (x^{2} - x + 1 \right )}}{3} - \frac{4 \sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} - \frac{\sqrt{3}}{3} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**3-x+1)/(x**3+1),x)

[Out]

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

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Giac [A]  time = 1.0711, size = 51, normalized size = 1.16 \begin{align*} -\frac{4}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 4 \, x + \frac{1}{3} \, \log \left (x^{2} - x + 1\right ) - \frac{2}{3} \, \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^3-x+1)/(x^3+1),x, algorithm="giac")

[Out]

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