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

Optimal. Leaf size=30 \[ -\frac{x^2}{2}+\frac{1}{3} \log \left (x^2-x+1\right )+x-\frac{2}{3} \log (x+1) \]

[Out]

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

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Rubi [A]  time = 0.0400098, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {1887, 1860, 31, 628} \[ -\frac{x^2}{2}+\frac{1}{3} \log \left (x^2-x+1\right )+x-\frac{2}{3} \log (x+1) \]

Antiderivative was successfully verified.

[In]

Int[((1 - x)*x^3)/(1 + x^3),x]

[Out]

x - x^2/2 - (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 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) x^3}{1+x^3} \, dx &=\int \left (1-x-\frac{1-x}{1+x^3}\right ) \, dx\\ &=x-\frac{x^2}{2}-\int \frac{1-x}{1+x^3} \, dx\\ &=x-\frac{x^2}{2}-\frac{1}{3} \int \frac{1-2 x}{1-x+x^2} \, dx-\frac{2}{3} \int \frac{1}{1+x} \, dx\\ &=x-\frac{x^2}{2}-\frac{2}{3} \log (1+x)+\frac{1}{3} \log \left (1-x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0043203, size = 30, normalized size = 1. \[ -\frac{x^2}{2}+\frac{1}{3} \log \left (x^2-x+1\right )+x-\frac{2}{3} \log (x+1) \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - x)*x^3)/(1 + x^3),x]

[Out]

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

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Maple [A]  time = 0.004, size = 25, normalized size = 0.8 \begin{align*} x-{\frac{{x}^{2}}{2}}-{\frac{2\,\ln \left ( 1+x \right ) }{3}}+{\frac{\ln \left ({x}^{2}-x+1 \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.42281, size = 32, normalized size = 1.07 \begin{align*} -\frac{1}{2} \, x^{2} + 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((1-x)*x^3/(x^3+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51494, size = 73, normalized size = 2.43 \begin{align*} -\frac{1}{2} \, x^{2} + 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((1-x)*x^3/(x^3+1),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.089936, size = 24, normalized size = 0.8 \begin{align*} - \frac{x^{2}}{2} + x - \frac{2 \log{\left (x + 1 \right )}}{3} + \frac{\log{\left (x^{2} - x + 1 \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2/2 + x - 2*log(x + 1)/3 + log(x**2 - x + 1)/3

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Giac [A]  time = 1.07455, size = 34, normalized size = 1.13 \begin{align*} -\frac{1}{2} \, x^{2} + 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((1-x)*x^3/(x^3+1),x, algorithm="giac")

[Out]

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