3.284 \(\int \frac{1}{1+x+x^2+x^3} \, dx\)

Optimal. Leaf size=25 \[ -\frac{1}{4} \log \left (x^2+1\right )+\frac{1}{2} \log (x+1)+\frac{1}{2} \tan ^{-1}(x) \]

[Out]

ArcTan[x]/2 + Log[1 + x]/2 - Log[1 + x^2]/4

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Rubi [A]  time = 0.0136644, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.364, Rules used = {2058, 635, 203, 260} \[ -\frac{1}{4} \log \left (x^2+1\right )+\frac{1}{2} \log (x+1)+\frac{1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x + x^2 + x^3)^(-1),x]

[Out]

ArcTan[x]/2 + Log[1 + x]/2 - Log[1 + x^2]/4

Rule 2058

Int[(P_)^(p_), x_Symbol] :> With[{u = Factor[P]}, Int[ExpandIntegrand[u^p, x], x] /;  !SumQ[NonfreeFactors[u,
x]]] /; PolyQ[P, x] && ILtQ[p, 0]

Rule 635

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

Rule 203

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

Rule 260

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

Rubi steps

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

Mathematica [A]  time = 0.0050561, size = 25, normalized size = 1. \[ -\frac{1}{4} \log \left (x^2+1\right )+\frac{1}{2} \log (x+1)+\frac{1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x + x^2 + x^3)^(-1),x]

[Out]

ArcTan[x]/2 + Log[1 + x]/2 - Log[1 + x^2]/4

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Maple [A]  time = 0.004, size = 20, normalized size = 0.8 \begin{align*}{\frac{\arctan \left ( x \right ) }{2}}+{\frac{\ln \left ( 1+x \right ) }{2}}-{\frac{\ln \left ({x}^{2}+1 \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3+x^2+x+1),x)

[Out]

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

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Maxima [A]  time = 1.44708, size = 26, normalized size = 1.04 \begin{align*} \frac{1}{2} \, \arctan \left (x\right ) - \frac{1}{4} \, \log \left (x^{2} + 1\right ) + \frac{1}{2} \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3+x^2+x+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.81518, size = 69, normalized size = 2.76 \begin{align*} \frac{1}{2} \, \arctan \left (x\right ) - \frac{1}{4} \, \log \left (x^{2} + 1\right ) + \frac{1}{2} \, \log \left (x + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3+x^2+x+1),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.11245, size = 19, normalized size = 0.76 \begin{align*} \frac{\log{\left (x + 1 \right )}}{2} - \frac{\log{\left (x^{2} + 1 \right )}}{4} + \frac{\operatorname{atan}{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**3+x**2+x+1),x)

[Out]

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

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Giac [A]  time = 1.06655, size = 27, normalized size = 1.08 \begin{align*} \frac{1}{2} \, \arctan \left (x\right ) - \frac{1}{4} \, \log \left (x^{2} + 1\right ) + \frac{1}{2} \, \log \left ({\left | x + 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^3+x^2+x+1),x, algorithm="giac")

[Out]

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