3.69 \(\int \frac {1}{1+x^2+x^3+x^5} \, dx\)

Optimal. Leaf size=38 \[ \frac {1}{4} \log \left (x^2+1\right )-\frac {1}{3} \log \left (x^2-x+1\right )+\frac {1}{6} \log (x+1)+\frac {1}{2} \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2058, 635, 203, 260, 628} \[ \frac {1}{4} \log \left (x^2+1\right )-\frac {1}{3} \log \left (x^2-x+1\right )+\frac {1}{6} \log (x+1)+\frac {1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^2 + x^3 + x^5)^(-1),x]

[Out]

ArcTan[x]/2 + Log[1 + x]/6 + Log[1 + x^2]/4 - Log[1 - x + x^2]/3

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]

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 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 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]

Rubi steps

\begin {align*} \int \frac {1}{1+x^2+x^3+x^5} \, dx &=\int \left (\frac {1}{6 (1+x)}+\frac {1+x}{2 \left (1+x^2\right )}+\frac {1-2 x}{3 \left (1-x+x^2\right )}\right ) \, dx\\ &=\frac {1}{6} \log (1+x)+\frac {1}{3} \int \frac {1-2 x}{1-x+x^2} \, dx+\frac {1}{2} \int \frac {1+x}{1+x^2} \, dx\\ &=\frac {1}{6} \log (1+x)-\frac {1}{3} \log \left (1-x+x^2\right )+\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}{6} \log (1+x)+\frac {1}{4} \log \left (1+x^2\right )-\frac {1}{3} \log \left (1-x+x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 38, normalized size = 1.00 \[ \frac {1}{4} \log \left (x^2+1\right )-\frac {1}{3} \log \left (x^2-x+1\right )+\frac {1}{6} \log (x+1)+\frac {1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^2 + x^3 + x^5)^(-1),x]

[Out]

ArcTan[x]/2 + Log[1 + x]/6 + Log[1 + x^2]/4 - Log[1 - x + x^2]/3

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fricas [A]  time = 0.79, size = 30, normalized size = 0.79 \[ \frac {1}{2} \, \arctan \relax (x) - \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^5+x^3+x^2+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.29, size = 31, normalized size = 0.82 \[ \frac {1}{2} \, \arctan \relax (x) - \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{6} \, \log \left ({\left | x + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^5+x^3+x^2+1),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 31, normalized size = 0.82 \[ \frac {\arctan \relax (x )}{2}+\frac {\ln \left (x +1\right )}{6}+\frac {\ln \left (x^{2}+1\right )}{4}-\frac {\ln \left (x^{2}-x +1\right )}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5+x^3+x^2+1),x)

[Out]

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

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maxima [A]  time = 1.11, size = 30, normalized size = 0.79 \[ \frac {1}{2} \, \arctan \relax (x) - \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) + \frac {1}{6} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x^5+x^3+x^2+1),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.16, size = 36, normalized size = 0.95 \[ \frac {\ln \left (x+1\right )}{6}-\frac {\ln \left (x^2-x+1\right )}{3}+\ln \left (x-\mathrm {i}\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2 + x^3 + x^5 + 1),x)

[Out]

log(x + 1)/6 + log(x - 1i)*(1/4 - 1i/4) + log(x + 1i)*(1/4 + 1i/4) - log(x^2 - x + 1)/3

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sympy [A]  time = 0.15, size = 29, normalized size = 0.76 \[ \frac {\log {\left (x + 1 \right )}}{6} + \frac {\log {\left (x^{2} + 1 \right )}}{4} - \frac {\log {\left (x^{2} - x + 1 \right )}}{3} + \frac {\operatorname {atan}{\relax (x )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(x**5+x**3+x**2+1),x)

[Out]

log(x + 1)/6 + log(x**2 + 1)/4 - log(x**2 - x + 1)/3 + atan(x)/2

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