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

Optimal. Leaf size=27 \[ -\frac{1}{4} \log \left (x^2+1\right )+\log (x)-\frac{1}{2} \log (x+1)-\frac{1}{2} \tan ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.0248596, antiderivative size = 27, 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 = {894, 635, 203, 260} \[ -\frac{1}{4} \log \left (x^2+1\right )+\log (x)-\frac{1}{2} \log (x+1)-\frac{1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 + x)*(1 + x^2)),x]

[Out]

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

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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}{x (1+x) \left (1+x^2\right )} \, dx &=\int \left (\frac{1}{x}-\frac{1}{2 (1+x)}+\frac{-1-x}{2 \left (1+x^2\right )}\right ) \, dx\\ &=\log (x)-\frac{1}{2} \log (1+x)+\frac{1}{2} \int \frac{-1-x}{1+x^2} \, dx\\ &=\log (x)-\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)+\log (x)-\frac{1}{2} \log (1+x)-\frac{1}{4} \log \left (1+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0067695, size = 27, normalized size = 1. \[ -\frac{1}{4} \log \left (x^2+1\right )+\log (x)-\frac{1}{2} \log (x+1)-\frac{1}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 + x)*(1 + x^2)),x]

[Out]

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

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Maple [A]  time = 0.004, size = 22, normalized size = 0.8 \begin{align*} -{\frac{\arctan \left ( x \right ) }{2}}+\ln \left ( x \right ) -{\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/(1+x)/(x^2+1),x)

[Out]

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

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Maxima [A]  time = 1.38402, size = 28, 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 ) + \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.02233, size = 82, normalized size = 3.04 \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 ) + \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.157123, size = 22, normalized size = 0.81 \begin{align*} \log{\left (x \right )} - \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/(1+x)/(x**2+1),x)

[Out]

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

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Giac [A]  time = 1.06179, size = 31, normalized size = 1.15 \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 ) + \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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