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

Optimal. Leaf size=24 \[ \frac{1}{2 \left (x^2+1\right )}-\frac{1}{2} \log \left (x^2+1\right )+\log (x) \]

[Out]

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

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Rubi [A]  time = 0.0095761, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {266, 44} \[ \frac{1}{2 \left (x^2+1\right )}-\frac{1}{2} \log \left (x^2+1\right )+\log (x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 + x^2)^2),x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0066393, size = 24, normalized size = 1. \[ \frac{1}{2 \left (x^2+1\right )}-\frac{1}{2} \log \left (x^2+1\right )+\log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 + x^2)^2),x]

[Out]

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

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Maple [A]  time = 0.009, size = 21, normalized size = 0.9 \begin{align*}{\frac{1}{2\,{x}^{2}+2}}+\ln \left ( x \right ) -{\frac{\ln \left ({x}^{2}+1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.972026, size = 32, normalized size = 1.33 \begin{align*} \frac{1}{2 \,{\left (x^{2} + 1\right )}} - \frac{1}{2} \, \log \left (x^{2} + 1\right ) + \frac{1}{2} \, \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.756444, size = 89, normalized size = 3.71 \begin{align*} -\frac{{\left (x^{2} + 1\right )} \log \left (x^{2} + 1\right ) - 2 \,{\left (x^{2} + 1\right )} \log \left (x\right ) - 1}{2 \,{\left (x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.095338, size = 19, normalized size = 0.79 \begin{align*} \log{\left (x \right )} - \frac{\log{\left (x^{2} + 1 \right )}}{2} + \frac{1}{2 x^{2} + 2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.09623, size = 39, normalized size = 1.62 \begin{align*} \frac{x^{2} + 2}{2 \,{\left (x^{2} + 1\right )}} - \frac{1}{2} \, \log \left (x^{2} + 1\right ) + \frac{1}{2} \, \log \left (x^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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