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

Optimal. Leaf size=28 \[ \frac {1}{2 \left (b x^2+1\right )}-\frac {1}{2} \log \left (b x^2+1\right )+\log (x) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {266, 44} \[ \frac {1}{2 \left (b x^2+1\right )}-\frac {1}{2} \log \left (b x^2+1\right )+\log (x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 + b*x^2)^2),x]

[Out]

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

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])

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]]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 25, normalized size = 0.89 \[ \frac {1}{2 b x^2+2}-\frac {1}{2} \log \left (b x^2+1\right )+\log (x) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 + b*x^2)^2),x]

[Out]

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

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fricas [A]  time = 0.72, size = 40, normalized size = 1.43 \[ -\frac {{\left (b x^{2} + 1\right )} \log \left (b x^{2} + 1\right ) - 2 \, {\left (b x^{2} + 1\right )} \log \relax (x) - 1}{2 \, {\left (b x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.63, size = 36, normalized size = 1.29 \[ \frac {b x^{2} + 2}{2 \, {\left (b x^{2} + 1\right )}} + \frac {1}{2} \, \log \left (x^{2}\right ) - \frac {1}{2} \, \log \left ({\left | b x^{2} + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 25, normalized size = 0.89 \[ \ln \relax (x )-\frac {\ln \left (b \,x^{2}+1\right )}{2}+\frac {1}{2 b \,x^{2}+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.35, size = 28, normalized size = 1.00 \[ \frac {1}{2 \, {\left (b x^{2} + 1\right )}} - \frac {1}{2} \, \log \left (b x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 24, normalized size = 0.86 \[ \ln \relax (x)-\frac {\ln \left (\frac {3\,b\,x^2}{2}+\frac {3}{2}\right )}{2}+\frac {1}{2\,\left (b\,x^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b*x^2 + 1)^2),x)

[Out]

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

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sympy [A]  time = 0.20, size = 22, normalized size = 0.79 \[ \log {\relax (x )} - \frac {\log {\left (x^{2} + \frac {1}{b} \right )}}{2} + \frac {1}{2 b x^{2} + 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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