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

Optimal. Leaf size=15 \[ \log (x)-\frac {1}{2} \log \left (b x^2+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 36, 29, 31} \[ \log (x)-\frac {1}{2} \log \left (b x^2+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(1 + b*x^2)),x]

[Out]

Log[x] - Log[1 + b*x^2]/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x (1+b x)} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} b \operatorname {Subst}\left (\int \frac {1}{1+b x} \, dx,x,x^2\right )\\ &=\log (x)-\frac {1}{2} \log \left (1+b x^2\right )\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 15, normalized size = 1.00 \[ \log (x)-\frac {1}{2} \log \left (b x^2+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 + b*x^2)),x]

[Out]

Log[x] - Log[1 + b*x^2]/2

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fricas [A]  time = 1.30, size = 13, normalized size = 0.87 \[ -\frac {1}{2} \, \log \left (b x^{2} + 1\right ) + \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(b*x^2 + 1) + log(x)

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giac [A]  time = 0.62, size = 18, normalized size = 1.20 \[ \frac {1}{2} \, \log \left (x^{2}\right ) - \frac {1}{2} \, \log \left ({\left | b x^{2} + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 14, normalized size = 0.93 \[ \ln \relax (x )-\frac {\ln \left (b \,x^{2}+1\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.35, size = 17, normalized size = 1.13 \[ -\frac {1}{2} \, \log \left (b x^{2} + 1\right ) + \frac {1}{2} \, \log \left (x^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(b*x^2 + 1) + 1/2*log(x^2)

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mupad [B]  time = 5.11, size = 14, normalized size = 0.93 \[ \ln \relax (x)-\frac {\ln \left (\frac {3\,b\,x^2}{2}+\frac {3}{2}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b*x^2 + 1)),x)

[Out]

log(x) - log((3*b*x^2)/2 + 3/2)/2

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sympy [A]  time = 0.13, size = 12, normalized size = 0.80 \[ \log {\relax (x )} - \frac {\log {\left (x^{2} + \frac {1}{b} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x) - log(x**2 + 1/b)/2

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