3.3.54 \(\int \frac {1}{x (-1+b x^2)} \, dx\) [254]

Optimal. Leaf size=18 \[ -\log (x)+\frac {1}{2} \log \left (1-b x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 18, 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 = {272, 36, 29, 31} \begin {gather*} \frac {1}{2} \log \left (1-b x^2\right )-\log (x) \end {gather*}

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 272

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} \text {Subst}\left (\int \frac {1}{x (-1+b x)} \, dx,x,x^2\right )\\ &=-\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )\right )+\frac {1}{2} b \text {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 = 18, normalized size = 1.00 \begin {gather*} -\log (x)+\frac {1}{2} \log \left (1-b x^2\right ) \end {gather*}

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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Maple [A]
time = 0.03, size = 16, normalized size = 0.89

method	result	size
default	\(\frac {\ln \left (b \,x^{2}-1\right )}{2}-\ln \left (x \right )\)	\(16\)
norman	\(\frac {\ln \left (b \,x^{2}-1\right )}{2}-\ln \left (x \right )\)	\(16\)
risch	\(-\ln \left (x \right )+\frac {\ln \left (-b \,x^{2}+1\right )}{2}\)	\(17\)
meijerg	\(\frac {\ln \left (-b \,x^{2}+1\right )}{2}-\ln \left (x \right )-\frac {\ln \left (-b \right )}{2}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b*x^2-1),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 17, normalized size = 0.94 \begin {gather*} \frac {1}{2} \, \log \left (b x^{2} - 1\right ) - \frac {1}{2} \, \log \left (x^{2}\right ) \end {gather*}

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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Fricas [A]
time = 0.71, size = 15, normalized size = 0.83 \begin {gather*} \frac {1}{2} \, \log \left (b x^{2} - 1\right ) - \log \left (x\right ) \end {gather*}

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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Sympy [A]
time = 0.05, size = 12, normalized size = 0.67 \begin {gather*} - \log {\left (x \right )} + \frac {\log {\left (x^{2} - \frac {1}{b} \right )}}{2} \end {gather*}

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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Giac [A]
time = 1.88, size = 18, normalized size = 1.00 \begin {gather*} -\frac {1}{2} \, \log \left (x^{2}\right ) + \frac {1}{2} \, \log \left ({\left | b x^{2} - 1 \right |}\right ) \end {gather*}

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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Mupad [B]
time = 0.06, size = 16, normalized size = 0.89 \begin {gather*} \frac {\ln \left (\frac {3}{2}-\frac {3\,b\,x^2}{2}\right )}{2}-\ln \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b*x^2 - 1)),x)

[Out]

log(3/2 - (3*b*x^2)/2)/2 - log(x)

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