3.3.60 \(\int \frac {1}{x (-1+b x^2)^2} \, dx\) [260]

Optimal. Leaf size=30 \[ \frac {1}{2 \left (1-b x^2\right )}+\log (x)-\frac {1}{2} \log \left (1-b x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 30, 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 = {272, 46} \begin {gather*} \frac {1}{2 \left (1-b x^2\right )}-\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)^2),x]

[Out]

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

Rule 46

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] && Lt
Q[m + n + 2, 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 )^2} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{x (-1+b x)^2} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {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 = 26, normalized size = 0.87 \begin {gather*} \frac {1}{2-2 b x^2}+\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)^2),x]

[Out]

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

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Maple [A]
time = 0.05, size = 33, normalized size = 1.10

method	result	size
risch	\(-\frac {1}{2 \left (b \,x^{2}-1\right )}-\frac {\ln \left (b \,x^{2}-1\right )}{2}+\ln \left (x \right )\)	\(25\)
norman	\(-\frac {b \,x^{2}}{2 \left (b \,x^{2}-1\right )}-\frac {\ln \left (b \,x^{2}-1\right )}{2}+\ln \left (x \right )\)	\(29\)
default	\(-\frac {b \left (\frac {\ln \left (b \,x^{2}-1\right )}{b}+\frac {1}{b \left (b \,x^{2}-1\right )}\right )}{2}+\ln \left (x \right )\)	\(33\)
meijerg	\(\frac {b \,x^{2}}{-2 b \,x^{2}+2}-\frac {\ln \left (-b \,x^{2}+1\right )}{2}+\frac {1}{2}+\ln \left (x \right )+\frac {\ln \left (-b \right )}{2}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 28, normalized size = 0.93 \begin {gather*} -\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 ) \end {gather*}

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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Fricas [A]
time = 1.19, size = 40, normalized size = 1.33 \begin {gather*} -\frac {{\left (b x^{2} - 1\right )} \log \left (b x^{2} - 1\right ) - 2 \, {\left (b x^{2} - 1\right )} \log \left (x\right ) + 1}{2 \, {\left (b x^{2} - 1\right )}} \end {gather*}

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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Sympy [A]
time = 0.08, size = 22, normalized size = 0.73 \begin {gather*} \log {\left (x \right )} - \frac {\log {\left (x^{2} - \frac {1}{b} \right )}}{2} - \frac {1}{2 b x^{2} - 2} \end {gather*}

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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Giac [A]
time = 1.33, size = 36, normalized size = 1.20 \begin {gather*} \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 ) \end {gather*}

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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Mupad [B]
time = 0.04, size = 26, normalized size = 0.87 \begin {gather*} \ln \left (x\right )-\frac {\ln \left (\frac {3\,b\,x^2}{2}-\frac {3}{2}\right )}{2}-\frac {1}{2\,\left (b\,x^2-1\right )} \end {gather*}

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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