3.18 \(\int \frac {\log (x)}{x^2 \sqrt {-1+x^2}} \, dx\)

Optimal. Leaf size=43 \[ \frac {\sqrt {x^2-1}}{x}+\frac {\sqrt {x^2-1} \log (x)}{x}-\tanh ^{-1}\left (\frac {x}{\sqrt {x^2-1}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2335, 277, 217, 206} \[ \frac {\sqrt {x^2-1}}{x}+\frac {\sqrt {x^2-1} \log (x)}{x}-\tanh ^{-1}\left (\frac {x}{\sqrt {x^2-1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Log[x]/(x^2*Sqrt[-1 + x^2]),x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2335

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n]))/(d*f*(m + 1)), x] - Dist[(b*n)/(d*(m + 1)), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 43, normalized size = 1.00 \[ \frac {\sqrt {x^2-1}}{x}+\frac {\sqrt {x^2-1} \log (x)}{x}-\log \left (\sqrt {x^2-1}+x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 32, normalized size = 0.74 \[ \frac {x \log \left (-x + \sqrt {x^{2} - 1}\right ) + \sqrt {x^{2} - 1} {\left (\log \relax (x) + 1\right )} + x}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.16, size = 62, normalized size = 1.44 \[ \frac {2 \, \log \relax (x)}{{\left (x - \sqrt {x^{2} - 1}\right )}^{2} + 1} + \frac {2}{{\left (x - \sqrt {x^{2} - 1}\right )}^{2} + 1} + \frac {1}{2} \, \log \left ({\left (x - \sqrt {x^{2} - 1}\right )}^{2}\right ) - \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(x)/((x - sqrt(x^2 - 1))^2 + 1) + 2/((x - sqrt(x^2 - 1))^2 + 1) + 1/2*log((x - sqrt(x^2 - 1))^2) - log(ab
s(x))

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maple [C]  time = 0.14, size = 89, normalized size = 2.07 \[ -\frac {\sqrt {-\mathrm {signum}\left (x^{2}-1\right )}\, \arcsin \relax (x )}{\sqrt {\mathrm {signum}\left (x^{2}-1\right )}}+\frac {-\frac {\sqrt {-\mathrm {signum}\left (x^{2}-1\right )}\, \sqrt {-x^{2}+1}\, \ln \relax (x )}{\sqrt {\mathrm {signum}\left (x^{2}-1\right )}}-\frac {\sqrt {-\mathrm {signum}\left (x^{2}-1\right )}\, \sqrt {-x^{2}+1}}{\sqrt {\mathrm {signum}\left (x^{2}-1\right )}}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(x)/x^2/(x^2-1)^(1/2),x)

[Out]

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

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maxima [A]  time = 0.96, size = 41, normalized size = 0.95 \[ \frac {\sqrt {x^{2} - 1} \log \relax (x)}{x} + \frac {\sqrt {x^{2} - 1}}{x} - \log \left (2 \, x + 2 \, \sqrt {x^{2} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\ln \relax (x)}{x^2\,\sqrt {x^2-1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(x)/(x^2*(x^2 - 1)^(1/2)),x)

[Out]

int(log(x)/(x^2*(x^2 - 1)^(1/2)), x)

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sympy [C]  time = 159.34, size = 37, normalized size = 0.86 \[ \left (\begin {cases} \frac {\sqrt {x^{2} - 1}}{x} & \text {for}\: x > -1 \wedge x < 1 \end {cases}\right ) \log {\relax (x )} - \begin {cases} \text {NaN} & \text {for}\: x < -1 \\\operatorname {acosh}{\relax (x )} - i \pi - \frac {\sqrt {x^{2} - 1}}{x} & \text {for}\: x < 1 \\\text {NaN} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(x)/x**2/(x**2-1)**(1/2),x)

[Out]

Piecewise((sqrt(x**2 - 1)/x, (x > -1) & (x < 1)))*log(x) - Piecewise((nan, x < -1), (acosh(x) - I*pi - sqrt(x*
*2 - 1)/x, x < 1), (nan, True))

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