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]

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

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Rubi [A]  time = 0.044734, antiderivative size = 43, normalized size of antiderivative = 1., 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.0225532, size = 43, normalized size = 1. \[ \frac{\sqrt{x^2-1}}{x}+\frac{\sqrt{x^2-1} \log (x)}{x}-\log \left (\sqrt{x^2-1}+x\right ) \]

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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Maple [C]  time = 0.053, size = 89, normalized size = 2.1 \begin{align*} -{\arcsin \left ( x \right ) \sqrt{-{\it signum} \left ({x}^{2}-1 \right ) }{\frac{1}{\sqrt{{\it signum} \left ({x}^{2}-1 \right ) }}}}+{\frac{1}{x} \left ( -{\sqrt{-{\it signum} \left ({x}^{2}-1 \right ) }\sqrt{-{x}^{2}+1}{\frac{1}{\sqrt{{\it signum} \left ({x}^{2}-1 \right ) }}}}-{\ln \left ( x \right ) \sqrt{-{\it signum} \left ({x}^{2}-1 \right ) }\sqrt{-{x}^{2}+1}{\frac{1}{\sqrt{{\it signum} \left ({x}^{2}-1 \right ) }}}} \right ) } \end{align*}

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 = 1.43524, size = 55, normalized size = 1.28 \begin{align*} \frac{\sqrt{x^{2} - 1} \log \left (x\right )}{x} + \frac{\sqrt{x^{2} - 1}}{x} - \log \left (2 \, x + 2 \, \sqrt{x^{2} - 1}\right ) \end{align*}

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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Fricas [A]  time = 2.14656, size = 86, normalized size = 2. \begin{align*} \frac{x \log \left (-x + \sqrt{x^{2} - 1}\right ) + \sqrt{x^{2} - 1}{\left (\log \left (x\right ) + 1\right )} + x}{x} \end{align*}

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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Sympy [C]  time = 177.985, size = 37, normalized size = 0.86 \begin{align*} \left (\begin{cases} \frac{\sqrt{x^{2} - 1}}{x} & \text{for}\: x > -1 \wedge x < 1 \end{cases}\right ) \log{\left (x \right )} - \begin{cases} \text{NaN} & \text{for}\: x < -1 \\\operatorname{acosh}{\left (x \right )} - i \pi - \frac{\sqrt{x^{2} - 1}}{x} & \text{for}\: x < 1 \\\text{NaN} & \text{otherwise} \end{cases} \end{align*}

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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Giac [A]  time = 1.11701, size = 84, normalized size = 1.95 \begin{align*} \frac{2 \, \log \left (x\right )}{{\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 ) \end{align*}

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))