\(\int \frac {x \log (x)}{\sqrt {-1+x^2}} \, dx\) [323]

Optimal result

Integrand size = 13, antiderivative size = 34 \[ \int \frac {x \log (x)}{\sqrt {-1+x^2}} \, dx=-\sqrt {-1+x^2}+\arctan \left (\sqrt {-1+x^2}\right )+\sqrt {-1+x^2} \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2376, 272, 52, 65, 209} \[ \int \frac {x \log (x)}{\sqrt {-1+x^2}} \, dx=\arctan \left (\sqrt {x^2-1}\right )-\sqrt {x^2-1}+\sqrt {x^2-1} \log (x) \]

[In]

Int[(x*Log[x])/Sqrt[-1 + x^2],x]

[Out]

-Sqrt[-1 + x^2] + ArcTan[Sqrt[-1 + x^2]] + Sqrt[-1 + x^2]*Log[x]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

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

Rule 2376

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

Rubi steps \begin{align*} \text {integral}& = \sqrt {-1+x^2} \log (x)-\int \frac {\sqrt {-1+x^2}}{x} \, dx \\ & = \sqrt {-1+x^2} \log (x)-\frac {1}{2} \text {Subst}\left (\int \frac {\sqrt {-1+x}}{x} \, dx,x,x^2\right ) \\ & = -\sqrt {-1+x^2}+\sqrt {-1+x^2} \log (x)+\frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {-1+x} x} \, dx,x,x^2\right ) \\ & = -\sqrt {-1+x^2}+\sqrt {-1+x^2} \log (x)+\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {-1+x^2}\right ) \\ & = -\sqrt {-1+x^2}+\arctan \left (\sqrt {-1+x^2}\right )+\sqrt {-1+x^2} \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {x \log (x)}{\sqrt {-1+x^2}} \, dx=-\arctan \left (\frac {1}{\sqrt {-1+x^2}}\right )+\sqrt {-1+x^2} (-1+\log (x)) \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 3.50

method	result	size
meijerg	\(-\frac {\sqrt {-\operatorname {signum}\left (x^{2}-1\right )}\, \left (2-2 \sqrt {-x^{2}+1}\right )}{4 \sqrt {\operatorname {signum}\left (x^{2}-1\right )}}+\frac {\sqrt {-\operatorname {signum}\left (x^{2}-1\right )}\, \ln \left (x \right ) \left (2-2 \sqrt {-x^{2}+1}\right )}{2 \sqrt {\operatorname {signum}\left (x^{2}-1\right )}}+\frac {\sqrt {-\operatorname {signum}\left (x^{2}-1\right )}\, \left (-16+16 \sqrt {-x^{2}+1}-32 \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{2}+1}}{2}\right )\right )}{32 \sqrt {\operatorname {signum}\left (x^{2}-1\right )}}\)	\(119\)

[In]

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

[Out]

-1/4/signum(x^2-1)^(1/2)*(-signum(x^2-1))^(1/2)*(2-2*(-x^2+1)^(1/2))+1/2/signum(x^2-1)^(1/2)*(-signum(x^2-1))^
(1/2)*ln(x)*(2-2*(-x^2+1)^(1/2))+1/32/signum(x^2-1)^(1/2)*(-signum(x^2-1))^(1/2)*(-16+16*(-x^2+1)^(1/2)-32*ln(
1/2+1/2*(-x^2+1)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {x \log (x)}{\sqrt {-1+x^2}} \, dx=\sqrt {x^{2} - 1} {\left (\log \left (x\right ) - 1\right )} + 2 \, \arctan \left (-x + \sqrt {x^{2} - 1}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {x \log (x)}{\sqrt {-1+x^2}} \, dx=\sqrt {x^{2} - 1} \log {\left (x \right )} - \begin {cases} \sqrt {x^{2} - 1} - \operatorname {acos}{\left (\frac {1}{x} \right )} & \text {for}\: x > -1 \wedge x < 1 \end {cases} \]

[In]

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

[Out]

sqrt(x**2 - 1)*log(x) - Piecewise((sqrt(x**2 - 1) - acos(1/x), (x > -1) & (x < 1)))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {x \log (x)}{\sqrt {-1+x^2}} \, dx=\sqrt {x^{2} - 1} \log \left (x\right ) - \sqrt {x^{2} - 1} - \arcsin \left (\frac {1}{{\left | x \right |}}\right ) \]

[In]

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

[Out]

sqrt(x^2 - 1)*log(x) - sqrt(x^2 - 1) - arcsin(1/abs(x))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {x \log (x)}{\sqrt {-1+x^2}} \, dx=\sqrt {x^{2} - 1} \log \left (x\right ) - \sqrt {x^{2} - 1} + \arctan \left (\sqrt {x^{2} - 1}\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x \log (x)}{\sqrt {-1+x^2}} \, dx=\int \frac {x\,\ln \left (x\right )}{\sqrt {x^2-1}} \,d x \]

[In]

int((x*log(x))/(x^2 - 1)^(1/2),x)

[Out]

int((x*log(x))/(x^2 - 1)^(1/2), x)