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

Optimal. Leaf size=34 \[ -\sqrt {x^2-1}+\sqrt {x^2-1} \log (x)+\tan ^{-1}\left (\sqrt {x^2-1}\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2338, 266, 50, 63, 203} \[ -\sqrt {x^2-1}+\sqrt {x^2-1} \log (x)+\tan ^{-1}\left (\sqrt {x^2-1}\right ) \]

Antiderivative was successfully verified.

[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 50

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 63

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 203

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

Rule 266

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 2338

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

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Mathematica [A]  time = 0.02, size = 27, normalized size = 0.79 \[ \sqrt {x^2-1} (\log (x)-1)-\tan ^{-1}\left (\frac {1}{\sqrt {x^2-1}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 27, normalized size = 0.79 \[ \sqrt {x^{2} - 1} {\left (\log \relax (x) - 1\right )} + 2 \, \arctan \left (-x + \sqrt {x^{2} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 1.01, size = 28, normalized size = 0.82 \[ \sqrt {x^{2} - 1} \log \relax (x) - \sqrt {x^{2} - 1} + \arctan \left (\sqrt {x^{2} - 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [C]  time = 0.14, size = 119, normalized size = 3.50 \[ \frac {\sqrt {-\mathrm {signum}\left (x^{2}-1\right )}\, \left (2-2 \sqrt {-x^{2}+1}\right ) \ln \relax (x )}{2 \sqrt {\mathrm {signum}\left (x^{2}-1\right )}}-\frac {\sqrt {-\mathrm {signum}\left (x^{2}-1\right )}\, \left (2-2 \sqrt {-x^{2}+1}\right )}{4 \sqrt {\mathrm {signum}\left (x^{2}-1\right )}}+\frac {\sqrt {-\mathrm {signum}\left (x^{2}-1\right )}\, \left (-32 \ln \left (\frac {1}{2}+\frac {\sqrt {-x^{2}+1}}{2}\right )-16+16 \sqrt {-x^{2}+1}\right )}{32 \sqrt {\mathrm {signum}\left (x^{2}-1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)))

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maxima [A]  time = 0.97, size = 27, normalized size = 0.79 \[ \sqrt {x^{2} - 1} \log \relax (x) - \sqrt {x^{2} - 1} - \arcsin \left (\frac {1}{{\left | x \right |}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {x\,\ln \relax (x)}{\sqrt {x^2-1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*log(x))/(x^2 - 1)^(1/2),x)

[Out]

int((x*log(x))/(x^2 - 1)^(1/2), x)

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sympy [A]  time = 2.74, size = 29, normalized size = 0.85 \[ \sqrt {x^{2} - 1} \log {\relax (x )} - \begin {cases} \sqrt {x^{2} - 1} - \operatorname {acos}{\left (\frac {1}{x} \right )} & \text {for}\: x > -1 \wedge x < 1 \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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