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

Optimal. Leaf size=78 \[ \sqrt{1-x^2}-\sqrt{1-x^2} \log \left (\sqrt{1-x^2}+x\right )-\frac{\tanh ^{-1}\left (\sqrt{2} \sqrt{1-x^2}\right )}{\sqrt{2}}+\frac{\tanh ^{-1}\left (\sqrt{2} x\right )}{\sqrt{2}} \]

[Out]

Sqrt[1 - x^2] + ArcTanh[Sqrt[2]*x]/Sqrt[2] - ArcTanh[Sqrt[2]*Sqrt[1 - x^2]]/Sqrt[2] - Sqrt[1 - x^2]*Log[x + Sq
rt[1 - x^2]]

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Rubi [A]  time = 0.2711, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 11, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.407, Rules used = {261, 2554, 6742, 2107, 321, 206, 444, 50, 63, 207, 388} \[ \sqrt{1-x^2}-\sqrt{1-x^2} \log \left (\sqrt{1-x^2}+x\right )-\frac{\tanh ^{-1}\left (\sqrt{2} \sqrt{1-x^2}\right )}{\sqrt{2}}+\frac{\tanh ^{-1}\left (\sqrt{2} x\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - x^2] + ArcTanh[Sqrt[2]*x]/Sqrt[2] - ArcTanh[Sqrt[2]*Sqrt[1 - x^2]]/Sqrt[2] - Sqrt[1 - x^2]*Log[x + Sq
rt[1 - x^2]]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 2107

Int[(x_)^(m_.)/((d_.)*(x_)^(n_.) + (c_.)*Sqrt[(a_.) + (b_.)*(x_)^(p_.)]), x_Symbol] :> -Dist[d, Int[x^(m + n)/
(a*c^2 + (b*c^2 - d^2)*x^(2*n)), x], x] + Dist[c, Int[(x^m*Sqrt[a + b*x^(2*n)])/(a*c^2 + (b*c^2 - d^2)*x^(2*n)
), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[p, 2*n] && NeQ[b*c^2 - d^2, 0]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, 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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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 207

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

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rubi steps

\begin{align*} \int \frac{x \log \left (x+\sqrt{1-x^2}\right )}{\sqrt{1-x^2}} \, dx &=-\sqrt{1-x^2} \log \left (x+\sqrt{1-x^2}\right )-\int \frac{x-\sqrt{1-x^2}}{x+\sqrt{1-x^2}} \, dx\\ &=-\sqrt{1-x^2} \log \left (x+\sqrt{1-x^2}\right )-\int \left (\frac{x}{x+\sqrt{1-x^2}}-\frac{\sqrt{1-x^2}}{x+\sqrt{1-x^2}}\right ) \, dx\\ &=-\sqrt{1-x^2} \log \left (x+\sqrt{1-x^2}\right )-\int \frac{x}{x+\sqrt{1-x^2}} \, dx+\int \frac{\sqrt{1-x^2}}{x+\sqrt{1-x^2}} \, dx\\ &=-\sqrt{1-x^2} \log \left (x+\sqrt{1-x^2}\right )+\int \frac{x^2}{1-2 x^2} \, dx-\int \frac{x \sqrt{1-x^2}}{1-2 x^2} \, dx+\int \left (\frac{x \sqrt{1-x^2}}{-1+2 x^2}-\frac{1-x^2}{-1+2 x^2}\right ) \, dx\\ &=-\frac{x}{2}-\sqrt{1-x^2} \log \left (x+\sqrt{1-x^2}\right )+\frac{1}{2} \int \frac{1}{1-2 x^2} \, dx-\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{1-x}}{1-2 x} \, dx,x,x^2\right )+\int \frac{x \sqrt{1-x^2}}{-1+2 x^2} \, dx-\int \frac{1-x^2}{-1+2 x^2} \, dx\\ &=\frac{\sqrt{1-x^2}}{2}+\frac{\tanh ^{-1}\left (\sqrt{2} x\right )}{2 \sqrt{2}}-\sqrt{1-x^2} \log \left (x+\sqrt{1-x^2}\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{(1-2 x) \sqrt{1-x}} \, dx,x,x^2\right )-\frac{1}{2} \int \frac{1}{-1+2 x^2} \, dx+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{1-x}}{-1+2 x} \, dx,x,x^2\right )\\ &=\sqrt{1-x^2}+\frac{\tanh ^{-1}\left (\sqrt{2} x\right )}{\sqrt{2}}-\sqrt{1-x^2} \log \left (x+\sqrt{1-x^2}\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x} (-1+2 x)} \, dx,x,x^2\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-1+2 x^2} \, dx,x,\sqrt{1-x^2}\right )\\ &=\sqrt{1-x^2}+\frac{\tanh ^{-1}\left (\sqrt{2} x\right )}{\sqrt{2}}-\frac{\tanh ^{-1}\left (\sqrt{2} \sqrt{1-x^2}\right )}{2 \sqrt{2}}-\sqrt{1-x^2} \log \left (x+\sqrt{1-x^2}\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-2 x^2} \, dx,x,\sqrt{1-x^2}\right )\\ &=\sqrt{1-x^2}+\frac{\tanh ^{-1}\left (\sqrt{2} x\right )}{\sqrt{2}}-\frac{\tanh ^{-1}\left (\sqrt{2} \sqrt{1-x^2}\right )}{\sqrt{2}}-\sqrt{1-x^2} \log \left (x+\sqrt{1-x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.0601325, size = 119, normalized size = 1.53 \[ \frac{1}{4} \left (4 \sqrt{1-x^2}-\sqrt{2} \log \left (\sqrt{2-2 x^2}-\sqrt{2} x+2\right )-\sqrt{2} \log \left (\sqrt{2-2 x^2}+\sqrt{2} x+2\right )-4 \sqrt{1-x^2} \log \left (\sqrt{1-x^2}+x\right )+2 \sqrt{2} \log \left (2 x+\sqrt{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[1 - x^2] + 2*Sqrt[2]*Log[Sqrt[2] + 2*x] - Sqrt[2]*Log[2 - Sqrt[2]*x + Sqrt[2 - 2*x^2]] - Sqrt[2]*Log[2
+ Sqrt[2]*x + Sqrt[2 - 2*x^2]] - 4*Sqrt[1 - x^2]*Log[x + Sqrt[1 - x^2]])/4

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Maple [F]  time = 0.019, size = 0, normalized size = 0. \begin{align*} \int{x\ln \left ( x+\sqrt{-{x}^{2}+1} \right ){\frac{1}{\sqrt{-{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (x^{2} - 1\right )} \log \left (x + \sqrt{x + 1} \sqrt{-x + 1}\right )}{\sqrt{x + 1} \sqrt{-x + 1}} - \int \frac{{\left (x^{2} - 1\right )} e^{\left (-\frac{1}{2} \, \log \left (x + 1\right ) - \frac{1}{2} \, \log \left (-x + 1\right )\right )}}{x}\,{d x} - \int \frac{1}{x^{2} + \sqrt{x + 1} x \sqrt{-x + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.1573, size = 293, normalized size = 3.76 \begin{align*} -\sqrt{-x^{2} + 1} \log \left (x + \sqrt{-x^{2} + 1}\right ) + \frac{1}{4} \, \sqrt{2} \log \left (\frac{6 \, x^{2} - 2 \, \sqrt{2}{\left (2 \, x^{2} - 3\right )} + 2 \, \sqrt{-x^{2} + 1}{\left (3 \, \sqrt{2} - 4\right )} - 9}{2 \, x^{2} - 1}\right ) + \frac{1}{4} \, \sqrt{2} \log \left (\frac{2 \, x^{2} + 2 \, \sqrt{2} x + 1}{2 \, x^{2} - 1}\right ) + \sqrt{-x^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-x^2 + 1)*log(x + sqrt(-x^2 + 1)) + 1/4*sqrt(2)*log((6*x^2 - 2*sqrt(2)*(2*x^2 - 3) + 2*sqrt(-x^2 + 1)*(3
*sqrt(2) - 4) - 9)/(2*x^2 - 1)) + 1/4*sqrt(2)*log((2*x^2 + 2*sqrt(2)*x + 1)/(2*x^2 - 1)) + sqrt(-x^2 + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \log{\left (x + \sqrt{1 - x^{2}} \right )}}{\sqrt{- \left (x - 1\right ) \left (x + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*log(x + sqrt(1 - x**2))/sqrt(-(x - 1)*(x + 1)), x)

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Giac [A]  time = 1.14819, size = 165, normalized size = 2.12 \begin{align*} -\sqrt{-x^{2} + 1} \log \left (x + \sqrt{-x^{2} + 1}\right ) - \frac{1}{4} \, \sqrt{2} \log \left (\frac{{\left | -4 \, \sqrt{2} + \frac{2 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 6 \right |}}{{\left | 4 \, \sqrt{2} + \frac{2 \,{\left (\sqrt{-x^{2} + 1} - 1\right )}^{2}}{x^{2}} - 6 \right |}}\right ) + \frac{1}{4} \, \sqrt{2} \log \left ({\left | x + \frac{1}{2} \, \sqrt{2} \right |}\right ) - \frac{1}{4} \, \sqrt{2} \log \left ({\left | x - \frac{1}{2} \, \sqrt{2} \right |}\right ) + \sqrt{-x^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-x^2 + 1)*log(x + sqrt(-x^2 + 1)) - 1/4*sqrt(2)*log(abs(-4*sqrt(2) + 2*(sqrt(-x^2 + 1) - 1)^2/x^2 - 6)/a
bs(4*sqrt(2) + 2*(sqrt(-x^2 + 1) - 1)^2/x^2 - 6)) + 1/4*sqrt(2)*log(abs(x + 1/2*sqrt(2))) - 1/4*sqrt(2)*log(ab
s(x - 1/2*sqrt(2))) + sqrt(-x^2 + 1)