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

Optimal. Leaf size=55 \[ \sqrt{1-x^2}-\sqrt{1-x^2} \log \left (\sqrt{1-x^2}+1\right )-\log \left (\sqrt{1-x^2}+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.0533209, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {261, 2554, 1591, 190, 43} \[ \sqrt{1-x^2}-\sqrt{1-x^2} \log \left (\sqrt{1-x^2}+1\right )-\log \left (\sqrt{1-x^2}+1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[1 - x^2] - Log[1 + Sqrt[1 - x^2]] - Sqrt[1 - x^2]*Log[1 + Sqrt[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 1591

Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[Pq, x], r = Expon[Qr, x]}, Dist[Coef
f[Qr, x, r]/(q*Coeff[Pq, x, q]), Subst[Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x,
r]*D[Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] && PolyQ[Qr, x]

Rule 190

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0207859, size = 41, normalized size = 0.75 \[ \sqrt{1-x^2}-\left (\sqrt{1-x^2}+1\right ) \log \left (\sqrt{1-x^2}+1\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 37, normalized size = 0.7 \begin{align*} -\ln \left ( 1+\sqrt{-{x}^{2}+1} \right ) \left ( 1+\sqrt{-{x}^{2}+1} \right ) +1+\sqrt{-{x}^{2}+1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.94243, size = 49, normalized size = 0.89 \begin{align*} -{\left (\sqrt{-x^{2} + 1} + 1\right )} \log \left (\sqrt{-x^{2} + 1} + 1\right ) + \sqrt{-x^{2} + 1} + 1 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.07777, size = 86, normalized size = 1.56 \begin{align*} -{\left (\sqrt{-x^{2} + 1} + 1\right )} \log \left (\sqrt{-x^{2} + 1} + 1\right ) + \sqrt{-x^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 7.99007, size = 31, normalized size = 0.56 \begin{align*} \sqrt{1 - x^{2}} - \left (\sqrt{1 - x^{2}} + 1\right ) \log{\left (\sqrt{1 - x^{2}} + 1 \right )} + 1 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.08964, size = 49, normalized size = 0.89 \begin{align*} -{\left (\sqrt{-x^{2} + 1} + 1\right )} \log \left (\sqrt{-x^{2} + 1} + 1\right ) + \sqrt{-x^{2} + 1} + 1 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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