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]

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

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Rubi [A]  time = 0.05, antiderivative size = 55, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.45, size = 35, normalized size = 0.64 \[ -{\left (\sqrt {-x^{2} + 1} + 1\right )} \log \left (\sqrt {-x^{2} + 1} + 1\right ) + \sqrt {-x^{2} + 1} \]

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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giac [A]  time = 1.03, size = 36, normalized size = 0.65 \[ -{\left (\sqrt {-x^{2} + 1} + 1\right )} \log \left (\sqrt {-x^{2} + 1} + 1\right ) + \sqrt {-x^{2} + 1} + 1 \]

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

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maple [A]  time = 0.01, size = 37, normalized size = 0.67 \[ -\left (1+\sqrt {-x^{2}+1}\right ) \ln \left (1+\sqrt {-x^{2}+1}\right )+1+\sqrt {-x^{2}+1} \]

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 = 0.41, size = 36, normalized size = 0.65 \[ -{\left (\sqrt {-x^{2} + 1} + 1\right )} \log \left (\sqrt {-x^{2} + 1} + 1\right ) + \sqrt {-x^{2} + 1} + 1 \]

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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mupad [B]  time = 0.34, size = 27, normalized size = 0.49 \[ -\left (\ln \left (\sqrt {1-x^2}+1\right )-1\right )\,\left (\sqrt {1-x^2}+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 5.99, size = 31, normalized size = 0.56 \[ \sqrt {1 - x^{2}} - \left (\sqrt {1 - x^{2}} + 1\right ) \log {\left (\sqrt {1 - x^{2}} + 1 \right )} + 1 \]

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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