3.881 \(\int \frac{x \sqrt{2-x^2}}{x-\sqrt{2-x^2}} \, dx\)

Optimal. Leaf size=60 \[ -\frac{x^2}{4}+\frac{1}{4} \sqrt{2-x^2} x-\frac{1}{2} \tanh ^{-1}\left (\frac{x}{\sqrt{2-x^2}}\right )+\frac{1}{4} \log (1-x)+\frac{1}{4} \log (x+1) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.299584, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {6742, 195, 216, 697, 402, 377, 207} \[ -\frac{x^2}{4}+\frac{1}{4} \sqrt{2-x^2} x-\frac{1}{2} \tanh ^{-1}\left (\frac{x}{\sqrt{2-x^2}}\right )+\frac{1}{4} \log (1-x)+\frac{1}{4} \log (x+1) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6742

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

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

Rule 697

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(a + c*
x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0]

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

Rule 377

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

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])

Rubi steps

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

Mathematica [A]  time = 0.0850813, size = 77, normalized size = 1.28 \[ \frac{1}{4} \left (-x^2+\sqrt{2-x^2} x+\log \left (1-x^2\right )-\log \left (\sqrt{2-x^2}-x+2\right )+\log \left (\sqrt{2-x^2}+x+2\right )+\log (1-x)-\log (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.005, size = 111, normalized size = 1.9 \begin{align*}{\frac{x}{4}\sqrt{-{x}^{2}+2}}+{\frac{1}{4}\sqrt{- \left ( x-1 \right ) ^{2}-2\,x+3}}-{\frac{1}{4}{\it Artanh} \left ({\frac{4-2\,x}{2}{\frac{1}{\sqrt{- \left ( x-1 \right ) ^{2}-2\,x+3}}}} \right ) }-{\frac{1}{4}\sqrt{- \left ( 1+x \right ) ^{2}+2\,x+3}}+{\frac{1}{4}{\it Artanh} \left ({\frac{4+2\,x}{2}{\frac{1}{\sqrt{- \left ( 1+x \right ) ^{2}+2\,x+3}}}} \right ) }-{\frac{{x}^{2}}{4}}+{\frac{\ln \left ( x-1 \right ) }{4}}+{\frac{\ln \left ( 1+x \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x*(-x^2+2)^(1/2)+1/4*(-(x-1)^2-2*x+3)^(1/2)-1/4*arctanh(1/2*(4-2*x)/(-(x-1)^2-2*x+3)^(1/2))-1/4*(-(1+x)^2+
2*x+3)^(1/2)+1/4*arctanh(1/2*(4+2*x)/(-(1+x)^2+2*x+3)^(1/2))-1/4*x^2+1/4*ln(x-1)+1/4*ln(1+x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.43754, size = 174, normalized size = 2.9 \begin{align*} -\frac{1}{4} \, x^{2} + \frac{1}{4} \, \sqrt{-x^{2} + 2} x + \frac{1}{4} \, \log \left (x^{2} - 1\right ) - \frac{1}{8} \, \log \left (-\frac{\sqrt{-x^{2} + 2} x + 1}{x^{2}}\right ) + \frac{1}{8} \, \log \left (\frac{\sqrt{-x^{2} + 2} x - 1}{x^{2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^2 + 1/4*sqrt(-x^2 + 2)*x + 1/4*log(x^2 - 1) - 1/8*log(-(sqrt(-x^2 + 2)*x + 1)/x^2) + 1/8*log((sqrt(-x^2
+ 2)*x - 1)/x^2)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \sqrt{2 - x^{2}}}{x - \sqrt{2 - x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(2 - x**2)/(x - sqrt(2 - x**2)), x)

________________________________________________________________________________________

Giac [B]  time = 1.24137, size = 158, normalized size = 2.63 \begin{align*} -\frac{1}{4} \, x^{2} + \frac{1}{4} \, \sqrt{-x^{2} + 2} x + \frac{1}{4} \, \log \left ({\left | x^{2} - 1 \right |}\right ) - \frac{1}{4} \, \log \left ({\left | \frac{x}{\sqrt{2} - \sqrt{-x^{2} + 2}} - \frac{\sqrt{2} - \sqrt{-x^{2} + 2}}{x} + 2 \right |}\right ) + \frac{1}{4} \, \log \left ({\left | \frac{x}{\sqrt{2} - \sqrt{-x^{2} + 2}} - \frac{\sqrt{2} - \sqrt{-x^{2} + 2}}{x} - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^2 + 1/4*sqrt(-x^2 + 2)*x + 1/4*log(abs(x^2 - 1)) - 1/4*log(abs(x/(sqrt(2) - sqrt(-x^2 + 2)) - (sqrt(2)
- sqrt(-x^2 + 2))/x + 2)) + 1/4*log(abs(x/(sqrt(2) - sqrt(-x^2 + 2)) - (sqrt(2) - sqrt(-x^2 + 2))/x - 2))