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

Optimal. Leaf size=54 \[ -\frac{x}{2}-\frac{1}{2} \sqrt{2-x} \sqrt{x}-\frac{1}{2} \log (1-x)+\frac{1}{2} \tanh ^{-1}\left (\sqrt{2-x} \sqrt{x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0491659, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2105, 101, 12, 92, 206, 43} \[ -\frac{x}{2}-\frac{1}{2} \sqrt{2-x} \sqrt{x}-\frac{1}{2} \log (1-x)+\frac{1}{2} \tanh ^{-1}\left (\sqrt{2-x} \sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2105

Int[(u_)/((e_.)*Sqrt[(a_.) + (b_.)*(x_)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[e, Int[(u*Sqrt[a
+ b*x])/(a*e^2 - c*f^2 + (b*e^2 - d*f^2)*x), x], x] - Dist[f, Int[(u*Sqrt[c + d*x])/(a*e^2 - c*f^2 + (b*e^2 -
d*f^2)*x), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a*e^2 - c*f^2, 0] && NeQ[b*e^2 - d*f^2, 0]

Rule 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

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

Mathematica [A]  time = 0.139623, size = 82, normalized size = 1.52 \[ \frac{1}{2} \left (-x-\sqrt{-(x-2) x}-\log \left (1-\sqrt{x}\right )-\log \left (\sqrt{x}+1\right )+\tanh ^{-1}\left (\frac{2-\sqrt{x}}{\sqrt{2-x}}\right )-\tanh ^{-1}\left (\frac{\sqrt{x}+2}{\sqrt{2-x}}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 51, normalized size = 0.9 \begin{align*} -{\frac{1}{2}\sqrt{2-x}\sqrt{x} \left ( \sqrt{-x \left ( -2+x \right ) }-{\it Artanh} \left ({\frac{1}{\sqrt{-x \left ( -2+x \right ) }}} \right ) \right ){\frac{1}{\sqrt{-x \left ( -2+x \right ) }}}}-{\frac{x}{2}}-{\frac{\ln \left ( x-1 \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate(sqrt(x)/(sqrt(x) - sqrt(-x + 2)), x)

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Fricas [A]  time = 1.44088, size = 180, normalized size = 3.33 \begin{align*} -\frac{1}{2} \, x - \frac{1}{2} \, \sqrt{x} \sqrt{-x + 2} - \frac{1}{2} \, \log \left (x - 1\right ) + \frac{1}{2} \, \log \left (\frac{x + \sqrt{x} \sqrt{-x + 2}}{x}\right ) - \frac{1}{2} \, \log \left (-\frac{x - \sqrt{x} \sqrt{-x + 2}}{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x - 1/2*sqrt(x)*sqrt(-x + 2) - 1/2*log(x - 1) + 1/2*log((x + sqrt(x)*sqrt(-x + 2))/x) - 1/2*log(-(x - sqr
t(x)*sqrt(-x + 2))/x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError