Integrand size = 25, antiderivative size = 50 \[ \int \frac {\sqrt {x}}{\sqrt {2-x}-\sqrt {x}} \, dx=-\frac {1}{2} \sqrt {2-x} \sqrt {x}-\frac {x}{2}+\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {2-x}}\right )-\frac {1}{2} \log (1-x) \] Output:
-1/2*(2-x)^(1/2)*x^(1/2)-1/2*x+arctanh(x^(1/2)/(2-x)^(1/2))-1/2*ln(1-x)
Time = 0.21 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {x}}{\sqrt {2-x}-\sqrt {x}} \, dx=-\frac {x}{2}-\frac {1}{2} \sqrt {-((-2+x) x)}+\log \left (-2+\sqrt {4-2 x}\right )-\log \left (-2+\sqrt {4-2 x}-\sqrt {2} \sqrt {x}+x+\sqrt {-((-2+x) x)}\right ) \] Input:
Integrate[Sqrt[x]/(Sqrt[2 - x] - Sqrt[x]),x]
Output:
-1/2*x - Sqrt[-((-2 + x)*x)]/2 + Log[-2 + Sqrt[4 - 2*x]] - Log[-2 + Sqrt[4 - 2*x] - Sqrt[2]*Sqrt[x] + x + Sqrt[-((-2 + x)*x)]]
Time = 0.47 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.16, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {2035, 25, 2532, 27, 243, 49, 380, 27, 291, 219, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {x}}{\sqrt {2-x}-\sqrt {x}} \, dx\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle 2 \int -\frac {x}{\sqrt {x}-\sqrt {2-x}}d\sqrt {x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 \int \frac {x}{\sqrt {x}-\sqrt {2-x}}d\sqrt {x}\) |
\(\Big \downarrow \) 2532 |
\(\displaystyle 2 \left (\int \frac {x^{3/2}}{2 (1-x)}d\sqrt {x}+\int \frac {\sqrt {2-x} x}{2 (1-x)}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \left (\frac {1}{2} \int \frac {x^{3/2}}{1-x}d\sqrt {x}+\frac {1}{2} \int \frac {\sqrt {2-x} x}{1-x}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle 2 \left (\frac {1}{4} \int \frac {x}{1-x}dx+\frac {1}{2} \int \frac {\sqrt {2-x} x}{1-x}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle 2 \left (\frac {1}{4} \int \left (\frac {1}{1-x}-1\right )dx+\frac {1}{2} \int \frac {\sqrt {2-x} x}{1-x}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 380 |
\(\displaystyle 2 \left (\frac {1}{4} \int \left (\frac {1}{1-x}-1\right )dx+\frac {1}{2} \left (\frac {1}{2} \int \frac {2}{(1-x) \sqrt {2-x}}d\sqrt {x}-\frac {1}{2} \sqrt {2-x} \sqrt {x}\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \left (\frac {1}{4} \int \left (\frac {1}{1-x}-1\right )dx+\frac {1}{2} \left (\int \frac {1}{(1-x) \sqrt {2-x}}d\sqrt {x}-\frac {1}{2} \sqrt {2-x} \sqrt {x}\right )\right )\) |
\(\Big \downarrow \) 291 |
\(\displaystyle 2 \left (\frac {1}{4} \int \left (\frac {1}{1-x}-1\right )dx+\frac {1}{2} \left (\int \frac {1}{1-x}d\frac {\sqrt {x}}{\sqrt {2-x}}-\frac {1}{2} \sqrt {2-x} \sqrt {x}\right )\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (\frac {1}{4} \int \left (\frac {1}{1-x}-1\right )dx+\frac {1}{2} \left (\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {2-x}}\right )-\frac {1}{2} \sqrt {2-x} \sqrt {x}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{2} \left (\text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {2-x}}\right )-\frac {1}{2} \sqrt {2-x} \sqrt {x}\right )+\frac {1}{4} (-x-\log (1-x))\right )\) |
Input:
Int[Sqrt[x]/(Sqrt[2 - x] - Sqrt[x]),x]
Output:
2*((-1/2*(Sqrt[2 - x]*Sqrt[x]) + ArcTanh[Sqrt[x]/Sqrt[2 - x]])/2 + (-x - L og[1 - x])/4)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b* (m + 2*(p + q) + 1))), x] - Simp[e^2/(b*(m + 2*(p + q) + 1)) Int[(e*x)^(m - 2)*(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[a*c*(m - 1) + (a*d*(m - 1) - 2 *q*(b*c - a*d))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 0] && GtQ[m, 1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(x_)^(m_.)/((d_.)*(x_)^(n_.) + (c_.)*Sqrt[(a_.) + (b_.)*(x_)^(p_.)]), x _Symbol] :> Simp[-d Int[x^(m + n)/(a*c^2 + (b*c^2 - d^2)*x^(2*n)), x], x] + Simp[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 ]
Time = 0.02 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.02
method | result | size |
default | \(-\frac {\sqrt {-x +2}\, \sqrt {x}\, \left (\sqrt {-x \left (x -2\right )}-\operatorname {arctanh}\left (\frac {1}{\sqrt {-x \left (x -2\right )}}\right )\right )}{2 \sqrt {-x \left (x -2\right )}}-\frac {x}{2}-\frac {\ln \left (-1+x \right )}{2}\) | \(51\) |
Input:
int(x^(1/2)/((-x+2)^(1/2)-x^(1/2)),x,method=_RETURNVERBOSE)
Output:
-1/2*(-x+2)^(1/2)*x^(1/2)/(-x*(x-2))^(1/2)*((-x*(x-2))^(1/2)-arctanh(1/(-x *(x-2))^(1/2)))-1/2*x-1/2*ln(-1+x)
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.28 \[ \int \frac {\sqrt {x}}{\sqrt {2-x}-\sqrt {x}} \, dx=-\frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x} \sqrt {-x + 2} - \frac {1}{2} \, \log \left (x - 1\right ) + \frac {1}{2} \, \log \left (\frac {x + \sqrt {x} \sqrt {-x + 2}}{x}\right ) - \frac {1}{2} \, \log \left (-\frac {x - \sqrt {x} \sqrt {-x + 2}}{x}\right ) \] Input:
integrate(x^(1/2)/((2-x)^(1/2)-x^(1/2)),x, algorithm="fricas")
Output:
-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 - sqrt(x)*sqrt(-x + 2))/x)
\[ \int \frac {\sqrt {x}}{\sqrt {2-x}-\sqrt {x}} \, dx=\int \frac {\sqrt {x}}{- \sqrt {x} + \sqrt {2 - x}}\, dx \] Input:
integrate(x**(1/2)/((2-x)**(1/2)-x**(1/2)),x)
Output:
Integral(sqrt(x)/(-sqrt(x) + sqrt(2 - x)), x)
\[ \int \frac {\sqrt {x}}{\sqrt {2-x}-\sqrt {x}} \, dx=\int { -\frac {\sqrt {x}}{\sqrt {x} - \sqrt {-x + 2}} \,d x } \] Input:
integrate(x^(1/2)/((2-x)^(1/2)-x^(1/2)),x, algorithm="maxima")
Output:
-integrate(sqrt(x)/(sqrt(x) - sqrt(-x + 2)), x)
Leaf count of result is larger than twice the leaf count of optimal. 109 vs. \(2 (36) = 72\).
Time = 0.17 (sec) , antiderivative size = 109, normalized size of antiderivative = 2.18 \[ \int \frac {\sqrt {x}}{\sqrt {2-x}-\sqrt {x}} \, dx=-\frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x} \sqrt {-x + 2} - \frac {1}{2} \, \log \left ({\left | x - 1 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 2}}{\sqrt {x}} + \frac {\sqrt {x}}{\sqrt {2} - \sqrt {-x + 2}} + 2 \right |}\right ) - \frac {1}{2} \, \log \left ({\left | -\frac {\sqrt {2} - \sqrt {-x + 2}}{\sqrt {x}} + \frac {\sqrt {x}}{\sqrt {2} - \sqrt {-x + 2}} - 2 \right |}\right ) \] Input:
integrate(x^(1/2)/((2-x)^(1/2)-x^(1/2)),x, algorithm="giac")
Output:
-1/2*x - 1/2*sqrt(x)*sqrt(-x + 2) - 1/2*log(abs(x - 1)) + 1/2*log(abs(-(sq rt(2) - sqrt(-x + 2))/sqrt(x) + sqrt(x)/(sqrt(2) - sqrt(-x + 2)) + 2)) - 1 /2*log(abs(-(sqrt(2) - sqrt(-x + 2))/sqrt(x) + sqrt(x)/(sqrt(2) - sqrt(-x + 2)) - 2))
Time = 22.68 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {x}}{\sqrt {2-x}-\sqrt {x}} \, dx=\mathrm {atanh}\left (\frac {\sqrt {x}\,\left (\sqrt {2}-\sqrt {2-x}\right )}{x+\sqrt {2}\,\sqrt {2-x}-2}\right )-\frac {\ln \left (x-1\right )}{2}-\frac {x}{2}-\frac {\sqrt {x}\,\sqrt {2-x}}{2} \] Input:
int(x^(1/2)/((2 - x)^(1/2) - x^(1/2)),x)
Output:
atanh((x^(1/2)*(2^(1/2) - (2 - x)^(1/2)))/(x + 2^(1/2)*(2 - x)^(1/2) - 2)) - log(x - 1)/2 - x/2 - (x^(1/2)*(2 - x)^(1/2))/2
Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {x}}{\sqrt {2-x}-\sqrt {x}} \, dx=-\frac {\sqrt {x}\, \sqrt {-x +2}}{2}+\mathrm {log}\left (-\sqrt {-x +2}\, i +\sqrt {-x +2}-\sqrt {x}\, i +\sqrt {x}\right )-\mathrm {log}\left (1-x \right )-\frac {x}{2}+1 \] Input:
int(x^(1/2)/((2-x)^(1/2)-x^(1/2)),x)
Output:
( - sqrt(x)*sqrt( - x + 2) + 2*log( - sqrt( - x + 2)*i + sqrt( - x + 2) - sqrt(x)*i + sqrt(x)) - 2*log( - x + 1) - x + 2)/2