Integrand size = 23, antiderivative size = 53 \[ \int \frac {\sqrt {x}}{\sqrt {2012-x}+\sqrt {x}} \, dx=-\frac {1}{2} \sqrt {2012-x} \sqrt {x}+\frac {x}{2}+503 \text {arctanh}\left (\frac {\sqrt {2012-x} \sqrt {x}}{1006}\right )+503 \log (1006-x) \]
Time = 0.21 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {x}}{\sqrt {2012-x}+\sqrt {x}} \, dx=\frac {1}{2} \left (x-\sqrt {-((-2012+x) x)}+2012 \log \left (2 \sqrt {503}-\sqrt {x}\right )+2012 \log \left (\sqrt {2012-x}+\sqrt {x}\right )-2012 \log \left (-1006+\sqrt {503} \sqrt {x}\right )\right ) \]
(x - Sqrt[-((-2012 + x)*x)] + 2012*Log[2*Sqrt[503] - Sqrt[x]] + 2012*Log[S qrt[2012 - x] + Sqrt[x]] - 2012*Log[-1006 + Sqrt[503]*Sqrt[x]])/2
Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {2035, 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.
\(\displaystyle \int \frac {\sqrt {x}}{\sqrt {2012-x}+\sqrt {x}} \, dx\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle 2 \int \frac {x}{\sqrt {2012-x}+\sqrt {x}}d\sqrt {x}\) |
\(\Big \downarrow \) 2532 |
\(\displaystyle 2 \left (\int \frac {\sqrt {2012-x} x}{2 (1006-x)}d\sqrt {x}-\int \frac {x^{3/2}}{2 (1006-x)}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \left (\frac {1}{2} \int \frac {\sqrt {2012-x} x}{1006-x}d\sqrt {x}-\frac {1}{2} \int \frac {x^{3/2}}{1006-x}d\sqrt {x}\right )\) |
\(\Big \downarrow \) 243 |
\(\displaystyle 2 \left (\frac {1}{2} \int \frac {\sqrt {2012-x} x}{1006-x}d\sqrt {x}-\frac {1}{4} \int \frac {x}{1006-x}dx\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle 2 \left (\frac {1}{2} \int \frac {\sqrt {2012-x} x}{1006-x}d\sqrt {x}-\frac {1}{4} \int \left (-1-\frac {1006}{x-1006}\right )dx\right )\) |
\(\Big \downarrow \) 380 |
\(\displaystyle 2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {2024072}{(1006-x) \sqrt {2012-x}}d\sqrt {x}-\frac {1}{2} \sqrt {2012-x} \sqrt {x}\right )-\frac {1}{4} \int \left (-1-\frac {1006}{x-1006}\right )dx\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \left (\frac {1}{2} \left (1012036 \int \frac {1}{(1006-x) \sqrt {2012-x}}d\sqrt {x}-\frac {1}{2} \sqrt {2012-x} \sqrt {x}\right )-\frac {1}{4} \int \left (-1-\frac {1006}{x-1006}\right )dx\right )\) |
\(\Big \downarrow \) 291 |
\(\displaystyle 2 \left (\frac {1}{2} \left (1012036 \int \frac {1}{1006-1006 x}d\frac {\sqrt {x}}{\sqrt {2012-x}}-\frac {1}{2} \sqrt {2012-x} \sqrt {x}\right )-\frac {1}{4} \int \left (-1-\frac {1006}{x-1006}\right )dx\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 \left (\frac {1}{2} \left (1006 \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {2012-x}}\right )-\frac {1}{2} \sqrt {2012-x} \sqrt {x}\right )-\frac {1}{4} \int \left (-1-\frac {1006}{x-1006}\right )dx\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (\frac {1}{2} \left (1006 \text {arctanh}\left (\frac {\sqrt {x}}{\sqrt {2012-x}}\right )-\frac {1}{2} \sqrt {2012-x} \sqrt {x}\right )+\frac {1}{4} (x+1006 \log (1006-x))\right )\) |
2*((-1/2*(Sqrt[2012 - x]*Sqrt[x]) + 1006*ArcTanh[Sqrt[x]/Sqrt[2012 - x]])/ 2 + (x + 1006*Log[1006 - x])/4)
3.1.58.3.1 Defintions of rubi rules used
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 = 53, normalized size of antiderivative = 1.00
method | result | size |
default | \(\frac {x}{2}+503 \ln \left (x -1006\right )-\frac {\sqrt {x}\, \sqrt {2012-x}\, \left (\sqrt {-x \left (-2012+x \right )}-1006 \,\operatorname {arctanh}\left (\frac {1006}{\sqrt {-x \left (-2012+x \right )}}\right )\right )}{2 \sqrt {-x \left (-2012+x \right )}}\) | \(53\) |
1/2*x+503*ln(x-1006)-1/2*x^(1/2)*(2012-x)^(1/2)*((-x*(-2012+x))^(1/2)-1006 *arctanh(1006/(-x*(-2012+x))^(1/2)))/(-x*(-2012+x))^(1/2)
Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {x}}{\sqrt {2012-x}+\sqrt {x}} \, dx=\frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x} \sqrt {-x + 2012} + 503 \, \log \left (x - 1006\right ) + 503 \, \log \left (\frac {x + \sqrt {x} \sqrt {-x + 2012}}{x}\right ) - 503 \, \log \left (-\frac {x - \sqrt {x} \sqrt {-x + 2012}}{x}\right ) \]
1/2*x - 1/2*sqrt(x)*sqrt(-x + 2012) + 503*log(x - 1006) + 503*log((x + sqr t(x)*sqrt(-x + 2012))/x) - 503*log(-(x - sqrt(x)*sqrt(-x + 2012))/x)
\[ \int \frac {\sqrt {x}}{\sqrt {2012-x}+\sqrt {x}} \, dx=\int \frac {\sqrt {x}}{\sqrt {x} + \sqrt {2012 - x}}\, dx \]
\[ \int \frac {\sqrt {x}}{\sqrt {2012-x}+\sqrt {x}} \, dx=\int { \frac {\sqrt {x}}{\sqrt {x} + \sqrt {-x + 2012}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (39) = 78\).
Time = 0.31 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.21 \[ \int \frac {\sqrt {x}}{\sqrt {2012-x}+\sqrt {x}} \, dx=\frac {1}{2} \, x - \frac {1}{2} \, \sqrt {x} \sqrt {-x + 2012} + 503 \, \log \left ({\left | x - 1006 \right |}\right ) + 503 \, \log \left ({\left | -\frac {2 \, \sqrt {503} - \sqrt {-x + 2012}}{\sqrt {x}} + \frac {\sqrt {x}}{2 \, \sqrt {503} - \sqrt {-x + 2012}} + 2 \right |}\right ) - 503 \, \log \left ({\left | -\frac {2 \, \sqrt {503} - \sqrt {-x + 2012}}{\sqrt {x}} + \frac {\sqrt {x}}{2 \, \sqrt {503} - \sqrt {-x + 2012}} - 2 \right |}\right ) \]
1/2*x - 1/2*sqrt(x)*sqrt(-x + 2012) + 503*log(abs(x - 1006)) + 503*log(abs (-(2*sqrt(503) - sqrt(-x + 2012))/sqrt(x) + sqrt(x)/(2*sqrt(503) - sqrt(-x + 2012)) + 2)) - 503*log(abs(-(2*sqrt(503) - sqrt(-x + 2012))/sqrt(x) + s qrt(x)/(2*sqrt(503) - sqrt(-x + 2012)) - 2))
Time = 16.55 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {x}}{\sqrt {2012-x}+\sqrt {x}} \, dx=\frac {x}{2}+1006\,\mathrm {atanh}\left (\frac {2\,\sqrt {503}\,\sqrt {x}-\sqrt {x}\,\sqrt {2012-x}}{x+2\,\sqrt {503}\,\sqrt {2012-x}-2012}\right )+503\,\ln \left (x-1006\right )-\frac {\sqrt {x}\,\sqrt {2012-x}}{2} \]