Integrand size = 11, antiderivative size = 50 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=-\frac {3}{2} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}+\sqrt {1+\frac {1}{\sqrt {x}}} x+\frac {3}{2} \text {arctanh}\left (\sqrt {1+\frac {1}{\sqrt {x}}}\right ) \] Output:
-3/2*(1+1/x^(1/2))^(1/2)*x^(1/2)+(1+1/x^(1/2))^(1/2)*x+3/2*arctanh((1+1/x^ (1/2))^(1/2))
Time = 0.06 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=\frac {1}{2} \left (\sqrt {1+\frac {1}{\sqrt {x}}} \left (-3+2 \sqrt {x}\right ) \sqrt {x}+3 \text {arctanh}\left (\sqrt {1+\frac {1}{\sqrt {x}}}\right )\right ) \] Input:
Integrate[1/Sqrt[1 + 1/Sqrt[x]],x]
Output:
(Sqrt[1 + 1/Sqrt[x]]*(-3 + 2*Sqrt[x])*Sqrt[x] + 3*ArcTanh[Sqrt[1 + 1/Sqrt[ x]]])/2
Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {774, 798, 52, 52, 73, 220}
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.
\(\displaystyle \int \frac {1}{\sqrt {\frac {1}{\sqrt {x}}+1}} \, dx\) |
\(\Big \downarrow \) 774 |
\(\displaystyle 2 \int \frac {\sqrt {x}}{\sqrt {1+\frac {1}{\sqrt {x}}}}d\sqrt {x}\) |
\(\Big \downarrow \) 798 |
\(\displaystyle -2 \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}} x^{3/2}}d\frac {1}{\sqrt {x}}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -2 \left (-\frac {3}{4} \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}} x}d\frac {1}{\sqrt {x}}-\frac {\sqrt {\frac {1}{\sqrt {x}}+1}}{2 x}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -2 \left (-\frac {3}{4} \left (-\frac {1}{2} \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}}d\frac {1}{\sqrt {x}}-\frac {\sqrt {\frac {1}{\sqrt {x}}+1}}{\sqrt {x}}\right )-\frac {\sqrt {\frac {1}{\sqrt {x}}+1}}{2 x}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -2 \left (-\frac {3}{4} \left (-\int \frac {1}{x-1}d\sqrt {1+\frac {1}{\sqrt {x}}}-\frac {\sqrt {\frac {1}{\sqrt {x}}+1}}{\sqrt {x}}\right )-\frac {\sqrt {\frac {1}{\sqrt {x}}+1}}{2 x}\right )\) |
\(\Big \downarrow \) 220 |
\(\displaystyle -2 \left (-\frac {3}{4} \left (\text {arctanh}\left (\sqrt {\frac {1}{\sqrt {x}}+1}\right )-\frac {\sqrt {\frac {1}{\sqrt {x}}+1}}{\sqrt {x}}\right )-\frac {\sqrt {\frac {1}{\sqrt {x}}+1}}{2 x}\right )\) |
Input:
Int[1/Sqrt[1 + 1/Sqrt[x]],x]
Output:
-2*(-1/2*Sqrt[1 + 1/Sqrt[x]]/x - (3*(-(Sqrt[1 + 1/Sqrt[x]]/Sqrt[x]) + ArcT anh[Sqrt[1 + 1/Sqrt[x]]]))/4)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Simp[k Subst[Int[x^(k - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; Fre eQ[{a, b, p}, x] && FractionQ[n]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Time = 0.45 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76
method | result | size |
meijerg | \(\frac {-\frac {\sqrt {\pi }\, x^{\frac {1}{4}} \left (-10 \sqrt {x}+15\right ) \sqrt {1+\sqrt {x}}}{10}+\frac {3 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{\frac {1}{4}}\right )}{2}}{\sqrt {\pi }}\) | \(38\) |
default | \(\frac {\sqrt {\frac {1+\sqrt {x}}{\sqrt {x}}}\, \sqrt {x}\, \left (4 \sqrt {x +\sqrt {x}}\, \sqrt {x}-6 \sqrt {x +\sqrt {x}}+3 \ln \left (\frac {1}{2}+\sqrt {x}+\sqrt {x +\sqrt {x}}\right )\right )}{4 \sqrt {\left (1+\sqrt {x}\right ) \sqrt {x}}}\) | \(65\) |
derivativedivides | \(\frac {\left (1+\sqrt {x}\right ) \left (4 \sqrt {x +\sqrt {x}}\, \sqrt {x}-6 \sqrt {x +\sqrt {x}}+3 \ln \left (\frac {1}{2}+\sqrt {x}+\sqrt {x +\sqrt {x}}\right )\right )}{4 \sqrt {\frac {1+\sqrt {x}}{\sqrt {x}}}\, \sqrt {\left (1+\sqrt {x}\right ) \sqrt {x}}}\) | \(67\) |
Input:
int(1/(1+1/x^(1/2))^(1/2),x,method=_RETURNVERBOSE)
Output:
2/Pi^(1/2)*(-1/20*Pi^(1/2)*x^(1/4)*(-10*x^(1/2)+15)*(1+x^(1/2))^(1/2)+3/4* Pi^(1/2)*arcsinh(x^(1/4)))
Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=\frac {1}{2} \, {\left (2 \, x - 3 \, \sqrt {x}\right )} \sqrt {\frac {x + \sqrt {x}}{x}} + \frac {3}{4} \, \log \left (\sqrt {\frac {x + \sqrt {x}}{x}} + 1\right ) - \frac {3}{4} \, \log \left (\sqrt {\frac {x + \sqrt {x}}{x}} - 1\right ) \] Input:
integrate(1/(1+1/x^(1/2))^(1/2),x, algorithm="fricas")
Output:
1/2*(2*x - 3*sqrt(x))*sqrt((x + sqrt(x))/x) + 3/4*log(sqrt((x + sqrt(x))/x ) + 1) - 3/4*log(sqrt((x + sqrt(x))/x) - 1)
Time = 2.68 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.20 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=\frac {x^{\frac {5}{4}}}{\sqrt {\sqrt {x} + 1}} - \frac {x^{\frac {3}{4}}}{2 \sqrt {\sqrt {x} + 1}} - \frac {3 \sqrt [4]{x}}{2 \sqrt {\sqrt {x} + 1}} + \frac {3 \operatorname {asinh}{\left (\sqrt [4]{x} \right )}}{2} \] Input:
integrate(1/(1+1/x**(1/2))**(1/2),x)
Output:
x**(5/4)/sqrt(sqrt(x) + 1) - x**(3/4)/(2*sqrt(sqrt(x) + 1)) - 3*x**(1/4)/( 2*sqrt(sqrt(x) + 1)) + 3*asinh(x**(1/4))/2
Time = 0.04 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=-\frac {3 \, {\left (\frac {1}{\sqrt {x}} + 1\right )}^{\frac {3}{2}} - 5 \, \sqrt {\frac {1}{\sqrt {x}} + 1}}{2 \, {\left ({\left (\frac {1}{\sqrt {x}} + 1\right )}^{2} - \frac {2}{\sqrt {x}} - 1\right )}} + \frac {3}{4} \, \log \left (\sqrt {\frac {1}{\sqrt {x}} + 1} + 1\right ) - \frac {3}{4} \, \log \left (\sqrt {\frac {1}{\sqrt {x}} + 1} - 1\right ) \] Input:
integrate(1/(1+1/x^(1/2))^(1/2),x, algorithm="maxima")
Output:
-1/2*(3*(1/sqrt(x) + 1)^(3/2) - 5*sqrt(1/sqrt(x) + 1))/((1/sqrt(x) + 1)^2 - 2/sqrt(x) - 1) + 3/4*log(sqrt(1/sqrt(x) + 1) + 1) - 3/4*log(sqrt(1/sqrt( x) + 1) - 1)
Time = 0.14 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=\frac {2 \, \sqrt {x + \sqrt {x}} {\left (2 \, \sqrt {x} - 3\right )} - 3 \, \log \left (-2 \, \sqrt {x + \sqrt {x}} + 2 \, \sqrt {x} + 1\right )}{4 \, \mathrm {sgn}\left (x\right )} \] Input:
integrate(1/(1+1/x^(1/2))^(1/2),x, algorithm="giac")
Output:
1/4*(2*sqrt(x + sqrt(x))*(2*sqrt(x) - 3) - 3*log(-2*sqrt(x + sqrt(x)) + 2* sqrt(x) + 1))/sgn(x)
Time = 0.41 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=\frac {4\,x\,\sqrt {\sqrt {x}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {5}{2};\ \frac {7}{2};\ -\sqrt {x}\right )}{5\,\sqrt {\frac {1}{\sqrt {x}}+1}} \] Input:
int(1/(1/x^(1/2) + 1)^(1/2),x)
Output:
(4*x*(x^(1/2) + 1)^(1/2)*hypergeom([1/2, 5/2], 7/2, -x^(1/2)))/(5*(1/x^(1/ 2) + 1)^(1/2))
Time = 0.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.64 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=x^{\frac {3}{4}} \sqrt {\sqrt {x}+1}-\frac {3 x^{\frac {1}{4}} \sqrt {\sqrt {x}+1}}{2}+\frac {3 \,\mathrm {log}\left (\sqrt {\sqrt {x}+1}+x^{\frac {1}{4}}\right )}{2} \] Input:
int(1/(1+1/x^(1/2))^(1/2),x)
Output:
(2*x**(3/4)*sqrt(sqrt(x) + 1) - 3*x**(1/4)*sqrt(sqrt(x) + 1) + 3*log(sqrt( sqrt(x) + 1) + x**(1/4)))/2