\(\int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx\) [2392]

Optimal result

Integrand size = 11, antiderivative size = 50 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=-\frac {3}{2} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}+\sqrt {1+\frac {1}{\sqrt {x}}} x+\frac {3}{2} \text {arctanh}\left (\sqrt {1+\frac {1}{\sqrt {x}}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {196, 44, 65, 213} \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=\frac {3}{2} \text {arctanh}\left (\sqrt {\frac {1}{\sqrt {x}}+1}\right )+\sqrt {\frac {1}{\sqrt {x}}+1} x-\frac {3}{2} \sqrt {\frac {1}{\sqrt {x}}+1} \sqrt {x} \]

[In]

Int[1/Sqrt[1 + 1/Sqrt[x]],x]

[Out]

(-3*Sqrt[1 + 1/Sqrt[x]]*Sqrt[x])/2 + Sqrt[1 + 1/Sqrt[x]]*x + (3*ArcTanh[Sqrt[1 + 1/Sqrt[x]]])/2

Rule 44

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] - Dist[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] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 196

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 213

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])

Rubi steps \begin{align*} \text {integral}& = -\left (2 \text {Subst}\left (\int \frac {1}{x^3 \sqrt {1+x}} \, dx,x,\frac {1}{\sqrt {x}}\right )\right ) \\ & = \sqrt {1+\frac {1}{\sqrt {x}}} x+\frac {3}{2} \text {Subst}\left (\int \frac {1}{x^2 \sqrt {1+x}} \, dx,x,\frac {1}{\sqrt {x}}\right ) \\ & = -\frac {3}{2} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}+\sqrt {1+\frac {1}{\sqrt {x}}} x-\frac {3}{4} \text {Subst}\left (\int \frac {1}{x \sqrt {1+x}} \, dx,x,\frac {1}{\sqrt {x}}\right ) \\ & = -\frac {3}{2} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}+\sqrt {1+\frac {1}{\sqrt {x}}} x-\frac {3}{2} \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sqrt {1+\frac {1}{\sqrt {x}}}\right ) \\ & = -\frac {3}{2} \sqrt {1+\frac {1}{\sqrt {x}}} \sqrt {x}+\sqrt {1+\frac {1}{\sqrt {x}}} x+\frac {3}{2} \tanh ^{-1}\left (\sqrt {1+\frac {1}{\sqrt {x}}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=\frac {1}{2} \left (\sqrt {1+\frac {1}{\sqrt {x}}} \left (-3+2 \sqrt {x}\right ) \sqrt {x}+3 \text {arctanh}\left (\sqrt {1+\frac {1}{\sqrt {x}}}\right )\right ) \]

[In]

Integrate[1/Sqrt[1 + 1/Sqrt[x]],x]

[Out]

(Sqrt[1 + 1/Sqrt[x]]*(-3 + 2*Sqrt[x])*Sqrt[x] + 3*ArcTanh[Sqrt[1 + 1/Sqrt[x]]])/2

Maple [A] (verified)

Time = 3.67 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.76

method	result	size
meijerg	\(\frac {-\frac {\sqrt {\pi }\, x^{\frac {1}{4}} \left (-10 \sqrt {x}+15\right ) \sqrt {\sqrt {x}+1}}{10}+\frac {3 \sqrt {\pi }\, \operatorname {arcsinh}\left (x^{\frac {1}{4}}\right )}{2}}{\sqrt {\pi }}\)	\(38\)
default	\(\frac {\sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \sqrt {x}\, \left (4 \sqrt {x +\sqrt {x}}\, \sqrt {x}+3 \ln \left (\frac {1}{2}+\sqrt {x}+\sqrt {x +\sqrt {x}}\right )-6 \sqrt {x +\sqrt {x}}\right )}{4 \sqrt {\left (\sqrt {x}+1\right ) \sqrt {x}}}\)	\(65\)
derivativedivides	\(\frac {\left (\sqrt {x}+1\right ) \left (4 \sqrt {x +\sqrt {x}}\, \sqrt {x}+3 \ln \left (\frac {1}{2}+\sqrt {x}+\sqrt {x +\sqrt {x}}\right )-6 \sqrt {x +\sqrt {x}}\right )}{4 \sqrt {\frac {\sqrt {x}+1}{\sqrt {x}}}\, \sqrt {\left (\sqrt {x}+1\right ) \sqrt {x}}}\)	\(67\)

[In]

int(1/(1+1/x^(1/2))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/Pi^(1/2)*(-1/20*Pi^(1/2)*x^(1/4)*(-10*x^(1/2)+15)*(x^(1/2)+1)^(1/2)+3/4*Pi^(1/2)*arcsinh(x^(1/4)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=\frac {1}{2} \, {\left (2 \, x - 3 \, \sqrt {x}\right )} \sqrt {\frac {x + \sqrt {x}}{x}} + \frac {3}{4} \, \log \left (\sqrt {\frac {x + \sqrt {x}}{x}} + 1\right ) - \frac {3}{4} \, \log \left (\sqrt {\frac {x + \sqrt {x}}{x}} - 1\right ) \]

[In]

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

[Out]

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)

Sympy [A] (verification not implemented)

Time = 2.22 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.20 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=\frac {x^{\frac {5}{4}}}{\sqrt {\sqrt {x} + 1}} - \frac {x^{\frac {3}{4}}}{2 \sqrt {\sqrt {x} + 1}} - \frac {3 \sqrt [4]{x}}{2 \sqrt {\sqrt {x} + 1}} + \frac {3 \operatorname {asinh}{\left (\sqrt [4]{x} \right )}}{2} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=-\frac {3 \, {\left (\frac {1}{\sqrt {x}} + 1\right )}^{\frac {3}{2}} - 5 \, \sqrt {\frac {1}{\sqrt {x}} + 1}}{2 \, {\left ({\left (\frac {1}{\sqrt {x}} + 1\right )}^{2} - \frac {2}{\sqrt {x}} - 1\right )}} + \frac {3}{4} \, \log \left (\sqrt {\frac {1}{\sqrt {x}} + 1} + 1\right ) - \frac {3}{4} \, \log \left (\sqrt {\frac {1}{\sqrt {x}} + 1} - 1\right ) \]

[In]

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

[Out]

-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/sq
rt(x) + 1) + 1) - 3/4*log(sqrt(1/sqrt(x) + 1) - 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=\frac {2 \, \sqrt {x + \sqrt {x}} {\left (2 \, \sqrt {x} - 3\right )} - 3 \, \log \left (-2 \, \sqrt {x + \sqrt {x}} + 2 \, \sqrt {x} + 1\right )}{4 \, \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

1/4*(2*sqrt(x + sqrt(x))*(2*sqrt(x) - 3) - 3*log(-2*sqrt(x + sqrt(x)) + 2*sqrt(x) + 1))/sgn(x)

Mupad [B] (verification not implemented)

Time = 5.86 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\sqrt {1+\frac {1}{\sqrt {x}}}} \, dx=\frac {4\,x\,\sqrt {\sqrt {x}+1}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {5}{2};\ \frac {7}{2};\ -\sqrt {x}\right )}{5\,\sqrt {\frac {1}{\sqrt {x}}+1}} \]

[In]

int(1/(1/x^(1/2) + 1)^(1/2),x)

[Out]

(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))