3.24.100 \(\int \frac {1}{\sqrt {1+\sqrt {\frac {1}{x}}}} \, dx\)

Optimal. Leaf size=58 \[ \sqrt {\sqrt {\frac {1}{x}}+1} x-\frac {3 \sqrt {\sqrt {\frac {1}{x}}+1}}{2 \sqrt {\frac {1}{x}}}+\frac {3}{2} \tanh ^{-1}\left (\sqrt {\sqrt {\frac {1}{x}}+1}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {255, 190, 51, 63, 207} \begin {gather*} \sqrt {\sqrt {\frac {1}{x}}+1} x-\frac {3 \sqrt {\sqrt {\frac {1}{x}}+1}}{2 \sqrt {\frac {1}{x}}}+\frac {3}{2} \tanh ^{-1}\left (\sqrt {\sqrt {\frac {1}{x}}+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 + Sqrt[x^(-1)]],x]

[Out]

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

Rule 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 190

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 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 255

Int[((a_) + (b_.)*((c_.)*(x_)^(q_.))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Subst[Int[(a + b*c^n
*x^(n*q))^p, x], x^(1/k), (c*x^q)^(1/k)/(c^(1/k)*(x^(1/k))^(q - 1))]] /; FreeQ[{a, b, c, p, q}, x] && Fraction
Q[n]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 70, normalized size = 1.21 \begin {gather*} \frac {1}{4} \left (2 \left (2-3 \sqrt {\frac {1}{x}}\right ) \sqrt {\sqrt {\frac {1}{x}}+1} x-3 \log \left (1-\frac {1}{\sqrt {\sqrt {\frac {1}{x}}+1}}\right )+3 \log \left (\frac {1}{\sqrt {\sqrt {\frac {1}{x}}+1}}+1\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[1 + Sqrt[x^(-1)]],x]

[Out]

(2*(2 - 3*Sqrt[x^(-1)])*Sqrt[1 + Sqrt[x^(-1)]]*x - 3*Log[1 - 1/Sqrt[1 + Sqrt[x^(-1)]]] + 3*Log[1 + 1/Sqrt[1 +
Sqrt[x^(-1)]]])/4

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IntegrateAlgebraic [A]  time = 0.04, size = 48, normalized size = 0.83 \begin {gather*} \frac {1}{2} \left (2-3 \sqrt {\frac {1}{x}}\right ) \sqrt {\sqrt {\frac {1}{x}}+1} x+\frac {3}{2} \tanh ^{-1}\left (\sqrt {\sqrt {\frac {1}{x}}+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/Sqrt[1 + Sqrt[x^(-1)]],x]

[Out]

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

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fricas [A]  time = 1.13, size = 55, normalized size = 0.95 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.24, size = 36, normalized size = 0.62 \begin {gather*} \frac {1}{2} \, \sqrt {x + \sqrt {x}} {\left (2 \, \sqrt {x} - 3\right )} - \frac {3}{4} \, \log \left (-2 \, \sqrt {x + \sqrt {x}} + 2 \, \sqrt {x} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.02, size = 92, normalized size = 1.59 \begin {gather*} -\frac {\sqrt {1+\sqrt {\frac {1}{x}}}\, \left (-3 \ln \left (\frac {\sqrt {\frac {1}{x}}\, \sqrt {x}}{2}+\sqrt {x}+\sqrt {\sqrt {\frac {1}{x}}\, x +x}\right )+6 \sqrt {\sqrt {\frac {1}{x}}\, x +x}\, \sqrt {\frac {1}{x}}\, \sqrt {x}-4 \sqrt {\sqrt {\frac {1}{x}}\, x +x}\, \sqrt {x}\right ) \sqrt {x}}{4 \sqrt {\left (1+\sqrt {\frac {1}{x}}\right ) x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(1+(1/x)^(1/2))^(1/2)*x^(1/2)*(6*((1/x)^(1/2)*x+x)^(1/2)*(1/x)^(1/2)*x^(1/2)-4*((1/x)^(1/2)*x+x)^(1/2)*x^
(1/2)-3*ln(1/2*(1/x)^(1/2)*x^(1/2)+x^(1/2)+((1/x)^(1/2)*x+x)^(1/2)))/(x*(1+(1/x)^(1/2)))^(1/2)

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maxima [A]  time = 0.55, size = 62, normalized size = 1.07 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.88, size = 37, normalized size = 0.64 \begin {gather*} \frac {3\,\mathrm {atanh}\left (\sqrt {\sqrt {\frac {1}{x}}+1}\right )}{2}+\frac {5\,x\,\sqrt {\sqrt {\frac {1}{x}}+1}}{2}-\frac {3\,x\,{\left (\sqrt {\frac {1}{x}}+1\right )}^{3/2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((1/x)^(1/2) + 1)^(1/2),x)

[Out]

(3*atanh(((1/x)^(1/2) + 1)^(1/2)))/2 + (5*x*((1/x)^(1/2) + 1)^(1/2))/2 - (3*x*((1/x)^(1/2) + 1)^(3/2))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {\sqrt {\frac {1}{x}} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(sqrt(1/x) + 1), x)

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