3.1.18 \(\int \sqrt {\sqrt {1+\frac {1}{x}}+\frac {1}{x}} \, dx\) [18]

Optimal. Leaf size=96 \[ \sqrt {\sqrt {1+\frac {1}{x}}+\frac {1}{x}} x+\frac {1}{4} \tan ^{-1}\left (\frac {3+\sqrt {1+\frac {1}{x}}}{2 \sqrt {\sqrt {1+\frac {1}{x}}+\frac {1}{x}}}\right )-\frac {3}{4} \tanh ^{-1}\left (\frac {1-3 \sqrt {1+\frac {1}{x}}}{2 \sqrt {\sqrt {1+\frac {1}{x}}+\frac {1}{x}}}\right ) \]

[Out]

1/4*arctan(1/2*(3+(1+1/x)^(1/2))/(1/x+(1+1/x)^(1/2))^(1/2))-3/4*arctanh(1/2*(1-3*(1+1/x)^(1/2))/(1/x+(1+1/x)^(
1/2))^(1/2))+x*(1/x+(1+1/x)^(1/2))^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {1028, 1047, 738, 212, 210} \begin {gather*} \sqrt {\sqrt {\frac {1}{x}+1}+\frac {1}{x}} x+\frac {1}{4} \tan ^{-1}\left (\frac {\sqrt {\frac {1}{x}+1}+3}{2 \sqrt {\sqrt {\frac {1}{x}+1}+\frac {1}{x}}}\right )-\frac {3}{4} \tanh ^{-1}\left (\frac {1-3 \sqrt {\frac {1}{x}+1}}{2 \sqrt {\sqrt {\frac {1}{x}+1}+\frac {1}{x}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[Sqrt[1 + x^(-1)] + x^(-1)],x]

[Out]

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

Rule 210

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 738

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 1028

Int[((g_.) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp
[(a*h - g*c*x)*(a + c*x^2)^(p + 1)*((d + e*x + f*x^2)^q/(2*a*c*(p + 1))), x] + Dist[2/(4*a*c*(p + 1)), Int[(a
+ c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q - 1)*Simp[g*c*d*(2*p + 3) - a*(h*e*q) + (g*c*e*(2*p + q + 3) - a*(2*h*f*
q))*x + g*c*f*(2*p + 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, g, h}, x] && NeQ[e^2 - 4*d*f, 0] && LtQ
[p, -1] && GtQ[q, 0]

Rule 1047

Int[((g_.) + (h_.)*(x_))/(((a_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[(-a)*c, 2]}, Dist[h/2 + c*(g/(2*q)), Int[1/((-q + c*x)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/2 - c*(g/
(2*q)), Int[1/((q + c*x)*Sqrt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, c, d, e, f, g, h}, x] && NeQ[e^2 - 4*d*f
, 0] && PosQ[(-a)*c]

Rubi steps

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

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Mathematica [A]
time = 0.36, size = 89, normalized size = 0.93 \begin {gather*} \frac {1}{2} \left (2 \sqrt {\sqrt {1+\frac {1}{x}}+\frac {1}{x}} x+\tan ^{-1}\left (1+\sqrt {1+\frac {1}{x}}-\sqrt {\sqrt {1+\frac {1}{x}}+\frac {1}{x}}\right )+3 \tanh ^{-1}\left (1-\sqrt {1+\frac {1}{x}}+\sqrt {\sqrt {1+\frac {1}{x}}+\frac {1}{x}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Sqrt[1 + x^(-1)] + x^(-1)],x]

[Out]

(2*Sqrt[Sqrt[1 + x^(-1)] + x^(-1)]*x + ArcTan[1 + Sqrt[1 + x^(-1)] - Sqrt[Sqrt[1 + x^(-1)] + x^(-1)]] + 3*ArcT
anh[1 - Sqrt[1 + x^(-1)] + Sqrt[Sqrt[1 + x^(-1)] + x^(-1)]])/2

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Mathics [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {cought exception: maximum recursion depth exceeded while calling a Python object} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[Sqrt[1/x + Sqrt[1 + 1/x]],x]')

[Out]

cought exception: maximum recursion depth exceeded while calling a Python object

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \sqrt {\frac {1}{x}+\sqrt {1+\frac {1}{x}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(sqrt(1/x + 1) + 1/x), x)

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Fricas [A]
time = 1.90, size = 122, normalized size = 1.27 \begin {gather*} x \sqrt {\frac {x \sqrt {\frac {x + 1}{x}} + 1}{x}} + \frac {1}{4} \, \arctan \left (\frac {2 \, {\left (x \sqrt {\frac {x + 1}{x}} - 3 \, x\right )} \sqrt {\frac {x \sqrt {\frac {x + 1}{x}} + 1}{x}}}{8 \, x - 1}\right ) + \frac {3}{4} \, \log \left (2 \, {\left (x \sqrt {\frac {x + 1}{x}} + x\right )} \sqrt {\frac {x \sqrt {\frac {x + 1}{x}} + 1}{x}} + 2 \, x \sqrt {\frac {x + 1}{x}} + 2 \, x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*sqrt((x*sqrt((x + 1)/x) + 1)/x) + 1/4*arctan(2*(x*sqrt((x + 1)/x) - 3*x)*sqrt((x*sqrt((x + 1)/x) + 1)/x)/(8*
x - 1)) + 3/4*log(2*(x*sqrt((x + 1)/x) + x)*sqrt((x*sqrt((x + 1)/x) + 1)/x) + 2*x*sqrt((x + 1)/x) + 2*x + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(sqrt(1 + 1/x) + 1/x), x)

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Giac [F] N/A
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Could not integrate

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sqrt {\sqrt {\frac {1}{x}+1}+\frac {1}{x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1/x + 1)^(1/2) + 1/x)^(1/2),x)

[Out]

int(((1/x + 1)^(1/2) + 1/x)^(1/2), x)

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