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

Optimal. Leaf size=24 \[ 2 \tanh ^{-1}\left (\sqrt {\frac {1}{x}+1}\right )-2 \sqrt {\frac {1}{x}+1} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1973, 266, 50, 63, 207} \[ 2 \tanh ^{-1}\left (\sqrt {\frac {1}{x}+1}\right )-2 \sqrt {\frac {1}{x}+1} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(1 + x)/x]/x,x]

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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]]

Rule 1973

Int[(u_)^(p_.)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[(c*x)^m*ExpandToSum[u, x]^p, x] /; FreeQ[{c, m, p}, x] &&
BinomialQ[u, x] &&  !BinomialMatchQ[u, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.00 \[ 2 \tanh ^{-1}\left (\sqrt {\frac {1}{x}+1}\right )-2 \sqrt {\frac {1}{x}+1} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(1 + x)/x]/x,x]

[Out]

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

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fricas [A]  time = 0.42, size = 38, normalized size = 1.58 \[ -2 \, \sqrt {\frac {x + 1}{x}} + \log \left (\sqrt {\frac {x + 1}{x}} + 1\right ) - \log \left (\sqrt {\frac {x + 1}{x}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.38, size = 38, normalized size = 1.58 \[ -\log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} + x} - 1 \right |}\right ) \mathrm {sgn}\relax (x) + \frac {2 \, \mathrm {sgn}\relax (x)}{x - \sqrt {x^{2} + x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(abs(-2*x + 2*sqrt(x^2 + x) - 1))*sgn(x) + 2*sgn(x)/(x - sqrt(x^2 + x))

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maple [B]  time = 0.01, size = 60, normalized size = 2.50 \[ -\frac {\sqrt {\frac {x +1}{x}}\, \left (-x^{2} \ln \left (x +\frac {1}{2}+\sqrt {x^{2}+x}\right )-2 \sqrt {x^{2}+x}\, x^{2}+2 \left (x^{2}+x \right )^{\frac {3}{2}}\right )}{\sqrt {\left (x +1\right ) x}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 38, normalized size = 1.58 \[ -2 \, \sqrt {\frac {x + 1}{x}} + \log \left (\sqrt {\frac {x + 1}{x}} + 1\right ) - \log \left (\sqrt {\frac {x + 1}{x}} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.04, size = 20, normalized size = 0.83 \[ 2\,\mathrm {atanh}\left (\sqrt {\frac {1}{x}+1}\right )-2\,\sqrt {\frac {1}{x}+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x + 1)/x)^(1/2)/x,x)

[Out]

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

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sympy [A]  time = 3.23, size = 32, normalized size = 1.33 \[ - 2 \sqrt {1 + \frac {1}{x}} - \log {\left (\sqrt {1 + \frac {1}{x}} - 1 \right )} + \log {\left (\sqrt {1 + \frac {1}{x}} + 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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