3.2232 \(\int \frac {1}{(2+b \sqrt {x}) x} \, dx\)

Optimal. Leaf size=19 \[ \frac {\log (x)}{2}-\log \left (b \sqrt {x}+2\right ) \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {266, 36, 29, 31} \[ \frac {\log (x)}{2}-\log \left (b \sqrt {x}+2\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/((2 + b*Sqrt[x])*x),x]

[Out]

-Log[2 + b*Sqrt[x]] + Log[x]/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \[ \frac {\log (x)}{2}-\log \left (b \sqrt {x}+2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[1/((2 + b*Sqrt[x])*x),x]

[Out]

-Log[2 + b*Sqrt[x]] + Log[x]/2

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fricas [A]  time = 1.14, size = 15, normalized size = 0.79 \[ -\log \left (b \sqrt {x} + 2\right ) + \log \left (\sqrt {x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(b*sqrt(x) + 2) + log(sqrt(x))

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giac [A]  time = 0.17, size = 17, normalized size = 0.89 \[ -\log \left ({\left | b \sqrt {x} + 2 \right |}\right ) + \frac {1}{2} \, \log \left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 16, normalized size = 0.84 \[ \frac {\ln \relax (x )}{2}-\ln \left (b \sqrt {x}+2\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.89, size = 15, normalized size = 0.79 \[ -\log \left (b \sqrt {x} + 2\right ) + \frac {1}{2} \, \log \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(b*sqrt(x) + 2) + 1/2*log(x)

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mupad [B]  time = 0.05, size = 10, normalized size = 0.53 \[ -2\,\mathrm {atanh}\left (b\,\sqrt {x}+1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(b*x^(1/2) + 2)),x)

[Out]

-2*atanh(b*x^(1/2) + 1)

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sympy [A]  time = 0.47, size = 19, normalized size = 1.00 \[ \begin {cases} \frac {\log {\relax (x )}}{2} - \log {\left (\sqrt {x} + \frac {2}{b} \right )} & \text {for}\: b \neq 0 \\\frac {\log {\relax (x )}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((log(x)/2 - log(sqrt(x) + 2/b), Ne(b, 0)), (log(x)/2, True))

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