3.2194 \(\int \frac{1}{a+b \sqrt{x}} \, dx\)

Optimal. Leaf size=27 \[ \frac{2 \sqrt{x}}{b}-\frac{2 a \log \left (a+b \sqrt{x}\right )}{b^2} \]

[Out]

(2*Sqrt[x])/b - (2*a*Log[a + b*Sqrt[x]])/b^2

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Rubi [A]  time = 0.0135099, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {190, 43} \[ \frac{2 \sqrt{x}}{b}-\frac{2 a \log \left (a+b \sqrt{x}\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^(-1),x]

[Out]

(2*Sqrt[x])/b - (2*a*Log[a + b*Sqrt[x]])/b^2

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0120507, size = 27, normalized size = 1. \[ \frac{2 \sqrt{x}}{b}-\frac{2 a \log \left (a+b \sqrt{x}\right )}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^(-1),x]

[Out]

(2*Sqrt[x])/b - (2*a*Log[a + b*Sqrt[x]])/b^2

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Maple [B]  time = 0.006, size = 57, normalized size = 2.1 \begin{align*} 2\,{\frac{\sqrt{x}}{b}}-{\frac{a}{{b}^{2}}\ln \left ( a+b\sqrt{x} \right ) }+{\frac{a}{{b}^{2}}\ln \left ( b\sqrt{x}-a \right ) }-{\frac{a\ln \left ({b}^{2}x-{a}^{2} \right ) }{{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.968798, size = 36, normalized size = 1.33 \begin{align*} -\frac{2 \, a \log \left (b \sqrt{x} + a\right )}{b^{2}} + \frac{2 \,{\left (b \sqrt{x} + a\right )}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*log(b*sqrt(x) + a)/b^2 + 2*(b*sqrt(x) + a)/b^2

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Fricas [A]  time = 1.31158, size = 58, normalized size = 2.15 \begin{align*} -\frac{2 \,{\left (a \log \left (b \sqrt{x} + a\right ) - b \sqrt{x}\right )}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(a*log(b*sqrt(x) + a) - b*sqrt(x))/b^2

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Sympy [A]  time = 0.154442, size = 27, normalized size = 1. \begin{align*} \begin{cases} - \frac{2 a \log{\left (\frac{a}{b} + \sqrt{x} \right )}}{b^{2}} + \frac{2 \sqrt{x}}{b} & \text{for}\: b \neq 0 \\\frac{x}{a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10129, size = 32, normalized size = 1.19 \begin{align*} -\frac{2 \, a \log \left ({\left | b \sqrt{x} + a \right |}\right )}{b^{2}} + \frac{2 \, \sqrt{x}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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