3.2822 \(\int \frac{1}{\sqrt{\frac{c}{a+b x}}} \, dx\)

Optimal. Leaf size=25 \[ \frac{2 (a+b x)}{3 b \sqrt{\frac{c}{a+b x}}} \]

[Out]

(2*(a + b*x))/(3*b*Sqrt[c/(a + b*x)])

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Rubi [A]  time = 0.006398, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {247, 15, 30} \[ \frac{2 (a+b x)}{3 b \sqrt{\frac{c}{a+b x}}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[c/(a + b*x)],x]

[Out]

(2*(a + b*x))/(3*b*Sqrt[c/(a + b*x)])

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{\frac{c}{a+b x}}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{\frac{c}{x}}} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \sqrt{x} \, dx,x,a+b x\right )}{b \sqrt{\frac{c}{a+b x}} \sqrt{a+b x}}\\ &=\frac{2 (a+b x)}{3 b \sqrt{\frac{c}{a+b x}}}\\ \end{align*}

Mathematica [A]  time = 0.0120305, size = 21, normalized size = 0.84 \[ \frac{2 c}{3 b \left (\frac{c}{a+b x}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[c/(a + b*x)],x]

[Out]

(2*c)/(3*b*(c/(a + b*x))^(3/2))

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Maple [A]  time = 0.001, size = 22, normalized size = 0.9 \begin{align*}{\frac{2\,bx+2\,a}{3\,b}{\frac{1}{\sqrt{{\frac{c}{bx+a}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b*x+a)/b/(c/(b*x+a))^(1/2)

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Maxima [A]  time = 1.2842, size = 23, normalized size = 0.92 \begin{align*} \frac{2 \, c}{3 \, b \left (\frac{c}{b x + a}\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*c/(b*(c/(b*x + a))^(3/2))

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Fricas [A]  time = 1.31992, size = 74, normalized size = 2.96 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{\frac{c}{b x + a}}}{3 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(b^2*x^2 + 2*a*b*x + a^2)*sqrt(c/(b*x + a))/(b*c)

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Sympy [A]  time = 0.914531, size = 49, normalized size = 1.96 \begin{align*} \begin{cases} \frac{2 a}{3 b \sqrt{c} \sqrt{\frac{1}{a + b x}}} + \frac{2 x}{3 \sqrt{c} \sqrt{\frac{1}{a + b x}}} & \text{for}\: b \neq 0 \\\frac{x}{\sqrt{\frac{c}{a}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*a/(3*b*sqrt(c)*sqrt(1/(a + b*x))) + 2*x/(3*sqrt(c)*sqrt(1/(a + b*x))), Ne(b, 0)), (x/sqrt(c/a), T
rue))

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Giac [B]  time = 1.14069, size = 84, normalized size = 3.36 \begin{align*} \frac{2 \,{\left (3 \, \sqrt{b c x + a c} a - \frac{3 \, \sqrt{b c x + a c} a c -{\left (b c x + a c\right )}^{\frac{3}{2}}}{c}\right )}}{3 \, b c \mathrm{sgn}\left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*sqrt(b*c*x + a*c)*a - (3*sqrt(b*c*x + a*c)*a*c - (b*c*x + a*c)^(3/2))/c)/(b*c*sgn(b*x + a))