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]

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

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Rubi [A]  time = 0.01, antiderivative size = 25, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.76, size = 35, normalized size = 1.40 \[ \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {\frac {c}{b x + a}}}{3 \, b c} \]

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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giac [B]  time = 0.16, size = 62, normalized size = 2.48 \[ \frac {2 \, {\left (3 \, \sqrt {b c x + a c} a - \frac {3 \, \sqrt {b c x + a c} a c - {\left (b c x + a c\right )}^{\frac {3}{2}}}{c}\right )}}{3 \, b c \mathrm {sgn}\left (b x + a\right )} \]

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))

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maple [A]  time = 0.00, size = 22, normalized size = 0.88 \[ \frac {\frac {2 b x}{3}+\frac {2 a}{3}}{\sqrt {\frac {c}{b x +a}}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.63, size = 17, normalized size = 0.68 \[ \frac {2 \, c}{3 \, b \left (\frac {c}{b x + a}\right )^{\frac {3}{2}}} \]

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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mupad [B]  time = 1.19, size = 40, normalized size = 1.60 \[ \sqrt {\frac {c}{a+b\,x}}\,\left (\frac {2\,b\,x^2}{3\,c}+\frac {2\,a^2}{3\,b\,c}+\frac {4\,a\,x}{3\,c}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c/(a + b*x))^(1/2),x)

[Out]

(c/(a + b*x))^(1/2)*((2*b*x^2)/(3*c) + (2*a^2)/(3*b*c) + (4*a*x)/(3*c))

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sympy [A]  time = 1.62, size = 49, normalized size = 1.96 \[ \begin {cases} \frac {2 a}{3 b \sqrt {c} \sqrt {\frac {1}{a + b x}}} + \frac {2 x}{3 \sqrt {c} \sqrt {\frac {1}{a + b x}}} & \text {for}\: b \neq 0 \\\frac {x}{\sqrt {\frac {c}{a}}} & \text {otherwise} \end {cases} \]

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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