3.2821 \(\int \sqrt {\frac {c}{a+b x}} \, dx\)

Optimal. Leaf size=23 \[ \frac {2 (a+b x) \sqrt {\frac {c}{a+b x}}}{b} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 23, 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} \[ \frac {2 (a+b x) \sqrt {\frac {c}{a+b x}}}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c/(a + b*x)],x]

[Out]

(2*Sqrt[c/(a + b*x)]*(a + b*x))/b

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

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.83 \[ \frac {2 c}{b \sqrt {\frac {c}{a+b x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c/(a + b*x)],x]

[Out]

(2*c)/(b*Sqrt[c/(a + b*x)])

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fricas [A]  time = 0.57, size = 21, normalized size = 0.91 \[ \frac {2 \, {\left (b x + a\right )} \sqrt {\frac {c}{b x + a}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b*x + a)*sqrt(c/(b*x + a))/b

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giac [A]  time = 0.16, size = 21, normalized size = 0.91 \[ \frac {2 \, \sqrt {b c x + a c} \mathrm {sgn}\left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b*c*x + a*c)*sgn(b*x + a)/b

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maple [A]  time = 0.00, size = 22, normalized size = 0.96 \[ \frac {2 \left (b x +a \right ) \sqrt {\frac {c}{b x +a}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.56, size = 17, normalized size = 0.74 \[ \frac {2 \, c}{b \sqrt {\frac {c}{b x + a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*c/(b*sqrt(c/(b*x + a)))

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mupad [B]  time = 1.16, size = 22, normalized size = 0.96 \[ \sqrt {\frac {c}{a+b\,x}}\,\left (2\,x+\frac {2\,a}{b}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/(a + b*x))^(1/2),x)

[Out]

(c/(a + b*x))^(1/2)*(2*x + (2*a)/b)

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sympy [A]  time = 0.34, size = 46, normalized size = 2.00 \[ \begin {cases} \frac {2 a \sqrt {c} \sqrt {\frac {1}{a + b x}}}{b} + 2 \sqrt {c} x \sqrt {\frac {1}{a + b x}} & \text {for}\: b \neq 0 \\x \sqrt {\frac {c}{a}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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