3.2828 \(\int \frac{1}{\sqrt{\frac{c}{(a+b x)^2}}} \, dx\)

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

[Out]

(a + b*x)/(2*b*Sqrt[c/(a + b*x)^2])

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Rubi [A]  time = 0.0073814, 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} \[ \frac{a+b x}{2 b \sqrt{\frac{c}{(a+b x)^2}}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[c/(a + b*x)^2],x]

[Out]

(a + b*x)/(2*b*Sqrt[c/(a + b*x)^2])

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

Mathematica [A]  time = 0.0128316, size = 32, normalized size = 1.28 \[ \frac{x (2 a+b x)}{2 (a+b x) \sqrt{\frac{c}{(a+b x)^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[c/(a + b*x)^2],x]

[Out]

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

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Maple [A]  time = 0., size = 29, normalized size = 1.2 \begin{align*}{\frac{x \left ( bx+2\,a \right ) }{2\,bx+2\,a}{\frac{1}{\sqrt{{\frac{c}{ \left ( bx+a \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.14642, size = 20, normalized size = 0.8 \begin{align*} \frac{b x^{2} + 2 \, a x}{2 \, \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x^2 + 2*a*x)/sqrt(c)

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Fricas [B]  time = 1.29815, size = 99, normalized size = 3.96 \begin{align*} \frac{{\left (b^{2} x^{3} + 3 \, a b x^{2} + 2 \, a^{2} x\right )} \sqrt{\frac{c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\frac{c}{\left (a + b x\right )^{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(c/(a + b*x)**2), x)

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Giac [A]  time = 1.39094, size = 39, normalized size = 1.56 \begin{align*} \frac{b \sqrt{c} x^{2} + 2 \, a \sqrt{c} x}{2 \, 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)^2)^(1/2),x, algorithm="giac")

[Out]

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