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]

1/2*(b*x+a)/b/(c/(b*x+a)^2)^(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} \[ \frac {a+b x}{2 b \sqrt {\frac {c}{(a+b x)^2}}} \]

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 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)^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*}

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Mathematica [A]  time = 0.01, 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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fricas [B]  time = 0.85, size = 48, normalized size = 1.92 \[ \frac {{\left (b^{2} x^{3} + 3 \, a b x^{2} + 2 \, a^{2} x\right )} \sqrt {\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{2 \, c} \]

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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giac [A]  time = 0.25, size = 23, normalized size = 0.92 \[ \frac {b x^{2} + 2 \, a x}{2 \, \sqrt {c} \mathrm {sgn}\left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 29, normalized size = 1.16 \[ \frac {\left (b x +2 a \right ) x}{2 \left (b x +a \right ) \sqrt {\frac {c}{\left (b x +a \right )^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.61, size = 15, normalized size = 0.60 \[ \frac {b x^{2} + 2 \, a x}{2 \, \sqrt {c}} \]

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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mupad [B]  time = 1.22, size = 42, normalized size = 1.68 \[ \sqrt {\frac {c}{{\left (a+b\,x\right )}^2}}\,\left (\frac {a^2\,x}{c}+\frac {b^2\,x^3}{2\,c}+\frac {3\,a\,b\,x^2}{2\,c}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c/(a + b*x)^2)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\frac {c}{\left (a + b x\right )^{2}}}}\, dx \]

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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