3.2827 \(\int \sqrt {\frac {c}{(a+b x)^2}} \, dx\)

Optimal. Leaf size=28 \[ \frac {(a+b x) \sqrt {\frac {c}{(a+b x)^2}} \log (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 28, 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, 29} \[ \frac {(a+b x) \sqrt {\frac {c}{(a+b x)^2}} \log (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c/(a + b*x)^2],x]

[Out]

(Sqrt[c/(a + b*x)^2]*(a + b*x)*Log[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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], 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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 28, normalized size = 1.00 \[ \frac {(a+b x) \sqrt {\frac {c}{(a+b x)^2}} \log (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c/(a + b*x)^2],x]

[Out]

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

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fricas [A]  time = 0.72, size = 37, normalized size = 1.32 \[ \frac {{\left (b x + a\right )} \sqrt {\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}} \log \left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 20, normalized size = 0.71 \[ \frac {\sqrt {c} \log \left ({\left | b x + a \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c)*log(abs(b*x + a))*sgn(b*x + a)/b

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maple [A]  time = 0.01, size = 27, normalized size = 0.96 \[ \frac {\left (b x +a \right ) \sqrt {\frac {c}{\left (b x +a \right )^{2}}}\, \ln \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.66, size = 13, normalized size = 0.46 \[ \frac {\sqrt {c} \log \left (b x + a\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c)*log(b*x + a)/b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \sqrt {\frac {c}{{\left (a+b\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/(a + b*x)^2)^(1/2),x)

[Out]

int((c/(a + b*x)^2)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(c/(a + b*x)**2), x)

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