3.2805 \(\int \frac{1}{\sqrt{c (a+b x)^2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0123486, antiderivative size = 28, 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, 29} \[ \frac{(a+b x) \log (a+b x)}{b \sqrt{c (a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[c*(a + b*x)^2],x]

[Out]

((a + b*x)*Log[a + b*x])/(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 29

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

Rubi steps

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

Mathematica [A]  time = 0.006314, size = 28, normalized size = 1. \[ \frac{(a+b x) \log (a+b x)}{b \sqrt{c (a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[c*(a + b*x)^2],x]

[Out]

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

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Maple [A]  time = 0.005, size = 27, normalized size = 1. \begin{align*}{\frac{ \left ( bx+a \right ) \ln \left ( bx+a \right ) }{b}{\frac{1}{\sqrt{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]

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

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Maxima [A]  time = 0.969728, size = 24, normalized size = 0.86 \begin{align*} \sqrt{\frac{1}{b^{2} c}} \log \left (x + \frac{a}{b}\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="maxima")

[Out]

sqrt(1/(b^2*c))*log(x + a/b)

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Fricas [A]  time = 1.29266, size = 92, normalized size = 3.29 \begin{align*} \frac{\sqrt{b^{2} c x^{2} + 2 \, a b c x + a^{2} c} \log \left (b x + a\right )}{b^{2} c x + a b 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]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{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 = 1.12227, size = 46, normalized size = 1.64 \begin{align*} \frac{\log \left ({\left | b x + a \right |} \sqrt{{\left | c \right |}}{\left | \mathrm{sgn}\left (b x + a\right ) \right |}\right )}{b \sqrt{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]

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