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

Optimal. Leaf size=24 \[ \frac{2 \tanh ^{-1}\left (\frac{b x}{\sqrt{b^2 x^2+b x}}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0077753, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {620, 206} \[ \frac{2 \tanh ^{-1}\left (\frac{b x}{\sqrt{b^2 x^2+b x}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*x + b^2*x^2],x]

[Out]

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

Rule 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0203353, size = 45, normalized size = 1.88 \[ \frac{2 \sqrt{x} \sqrt{b x+1} \sinh ^{-1}\left (\sqrt{b} \sqrt{x}\right )}{\sqrt{b} \sqrt{b x (b x+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*x + b^2*x^2],x]

[Out]

(2*Sqrt[x]*Sqrt[1 + b*x]*ArcSinh[Sqrt[b]*Sqrt[x]])/(Sqrt[b]*Sqrt[b*x*(1 + b*x)])

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Maple [A]  time = 0.057, size = 37, normalized size = 1.5 \begin{align*}{\ln \left ({ \left ({\frac{b}{2}}+{b}^{2}x \right ){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+bx} \right ){\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.15796, size = 59, normalized size = 2.46 \begin{align*} -\frac{\log \left (-2 \, b x + 2 \, \sqrt{b^{2} x^{2} + b x} - 1\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-2*b*x + 2*sqrt(b^2*x^2 + b*x) - 1)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(b**2*x**2 + b*x), x)

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Giac [A]  time = 1.80615, size = 49, normalized size = 2.04 \begin{align*} -\frac{\log \left ({\left | -2 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + b x}\right )}{\left | b \right |} - b \right |}\right )}{{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(abs(-2*(x*abs(b) - sqrt(b^2*x^2 + b*x))*abs(b) - b))/abs(b)