3.1545 \(\int \frac{1}{\sqrt{1-b x} \sqrt{2-b x}} \, dx\)

Optimal. Leaf size=16 \[ -\frac{2 \sinh ^{-1}\left (\sqrt{1-b x}\right )}{b} \]

[Out]

(-2*ArcSinh[Sqrt[1 - b*x]])/b

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Rubi [A]  time = 0.0052831, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {63, 215} \[ -\frac{2 \sinh ^{-1}\left (\sqrt{1-b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 - b*x]*Sqrt[2 - b*x]),x]

[Out]

(-2*ArcSinh[Sqrt[1 - b*x]])/b

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0065096, size = 16, normalized size = 1. \[ -\frac{2 \sinh ^{-1}\left (\sqrt{1-b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[1 - b*x]*Sqrt[2 - b*x]),x]

[Out]

(-2*ArcSinh[Sqrt[1 - b*x]])/b

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Maple [B]  time = 0.007, size = 70, normalized size = 4.4 \begin{align*}{\sqrt{ \left ( -bx+1 \right ) \left ( -bx+2 \right ) }\ln \left ({ \left ( -{\frac{3\,b}{2}}+{b}^{2}x \right ){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}-3\,bx+2} \right ){\frac{1}{\sqrt{-bx+1}}}{\frac{1}{\sqrt{-bx+2}}}{\frac{1}{\sqrt{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((-b*x+1)*(-b*x+2))^(1/2)/(-b*x+1)^(1/2)/(-b*x+2)^(1/2)*ln((-3/2*b+b^2*x)/(b^2)^(1/2)+(b^2*x^2-3*b*x+2)^(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*x+1)^(1/2)/(-b*x+2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.94618, size = 73, normalized size = 4.56 \begin{align*} -\frac{\log \left (-2 \, b x + 2 \, \sqrt{-b x + 2} \sqrt{-b x + 1} + 3\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(-2*b*x + 2*sqrt(-b*x + 2)*sqrt(-b*x + 1) + 3)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt(-b*x + 1)*sqrt(-b*x + 2)), x)

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Giac [A]  time = 1.14366, size = 35, normalized size = 2.19 \begin{align*} \frac{2 \, \log \left ({\left | -\sqrt{-b x + 2} + \sqrt{-b x + 1} \right |}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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