3.1546 \(\int \frac{1}{\sqrt{-b x} \sqrt{2-b x}} \, dx\)

Optimal. Leaf size=20 \[ -\frac{2 \sinh ^{-1}\left (\frac{\sqrt{-b x}}{\sqrt{2}}\right )}{b} \]

[Out]

(-2*ArcSinh[Sqrt[-(b*x)]/Sqrt[2]])/b

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Rubi [A]  time = 0.0053592, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {63, 215} \[ -\frac{2 \sinh ^{-1}\left (\frac{\sqrt{-b x}}{\sqrt{2}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[-(b*x)]*Sqrt[2 - b*x]),x]

[Out]

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

Mathematica [A]  time = 0.0085961, size = 37, normalized size = 1.85 \[ \frac{2 \sqrt{x} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{\sqrt{b} \sqrt{-b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(Sqrt[b]*Sqrt[-(b*x)])

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Maple [B]  time = 0.004, size = 64, normalized size = 3.2 \begin{align*}{\sqrt{-bx \left ( -bx+2 \right ) }\ln \left ({({b}^{2}x-b){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}-2\,bx} \right ){\frac{1}{\sqrt{-bx}}}{\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)^(1/2)/(-b*x+2)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.99129, size = 62, normalized size = 3.1 \begin{align*} -\frac{\log \left (-b x + \sqrt{-b x + 2} \sqrt{-b x} + 1\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.48838, size = 53, normalized size = 2.65 \begin{align*} \begin{cases} - \frac{2 \operatorname{acosh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{b} & \text{for}\: \frac{\left |{b x}\right |}{2} > 1 \\- \frac{2 i \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{b} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b, Abs(b*x)/2 > 1), (-2*I*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/b, Tr
ue))

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Giac [A]  time = 1.1284, size = 32, normalized size = 1.6 \begin{align*} \frac{2 \, \log \left ({\left | -\sqrt{-b x + 2} + \sqrt{-b x} \right |}\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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