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

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 1.00 \[ -\frac {2 \sinh ^{-1}\left (\sqrt {1-b x}\right )}{b} \]

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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fricas [B]  time = 0.42, size = 30, normalized size = 1.88 \[ -\frac {\log \left (-2 \, b x + 2 \, \sqrt {-b x + 2} \sqrt {-b x + 1} + 3\right )}{b} \]

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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giac [A]  time = 0.85, size = 25, normalized size = 1.56 \[ \frac {2 \, \log \left (\sqrt {-b x + 2} - \sqrt {-b x + 1}\right )}{b} \]

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(sqrt(-b*x + 2) - sqrt(-b*x + 1))/b

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maple [B]  time = 0.01, size = 70, normalized size = 4.38 \[ \frac {\sqrt {\left (-b x +1\right ) \left (-b x +2\right )}\, \ln \left (\frac {b^{2} x -\frac {3}{2} b}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}-3 b x +2}\right )}{\sqrt {-b x +1}\, \sqrt {-b x +2}\, \sqrt {b^{2}}} \]

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 [B]  time = 1.33, size = 33, normalized size = 2.06 \[ \frac {\log \left (2 \, b^{2} x + 2 \, \sqrt {b^{2} x^{2} - 3 \, b x + 2} b - 3 \, b\right )}{b} \]

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]

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

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mupad [B]  time = 0.31, size = 45, normalized size = 2.81 \[ -\frac {4\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {2}-\sqrt {2-b\,x}\right )}{\left (\sqrt {1-b\,x}-1\right )\,\sqrt {-b^2}}\right )}{\sqrt {-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*atan((b*(2^(1/2) - (2 - b*x)^(1/2)))/(((1 - b*x)^(1/2) - 1)*(-b^2)^(1/2))))/(-b^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {- b x + 1} \sqrt {- b x + 2}}\, dx \]

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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