\(\int \frac {1}{\sqrt {2-b x} \sqrt {3-b x}} \, dx\) [1543]

Optimal result

Integrand size = 21, antiderivative size = 16 \[ \int \frac {1}{\sqrt {2-b x} \sqrt {3-b x}} \, dx=-\frac {2 \text {arcsinh}\left (\sqrt {2-b x}\right )}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {65, 221} \[ \int \frac {1}{\sqrt {2-b x} \sqrt {3-b x}} \, dx=-\frac {2 \text {arcsinh}\left (\sqrt {2-b x}\right )}{b} \]

[In]

Int[1/(Sqrt[2 - b*x]*Sqrt[3 - b*x]),x]

[Out]

(-2*ArcSinh[Sqrt[2 - b*x]])/b

Rule 65

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 221

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(60\) vs. \(2(16)=32\).

Time = 0.05 (sec) , antiderivative size = 60, normalized size of antiderivative = 3.75 \[ \int \frac {1}{\sqrt {2-b x} \sqrt {3-b x}} \, dx=\frac {\log \left (\sqrt {2-b x}-\sqrt {3-b x}\right )}{b}-\frac {\log \left (b \sqrt {2-b x}+b \sqrt {3-b x}\right )}{b} \]

[In]

Integrate[1/(Sqrt[2 - b*x]*Sqrt[3 - b*x]),x]

[Out]

Log[Sqrt[2 - b*x] - Sqrt[3 - b*x]]/b - Log[b*Sqrt[2 - b*x] + b*Sqrt[3 - b*x]]/b

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(69\) vs. \(2(14)=28\).

Time = 0.54 (sec) , antiderivative size = 70, normalized size of antiderivative = 4.38

method	result	size
default	\(\frac {\sqrt {\left (-b x +2\right ) \left (-b x +3\right )}\, \ln \left (\frac {-\frac {5}{2} b +b^{2} x}{\sqrt {b^{2}}}+\sqrt {b^{2} x^{2}-5 b x +6}\right )}{\sqrt {-b x +2}\, \sqrt {-b x +3}\, \sqrt {b^{2}}}\)	\(70\)

[In]

int(1/(-b*x+2)^(1/2)/(-b*x+3)^(1/2),x,method=_RETURNVERBOSE)

[Out]

((-b*x+2)*(-b*x+3))^(1/2)/(-b*x+2)^(1/2)/(-b*x+3)^(1/2)*ln((-5/2*b+b^2*x)/(b^2)^(1/2)+(b^2*x^2-5*b*x+6)^(1/2))
/(b^2)^(1/2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 30 vs. \(2 (14) = 28\).

Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.88 \[ \int \frac {1}{\sqrt {2-b x} \sqrt {3-b x}} \, dx=-\frac {\log \left (-2 \, b x + 2 \, \sqrt {-b x + 3} \sqrt {-b x + 2} + 5\right )}{b} \]

[In]

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

[Out]

-log(-2*b*x + 2*sqrt(-b*x + 3)*sqrt(-b*x + 2) + 5)/b

Sympy [F]

\[ \int \frac {1}{\sqrt {2-b x} \sqrt {3-b x}} \, dx=\int \frac {1}{\sqrt {- b x + 2} \sqrt {- b x + 3}}\, dx \]

[In]

integrate(1/(-b*x+2)**(1/2)/(-b*x+3)**(1/2),x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 33 vs. \(2 (14) = 28\).

Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 2.06 \[ \int \frac {1}{\sqrt {2-b x} \sqrt {3-b x}} \, dx=\frac {\log \left (2 \, b^{2} x + 2 \, \sqrt {b^{2} x^{2} - 5 \, b x + 6} b - 5 \, b\right )}{b} \]

[In]

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

[Out]

log(2*b^2*x + 2*sqrt(b^2*x^2 - 5*b*x + 6)*b - 5*b)/b

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56 \[ \int \frac {1}{\sqrt {2-b x} \sqrt {3-b x}} \, dx=\frac {2 \, \log \left (\sqrt {-b x + 3} - \sqrt {-b x + 2}\right )}{b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 49, normalized size of antiderivative = 3.06 \[ \int \frac {1}{\sqrt {2-b x} \sqrt {3-b x}} \, dx=\frac {4\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {3}-\sqrt {3-b\,x}\right )}{\left (\sqrt {2}-\sqrt {2-b\,x}\right )\,\sqrt {-b^2}}\right )}{\sqrt {-b^2}} \]

[In]

int(1/((2 - b*x)^(1/2)*(3 - b*x)^(1/2)),x)

[Out]

(4*atan((b*(3^(1/2) - (3 - b*x)^(1/2)))/((2^(1/2) - (2 - b*x)^(1/2))*(-b^2)^(1/2))))/(-b^2)^(1/2)