\(\int \frac {1}{\sqrt {x} \sqrt {1-b x}} \, dx\) [649]

Optimal result

Integrand size = 16, antiderivative size = 19 \[ \int \frac {1}{\sqrt {x} \sqrt {1-b x}} \, dx=\frac {2 \arcsin \left (\sqrt {b} \sqrt {x}\right )}{\sqrt {b}} \]

[Out]

2*arcsin(b^(1/2)*x^(1/2))/b^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {56, 222} \[ \int \frac {1}{\sqrt {x} \sqrt {1-b x}} \, dx=\frac {2 \arcsin \left (\sqrt {b} \sqrt {x}\right )}{\sqrt {b}} \]

[In]

Int[1/(Sqrt[x]*Sqrt[1 - b*x]),x]

[Out]

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

Rule 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 222

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.74 \[ \int \frac {1}{\sqrt {x} \sqrt {1-b x}} \, dx=\frac {4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{-1+\sqrt {1-b x}}\right )}{\sqrt {b}} \]

[In]

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

[Out]

(4*ArcTan[(Sqrt[b]*Sqrt[x])/(-1 + Sqrt[1 - b*x])])/Sqrt[b]

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
meijerg	\(\frac {2 \arcsin \left (\sqrt {b}\, \sqrt {x}\right )}{\sqrt {b}}\)	\(14\)
default	\(\frac {\sqrt {x \left (-b x +1\right )}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{2 b}\right )}{\sqrt {-b \,x^{2}+x}}\right )}{\sqrt {x}\, \sqrt {-b x +1}\, \sqrt {b}}\)	\(48\)

[In]

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

[Out]

2*arcsin(b^(1/2)*x^(1/2))/b^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 57, normalized size of antiderivative = 3.00 \[ \int \frac {1}{\sqrt {x} \sqrt {1-b x}} \, dx=\left [-\frac {\sqrt {-b} \log \left (-2 \, b x + 2 \, \sqrt {-b x + 1} \sqrt {-b} \sqrt {x} + 1\right )}{b}, -\frac {2 \, \arctan \left (\frac {\sqrt {-b x + 1}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}}\right ] \]

[In]

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

[Out]

[-sqrt(-b)*log(-2*b*x + 2*sqrt(-b*x + 1)*sqrt(-b)*sqrt(x) + 1)/b, -2*arctan(sqrt(-b*x + 1)/(sqrt(b)*sqrt(x)))/
sqrt(b)]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.68 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.21 \[ \int \frac {1}{\sqrt {x} \sqrt {1-b x}} \, dx=\begin {cases} - \frac {2 i \operatorname {acosh}{\left (\sqrt {b} \sqrt {x} \right )}}{\sqrt {b}} & \text {for}\: \left |{b x}\right | > 1 \\\frac {2 \operatorname {asin}{\left (\sqrt {b} \sqrt {x} \right )}}{\sqrt {b}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.11 \[ \int \frac {1}{\sqrt {x} \sqrt {1-b x}} \, dx=-\frac {2 \, \arctan \left (\frac {\sqrt {-b x + 1}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (13) = 26\).

Time = 5.68 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.16 \[ \int \frac {1}{\sqrt {x} \sqrt {1-b x}} \, dx=\frac {2 \, b \log \left ({\left | -\sqrt {-b x + 1} \sqrt {-b} + \sqrt {{\left (b x - 1\right )} b + b} \right |}\right )}{\sqrt {-b} {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\sqrt {x} \sqrt {1-b x}} \, dx=-\frac {4\,\mathrm {atan}\left (\frac {\sqrt {1-b\,x}-1}{\sqrt {b}\,\sqrt {x}}\right )}{\sqrt {b}} \]

[In]

int(1/(x^(1/2)*(1 - b*x)^(1/2)),x)

[Out]

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