\(\int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx\) [632]

Optimal result

Integrand size = 16, antiderivative size = 24 \[ \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx=\frac {2 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, 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} \[ \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx=\frac {2 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \]

[In]

Int[1/(Sqrt[x]*Sqrt[2 - b*x]),x]

[Out]

(2*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/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 {2-b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = \frac {2 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.62 \[ \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx=-\frac {4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{\sqrt {b}} \]

[In]

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

[Out]

(-4*ArcTan[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 - b*x])])/Sqrt[b]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75

method	result	size
meijerg	\(\frac {2 \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{\sqrt {b}}\)	\(18\)
default	\(\frac {\sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{\sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx=\left [-\frac {\sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{b}, -\frac {2 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}}\right ] \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.33 \[ \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx=\begin {cases} - \frac {2 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} & \text {for}\: \left |{b x}\right | > 2 \\\frac {2 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx=-\frac {2 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 43 vs. \(2 (17) = 34\).

Time = 6.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx=\frac {2 \, b \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx=\frac {4\,\mathrm {atan}\left (\frac {\sqrt {2}-\sqrt {2-b\,x}}{\sqrt {b}\,\sqrt {x}}\right )}{\sqrt {b}} \]

[In]

int(1/(x^(1/2)*(2 - b*x)^(1/2)),x)

[Out]

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