\(\int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx\) [516]

Optimal result

Integrand size = 16, antiderivative size = 41 \[ \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx=\sqrt {x} \sqrt {2-b x}+\frac {2 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {52, 56, 222} \[ \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx=\frac {2 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}}+\sqrt {x} \sqrt {2-b x} \]

[In]

Int[Sqrt[2 - b*x]/Sqrt[x],x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.37 \[ \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx=\sqrt {x} \sqrt {2-b x}-\frac {4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{\sqrt {b}} \]

[In]

Integrate[Sqrt[2 - b*x]/Sqrt[x],x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(62\) vs. \(2(30)=60\).

Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.54

method	result	size
default	\(\sqrt {x}\, \sqrt {-b x +2}+\frac {\sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{\sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(63\)
meijerg	\(\frac {\sqrt {-b}\, \left (-\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {-b}\, \sqrt {-\frac {b x}{2}+1}-\frac {2 \sqrt {\pi }\, \sqrt {-b}\, \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{\sqrt {b}}\right )}{\sqrt {\pi }\, b}\)	\(63\)
risch	\(-\frac {\sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{\sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {\sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{\sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(89\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 89, normalized size of antiderivative = 2.17 \[ \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx=\left [\frac {\sqrt {-b x + 2} b \sqrt {x} - \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{b}, \frac {\sqrt {-b x + 2} b \sqrt {x} - 2 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b}\right ] \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.90 \[ \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx=\begin {cases} \frac {i b x^{\frac {3}{2}}}{\sqrt {b x - 2}} - \frac {2 i \sqrt {x}}{\sqrt {b x - 2}} - \frac {2 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {b x^{\frac {3}{2}}}{\sqrt {- b x + 2}} + \frac {2 \sqrt {x}}{\sqrt {- b x + 2}} + \frac {2 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((I*b*x**(3/2)/sqrt(b*x - 2) - 2*I*sqrt(x)/sqrt(b*x - 2) - 2*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/sqrt(
b), Abs(b*x) > 2), (-b*x**(3/2)/sqrt(-b*x + 2) + 2*sqrt(x)/sqrt(-b*x + 2) + 2*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/
sqrt(b), True))

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx=-\frac {2 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{\sqrt {b}} + \frac {2 \, \sqrt {-b x + 2}}{{\left (b - \frac {b x - 2}{x}\right )} \sqrt {x}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (30) = 60\).

Time = 6.13 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.71 \[ \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx=\frac {b {\left (\frac {2 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}} + \frac {\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2}}{b}\right )}}{{\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.81 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx=\sqrt {x}\,\sqrt {2-b\,x}-\frac {4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {2}-\sqrt {2-b\,x}}\right )}{\sqrt {b}} \]

[In]

int((2 - b*x)^(1/2)/x^(1/2),x)

[Out]

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