\(\int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx\) [631]

Optimal result

Integrand size = 16, antiderivative size = 45 \[ \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx=-\frac {\sqrt {x} \sqrt {2-b x}}{b}+\frac {2 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \]

[Out]

2*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(3/2)-x^(1/2)*(-b*x+2)^(1/2)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, 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} \[ \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx=\frac {2 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}-\frac {\sqrt {x} \sqrt {2-b x}}{b} \]

[In]

Int[Sqrt[x]/Sqrt[2 - b*x],x]

[Out]

-((Sqrt[x]*Sqrt[2 - b*x])/b) + (2*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(3/2)

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}& = -\frac {\sqrt {x} \sqrt {2-b x}}{b}+\frac {\int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{b} \\ & = -\frac {\sqrt {x} \sqrt {2-b x}}{b}+\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{b} \\ & = -\frac {\sqrt {x} \sqrt {2-b x}}{b}+\frac {2 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.33 \[ \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx=-\frac {\sqrt {x} \sqrt {2-b x}}{b}-\frac {4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{b^{3/2}} \]

[In]

Integrate[Sqrt[x]/Sqrt[2 - b*x],x]

[Out]

-((Sqrt[x]*Sqrt[2 - b*x])/b) - (4*ArcTan[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 - b*x])])/b^(3/2)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.47

method	result	size
meijerg	\(-\frac {2 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {3}{2}} \sqrt {-\frac {b x}{2}+1}}{2 b}+\frac {\sqrt {\pi }\, \left (-b \right )^{\frac {3}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {3}{2}}}\right )}{\sqrt {-b}\, \sqrt {\pi }\, b}\)	\(66\)
default	\(-\frac {\sqrt {x}\, \sqrt {-b x +2}}{b}+\frac {\arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +2}}\)	\(67\)
risch	\(\frac {\sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{b \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {\arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +2}}\)	\(91\)

[In]

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

[Out]

-2/(-b)^(1/2)/Pi^(1/2)/b*(-1/2*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(3/2)/b*(-1/2*b*x+1)^(1/2)+Pi^(1/2)*(-b)^(3/2)/b^
(3/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.00 \[ \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx=\left [-\frac {\sqrt {-b x + 2} b \sqrt {x} + \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{b^{2}}, -\frac {\sqrt {-b x + 2} b \sqrt {x} + 2 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{2}}\right ] \]

[In]

integrate(x^(1/2)/(-b*x+2)^(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^2, -(sqrt(-b*x + 2)*
b*sqrt(x) + 2*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))))/b^2]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.67 (sec) , antiderivative size = 119, normalized size of antiderivative = 2.64 \[ \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx=\begin {cases} - \frac {i x^{\frac {3}{2}}}{\sqrt {b x - 2}} + \frac {2 i \sqrt {x}}{b \sqrt {b x - 2}} - \frac {2 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\\frac {x^{\frac {3}{2}}}{\sqrt {- b x + 2}} - \frac {2 \sqrt {x}}{b \sqrt {- b x + 2}} + \frac {2 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.16 \[ \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx=-\frac {2 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {3}{2}}} - \frac {2 \, \sqrt {-b x + 2}}{{\left (b^{2} - \frac {{\left (b x - 2\right )} b}{x}\right )} \sqrt {x}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 69 vs. \(2 (34) = 68\).

Time = 6.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.53 \[ \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx=\frac {{\left (\frac {2 \, b \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}} - \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2}\right )} {\left | b \right |}}{b^{3}} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx=-\frac {4\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {2}-\sqrt {2-b\,x}}\right )}{b^{3/2}}-\frac {\sqrt {x}\,\sqrt {2-b\,x}}{b} \]

[In]

int(x^(1/2)/(2 - b*x)^(1/2),x)

[Out]

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