\(\int \frac {\sqrt {x}}{(2-b x)^{3/2}} \, dx\) [637]

Optimal result

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

[Out]

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

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 = {49, 56, 222} \[ \int \frac {\sqrt {x}}{(2-b x)^{3/2}} \, dx=\frac {2 \sqrt {x}}{b \sqrt {2-b x}}-\frac {2 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \]

[In]

Int[Sqrt[x]/(2 - b*x)^(3/2),x]

[Out]

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

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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 {2 \sqrt {x}}{b \sqrt {2-b x}}-\frac {\int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{b} \\ & = \frac {2 \sqrt {x}}{b \sqrt {2-b x}}-\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{b} \\ & = \frac {2 \sqrt {x}}{b \sqrt {2-b x}}-\frac {2 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.33 \[ \int \frac {\sqrt {x}}{(2-b x)^{3/2}} \, dx=\frac {2 \sqrt {x}}{b \sqrt {2-b x}}+\frac {4 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{b^{3/2}} \]

[In]

Integrate[Sqrt[x]/(2 - b*x)^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.49

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

[In]

int(x^(1/2)/(-b*x+2)^(3/2),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 = 122, normalized size of antiderivative = 2.71 \[ \int \frac {\sqrt {x}}{(2-b x)^{3/2}} \, dx=\left [-\frac {{\left (b x - 2\right )} \sqrt {-b} \log \left (-b x - \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right ) + 2 \, \sqrt {-b x + 2} b \sqrt {x}}{b^{3} x - 2 \, b^{2}}, \frac {2 \, {\left ({\left (b x - 2\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) - \sqrt {-b x + 2} b \sqrt {x}\right )}}{b^{3} x - 2 \, b^{2}}\right ] \]

[In]

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

[Out]

[-((b*x - 2)*sqrt(-b)*log(-b*x - sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) + 2*sqrt(-b*x + 2)*b*sqrt(x))/(b^3*x - 2
*b^2), 2*((b*x - 2)*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) - sqrt(-b*x + 2)*b*sqrt(x))/(b^3*x - 2*b^
2)]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.67 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.00 \[ \int \frac {\sqrt {x}}{(2-b x)^{3/2}} \, dx=\begin {cases} - \frac {2 i \sqrt {x}}{b \sqrt {b x - 2}} + \frac {2 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\\frac {2 \sqrt {x}}{b \sqrt {- b x + 2}} - \frac {2 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-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), (2*s
qrt(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 = 38, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {x}}{(2-b x)^{3/2}} \, dx=\frac {2 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {3}{2}}} + \frac {2 \, \sqrt {x}}{\sqrt {-b x + 2} b} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (34) = 68\).

Time = 1.53 (sec) , antiderivative size = 92, normalized size of antiderivative = 2.04 \[ \int \frac {\sqrt {x}}{(2-b x)^{3/2}} \, dx=-\frac {{\left (\frac {\log \left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2}\right )}{\sqrt {-b}} + \frac {8 \, \sqrt {-b}}{{\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b}\right )} {\left | b \right |}}{b^{2}} \]

[In]

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

[Out]

-(log((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2)/sqrt(-b) + 8*sqrt(-b)/((sqrt(-b*x + 2)*sqrt(-b) -
sqrt((b*x - 2)*b + 2*b))^2 - 2*b))*abs(b)/b^2

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{(2-b x)^{3/2}} \, dx=\int \frac {\sqrt {x}}{{\left (2-b\,x\right )}^{3/2}} \,d x \]

[In]

int(x^(1/2)/(2 - b*x)^(3/2),x)

[Out]

int(x^(1/2)/(2 - b*x)^(3/2), x)