\(\int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx\) [630]

Optimal result

Integrand size = 16, antiderivative size = 69 \[ \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx=-\frac {3 \sqrt {x} \sqrt {2-b x}}{2 b^2}-\frac {x^{3/2} \sqrt {2-b x}}{2 b}+\frac {3 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}} \]

[Out]

3*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(5/2)-1/2*x^(3/2)*(-b*x+2)^(1/2)/b-3/2*x^(1/2)*(-b*x+2)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {52, 56, 222} \[ \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx=\frac {3 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}-\frac {3 \sqrt {x} \sqrt {2-b x}}{2 b^2}-\frac {x^{3/2} \sqrt {2-b x}}{2 b} \]

[In]

Int[x^(3/2)/Sqrt[2 - b*x],x]

[Out]

(-3*Sqrt[x]*Sqrt[2 - b*x])/(2*b^2) - (x^(3/2)*Sqrt[2 - b*x])/(2*b) + (3*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(
5/2)

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 {x^{3/2} \sqrt {2-b x}}{2 b}+\frac {3 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{2 b} \\ & = -\frac {3 \sqrt {x} \sqrt {2-b x}}{2 b^2}-\frac {x^{3/2} \sqrt {2-b x}}{2 b}+\frac {3 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{2 b^2} \\ & = -\frac {3 \sqrt {x} \sqrt {2-b x}}{2 b^2}-\frac {x^{3/2} \sqrt {2-b x}}{2 b}+\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{b^2} \\ & = -\frac {3 \sqrt {x} \sqrt {2-b x}}{2 b^2}-\frac {x^{3/2} \sqrt {2-b x}}{2 b}+\frac {3 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97 \[ \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx=-\frac {\sqrt {x} \sqrt {2-b x} (3+b x)}{2 b^2}-\frac {6 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{b^{5/2}} \]

[In]

Integrate[x^(3/2)/Sqrt[2 - b*x],x]

[Out]

-1/2*(Sqrt[x]*Sqrt[2 - b*x]*(3 + b*x))/b^2 - (6*ArcTan[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 - b*x])])/b^(5/2)

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.06

method	result	size
meijerg	\(-\frac {4 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {5}{2}} \left (5 b x +15\right ) \sqrt {-\frac {b x}{2}+1}}{40 b^{2}}+\frac {3 \sqrt {\pi }\, \left (-b \right )^{\frac {5}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{4 b^{\frac {5}{2}}}\right )}{\left (-b \right )^{\frac {3}{2}} \sqrt {\pi }\, b}\)	\(73\)
default	\(-\frac {x^{\frac {3}{2}} \sqrt {-b x +2}}{2 b}+\frac {-\frac {3 \sqrt {x}\, \sqrt {-b x +2}}{2 b}+\frac {3 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{2 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +2}}}{b}\)	\(89\)
risch	\(\frac {\left (b x +3\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{2 b^{2} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {3 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{2 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {-b x +2}}\)	\(98\)

[In]

int(x^(3/2)/(-b*x+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-4/(-b)^(3/2)/Pi^(1/2)/b*(-1/40*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(5/2)*(5*b*x+15)/b^2*(-1/2*b*x+1)^(1/2)+3/4*Pi^(
1/2)*(-b)^(5/2)/b^(5/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.55 \[ \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx=\left [-\frac {{\left (b^{2} x + 3 \, b\right )} \sqrt {-b x + 2} \sqrt {x} + 3 \, \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{2 \, b^{3}}, -\frac {{\left (b^{2} x + 3 \, b\right )} \sqrt {-b x + 2} \sqrt {x} + 6 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{2 \, b^{3}}\right ] \]

[In]

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

[Out]

[-1/2*((b^2*x + 3*b)*sqrt(-b*x + 2)*sqrt(x) + 3*sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1))/b^3,
-1/2*((b^2*x + 3*b)*sqrt(-b*x + 2)*sqrt(x) + 6*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))))/b^3]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.64 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.35 \[ \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx=\begin {cases} - \frac {i x^{\frac {5}{2}}}{2 \sqrt {b x - 2}} - \frac {i x^{\frac {3}{2}}}{2 b \sqrt {b x - 2}} + \frac {3 i \sqrt {x}}{b^{2} \sqrt {b x - 2}} - \frac {3 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {5}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\\frac {x^{\frac {5}{2}}}{2 \sqrt {- b x + 2}} + \frac {x^{\frac {3}{2}}}{2 b \sqrt {- b x + 2}} - \frac {3 \sqrt {x}}{b^{2} \sqrt {- b x + 2}} + \frac {3 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-I*x**(5/2)/(2*sqrt(b*x - 2)) - I*x**(3/2)/(2*b*sqrt(b*x - 2)) + 3*I*sqrt(x)/(b**2*sqrt(b*x - 2)) -
3*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(5/2), Abs(b*x) > 2), (x**(5/2)/(2*sqrt(-b*x + 2)) + x**(3/2)/(2*b*sq
rt(-b*x + 2)) - 3*sqrt(x)/(b**2*sqrt(-b*x + 2)) + 3*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.23 \[ \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx=-\frac {\frac {5 \, \sqrt {-b x + 2} b}{\sqrt {x}} + \frac {3 \, {\left (-b x + 2\right )}^{\frac {3}{2}}}{x^{\frac {3}{2}}}}{b^{4} - \frac {2 \, {\left (b x - 2\right )} b^{3}}{x} + \frac {{\left (b x - 2\right )}^{2} b^{2}}{x^{2}}} - \frac {3 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {5}{2}}} \]

[In]

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

[Out]

-(5*sqrt(-b*x + 2)*b/sqrt(x) + 3*(-b*x + 2)^(3/2)/x^(3/2))/(b^4 - 2*(b*x - 2)*b^3/x + (b*x - 2)^2*b^2/x^2) - 3
*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(5/2)

Giac [A] (verification not implemented)

none

Time = 5.92 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07 \[ \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx=-\frac {{\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} {\left (b x + 3\right )} \sqrt {-b x + 2} - \frac {6 \, b \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}}\right )} {\left | b \right |}}{2 \, b^{4}} \]

[In]

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

[Out]

-1/2*(sqrt((b*x - 2)*b + 2*b)*(b*x + 3)*sqrt(-b*x + 2) - 6*b*log(abs(-sqrt(-b*x + 2)*sqrt(-b) + sqrt((b*x - 2)
*b + 2*b)))/sqrt(-b))*abs(b)/b^4

Mupad [F(-1)]

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

[In]

int(x^(3/2)/(2 - b*x)^(1/2),x)

[Out]

int(x^(3/2)/(2 - b*x)^(1/2), x)