\(\int x^{3/2} \sqrt {2-b x} \, dx\) [514]

Optimal result

Integrand size = 16, antiderivative size = 87 \[ \int x^{3/2} \sqrt {2-b x} \, dx=-\frac {\sqrt {x} \sqrt {2-b x}}{2 b^2}-\frac {x^{3/2} \sqrt {2-b x}}{6 b}+\frac {1}{3} x^{5/2} \sqrt {2-b x}+\frac {\arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}} \]

[Out]

arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(5/2)-1/6*x^(3/2)*(-b*x+2)^(1/2)/b+1/3*x^(5/2)*(-b*x+2)^(1/2)-1/2*x^(1/2
)*(-b*x+2)^(1/2)/b^2

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {52, 56, 222} \[ \int x^{3/2} \sqrt {2-b x} \, dx=\frac {\arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}}-\frac {\sqrt {x} \sqrt {2-b x}}{2 b^2}+\frac {1}{3} x^{5/2} \sqrt {2-b x}-\frac {x^{3/2} \sqrt {2-b x}}{6 b} \]

[In]

Int[x^(3/2)*Sqrt[2 - b*x],x]

[Out]

-1/2*(Sqrt[x]*Sqrt[2 - b*x])/b^2 - (x^(3/2)*Sqrt[2 - b*x])/(6*b) + (x^(5/2)*Sqrt[2 - b*x])/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 {1}{3} x^{5/2} \sqrt {2-b x}+\frac {1}{3} \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx \\ & = -\frac {x^{3/2} \sqrt {2-b x}}{6 b}+\frac {1}{3} x^{5/2} \sqrt {2-b x}+\frac {\int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{2 b} \\ & = -\frac {\sqrt {x} \sqrt {2-b x}}{2 b^2}-\frac {x^{3/2} \sqrt {2-b x}}{6 b}+\frac {1}{3} x^{5/2} \sqrt {2-b x}+\frac {\int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{2 b^2} \\ & = -\frac {\sqrt {x} \sqrt {2-b x}}{2 b^2}-\frac {x^{3/2} \sqrt {2-b x}}{6 b}+\frac {1}{3} x^{5/2} \sqrt {2-b x}+\frac {\text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{b^2} \\ & = -\frac {\sqrt {x} \sqrt {2-b x}}{2 b^2}-\frac {x^{3/2} \sqrt {2-b x}}{6 b}+\frac {1}{3} x^{5/2} \sqrt {2-b x}+\frac {\sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.87 \[ \int x^{3/2} \sqrt {2-b x} \, dx=\frac {\sqrt {x} \sqrt {2-b x} \left (-3-b x+2 b^2 x^2\right )}{6 b^2}-\frac {2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{b^{5/2}} \]

[In]

Integrate[x^(3/2)*Sqrt[2 - b*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.93

method	result	size
meijerg	\(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {5}{2}} \left (-10 b^{2} x^{2}+5 b x +15\right ) \sqrt {-\frac {b x}{2}+1}}{30 b^{2}}-\frac {\sqrt {\pi }\, \left (-b \right )^{\frac {5}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {5}{2}}}}{\left (-b \right )^{\frac {3}{2}} \sqrt {\pi }\, b}\)	\(81\)
default	\(-\frac {x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {3}{2}}}{3 b}+\frac {-\frac {\sqrt {x}\, \left (-b x +2\right )^{\frac {3}{2}}}{2 b}+\frac {\sqrt {x}\, \sqrt {-b x +2}+\frac {\sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{\sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}}{2 b}}{b}\)	\(106\)
risch	\(-\frac {\left (2 b^{2} x^{2}-b x -3\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{6 b^{2} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {\arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{2 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {-b x +2}}\)	\(107\)

[In]

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

[Out]

4/(-b)^(3/2)/Pi^(1/2)/b*(1/120*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(5/2)*(-10*b^2*x^2+5*b*x+15)/b^2*(-1/2*b*x+1)^(1/
2)-1/4*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 = 0.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.44 \[ \int x^{3/2} \sqrt {2-b x} \, dx=\left [\frac {{\left (2 \, b^{3} x^{2} - b^{2} x - 3 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 3 \, \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{6 \, b^{3}}, \frac {{\left (2 \, b^{3} x^{2} - b^{2} x - 3 \, b\right )} \sqrt {-b x + 2} \sqrt {x} - 6 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{6 \, b^{3}}\right ] \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 7.60 (sec) , antiderivative size = 194, normalized size of antiderivative = 2.23 \[ \int x^{3/2} \sqrt {2-b x} \, dx=\begin {cases} \frac {i b x^{\frac {7}{2}}}{3 \sqrt {b x - 2}} - \frac {5 i x^{\frac {5}{2}}}{6 \sqrt {b x - 2}} - \frac {i x^{\frac {3}{2}}}{6 b \sqrt {b x - 2}} + \frac {i \sqrt {x}}{b^{2} \sqrt {b x - 2}} - \frac {i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {5}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {b x^{\frac {7}{2}}}{3 \sqrt {- b x + 2}} + \frac {5 x^{\frac {5}{2}}}{6 \sqrt {- b x + 2}} + \frac {x^{\frac {3}{2}}}{6 b \sqrt {- b x + 2}} - \frac {\sqrt {x}}{b^{2} \sqrt {- b x + 2}} + \frac {\operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((I*b*x**(7/2)/(3*sqrt(b*x - 2)) - 5*I*x**(5/2)/(6*sqrt(b*x - 2)) - I*x**(3/2)/(6*b*sqrt(b*x - 2)) +
I*sqrt(x)/(b**2*sqrt(b*x - 2)) - I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(5/2), Abs(b*x) > 2), (-b*x**(7/2)/(3*s
qrt(-b*x + 2)) + 5*x**(5/2)/(6*sqrt(-b*x + 2)) + x**(3/2)/(6*b*sqrt(-b*x + 2)) - sqrt(x)/(b**2*sqrt(-b*x + 2))
+ asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(5/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.34 \[ \int x^{3/2} \sqrt {2-b x} \, dx=\frac {\frac {3 \, \sqrt {-b x + 2} b^{2}}{\sqrt {x}} - \frac {8 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b}{x^{\frac {3}{2}}} - \frac {3 \, {\left (-b x + 2\right )}^{\frac {5}{2}}}{x^{\frac {5}{2}}}}{3 \, {\left (b^{5} - \frac {3 \, {\left (b x - 2\right )} b^{4}}{x} + \frac {3 \, {\left (b x - 2\right )}^{2} b^{3}}{x^{2}} - \frac {{\left (b x - 2\right )}^{3} b^{2}}{x^{3}}\right )}} - \frac {\arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {5}{2}}} \]

[In]

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

[Out]

1/3*(3*sqrt(-b*x + 2)*b^2/sqrt(x) - 8*(-b*x + 2)^(3/2)*b/x^(3/2) - 3*(-b*x + 2)^(5/2)/x^(5/2))/(b^5 - 3*(b*x -
2)*b^4/x + 3*(b*x - 2)^2*b^3/x^2 - (b*x - 2)^3*b^2/x^3) - arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(5/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (62) = 124\).

Time = 12.02 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.05 \[ \int x^{3/2} \sqrt {2-b x} \, dx=\frac {\frac {{\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} {\left ({\left (b x - 2\right )} {\left (\frac {2 \, {\left (b x - 2\right )}}{b^{2}} + \frac {13}{b^{2}}\right )} + \frac {33}{b^{2}}\right )} - \frac {30 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b}\right )} {\left | b \right |}}{b} - \frac {6 \, {\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} {\left (b x + 3\right )} \sqrt {-b x + 2} - \frac {6 \, b \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}}\right )} {\left | b \right |}}{b^{3}}}{6 \, b} \]

[In]

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

[Out]

1/6*((sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2)*((b*x - 2)*(2*(b*x - 2)/b^2 + 13/b^2) + 33/b^2) - 30*log(abs(-sqr
t(-b*x + 2)*sqrt(-b) + sqrt((b*x - 2)*b + 2*b)))/(sqrt(-b)*b))*abs(b)/b - 6*(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^3)/b

Mupad [F(-1)]

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

[In]

int(x^(3/2)*(2 - b*x)^(1/2),x)

[Out]

int(x^(3/2)*(2 - b*x)^(1/2), x)