\(\int \sqrt {x} \sqrt {2-b x} \, dx\) [515]

Optimal result

Integrand size = 16, antiderivative size = 65 \[ \int \sqrt {x} \sqrt {2-b x} \, dx=-\frac {\sqrt {x} \sqrt {2-b x}}{2 b}+\frac {1}{2} x^{3/2} \sqrt {2-b x}+\frac {\arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 65, 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} \[ \int \sqrt {x} \sqrt {2-b x} \, dx=\frac {\arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}+\frac {1}{2} x^{3/2} \sqrt {2-b x}-\frac {\sqrt {x} \sqrt {2-b x}}{2 b} \]

[In]

Int[Sqrt[x]*Sqrt[2 - b*x],x]

[Out]

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int \sqrt {x} \sqrt {2-b x} \, dx=\frac {\sqrt {x} \sqrt {2-b x} (-1+b x)}{2 b}-\frac {2 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{b^{3/2}} \]

[In]

Integrate[Sqrt[x]*Sqrt[2 - b*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.12

method	result	size
meijerg	\(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {3}{2}} \left (-3 b x +3\right ) \sqrt {-\frac {b x}{2}+1}}{6 b}-\frac {\sqrt {\pi }\, \left (-b \right )^{\frac {3}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {3}{2}}}}{\sqrt {-b}\, \sqrt {\pi }\, b}\)	\(73\)
default	\(-\frac {\sqrt {x}\, \left (-b x +2\right )^{\frac {3}{2}}}{2 b}+\frac {\sqrt {x}\, \sqrt {-b x +2}+\frac {\sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right )}{\sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}}{2 b}\)	\(85\)
risch	\(-\frac {\left (b x -1\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{2 b \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {\arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{2 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +2}}\)	\(98\)

[In]

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

[Out]

2/(-b)^(1/2)/Pi^(1/2)/b*(1/12*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(3/2)*(-3*b*x+3)/b*(-1/2*b*x+1)^(1/2)-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 = 0.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.65 \[ \int \sqrt {x} \sqrt {2-b x} \, dx=\left [\frac {{\left (b^{2} x - b\right )} \sqrt {-b x + 2} \sqrt {x} - \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{2 \, b^{2}}, \frac {{\left (b^{2} x - b\right )} \sqrt {-b x + 2} \sqrt {x} - 2 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{2 \, b^{2}}\right ] \]

[In]

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

[Out]

[1/2*((b^2*x - b)*sqrt(-b*x + 2)*sqrt(x) - sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1))/b^2, 1/2*
((b^2*x - b)*sqrt(-b*x + 2)*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 = 3.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.38 \[ \int \sqrt {x} \sqrt {2-b x} \, dx=\begin {cases} \frac {i b x^{\frac {5}{2}}}{2 \sqrt {b x - 2}} - \frac {3 i x^{\frac {3}{2}}}{2 \sqrt {b x - 2}} + \frac {i \sqrt {x}}{b \sqrt {b x - 2}} - \frac {i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {b x^{\frac {5}{2}}}{2 \sqrt {- b x + 2}} + \frac {3 x^{\frac {3}{2}}}{2 \sqrt {- b x + 2}} - \frac {\sqrt {x}}{b \sqrt {- b x + 2}} + \frac {\operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((I*b*x**(5/2)/(2*sqrt(b*x - 2)) - 3*I*x**(3/2)/(2*sqrt(b*x - 2)) + I*sqrt(x)/(b*sqrt(b*x - 2)) - I*a
cosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(3/2), Abs(b*x) > 2), (-b*x**(5/2)/(2*sqrt(-b*x + 2)) + 3*x**(3/2)/(2*sqrt(
-b*x + 2)) - sqrt(x)/(b*sqrt(-b*x + 2)) + asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(3/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.25 \[ \int \sqrt {x} \sqrt {2-b x} \, dx=\frac {\frac {\sqrt {-b x + 2} b}{\sqrt {x}} - \frac {{\left (-b x + 2\right )}^{\frac {3}{2}}}{x^{\frac {3}{2}}}}{b^{3} - \frac {2 \, {\left (b x - 2\right )} b^{2}}{x} + \frac {{\left (b x - 2\right )}^{2} b}{x^{2}}} - \frac {\arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (46) = 92\).

Time = 12.15 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.29 \[ \int \sqrt {x} \sqrt {2-b x} \, dx=\frac {\frac {{\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} {\left (b x + 3\right )} \sqrt {-b x + 2} - \frac {6 \, b \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}}\right )} {\left | b \right |}}{b^{2}} + \frac {4 \, {\left (\frac {2 \, b \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}} - \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2}\right )} {\left | b \right |}}{b^{2}}}{2 \, b} \]

[In]

integrate(x^(1/2)*(-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^2 + 4*(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^2)/b

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82 \[ \int \sqrt {x} \sqrt {2-b x} \, dx=\sqrt {x}\,\left (\frac {x}{2}-\frac {1}{2\,b}\right )\,\sqrt {2-b\,x}-\frac {\ln \left (\sqrt {-b}\,\sqrt {x}\,\sqrt {2-b\,x}-b\,x+1\right )}{2\,{\left (-b\right )}^{3/2}} \]

[In]

int(x^(1/2)*(2 - b*x)^(1/2),x)

[Out]

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