3.515 \(\int \sqrt{x} \sqrt{2-b x} \, dx\)

Optimal. Leaf size=65 \[ \frac{\sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{3/2}}+\frac{1}{2} x^{3/2} \sqrt{2-b x}-\frac{\sqrt{x} \sqrt{2-b x}}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.0149726, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {50, 54, 216} \[ \frac{\sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{3/2}}+\frac{1}{2} x^{3/2} \sqrt{2-b x}-\frac{\sqrt{x} \sqrt{2-b x}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*Sqrt[2 - b*x],x]

[Out]

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

Rule 50

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 54

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 216

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*} \int \sqrt{x} \sqrt{2-b x} \, dx &=\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{\operatorname{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]  time = 0.037953, size = 51, normalized size = 0.78 \[ \frac{\sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{3/2}}+\frac{\sqrt{x} \sqrt{2-b x} (b x-1)}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*Sqrt[2 - b*x],x]

[Out]

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

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Maple [A]  time = 0.002, size = 81, normalized size = 1.3 \begin{align*}{\frac{1}{2}{x}^{{\frac{3}{2}}}\sqrt{-bx+2}}-{\frac{1}{2\,b}\sqrt{x}\sqrt{-bx+2}}+{\frac{1}{2}\sqrt{ \left ( -bx+2 \right ) x}\arctan \left ({\sqrt{b} \left ( x-{b}^{-1} \right ){\frac{1}{\sqrt{-b{x}^{2}+2\,x}}}} \right ){b}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{-bx+2}}}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60081, size = 281, normalized size = 4.32 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]

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Sympy [A]  time = 3.31303, size = 156, normalized size = 2.4 \begin{align*} \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}\: \frac{\left |{b x}\right |}{2} > 1 \\- \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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 > 1), (-b*x**(5/2)/(2*sqrt(-b*x + 2)) + 3*x**(3/2)/(2*sqr
t(-b*x + 2)) - sqrt(x)/(b*sqrt(-b*x + 2)) + asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(3/2), True))

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError