3.513 \(\int x^{5/2} \sqrt{2-b x} \, dx\)

Optimal. Leaf size=112 \[ -\frac{5 x^{3/2} \sqrt{2-b x}}{24 b^2}-\frac{5 \sqrt{x} \sqrt{2-b x}}{8 b^3}+\frac{5 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{7/2}}+\frac{1}{4} x^{7/2} \sqrt{2-b x}-\frac{x^{5/2} \sqrt{2-b x}}{12 b} \]

[Out]

(-5*Sqrt[x]*Sqrt[2 - b*x])/(8*b^3) - (5*x^(3/2)*Sqrt[2 - b*x])/(24*b^2) - (x^(5/2)*Sqrt[2 - b*x])/(12*b) + (x^
(7/2)*Sqrt[2 - b*x])/4 + (5*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(4*b^(7/2))

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Rubi [A]  time = 0.0286051, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {50, 54, 216} \[ -\frac{5 x^{3/2} \sqrt{2-b x}}{24 b^2}-\frac{5 \sqrt{x} \sqrt{2-b x}}{8 b^3}+\frac{5 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{7/2}}+\frac{1}{4} x^{7/2} \sqrt{2-b x}-\frac{x^{5/2} \sqrt{2-b x}}{12 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*Sqrt[2 - b*x],x]

[Out]

(-5*Sqrt[x]*Sqrt[2 - b*x])/(8*b^3) - (5*x^(3/2)*Sqrt[2 - b*x])/(24*b^2) - (x^(5/2)*Sqrt[2 - b*x])/(12*b) + (x^
(7/2)*Sqrt[2 - b*x])/4 + (5*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(4*b^(7/2))

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 x^{5/2} \sqrt{2-b x} \, dx &=\frac{1}{4} x^{7/2} \sqrt{2-b x}+\frac{1}{4} \int \frac{x^{5/2}}{\sqrt{2-b x}} \, dx\\ &=-\frac{x^{5/2} \sqrt{2-b x}}{12 b}+\frac{1}{4} x^{7/2} \sqrt{2-b x}+\frac{5 \int \frac{x^{3/2}}{\sqrt{2-b x}} \, dx}{12 b}\\ &=-\frac{5 x^{3/2} \sqrt{2-b x}}{24 b^2}-\frac{x^{5/2} \sqrt{2-b x}}{12 b}+\frac{1}{4} x^{7/2} \sqrt{2-b x}+\frac{5 \int \frac{\sqrt{x}}{\sqrt{2-b x}} \, dx}{8 b^2}\\ &=-\frac{5 \sqrt{x} \sqrt{2-b x}}{8 b^3}-\frac{5 x^{3/2} \sqrt{2-b x}}{24 b^2}-\frac{x^{5/2} \sqrt{2-b x}}{12 b}+\frac{1}{4} x^{7/2} \sqrt{2-b x}+\frac{5 \int \frac{1}{\sqrt{x} \sqrt{2-b x}} \, dx}{8 b^3}\\ &=-\frac{5 \sqrt{x} \sqrt{2-b x}}{8 b^3}-\frac{5 x^{3/2} \sqrt{2-b x}}{24 b^2}-\frac{x^{5/2} \sqrt{2-b x}}{12 b}+\frac{1}{4} x^{7/2} \sqrt{2-b x}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-b x^2}} \, dx,x,\sqrt{x}\right )}{4 b^3}\\ &=-\frac{5 \sqrt{x} \sqrt{2-b x}}{8 b^3}-\frac{5 x^{3/2} \sqrt{2-b x}}{24 b^2}-\frac{x^{5/2} \sqrt{2-b x}}{12 b}+\frac{1}{4} x^{7/2} \sqrt{2-b x}+\frac{5 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0450378, size = 71, normalized size = 0.63 \[ \frac{\sqrt{x} \sqrt{2-b x} \left (6 b^3 x^3-2 b^2 x^2-5 b x-15\right )}{24 b^3}+\frac{5 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*Sqrt[2 - b*x],x]

[Out]

(Sqrt[x]*Sqrt[2 - b*x]*(-15 - 5*b*x - 2*b^2*x^2 + 6*b^3*x^3))/(24*b^3) + (5*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])
/(4*b^(7/2))

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Maple [A]  time = 0.007, size = 116, normalized size = 1. \begin{align*} -{\frac{1}{4\,b}{x}^{{\frac{5}{2}}} \left ( -bx+2 \right ) ^{{\frac{3}{2}}}}-{\frac{5}{12\,{b}^{2}}{x}^{{\frac{3}{2}}} \left ( -bx+2 \right ) ^{{\frac{3}{2}}}}-{\frac{5}{8\,{b}^{3}} \left ( -bx+2 \right ) ^{{\frac{3}{2}}}\sqrt{x}}+{\frac{5}{8\,{b}^{3}}\sqrt{x}\sqrt{-bx+2}}+{\frac{5}{8}\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{7}{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^(5/2)*(-b*x+2)^(1/2),x)

[Out]

-1/4/b*x^(5/2)*(-b*x+2)^(3/2)-5/12/b^2*x^(3/2)*(-b*x+2)^(3/2)-5/8/b^3*x^(1/2)*(-b*x+2)^(3/2)+5/8*x^(1/2)*(-b*x
+2)^(1/2)/b^3+5/8/b^(7/2)*((-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^(5/2)*(-b*x+2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51176, size = 367, normalized size = 3.28 \begin{align*} \left [\frac{{\left (6 \, b^{4} x^{3} - 2 \, b^{3} x^{2} - 5 \, b^{2} x - 15 \, b\right )} \sqrt{-b x + 2} \sqrt{x} - 15 \, \sqrt{-b} \log \left (-b x + \sqrt{-b x + 2} \sqrt{-b} \sqrt{x} + 1\right )}{24 \, b^{4}}, \frac{{\left (6 \, b^{4} x^{3} - 2 \, b^{3} x^{2} - 5 \, b^{2} x - 15 \, b\right )} \sqrt{-b x + 2} \sqrt{x} - 30 \, \sqrt{b} \arctan \left (\frac{\sqrt{-b x + 2}}{\sqrt{b} \sqrt{x}}\right )}{24 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(-b*x+2)^(1/2),x, algorithm="fricas")

[Out]

[1/24*((6*b^4*x^3 - 2*b^3*x^2 - 5*b^2*x - 15*b)*sqrt(-b*x + 2)*sqrt(x) - 15*sqrt(-b)*log(-b*x + sqrt(-b*x + 2)
*sqrt(-b)*sqrt(x) + 1))/b^4, 1/24*((6*b^4*x^3 - 2*b^3*x^2 - 5*b^2*x - 15*b)*sqrt(-b*x + 2)*sqrt(x) - 30*sqrt(b
)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))))/b^4]

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Sympy [A]  time = 17.6091, size = 252, normalized size = 2.25 \begin{align*} \begin{cases} \frac{i b x^{\frac{9}{2}}}{4 \sqrt{b x - 2}} - \frac{7 i x^{\frac{7}{2}}}{12 \sqrt{b x - 2}} - \frac{i x^{\frac{5}{2}}}{24 b \sqrt{b x - 2}} - \frac{5 i x^{\frac{3}{2}}}{24 b^{2} \sqrt{b x - 2}} + \frac{5 i \sqrt{x}}{4 b^{3} \sqrt{b x - 2}} - \frac{5 i \operatorname{acosh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{4 b^{\frac{7}{2}}} & \text{for}\: \frac{\left |{b x}\right |}{2} > 1 \\- \frac{b x^{\frac{9}{2}}}{4 \sqrt{- b x + 2}} + \frac{7 x^{\frac{7}{2}}}{12 \sqrt{- b x + 2}} + \frac{x^{\frac{5}{2}}}{24 b \sqrt{- b x + 2}} + \frac{5 x^{\frac{3}{2}}}{24 b^{2} \sqrt{- b x + 2}} - \frac{5 \sqrt{x}}{4 b^{3} \sqrt{- b x + 2}} + \frac{5 \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{4 b^{\frac{7}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(-b*x+2)**(1/2),x)

[Out]

Piecewise((I*b*x**(9/2)/(4*sqrt(b*x - 2)) - 7*I*x**(7/2)/(12*sqrt(b*x - 2)) - I*x**(5/2)/(24*b*sqrt(b*x - 2))
- 5*I*x**(3/2)/(24*b**2*sqrt(b*x - 2)) + 5*I*sqrt(x)/(4*b**3*sqrt(b*x - 2)) - 5*I*acosh(sqrt(2)*sqrt(b)*sqrt(x
)/2)/(4*b**(7/2)), Abs(b*x)/2 > 1), (-b*x**(9/2)/(4*sqrt(-b*x + 2)) + 7*x**(7/2)/(12*sqrt(-b*x + 2)) + x**(5/2
)/(24*b*sqrt(-b*x + 2)) + 5*x**(3/2)/(24*b**2*sqrt(-b*x + 2)) - 5*sqrt(x)/(4*b**3*sqrt(-b*x + 2)) + 5*asin(sqr
t(2)*sqrt(b)*sqrt(x)/2)/(4*b**(7/2)), 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^(5/2)*(-b*x+2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError