3.540 \(\int x^{3/2} (2-b x)^{3/2} \, dx\)

Optimal. Leaf size=109 \[ -\frac{3 \sqrt{x} \sqrt{2-b x}}{8 b^2}+\frac{3 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{5/2}}+\frac{1}{4} x^{5/2} (2-b x)^{3/2}+\frac{1}{4} x^{5/2} \sqrt{2-b x}-\frac{x^{3/2} \sqrt{2-b x}}{8 b} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0279175, antiderivative size = 109, 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} \[ -\frac{3 \sqrt{x} \sqrt{2-b x}}{8 b^2}+\frac{3 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{5/2}}+\frac{1}{4} x^{5/2} (2-b x)^{3/2}+\frac{1}{4} x^{5/2} \sqrt{2-b x}-\frac{x^{3/2} \sqrt{2-b x}}{8 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(2 - b*x)^(3/2),x]

[Out]

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

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

Mathematica [A]  time = 0.0539062, size = 70, normalized size = 0.64 \[ \frac{3 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{5/2}}-\frac{\sqrt{x} \sqrt{2-b x} \left (2 b^3 x^3-6 b^2 x^2+b x+3\right )}{8 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(2 - b*x)^(3/2),x]

[Out]

-(Sqrt[x]*Sqrt[2 - b*x]*(3 + b*x - 6*b^2*x^2 + 2*b^3*x^3))/(8*b^2) + (3*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(4*
b^(5/2))

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 116, normalized size = 1.1 \begin{align*} -{\frac{1}{4\,b}{x}^{{\frac{3}{2}}} \left ( -bx+2 \right ) ^{{\frac{5}{2}}}}-{\frac{1}{4\,{b}^{2}} \left ( -bx+2 \right ) ^{{\frac{5}{2}}}\sqrt{x}}+{\frac{1}{8\,{b}^{2}} \left ( -bx+2 \right ) ^{{\frac{3}{2}}}\sqrt{x}}+{\frac{3}{8\,{b}^{2}}\sqrt{x}\sqrt{-bx+2}}+{\frac{3}{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{5}{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^(3/2)*(-b*x+2)^(3/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.83026, size = 356, normalized size = 3.27 \begin{align*} \left [-\frac{{\left (2 \, b^{4} x^{3} - 6 \, 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 )}{8 \, b^{3}}, -\frac{{\left (2 \, b^{4} x^{3} - 6 \, 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 )}{8 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*((2*b^4*x^3 - 6*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)*sqr
t(-b)*sqrt(x) + 1))/b^3, -1/8*((2*b^4*x^3 - 6*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)*sqrt(x))))/b^3]

________________________________________________________________________________________

Sympy [A]  time = 10.743, size = 252, normalized size = 2.31 \begin{align*} \begin{cases} - \frac{i b^{2} x^{\frac{9}{2}}}{4 \sqrt{b x - 2}} + \frac{5 i b x^{\frac{7}{2}}}{4 \sqrt{b x - 2}} - \frac{13 i x^{\frac{5}{2}}}{8 \sqrt{b x - 2}} - \frac{i x^{\frac{3}{2}}}{8 b \sqrt{b x - 2}} + \frac{3 i \sqrt{x}}{4 b^{2} \sqrt{b x - 2}} - \frac{3 i \operatorname{acosh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{4 b^{\frac{5}{2}}} & \text{for}\: \frac{\left |{b x}\right |}{2} > 1 \\\frac{b^{2} x^{\frac{9}{2}}}{4 \sqrt{- b x + 2}} - \frac{5 b x^{\frac{7}{2}}}{4 \sqrt{- b x + 2}} + \frac{13 x^{\frac{5}{2}}}{8 \sqrt{- b x + 2}} + \frac{x^{\frac{3}{2}}}{8 b \sqrt{- b x + 2}} - \frac{3 \sqrt{x}}{4 b^{2} \sqrt{- b x + 2}} + \frac{3 \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{4 b^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I*b**2*x**(9/2)/(4*sqrt(b*x - 2)) + 5*I*b*x**(7/2)/(4*sqrt(b*x - 2)) - 13*I*x**(5/2)/(8*sqrt(b*x -
2)) - I*x**(3/2)/(8*b*sqrt(b*x - 2)) + 3*I*sqrt(x)/(4*b**2*sqrt(b*x - 2)) - 3*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)
/2)/(4*b**(5/2)), Abs(b*x)/2 > 1), (b**2*x**(9/2)/(4*sqrt(-b*x + 2)) - 5*b*x**(7/2)/(4*sqrt(-b*x + 2)) + 13*x*
*(5/2)/(8*sqrt(-b*x + 2)) + x**(3/2)/(8*b*sqrt(-b*x + 2)) - 3*sqrt(x)/(4*b**2*sqrt(-b*x + 2)) + 3*asin(sqrt(2)
*sqrt(b)*sqrt(x)/2)/(4*b**(5/2)), True))

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out