3.539 \(\int x^{5/2} (2-b x)^{3/2} \, dx\)

Optimal. Leaf size=131 \[ -\frac{x^{3/2} \sqrt{2-b x}}{8 b^2}-\frac{3 \sqrt{x} \sqrt{2-b x}}{8 b^3}+\frac{3 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{7/2}}+\frac{1}{5} x^{7/2} (2-b x)^{3/2}+\frac{3}{20} x^{7/2} \sqrt{2-b x}-\frac{x^{5/2} \sqrt{2-b x}}{20 b} \]

[Out]

(-3*Sqrt[x]*Sqrt[2 - b*x])/(8*b^3) - (x^(3/2)*Sqrt[2 - b*x])/(8*b^2) - (x^(5/2)*Sqrt[2 - b*x])/(20*b) + (3*x^(
7/2)*Sqrt[2 - b*x])/20 + (x^(7/2)*(2 - b*x)^(3/2))/5 + (3*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(4*b^(7/2))

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Rubi [A]  time = 0.0399673, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {50, 54, 216} \[ -\frac{x^{3/2} \sqrt{2-b x}}{8 b^2}-\frac{3 \sqrt{x} \sqrt{2-b x}}{8 b^3}+\frac{3 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{7/2}}+\frac{1}{5} x^{7/2} (2-b x)^{3/2}+\frac{3}{20} x^{7/2} \sqrt{2-b x}-\frac{x^{5/2} \sqrt{2-b x}}{20 b} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(2 - b*x)^(3/2),x]

[Out]

(-3*Sqrt[x]*Sqrt[2 - b*x])/(8*b^3) - (x^(3/2)*Sqrt[2 - b*x])/(8*b^2) - (x^(5/2)*Sqrt[2 - b*x])/(20*b) + (3*x^(
7/2)*Sqrt[2 - b*x])/20 + (x^(7/2)*(2 - b*x)^(3/2))/5 + (3*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} (2-b x)^{3/2} \, dx &=\frac{1}{5} x^{7/2} (2-b x)^{3/2}+\frac{3}{5} \int x^{5/2} \sqrt{2-b x} \, dx\\ &=\frac{3}{20} x^{7/2} \sqrt{2-b x}+\frac{1}{5} x^{7/2} (2-b x)^{3/2}+\frac{3}{20} \int \frac{x^{5/2}}{\sqrt{2-b x}} \, dx\\ &=-\frac{x^{5/2} \sqrt{2-b x}}{20 b}+\frac{3}{20} x^{7/2} \sqrt{2-b x}+\frac{1}{5} x^{7/2} (2-b x)^{3/2}+\frac{\int \frac{x^{3/2}}{\sqrt{2-b x}} \, dx}{4 b}\\ &=-\frac{x^{3/2} \sqrt{2-b x}}{8 b^2}-\frac{x^{5/2} \sqrt{2-b x}}{20 b}+\frac{3}{20} x^{7/2} \sqrt{2-b x}+\frac{1}{5} x^{7/2} (2-b x)^{3/2}+\frac{3 \int \frac{\sqrt{x}}{\sqrt{2-b x}} \, dx}{8 b^2}\\ &=-\frac{3 \sqrt{x} \sqrt{2-b x}}{8 b^3}-\frac{x^{3/2} \sqrt{2-b x}}{8 b^2}-\frac{x^{5/2} \sqrt{2-b x}}{20 b}+\frac{3}{20} x^{7/2} \sqrt{2-b x}+\frac{1}{5} x^{7/2} (2-b x)^{3/2}+\frac{3 \int \frac{1}{\sqrt{x} \sqrt{2-b x}} \, dx}{8 b^3}\\ &=-\frac{3 \sqrt{x} \sqrt{2-b x}}{8 b^3}-\frac{x^{3/2} \sqrt{2-b x}}{8 b^2}-\frac{x^{5/2} \sqrt{2-b x}}{20 b}+\frac{3}{20} x^{7/2} \sqrt{2-b x}+\frac{1}{5} x^{7/2} (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^3}\\ &=-\frac{3 \sqrt{x} \sqrt{2-b x}}{8 b^3}-\frac{x^{3/2} \sqrt{2-b x}}{8 b^2}-\frac{x^{5/2} \sqrt{2-b x}}{20 b}+\frac{3}{20} x^{7/2} \sqrt{2-b x}+\frac{1}{5} x^{7/2} (2-b x)^{3/2}+\frac{3 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0616042, size = 79, normalized size = 0.6 \[ \frac{3 \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{4 b^{7/2}}-\frac{\sqrt{x} \sqrt{2-b x} \left (8 b^4 x^4-22 b^3 x^3+2 b^2 x^2+5 b x+15\right )}{40 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(2 - b*x)^(3/2),x]

[Out]

-(Sqrt[x]*Sqrt[2 - b*x]*(15 + 5*b*x + 2*b^2*x^2 - 22*b^3*x^3 + 8*b^4*x^4))/(40*b^3) + (3*ArcSin[(Sqrt[b]*Sqrt[
x])/Sqrt[2]])/(4*b^(7/2))

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Maple [A]  time = 0.005, size = 132, normalized size = 1. \begin{align*} -{\frac{1}{5\,b}{x}^{{\frac{5}{2}}} \left ( -bx+2 \right ) ^{{\frac{5}{2}}}}-{\frac{1}{4\,{b}^{2}}{x}^{{\frac{3}{2}}} \left ( -bx+2 \right ) ^{{\frac{5}{2}}}}-{\frac{1}{4\,{b}^{3}} \left ( -bx+2 \right ) ^{{\frac{5}{2}}}\sqrt{x}}+{\frac{1}{8\,{b}^{3}} \left ( -bx+2 \right ) ^{{\frac{3}{2}}}\sqrt{x}}+{\frac{3}{8\,{b}^{3}}\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{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)^(3/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88604, size = 405, normalized size = 3.09 \begin{align*} \left [-\frac{{\left (8 \, b^{5} x^{4} - 22 \, 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 )}{40 \, b^{4}}, -\frac{{\left (8 \, b^{5} x^{4} - 22 \, 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 )}{40 \, b^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(-b*x+2)^(3/2),x, algorithm="fricas")

[Out]

[-1/40*((8*b^5*x^4 - 22*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, -1/40*((8*b^5*x^4 - 22*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 = 35.1992, size = 291, normalized size = 2.22 \begin{align*} \begin{cases} - \frac{i b^{2} x^{\frac{11}{2}}}{5 \sqrt{b x - 2}} + \frac{19 i b x^{\frac{9}{2}}}{20 \sqrt{b x - 2}} - \frac{23 i x^{\frac{7}{2}}}{20 \sqrt{b x - 2}} - \frac{i x^{\frac{5}{2}}}{40 b \sqrt{b x - 2}} - \frac{i x^{\frac{3}{2}}}{8 b^{2} \sqrt{b x - 2}} + \frac{3 i \sqrt{x}}{4 b^{3} \sqrt{b x - 2}} - \frac{3 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^{2} x^{\frac{11}{2}}}{5 \sqrt{- b x + 2}} - \frac{19 b x^{\frac{9}{2}}}{20 \sqrt{- b x + 2}} + \frac{23 x^{\frac{7}{2}}}{20 \sqrt{- b x + 2}} + \frac{x^{\frac{5}{2}}}{40 b \sqrt{- b x + 2}} + \frac{x^{\frac{3}{2}}}{8 b^{2} \sqrt{- b x + 2}} - \frac{3 \sqrt{x}}{4 b^{3} \sqrt{- b x + 2}} + \frac{3 \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)**(3/2),x)

[Out]

Piecewise((-I*b**2*x**(11/2)/(5*sqrt(b*x - 2)) + 19*I*b*x**(9/2)/(20*sqrt(b*x - 2)) - 23*I*x**(7/2)/(20*sqrt(b
*x - 2)) - I*x**(5/2)/(40*b*sqrt(b*x - 2)) - I*x**(3/2)/(8*b**2*sqrt(b*x - 2)) + 3*I*sqrt(x)/(4*b**3*sqrt(b*x
- 2)) - 3*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(4*b**(7/2)), Abs(b*x)/2 > 1), (b**2*x**(11/2)/(5*sqrt(-b*x + 2))
- 19*b*x**(9/2)/(20*sqrt(-b*x + 2)) + 23*x**(7/2)/(20*sqrt(-b*x + 2)) + x**(5/2)/(40*b*sqrt(-b*x + 2)) + x**(
3/2)/(8*b**2*sqrt(-b*x + 2)) - 3*sqrt(x)/(4*b**3*sqrt(-b*x + 2)) + 3*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/(4*b**(7/
2)), True))

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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^(5/2)*(-b*x+2)^(3/2),x, algorithm="giac")

[Out]

Timed out