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

Optimal. Leaf size=62 \[ 2 b^{3/2} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )-\frac{2 (2-b x)^{3/2}}{3 x^{3/2}}+\frac{2 b \sqrt{2-b x}}{\sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.0132896, antiderivative size = 62, 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 = {47, 54, 216} \[ 2 b^{3/2} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )-\frac{2 (2-b x)^{3/2}}{3 x^{3/2}}+\frac{2 b \sqrt{2-b x}}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[(2 - b*x)^(3/2)/x^(5/2),x]

[Out]

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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 \frac{(2-b x)^{3/2}}{x^{5/2}} \, dx &=-\frac{2 (2-b x)^{3/2}}{3 x^{3/2}}-b \int \frac{\sqrt{2-b x}}{x^{3/2}} \, dx\\ &=\frac{2 b \sqrt{2-b x}}{\sqrt{x}}-\frac{2 (2-b x)^{3/2}}{3 x^{3/2}}+b^2 \int \frac{1}{\sqrt{x} \sqrt{2-b x}} \, dx\\ &=\frac{2 b \sqrt{2-b x}}{\sqrt{x}}-\frac{2 (2-b x)^{3/2}}{3 x^{3/2}}+\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-b x^2}} \, dx,x,\sqrt{x}\right )\\ &=\frac{2 b \sqrt{2-b x}}{\sqrt{x}}-\frac{2 (2-b x)^{3/2}}{3 x^{3/2}}+2 b^{3/2} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0053345, size = 30, normalized size = 0.48 \[ -\frac{4 \sqrt{2} \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};\frac{b x}{2}\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 - b*x)^(3/2)/x^(5/2),x]

[Out]

(-4*Sqrt[2]*Hypergeometric2F1[-3/2, -3/2, -1/2, (b*x)/2])/(3*x^(3/2))

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Maple [B]  time = 0.018, size = 98, normalized size = 1.6 \begin{align*} -{\frac{8\,{b}^{2}{x}^{2}-20\,bx+8}{3}\sqrt{ \left ( -bx+2 \right ) x}{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{-x \left ( bx-2 \right ) }}}{\frac{1}{\sqrt{-bx+2}}}}+{{b}^{{\frac{3}{2}}}\arctan \left ({\sqrt{b} \left ( x-{b}^{-1} \right ){\frac{1}{\sqrt{-b{x}^{2}+2\,x}}}} \right ) \sqrt{ \left ( -bx+2 \right ) x}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{-bx+2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x+2)^(3/2)/x^(5/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92141, size = 304, normalized size = 4.9 \begin{align*} \left [\frac{3 \, \sqrt{-b} b x^{2} \log \left (-b x - \sqrt{-b x + 2} \sqrt{-b} \sqrt{x} + 1\right ) + 4 \,{\left (2 \, b x - 1\right )} \sqrt{-b x + 2} \sqrt{x}}{3 \, x^{2}}, -\frac{2 \,{\left (3 \, b^{\frac{3}{2}} x^{2} \arctan \left (\frac{\sqrt{-b x + 2}}{\sqrt{b} \sqrt{x}}\right ) - 2 \,{\left (2 \, b x - 1\right )} \sqrt{-b x + 2} \sqrt{x}\right )}}{3 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*sqrt(-b)*b*x^2*log(-b*x - sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) + 4*(2*b*x - 1)*sqrt(-b*x + 2)*sqrt(x))
/x^2, -2/3*(3*b^(3/2)*x^2*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) - 2*(2*b*x - 1)*sqrt(-b*x + 2)*sqrt(x))/x^2
]

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Sympy [C]  time = 4.92877, size = 182, normalized size = 2.94 \begin{align*} \begin{cases} \frac{8 b^{\frac{3}{2}} \sqrt{-1 + \frac{2}{b x}}}{3} + i b^{\frac{3}{2}} \log{\left (\frac{1}{b x} \right )} - 2 i b^{\frac{3}{2}} \log{\left (\frac{1}{\sqrt{b} \sqrt{x}} \right )} + 2 b^{\frac{3}{2}} \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )} - \frac{4 \sqrt{b} \sqrt{-1 + \frac{2}{b x}}}{3 x} & \text{for}\: \frac{2}{\left |{b x}\right |} > 1 \\\frac{8 i b^{\frac{3}{2}} \sqrt{1 - \frac{2}{b x}}}{3} + i b^{\frac{3}{2}} \log{\left (\frac{1}{b x} \right )} - 2 i b^{\frac{3}{2}} \log{\left (\sqrt{1 - \frac{2}{b x}} + 1 \right )} - \frac{4 i \sqrt{b} \sqrt{1 - \frac{2}{b x}}}{3 x} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+2)**(3/2)/x**(5/2),x)

[Out]

Piecewise((8*b**(3/2)*sqrt(-1 + 2/(b*x))/3 + I*b**(3/2)*log(1/(b*x)) - 2*I*b**(3/2)*log(1/(sqrt(b)*sqrt(x))) +
2*b**(3/2)*asin(sqrt(2)*sqrt(b)*sqrt(x)/2) - 4*sqrt(b)*sqrt(-1 + 2/(b*x))/(3*x), 2/Abs(b*x) > 1), (8*I*b**(3/
2)*sqrt(1 - 2/(b*x))/3 + I*b**(3/2)*log(1/(b*x)) - 2*I*b**(3/2)*log(sqrt(1 - 2/(b*x)) + 1) - 4*I*sqrt(b)*sqrt(
1 - 2/(b*x))/(3*x), 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((-b*x+2)^(3/2)/x^(5/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError