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

Optimal. Leaf size=60 \[ -\frac{2 (2-b x)^{3/2}}{\sqrt{x}}-3 b \sqrt{x} \sqrt{2-b x}-6 \sqrt{b} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right ) \]

[Out]

-3*b*Sqrt[x]*Sqrt[2 - b*x] - (2*(2 - b*x)^(3/2))/Sqrt[x] - 6*Sqrt[b]*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]]

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Rubi [A]  time = 0.0115209, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {47, 50, 54, 216} \[ -\frac{2 (2-b x)^{3/2}}{\sqrt{x}}-3 b \sqrt{x} \sqrt{2-b x}-6 \sqrt{b} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(2 - b*x)^(3/2)/x^(3/2),x]

[Out]

-3*b*Sqrt[x]*Sqrt[2 - b*x] - (2*(2 - b*x)^(3/2))/Sqrt[x] - 6*Sqrt[b]*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 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 \frac{(2-b x)^{3/2}}{x^{3/2}} \, dx &=-\frac{2 (2-b x)^{3/2}}{\sqrt{x}}-(3 b) \int \frac{\sqrt{2-b x}}{\sqrt{x}} \, dx\\ &=-3 b \sqrt{x} \sqrt{2-b x}-\frac{2 (2-b x)^{3/2}}{\sqrt{x}}-(3 b) \int \frac{1}{\sqrt{x} \sqrt{2-b x}} \, dx\\ &=-3 b \sqrt{x} \sqrt{2-b x}-\frac{2 (2-b x)^{3/2}}{\sqrt{x}}-(6 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-b x^2}} \, dx,x,\sqrt{x}\right )\\ &=-3 b \sqrt{x} \sqrt{2-b x}-\frac{2 (2-b x)^{3/2}}{\sqrt{x}}-6 \sqrt{b} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0051878, size = 28, normalized size = 0.47 \[ -\frac{4 \sqrt{2} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};\frac{b x}{2}\right )}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 - b*x)^(3/2)/x^(3/2),x]

[Out]

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

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Maple [B]  time = 0.018, size = 97, normalized size = 1.6 \begin{align*}{({b}^{2}{x}^{2}+2\,bx-8)\sqrt{ \left ( -bx+2 \right ) x}{\frac{1}{\sqrt{-x \left ( bx-2 \right ) }}}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{-bx+2}}}}-3\,{\frac{\sqrt{b}\sqrt{ \left ( -bx+2 \right ) x}}{\sqrt{x}\sqrt{-bx+2}}\arctan \left ({\frac{\sqrt{b}}{\sqrt{-b{x}^{2}+2\,x}} \left ( x-{b}^{-1} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85379, size = 267, normalized size = 4.45 \begin{align*} \left [\frac{3 \, \sqrt{-b} x \log \left (-b x + \sqrt{-b x + 2} \sqrt{-b} \sqrt{x} + 1\right ) -{\left (b x + 4\right )} \sqrt{-b x + 2} \sqrt{x}}{x}, \frac{6 \, \sqrt{b} x \arctan \left (\frac{\sqrt{-b x + 2}}{\sqrt{b} \sqrt{x}}\right ) -{\left (b x + 4\right )} \sqrt{-b x + 2} \sqrt{x}}{x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(3*sqrt(-b)*x*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) - (b*x + 4)*sqrt(-b*x + 2)*sqrt(x))/x, (6*sqrt(
b)*x*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) - (b*x + 4)*sqrt(-b*x + 2)*sqrt(x))/x]

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Sympy [A]  time = 2.92056, size = 160, normalized size = 2.67 \begin{align*} \begin{cases} 6 i \sqrt{b} \operatorname{acosh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )} - \frac{i b^{2} x^{\frac{3}{2}}}{\sqrt{b x - 2}} - \frac{2 i b \sqrt{x}}{\sqrt{b x - 2}} + \frac{8 i}{\sqrt{x} \sqrt{b x - 2}} & \text{for}\: \frac{\left |{b x}\right |}{2} > 1 \\- 6 \sqrt{b} \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )} + \frac{b^{2} x^{\frac{3}{2}}}{\sqrt{- b x + 2}} + \frac{2 b \sqrt{x}}{\sqrt{- b x + 2}} - \frac{8}{\sqrt{x} \sqrt{- b x + 2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((6*I*sqrt(b)*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2) - I*b**2*x**(3/2)/sqrt(b*x - 2) - 2*I*b*sqrt(x)/sqrt(b
*x - 2) + 8*I/(sqrt(x)*sqrt(b*x - 2)), Abs(b*x)/2 > 1), (-6*sqrt(b)*asin(sqrt(2)*sqrt(b)*sqrt(x)/2) + b**2*x**
(3/2)/sqrt(-b*x + 2) + 2*b*sqrt(x)/sqrt(-b*x + 2) - 8/(sqrt(x)*sqrt(-b*x + 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((-b*x+2)^(3/2)/x^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError