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

Optimal. Leaf size=82 \[ -\frac{2 (2-b x)^{5/2}}{\sqrt{x}}-\frac{5}{2} b \sqrt{x} (2-b x)^{3/2}-\frac{15}{2} b \sqrt{x} \sqrt{2-b x}-15 \sqrt{b} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right ) \]

[Out]

(-15*b*Sqrt[x]*Sqrt[2 - b*x])/2 - (5*b*Sqrt[x]*(2 - b*x)^(3/2))/2 - (2*(2 - b*x)^(5/2))/Sqrt[x] - 15*Sqrt[b]*A
rcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]]

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Rubi [A]  time = 0.0169737, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {47, 50, 54, 216} \[ -\frac{2 (2-b x)^{5/2}}{\sqrt{x}}-\frac{5}{2} b \sqrt{x} (2-b x)^{3/2}-\frac{15}{2} b \sqrt{x} \sqrt{2-b x}-15 \sqrt{b} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(2 - b*x)^(5/2)/x^(3/2),x]

[Out]

(-15*b*Sqrt[x]*Sqrt[2 - b*x])/2 - (5*b*Sqrt[x]*(2 - b*x)^(3/2))/2 - (2*(2 - b*x)^(5/2))/Sqrt[x] - 15*Sqrt[b]*A
rcSin[(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)^{5/2}}{x^{3/2}} \, dx &=-\frac{2 (2-b x)^{5/2}}{\sqrt{x}}-(5 b) \int \frac{(2-b x)^{3/2}}{\sqrt{x}} \, dx\\ &=-\frac{5}{2} b \sqrt{x} (2-b x)^{3/2}-\frac{2 (2-b x)^{5/2}}{\sqrt{x}}-\frac{1}{2} (15 b) \int \frac{\sqrt{2-b x}}{\sqrt{x}} \, dx\\ &=-\frac{15}{2} b \sqrt{x} \sqrt{2-b x}-\frac{5}{2} b \sqrt{x} (2-b x)^{3/2}-\frac{2 (2-b x)^{5/2}}{\sqrt{x}}-\frac{1}{2} (15 b) \int \frac{1}{\sqrt{x} \sqrt{2-b x}} \, dx\\ &=-\frac{15}{2} b \sqrt{x} \sqrt{2-b x}-\frac{5}{2} b \sqrt{x} (2-b x)^{3/2}-\frac{2 (2-b x)^{5/2}}{\sqrt{x}}-(15 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-b x^2}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{15}{2} b \sqrt{x} \sqrt{2-b x}-\frac{5}{2} b \sqrt{x} (2-b x)^{3/2}-\frac{2 (2-b x)^{5/2}}{\sqrt{x}}-15 \sqrt{b} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0053384, size = 28, normalized size = 0.34 \[ -\frac{8 \sqrt{2} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};\frac{b x}{2}\right )}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 - b*x)^(5/2)/x^(3/2),x]

[Out]

(-8*Sqrt[2]*Hypergeometric2F1[-5/2, -1/2, 1/2, (b*x)/2])/Sqrt[x]

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Maple [A]  time = 0.018, size = 106, normalized size = 1.3 \begin{align*} -{\frac{{b}^{3}{x}^{3}-11\,{b}^{2}{x}^{2}+2\,bx+32}{2}\sqrt{ \left ( -bx+2 \right ) x}{\frac{1}{\sqrt{-x \left ( bx-2 \right ) }}}{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{-bx+2}}}}-{\frac{15}{2}\sqrt{b}\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)^(5/2)/x^(3/2),x)

[Out]

-1/2*(b^3*x^3-11*b^2*x^2+2*b*x+32)/(-x*(b*x-2))^(1/2)*((-b*x+2)*x)^(1/2)/x^(1/2)/(-b*x+2)^(1/2)-15/2*b^(1/2)*a
rctan(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)^(5/2)/x^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.95462, size = 316, normalized size = 3.85 \begin{align*} \left [\frac{15 \, \sqrt{-b} x \log \left (-b x + \sqrt{-b x + 2} \sqrt{-b} \sqrt{x} + 1\right ) +{\left (b^{2} x^{2} - 9 \, b x - 16\right )} \sqrt{-b x + 2} \sqrt{x}}{2 \, x}, \frac{30 \, \sqrt{b} x \arctan \left (\frac{\sqrt{-b x + 2}}{\sqrt{b} \sqrt{x}}\right ) +{\left (b^{2} x^{2} - 9 \, b x - 16\right )} \sqrt{-b x + 2} \sqrt{x}}{2 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(15*sqrt(-b)*x*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) + (b^2*x^2 - 9*b*x - 16)*sqrt(-b*x + 2)*sq
rt(x))/x, 1/2*(30*sqrt(b)*x*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) + (b^2*x^2 - 9*b*x - 16)*sqrt(-b*x + 2)*s
qrt(x))/x]

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Sympy [A]  time = 11.0837, size = 202, normalized size = 2.46 \begin{align*} \begin{cases} 15 i \sqrt{b} \operatorname{acosh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )} + \frac{i b^{3} x^{\frac{5}{2}}}{2 \sqrt{b x - 2}} - \frac{11 i b^{2} x^{\frac{3}{2}}}{2 \sqrt{b x - 2}} + \frac{i b \sqrt{x}}{\sqrt{b x - 2}} + \frac{16 i}{\sqrt{x} \sqrt{b x - 2}} & \text{for}\: \frac{\left |{b x}\right |}{2} > 1 \\- 15 \sqrt{b} \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )} - \frac{b^{3} x^{\frac{5}{2}}}{2 \sqrt{- b x + 2}} + \frac{11 b^{2} x^{\frac{3}{2}}}{2 \sqrt{- b x + 2}} - \frac{b \sqrt{x}}{\sqrt{- b x + 2}} - \frac{16}{\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)**(5/2)/x**(3/2),x)

[Out]

Piecewise((15*I*sqrt(b)*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2) + I*b**3*x**(5/2)/(2*sqrt(b*x - 2)) - 11*I*b**2*x**(3
/2)/(2*sqrt(b*x - 2)) + I*b*sqrt(x)/sqrt(b*x - 2) + 16*I/(sqrt(x)*sqrt(b*x - 2)), Abs(b*x)/2 > 1), (-15*sqrt(b
)*asin(sqrt(2)*sqrt(b)*sqrt(x)/2) - b**3*x**(5/2)/(2*sqrt(-b*x + 2)) + 11*b**2*x**(3/2)/(2*sqrt(-b*x + 2)) - b
*sqrt(x)/sqrt(-b*x + 2) - 16/(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)^(5/2)/x^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError