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

Optimal. Leaf size=84 \[ 5 b^2 \sqrt{x} \sqrt{2-b x}+10 b^{3/2} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )-\frac{2 (2-b x)^{5/2}}{3 x^{3/2}}+\frac{10 b (2-b x)^{3/2}}{3 \sqrt{x}} \]

[Out]

5*b^2*Sqrt[x]*Sqrt[2 - b*x] + (10*b*(2 - b*x)^(3/2))/(3*Sqrt[x]) - (2*(2 - b*x)^(5/2))/(3*x^(3/2)) + 10*b^(3/2
)*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]]

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Rubi [A]  time = 0.0172905, antiderivative size = 84, 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} \[ 5 b^2 \sqrt{x} \sqrt{2-b x}+10 b^{3/2} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )-\frac{2 (2-b x)^{5/2}}{3 x^{3/2}}+\frac{10 b (2-b x)^{3/2}}{3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[(2 - b*x)^(5/2)/x^(5/2),x]

[Out]

5*b^2*Sqrt[x]*Sqrt[2 - b*x] + (10*b*(2 - b*x)^(3/2))/(3*Sqrt[x]) - (2*(2 - b*x)^(5/2))/(3*x^(3/2)) + 10*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 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^{5/2}} \, dx &=-\frac{2 (2-b x)^{5/2}}{3 x^{3/2}}-\frac{1}{3} (5 b) \int \frac{(2-b x)^{3/2}}{x^{3/2}} \, dx\\ &=\frac{10 b (2-b x)^{3/2}}{3 \sqrt{x}}-\frac{2 (2-b x)^{5/2}}{3 x^{3/2}}+\left (5 b^2\right ) \int \frac{\sqrt{2-b x}}{\sqrt{x}} \, dx\\ &=5 b^2 \sqrt{x} \sqrt{2-b x}+\frac{10 b (2-b x)^{3/2}}{3 \sqrt{x}}-\frac{2 (2-b x)^{5/2}}{3 x^{3/2}}+\left (5 b^2\right ) \int \frac{1}{\sqrt{x} \sqrt{2-b x}} \, dx\\ &=5 b^2 \sqrt{x} \sqrt{2-b x}+\frac{10 b (2-b x)^{3/2}}{3 \sqrt{x}}-\frac{2 (2-b x)^{5/2}}{3 x^{3/2}}+\left (10 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2-b x^2}} \, dx,x,\sqrt{x}\right )\\ &=5 b^2 \sqrt{x} \sqrt{2-b x}+\frac{10 b (2-b x)^{3/2}}{3 \sqrt{x}}-\frac{2 (2-b x)^{5/2}}{3 x^{3/2}}+10 b^{3/2} \sin ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0069384, size = 30, normalized size = 0.36 \[ -\frac{8 \sqrt{2} \, _2F_1\left (-\frac{5}{2},-\frac{3}{2};-\frac{1}{2};\frac{b x}{2}\right )}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(2 - b*x)^(5/2)/x^(5/2),x]

[Out]

(-8*Sqrt[2]*Hypergeometric2F1[-5/2, -3/2, -1/2, (b*x)/2])/(3*x^(3/2))

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Maple [A]  time = 0.017, size = 107, normalized size = 1.3 \begin{align*} -{\frac{3\,{b}^{3}{x}^{3}+22\,{b}^{2}{x}^{2}-64\,bx+16}{3}\sqrt{ \left ( -bx+2 \right ) x}{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{-x \left ( bx-2 \right ) }}}{\frac{1}{\sqrt{-bx+2}}}}+5\,{\frac{{b}^{3/2}\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)^(5/2)/x^(5/2),x)

[Out]

-1/3*(3*b^3*x^3+22*b^2*x^2-64*b*x+16)/x^(3/2)/(-x*(b*x-2))^(1/2)*((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)+5*b^(3/2)*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^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92463, size = 336, normalized size = 4. \begin{align*} \left [\frac{15 \, \sqrt{-b} b x^{2} \log \left (-b x - \sqrt{-b x + 2} \sqrt{-b} \sqrt{x} + 1\right ) +{\left (3 \, b^{2} x^{2} + 28 \, b x - 8\right )} \sqrt{-b x + 2} \sqrt{x}}{3 \, x^{2}}, -\frac{30 \, b^{\frac{3}{2}} x^{2} \arctan \left (\frac{\sqrt{-b x + 2}}{\sqrt{b} \sqrt{x}}\right ) -{\left (3 \, b^{2} x^{2} + 28 \, b x - 8\right )} \sqrt{-b x + 2} \sqrt{x}}{3 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(15*sqrt(-b)*b*x^2*log(-b*x - sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) + (3*b^2*x^2 + 28*b*x - 8)*sqrt(-b*x +
2)*sqrt(x))/x^2, -1/3*(30*b^(3/2)*x^2*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) - (3*b^2*x^2 + 28*b*x - 8)*sqr
t(-b*x + 2)*sqrt(x))/x^2]

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Sympy [C]  time = 10.6367, size = 221, normalized size = 2.63 \begin{align*} \begin{cases} b^{\frac{5}{2}} x \sqrt{-1 + \frac{2}{b x}} + \frac{28 b^{\frac{3}{2}} \sqrt{-1 + \frac{2}{b x}}}{3} + 5 i b^{\frac{3}{2}} \log{\left (\frac{1}{b x} \right )} - 10 i b^{\frac{3}{2}} \log{\left (\frac{1}{\sqrt{b} \sqrt{x}} \right )} + 10 b^{\frac{3}{2}} \operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )} - \frac{8 \sqrt{b} \sqrt{-1 + \frac{2}{b x}}}{3 x} & \text{for}\: \frac{2}{\left |{b x}\right |} > 1 \\i b^{\frac{5}{2}} x \sqrt{1 - \frac{2}{b x}} + \frac{28 i b^{\frac{3}{2}} \sqrt{1 - \frac{2}{b x}}}{3} + 5 i b^{\frac{3}{2}} \log{\left (\frac{1}{b x} \right )} - 10 i b^{\frac{3}{2}} \log{\left (\sqrt{1 - \frac{2}{b x}} + 1 \right )} - \frac{8 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)**(5/2)/x**(5/2),x)

[Out]

Piecewise((b**(5/2)*x*sqrt(-1 + 2/(b*x)) + 28*b**(3/2)*sqrt(-1 + 2/(b*x))/3 + 5*I*b**(3/2)*log(1/(b*x)) - 10*I
*b**(3/2)*log(1/(sqrt(b)*sqrt(x))) + 10*b**(3/2)*asin(sqrt(2)*sqrt(b)*sqrt(x)/2) - 8*sqrt(b)*sqrt(-1 + 2/(b*x)
)/(3*x), 2/Abs(b*x) > 1), (I*b**(5/2)*x*sqrt(1 - 2/(b*x)) + 28*I*b**(3/2)*sqrt(1 - 2/(b*x))/3 + 5*I*b**(3/2)*l
og(1/(b*x)) - 10*I*b**(3/2)*log(sqrt(1 - 2/(b*x)) + 1) - 8*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)^(5/2)/x^(5/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError