3.646 \(\int \frac{1}{x^{5/2} (2-b x)^{5/2}} \, dx\)

Optimal. Leaf size=75 \[ -\frac{2 \sqrt{2-b x}}{3 x^{3/2}}+\frac{1}{x^{3/2} \sqrt{2-b x}}+\frac{1}{3 x^{3/2} (2-b x)^{3/2}}-\frac{2 b \sqrt{2-b x}}{3 \sqrt{x}} \]

[Out]

1/(3*x^(3/2)*(2 - b*x)^(3/2)) + 1/(x^(3/2)*Sqrt[2 - b*x]) - (2*Sqrt[2 - b*x])/(3*x^(3/2)) - (2*b*Sqrt[2 - b*x]
)/(3*Sqrt[x])

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Rubi [A]  time = 0.0095324, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {45, 37} \[ -\frac{2 \sqrt{2-b x}}{3 x^{3/2}}+\frac{1}{x^{3/2} \sqrt{2-b x}}+\frac{1}{3 x^{3/2} (2-b x)^{3/2}}-\frac{2 b \sqrt{2-b x}}{3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(2 - b*x)^(5/2)),x]

[Out]

1/(3*x^(3/2)*(2 - b*x)^(3/2)) + 1/(x^(3/2)*Sqrt[2 - b*x]) - (2*Sqrt[2 - b*x])/(3*x^(3/2)) - (2*b*Sqrt[2 - b*x]
)/(3*Sqrt[x])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{x^{5/2} (2-b x)^{5/2}} \, dx &=\frac{1}{3 x^{3/2} (2-b x)^{3/2}}+\int \frac{1}{x^{5/2} (2-b x)^{3/2}} \, dx\\ &=\frac{1}{3 x^{3/2} (2-b x)^{3/2}}+\frac{1}{x^{3/2} \sqrt{2-b x}}+2 \int \frac{1}{x^{5/2} \sqrt{2-b x}} \, dx\\ &=\frac{1}{3 x^{3/2} (2-b x)^{3/2}}+\frac{1}{x^{3/2} \sqrt{2-b x}}-\frac{2 \sqrt{2-b x}}{3 x^{3/2}}+\frac{1}{3} (2 b) \int \frac{1}{x^{3/2} \sqrt{2-b x}} \, dx\\ &=\frac{1}{3 x^{3/2} (2-b x)^{3/2}}+\frac{1}{x^{3/2} \sqrt{2-b x}}-\frac{2 \sqrt{2-b x}}{3 x^{3/2}}-\frac{2 b \sqrt{2-b x}}{3 \sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0127586, size = 41, normalized size = 0.55 \[ -\frac{2 b^3 x^3-6 b^2 x^2+3 b x+1}{3 x^{3/2} (2-b x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(2 - b*x)^(5/2)),x]

[Out]

-(1 + 3*b*x - 6*b^2*x^2 + 2*b^3*x^3)/(3*x^(3/2)*(2 - b*x)^(3/2))

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Maple [A]  time = 0.003, size = 36, normalized size = 0.5 \begin{align*} -{\frac{2\,{b}^{3}{x}^{3}-6\,{b}^{2}{x}^{2}+3\,bx+1}{3}{x}^{-{\frac{3}{2}}} \left ( -bx+2 \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(5/2)/(-b*x+2)^(5/2),x)

[Out]

-1/3*(2*b^3*x^3-6*b^2*x^2+3*b*x+1)/x^(3/2)/(-b*x+2)^(3/2)

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Maxima [A]  time = 0.990595, size = 78, normalized size = 1.04 \begin{align*} -\frac{3 \, \sqrt{-b x + 2} b}{8 \, \sqrt{x}} + \frac{{\left (b^{3} - \frac{9 \,{\left (b x - 2\right )} b^{2}}{x}\right )} x^{\frac{3}{2}}}{24 \,{\left (-b x + 2\right )}^{\frac{3}{2}}} - \frac{{\left (-b x + 2\right )}^{\frac{3}{2}}}{24 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*sqrt(-b*x + 2)*b/sqrt(x) + 1/24*(b^3 - 9*(b*x - 2)*b^2/x)*x^(3/2)/(-b*x + 2)^(3/2) - 1/24*(-b*x + 2)^(3/2
)/x^(3/2)

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Fricas [A]  time = 1.63185, size = 126, normalized size = 1.68 \begin{align*} -\frac{{\left (2 \, b^{3} x^{3} - 6 \, b^{2} x^{2} + 3 \, b x + 1\right )} \sqrt{-b x + 2} \sqrt{x}}{3 \,{\left (b^{2} x^{4} - 4 \, b x^{3} + 4 \, x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(-b*x+2)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(2*b^3*x^3 - 6*b^2*x^2 + 3*b*x + 1)*sqrt(-b*x + 2)*sqrt(x)/(b^2*x^4 - 4*b*x^3 + 4*x^2)

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Sympy [B]  time = 30.8685, size = 529, normalized size = 7.05 \begin{align*} \begin{cases} - \frac{2 b^{\frac{27}{2}} x^{4} \sqrt{-1 + \frac{2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac{10 b^{\frac{25}{2}} x^{3} \sqrt{-1 + \frac{2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} - \frac{15 b^{\frac{23}{2}} x^{2} \sqrt{-1 + \frac{2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac{5 b^{\frac{21}{2}} x \sqrt{-1 + \frac{2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac{2 b^{\frac{19}{2}} \sqrt{-1 + \frac{2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} & \text{for}\: \frac{2}{\left |{b x}\right |} > 1 \\- \frac{2 i b^{\frac{27}{2}} x^{4} \sqrt{1 - \frac{2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac{10 i b^{\frac{25}{2}} x^{3} \sqrt{1 - \frac{2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} - \frac{15 i b^{\frac{23}{2}} x^{2} \sqrt{1 - \frac{2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac{5 i b^{\frac{21}{2}} x \sqrt{1 - \frac{2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac{2 i b^{\frac{19}{2}} \sqrt{1 - \frac{2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(5/2)/(-b*x+2)**(5/2),x)

[Out]

Piecewise((-2*b**(27/2)*x**4*sqrt(-1 + 2/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) + 1
0*b**(25/2)*x**3*sqrt(-1 + 2/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) - 15*b**(23/2)*
x**2*sqrt(-1 + 2/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) + 5*b**(21/2)*x*sqrt(-1 + 2
/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) + 2*b**(19/2)*sqrt(-1 + 2/(b*x))/(3*b**12*x
**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x), 2/Abs(b*x) > 1), (-2*I*b**(27/2)*x**4*sqrt(1 - 2/(b*x))/(3*b
**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) + 10*I*b**(25/2)*x**3*sqrt(1 - 2/(b*x))/(3*b**12*x**4 -
18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) - 15*I*b**(23/2)*x**2*sqrt(1 - 2/(b*x))/(3*b**12*x**4 - 18*b**11*x
**3 + 36*b**10*x**2 - 24*b**9*x) + 5*I*b**(21/2)*x*sqrt(1 - 2/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**10*
x**2 - 24*b**9*x) + 2*I*b**(19/2)*sqrt(1 - 2/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x)
, True))

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Giac [B]  time = 1.16395, size = 247, normalized size = 3.29 \begin{align*} -\frac{{\left (4 \,{\left (b x - 2\right )} b^{2}{\left | b \right |} + 9 \, b^{2}{\left | b \right |}\right )} \sqrt{-b x + 2}}{12 \,{\left ({\left (b x - 2\right )} b + 2 \, b\right )}^{\frac{3}{2}}} - \frac{3 \,{\left (\sqrt{-b x + 2} \sqrt{-b} - \sqrt{{\left (b x - 2\right )} b + 2 \, b}\right )}^{4} \sqrt{-b} b^{3} - 18 \,{\left (\sqrt{-b x + 2} \sqrt{-b} - \sqrt{{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} \sqrt{-b} b^{4} + 16 \, \sqrt{-b} b^{5}}{3 \,{\left ({\left (\sqrt{-b x + 2} \sqrt{-b} - \sqrt{{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b\right )}^{3}{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(-b*x+2)^(5/2),x, algorithm="giac")

[Out]

-1/12*(4*(b*x - 2)*b^2*abs(b) + 9*b^2*abs(b))*sqrt(-b*x + 2)/((b*x - 2)*b + 2*b)^(3/2) - 1/3*(3*(sqrt(-b*x + 2
)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^4*sqrt(-b)*b^3 - 18*(sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^
2*sqrt(-b)*b^4 + 16*sqrt(-b)*b^5)/(((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2 - 2*b)^3*abs(b))