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

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

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Rubi [A]  time = 0.01, antiderivative size = 75, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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]

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])

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*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 0.55 \begin {gather*} -\frac {2 b^3 x^3-6 b^2 x^2+3 b x+1}{3 x^{3/2} (2-b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 48, normalized size = 0.64 \begin {gather*} \frac {\sqrt {2-b x} \left (-2 b^3 x^3+6 b^2 x^2-3 b x-1\right )}{3 x^{3/2} (b x-2)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.10, size = 56, normalized size = 0.75 \begin {gather*} -\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 {gather*}

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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giac [B]  time = 1.25, size = 183, normalized size = 2.44 \begin {gather*} -\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 {gather*}

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))

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maple [A]  time = 0.00, size = 36, normalized size = 0.48 \begin {gather*} -\frac {2 b^{3} x^{3}-6 b^{2} x^{2}+3 b x +1}{3 \left (-b x +2\right )^{\frac {3}{2}} x^{\frac {3}{2}}} \end {gather*}

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 = 1.35, size = 58, normalized size = 0.77 \begin {gather*} -\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 {gather*}

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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mupad [B]  time = 0.44, size = 73, normalized size = 0.97 \begin {gather*} \frac {\sqrt {2-b\,x}+3\,b\,x\,\sqrt {2-b\,x}-6\,b^2\,x^2\,\sqrt {2-b\,x}+2\,b^3\,x^3\,\sqrt {2-b\,x}}{x^{3/2}\,\left (x\,\left (12\,b-3\,b^2\,x\right )-12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(2 - b*x)^(5/2)),x)

[Out]

((2 - b*x)^(1/2) + 3*b*x*(2 - b*x)^(1/2) - 6*b^2*x^2*(2 - b*x)^(1/2) + 2*b^3*x^3*(2 - b*x)^(1/2))/(x^(3/2)*(x*
(12*b - 3*b^2*x) - 12))

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sympy [B]  time = 12.40, size = 529, normalized size = 7.05 \begin {gather*} \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 {gather*}

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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