3.7.40 \(\int \frac {1}{x^{5/2} (2-b x)^{3/2}} \, dx\)

Optimal. Leaf size=56 \[ -\frac {2 \sqrt {2-b x}}{3 x^{3/2}}+\frac {1}{x^{3/2} \sqrt {2-b x}}-\frac {2 b \sqrt {2-b x}}{3 \sqrt {x}} \]

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Rubi [A]  time = 0.01, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, 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 {2 b \sqrt {2-b x}}{3 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(5/2)*(2 - b*x)^(3/2)),x]

[Out]

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)^{3/2}} \, dx &=\frac {1}{x^{3/2} \sqrt {2-b x}}+2 \int \frac {1}{x^{5/2} \sqrt {2-b x}} \, dx\\ &=\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}{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 = 33, normalized size = 0.59 \begin {gather*} \frac {2 b^2 x^2-2 b x-1}{3 x^{3/2} \sqrt {2-b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(5/2)*(2 - b*x)^(3/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.13, size = 40, normalized size = 0.71 \begin {gather*} \frac {\sqrt {2-b x} \left (-2 b^2 x^2+2 b x+1\right )}{3 x^{3/2} (b x-2)} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^(5/2)*(2 - b*x)^(3/2)),x]

[Out]

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

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fricas [A]  time = 1.26, size = 40, normalized size = 0.71 \begin {gather*} -\frac {{\left (2 \, b^{2} x^{2} - 2 \, b x - 1\right )} \sqrt {-b x + 2} \sqrt {x}}{3 \, {\left (b x^{3} - 2 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(-b*x+2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.17, size = 96, normalized size = 1.71 \begin {gather*} -\frac {\sqrt {-b} b^{3}}{{\left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b\right )} {\left | b \right |}} - \frac {{\left (5 \, {\left (b x - 2\right )} b^{2} {\left | b \right |} + 12 \, b^{2} {\left | b \right |}\right )} \sqrt {-b x + 2}}{12 \, {\left ({\left (b x - 2\right )} b + 2 \, b\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(5/2)/(-b*x+2)^(3/2),x, algorithm="giac")

[Out]

-sqrt(-b)*b^3/(((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2 - 2*b)*abs(b)) - 1/12*(5*(b*x - 2)*b^2*a
bs(b) + 12*b^2*abs(b))*sqrt(-b*x + 2)/((b*x - 2)*b + 2*b)^(3/2)

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maple [A]  time = 0.01, size = 28, normalized size = 0.50 \begin {gather*} \frac {2 b^{2} x^{2}-2 b x -1}{3 \sqrt {-b x +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)^(3/2),x)

[Out]

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

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maxima [A]  time = 1.28, size = 44, normalized size = 0.79 \begin {gather*} \frac {b^{2} \sqrt {x}}{4 \, \sqrt {-b x + 2}} - \frac {\sqrt {-b x + 2} b}{2 \, \sqrt {x}} - \frac {{\left (-b x + 2\right )}^{\frac {3}{2}}}{12 \, 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)^(3/2),x, algorithm="maxima")

[Out]

1/4*b^2*sqrt(x)/sqrt(-b*x + 2) - 1/2*sqrt(-b*x + 2)*b/sqrt(x) - 1/12*(-b*x + 2)^(3/2)/x^(3/2)

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mupad [B]  time = 0.36, size = 38, normalized size = 0.68 \begin {gather*} \frac {\sqrt {2-b\,x}\,\left (\frac {2\,x}{3}-\frac {2\,b\,x^2}{3}+\frac {1}{3\,b}\right )}{x^{5/2}-\frac {2\,x^{3/2}}{b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(2 - b*x)^(3/2)),x)

[Out]

((2 - b*x)^(1/2)*((2*x)/3 - (2*b*x^2)/3 + 1/(3*b)))/(x^(5/2) - (2*x^(3/2))/b)

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sympy [B]  time = 4.29, size = 354, normalized size = 6.32 \begin {gather*} \begin {cases} \frac {2 b^{\frac {15}{2}} x^{3} \sqrt {-1 + \frac {2}{b x}}}{- 3 b^{6} x^{3} + 12 b^{5} x^{2} - 12 b^{4} x} - \frac {6 b^{\frac {13}{2}} x^{2} \sqrt {-1 + \frac {2}{b x}}}{- 3 b^{6} x^{3} + 12 b^{5} x^{2} - 12 b^{4} x} + \frac {3 b^{\frac {11}{2}} x \sqrt {-1 + \frac {2}{b x}}}{- 3 b^{6} x^{3} + 12 b^{5} x^{2} - 12 b^{4} x} + \frac {2 b^{\frac {9}{2}} \sqrt {-1 + \frac {2}{b x}}}{- 3 b^{6} x^{3} + 12 b^{5} x^{2} - 12 b^{4} x} & \text {for}\: \frac {2}{\left |{b x}\right |} > 1 \\\frac {2 i b^{\frac {15}{2}} x^{3} \sqrt {1 - \frac {2}{b x}}}{- 3 b^{6} x^{3} + 12 b^{5} x^{2} - 12 b^{4} x} - \frac {6 i b^{\frac {13}{2}} x^{2} \sqrt {1 - \frac {2}{b x}}}{- 3 b^{6} x^{3} + 12 b^{5} x^{2} - 12 b^{4} x} + \frac {3 i b^{\frac {11}{2}} x \sqrt {1 - \frac {2}{b x}}}{- 3 b^{6} x^{3} + 12 b^{5} x^{2} - 12 b^{4} x} + \frac {2 i b^{\frac {9}{2}} \sqrt {1 - \frac {2}{b x}}}{- 3 b^{6} x^{3} + 12 b^{5} x^{2} - 12 b^{4} 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)**(3/2),x)

[Out]

Piecewise((2*b**(15/2)*x**3*sqrt(-1 + 2/(b*x))/(-3*b**6*x**3 + 12*b**5*x**2 - 12*b**4*x) - 6*b**(13/2)*x**2*sq
rt(-1 + 2/(b*x))/(-3*b**6*x**3 + 12*b**5*x**2 - 12*b**4*x) + 3*b**(11/2)*x*sqrt(-1 + 2/(b*x))/(-3*b**6*x**3 +
12*b**5*x**2 - 12*b**4*x) + 2*b**(9/2)*sqrt(-1 + 2/(b*x))/(-3*b**6*x**3 + 12*b**5*x**2 - 12*b**4*x), 2/Abs(b*x
) > 1), (2*I*b**(15/2)*x**3*sqrt(1 - 2/(b*x))/(-3*b**6*x**3 + 12*b**5*x**2 - 12*b**4*x) - 6*I*b**(13/2)*x**2*s
qrt(1 - 2/(b*x))/(-3*b**6*x**3 + 12*b**5*x**2 - 12*b**4*x) + 3*I*b**(11/2)*x*sqrt(1 - 2/(b*x))/(-3*b**6*x**3 +
12*b**5*x**2 - 12*b**4*x) + 2*I*b**(9/2)*sqrt(1 - 2/(b*x))/(-3*b**6*x**3 + 12*b**5*x**2 - 12*b**4*x), True))

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