3.6.100 \(\int \frac {1}{x^{3/2} (a-b x)^{3/2}} \, dx\) [600]

Optimal. Leaf size=41 \[ \frac {2}{a \sqrt {x} \sqrt {a-b x}}-\frac {4 \sqrt {a-b x}}{a^2 \sqrt {x}} \]

[Out]

2/a/x^(1/2)/(-b*x+a)^(1/2)-4*(-b*x+a)^(1/2)/a^2/x^(1/2)

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Rubi [A]
time = 0.00, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {47, 37} \begin {gather*} \frac {2}{a \sqrt {x} \sqrt {a-b x}}-\frac {4 \sqrt {a-b x}}{a^2 \sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(a - b*x)^(3/2)),x]

[Out]

2/(a*Sqrt[x]*Sqrt[a - b*x]) - (4*Sqrt[a - b*x])/(a^2*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 47

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^{3/2} (a-b x)^{3/2}} \, dx &=\frac {2}{a \sqrt {x} \sqrt {a-b x}}+\frac {2 \int \frac {1}{x^{3/2} \sqrt {a-b x}} \, dx}{a}\\ &=\frac {2}{a \sqrt {x} \sqrt {a-b x}}-\frac {4 \sqrt {a-b x}}{a^2 \sqrt {x}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 26, normalized size = 0.63 \begin {gather*} -\frac {2 (a-2 b x)}{a^2 \sqrt {x} \sqrt {a-b x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a - 2*b*x))/(a^2*Sqrt[x]*Sqrt[a - b*x])

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 2 in optimal.
time = 2.73, size = 126, normalized size = 3.07 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {2 \sqrt {b} \left (-a+2 b x\right ) \sqrt {\frac {a-b x}{b x}}}{a^2 \left (a-b x\right )},\text {Abs}\left [\frac {a}{b x}\right ]>1\right \}\right \},\frac {-2 I a b^{\frac {3}{2}} \sqrt {1-\frac {a}{b x}}}{a^3 b-a^2 b^2 x}+\frac {I 4 b^{\frac {5}{2}} x \sqrt {1-\frac {a}{b x}}}{a^3 b-a^2 b^2 x}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[1/(x^(3/2)*(a - b*x)^(3/2)),x]')

[Out]

Piecewise[{{2 Sqrt[b] (-a + 2 b x) Sqrt[(a - b x) / (b x)] / (a ^ 2 (a - b x)), Abs[a / (b x)] > 1}}, -2 I a b
 ^ (3 / 2) Sqrt[1 - a / (b x)] / (a ^ 3 b - a ^ 2 b ^ 2 x) + I 4 b ^ (5 / 2) x Sqrt[1 - a / (b x)] / (a ^ 3 b
- a ^ 2 b ^ 2 x)]

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Maple [A]
time = 0.13, size = 35, normalized size = 0.85

method result size
gosper \(-\frac {2 \left (-2 b x +a \right )}{\sqrt {x}\, \sqrt {-b x +a}\, a^{2}}\) \(23\)
default \(-\frac {2}{a \sqrt {x}\, \sqrt {-b x +a}}+\frac {4 b \sqrt {x}}{a^{2} \sqrt {-b x +a}}\) \(35\)
risch \(-\frac {2 \sqrt {-b x +a}}{a^{2} \sqrt {x}}+\frac {2 b \sqrt {x}}{a^{2} \sqrt {-b x +a}}\) \(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^(3/2)/(-b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/a/x^(1/2)/(-b*x+a)^(1/2)+4*b/a^2*x^(1/2)/(-b*x+a)^(1/2)

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Maxima [A]
time = 0.25, size = 34, normalized size = 0.83 \begin {gather*} \frac {2 \, b \sqrt {x}}{\sqrt {-b x + a} a^{2}} - \frac {2 \, \sqrt {-b x + a}}{a^{2} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*sqrt(x)/(sqrt(-b*x + a)*a^2) - 2*sqrt(-b*x + a)/(a^2*sqrt(x))

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Fricas [A]
time = 0.55, size = 38, normalized size = 0.93 \begin {gather*} -\frac {2 \, {\left (2 \, b x - a\right )} \sqrt {-b x + a} \sqrt {x}}{a^{2} b x^{2} - a^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.92, size = 112, normalized size = 2.73 \begin {gather*} \begin {cases} - \frac {2}{a \sqrt {b} x \sqrt {\frac {a}{b x} - 1}} + \frac {4 \sqrt {b}}{a^{2} \sqrt {\frac {a}{b x} - 1}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {2 i a b^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}}{- a^{3} b + a^{2} b^{2} x} - \frac {4 i b^{\frac {5}{2}} x \sqrt {- \frac {a}{b x} + 1}}{- a^{3} b + a^{2} b^{2} x} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(3/2)/(-b*x+a)**(3/2),x)

[Out]

Piecewise((-2/(a*sqrt(b)*x*sqrt(a/(b*x) - 1)) + 4*sqrt(b)/(a**2*sqrt(a/(b*x) - 1)), Abs(a/(b*x)) > 1), (2*I*a*
b**(3/2)*sqrt(-a/(b*x) + 1)/(-a**3*b + a**2*b**2*x) - 4*I*b**(5/2)*x*sqrt(-a/(b*x) + 1)/(-a**3*b + a**2*b**2*x
), True))

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Giac [A]
time = 0.00, size = 77, normalized size = 1.88 \begin {gather*} 2 \left (\frac {\frac {1}{2}\cdot 2 b \sqrt {x} \sqrt {a-b x}}{a^{2} \left (a-b x\right )}+\frac {4 \sqrt {-b}}{2 a \left (\left (\sqrt {a-b x}-\sqrt {-b} \sqrt {x}\right )^{2}-a\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(-b*x + a)*b*sqrt(x)/((b*x - a)*a^2) + 4*sqrt(-b)/(((sqrt(-b)*sqrt(x) - sqrt(-b*x + a))^2 - a)*a)

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Mupad [B]
time = 0.40, size = 42, normalized size = 1.02 \begin {gather*} -\frac {2\,a\,\sqrt {a-b\,x}-4\,b\,x\,\sqrt {a-b\,x}}{\sqrt {x}\,\left (a^3-a^2\,b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(3/2)*(a - b*x)^(3/2)),x)

[Out]

-(2*a*(a - b*x)^(1/2) - 4*b*x*(a - b*x)^(1/2))/(x^(1/2)*(a^3 - a^2*b*x))

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