3.503 \(\int \frac {\sqrt {a-b x}}{x^{7/2}} \, dx\)

Optimal. Leaf size=46 \[ -\frac {4 b (a-b x)^{3/2}}{15 a^2 x^{3/2}}-\frac {2 (a-b x)^{3/2}}{5 a x^{5/2}} \]

[Out]

-2/5*(-b*x+a)^(3/2)/a/x^(5/2)-4/15*b*(-b*x+a)^(3/2)/a^2/x^(3/2)

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Rubi [A]  time = 0.01, antiderivative size = 46, 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 = {45, 37} \[ -\frac {4 b (a-b x)^{3/2}}{15 a^2 x^{3/2}}-\frac {2 (a-b x)^{3/2}}{5 a x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a - b*x]/x^(7/2),x]

[Out]

(-2*(a - b*x)^(3/2))/(5*a*x^(5/2)) - (4*b*(a - b*x)^(3/2))/(15*a^2*x^(3/2))

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 {\sqrt {a-b x}}{x^{7/2}} \, dx &=-\frac {2 (a-b x)^{3/2}}{5 a x^{5/2}}+\frac {(2 b) \int \frac {\sqrt {a-b x}}{x^{5/2}} \, dx}{5 a}\\ &=-\frac {2 (a-b x)^{3/2}}{5 a x^{5/2}}-\frac {4 b (a-b x)^{3/2}}{15 a^2 x^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.65 \[ -\frac {2 (a-b x)^{3/2} (3 a+2 b x)}{15 a^2 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a - b*x]/x^(7/2),x]

[Out]

(-2*(a - b*x)^(3/2)*(3*a + 2*b*x))/(15*a^2*x^(5/2))

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fricas [A]  time = 0.45, size = 34, normalized size = 0.74 \[ \frac {2 \, {\left (2 \, b^{2} x^{2} + a b x - 3 \, a^{2}\right )} \sqrt {-b x + a}}{15 \, a^{2} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+a)^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

2/15*(2*b^2*x^2 + a*b*x - 3*a^2)*sqrt(-b*x + a)/(a^2*x^(5/2))

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giac [A]  time = 1.38, size = 61, normalized size = 1.33 \[ \frac {2 \, {\left (\frac {2 \, {\left (b x - a\right )} b^{5}}{a^{2}} + \frac {5 \, b^{5}}{a}\right )} {\left (b x - a\right )} \sqrt {-b x + a} b}{15 \, {\left ({\left (b x - a\right )} b + a b\right )}^{\frac {5}{2}} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+a)^(1/2)/x^(7/2),x, algorithm="giac")

[Out]

2/15*(2*(b*x - a)*b^5/a^2 + 5*b^5/a)*(b*x - a)*sqrt(-b*x + a)*b/(((b*x - a)*b + a*b)^(5/2)*abs(b))

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maple [A]  time = 0.00, size = 25, normalized size = 0.54 \[ -\frac {2 \left (-b x +a \right )^{\frac {3}{2}} \left (2 b x +3 a \right )}{15 a^{2} x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x+a)^(1/2)/x^(7/2),x)

[Out]

-2/15*(-b*x+a)^(3/2)*(2*b*x+3*a)/x^(5/2)/a^2

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maxima [A]  time = 1.34, size = 33, normalized size = 0.72 \[ -\frac {2 \, {\left (\frac {5 \, {\left (-b x + a\right )}^{\frac {3}{2}} b}{x^{\frac {3}{2}}} + \frac {3 \, {\left (-b x + a\right )}^{\frac {5}{2}}}{x^{\frac {5}{2}}}\right )}}{15 \, a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+a)^(1/2)/x^(7/2),x, algorithm="maxima")

[Out]

-2/15*(5*(-b*x + a)^(3/2)*b/x^(3/2) + 3*(-b*x + a)^(5/2)/x^(5/2))/a^2

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mupad [B]  time = 0.25, size = 32, normalized size = 0.70 \[ \frac {\sqrt {a-b\,x}\,\left (\frac {4\,b^2\,x^2}{15\,a^2}+\frac {2\,b\,x}{15\,a}-\frac {2}{5}\right )}{x^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x)^(1/2)/x^(7/2),x)

[Out]

((a - b*x)^(1/2)*((4*b^2*x^2)/(15*a^2) + (2*b*x)/(15*a) - 2/5))/x^(5/2)

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sympy [A]  time = 5.01, size = 241, normalized size = 5.24 \[ \begin {cases} - \frac {2 \sqrt {b} \sqrt {\frac {a}{b x} - 1}}{5 x^{2}} + \frac {2 b^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}}{15 a x} + \frac {4 b^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}}{15 a^{2}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {6 i a^{3} b^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}}{x \left (- 15 a^{3} b x + 15 a^{2} b^{2} x^{2}\right )} - \frac {8 i a^{2} b^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}}{- 15 a^{3} b x + 15 a^{2} b^{2} x^{2}} - \frac {2 i a b^{\frac {7}{2}} x \sqrt {- \frac {a}{b x} + 1}}{- 15 a^{3} b x + 15 a^{2} b^{2} x^{2}} + \frac {4 i b^{\frac {9}{2}} x^{2} \sqrt {- \frac {a}{b x} + 1}}{- 15 a^{3} b x + 15 a^{2} b^{2} x^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+a)**(1/2)/x**(7/2),x)

[Out]

Piecewise((-2*sqrt(b)*sqrt(a/(b*x) - 1)/(5*x**2) + 2*b**(3/2)*sqrt(a/(b*x) - 1)/(15*a*x) + 4*b**(5/2)*sqrt(a/(
b*x) - 1)/(15*a**2), Abs(a/(b*x)) > 1), (6*I*a**3*b**(3/2)*sqrt(-a/(b*x) + 1)/(x*(-15*a**3*b*x + 15*a**2*b**2*
x**2)) - 8*I*a**2*b**(5/2)*sqrt(-a/(b*x) + 1)/(-15*a**3*b*x + 15*a**2*b**2*x**2) - 2*I*a*b**(7/2)*x*sqrt(-a/(b
*x) + 1)/(-15*a**3*b*x + 15*a**2*b**2*x**2) + 4*I*b**(9/2)*x**2*sqrt(-a/(b*x) + 1)/(-15*a**3*b*x + 15*a**2*b**
2*x**2), True))

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