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

Optimal. Leaf size=19 \[ -\frac {(2-b x)^{3/2}}{3 x^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.00, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {37} \[ -\frac {(2-b x)^{3/2}}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 - b*x]/x^(5/2),x]

[Out]

-(2 - b*x)^(3/2)/(3*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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 19, normalized size = 1.00 \[ -\frac {(2-b x)^{3/2}}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 - b*x]/x^(5/2),x]

[Out]

-1/3*(2 - b*x)^(3/2)/x^(3/2)

________________________________________________________________________________________

fricas [A]  time = 0.44, size = 18, normalized size = 0.95 \[ \frac {{\left (b x - 2\right )} \sqrt {-b x + 2}}{3 \, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(b*x - 2)*sqrt(-b*x + 2)/x^(3/2)

________________________________________________________________________________________

giac [B]  time = 1.12, size = 35, normalized size = 1.84 \[ \frac {{\left (b x - 2\right )} \sqrt {-b x + 2} b^{4}}{3 \, {\left ({\left (b x - 2\right )} b + 2 \, b\right )}^{\frac {3}{2}} {\left | b \right |}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.00, size = 14, normalized size = 0.74 \[ -\frac {\left (-b x +2\right )^{\frac {3}{2}}}{3 x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.28, size = 13, normalized size = 0.68 \[ -\frac {{\left (-b x + 2\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-b*x + 2)^(3/2)/x^(3/2)

________________________________________________________________________________________

mupad [B]  time = 0.22, size = 18, normalized size = 0.95 \[ \frac {\sqrt {2-b\,x}\,\left (\frac {b\,x}{3}-\frac {2}{3}\right )}{x^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2 - b*x)^(1/2)/x^(5/2),x)

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 1.51, size = 82, normalized size = 4.32 \[ \begin {cases} \frac {b^{\frac {3}{2}} \sqrt {-1 + \frac {2}{b x}}}{3} - \frac {2 \sqrt {b} \sqrt {-1 + \frac {2}{b x}}}{3 x} & \text {for}\: \frac {2}{\left |{b x}\right |} > 1 \\\frac {i b^{\frac {3}{2}} \sqrt {1 - \frac {2}{b x}}}{3} - \frac {2 i \sqrt {b} \sqrt {1 - \frac {2}{b x}}}{3 x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((b**(3/2)*sqrt(-1 + 2/(b*x))/3 - 2*sqrt(b)*sqrt(-1 + 2/(b*x))/(3*x), 2/Abs(b*x) > 1), (I*b**(3/2)*sq
rt(1 - 2/(b*x))/3 - 2*I*sqrt(b)*sqrt(1 - 2/(b*x))/(3*x), True))

________________________________________________________________________________________