\(\int \frac {\sqrt {2-b x}}{x^{5/2}} \, dx\) [518]

Optimal result

Integrand size = 16, antiderivative size = 19 \[ \int \frac {\sqrt {2-b x}}{x^{5/2}} \, dx=-\frac {(2-b x)^{3/2}}{3 x^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {37} \[ \int \frac {\sqrt {2-b x}}{x^{5/2}} \, dx=-\frac {(2-b x)^{3/2}}{3 x^{3/2}} \]

[In]

Int[Sqrt[2 - b*x]/x^(5/2),x]

[Out]

-1/3*(2 - b*x)^(3/2)/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]

Rubi steps \begin{align*} \text {integral}& = -\frac {(2-b x)^{3/2}}{3 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {2-b x}}{x^{5/2}} \, dx=-\frac {(2-b x)^{3/2}}{3 x^{3/2}} \]

[In]

Integrate[Sqrt[2 - b*x]/x^(5/2),x]

[Out]

-1/3*(2 - b*x)^(3/2)/x^(3/2)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74

method	result	size
gosper	\(-\frac {\left (-b x +2\right )^{\frac {3}{2}}}{3 x^{\frac {3}{2}}}\)	\(14\)
meijerg	\(-\frac {2 \sqrt {2}\, \left (-\frac {b x}{2}+1\right )^{\frac {3}{2}}}{3 x^{\frac {3}{2}}}\)	\(17\)
default	\(-\frac {2 \sqrt {-b x +2}}{3 x^{\frac {3}{2}}}+\frac {b \sqrt {-b x +2}}{3 \sqrt {x}}\)	\(29\)
risch	\(-\frac {\sqrt {\left (-b x +2\right ) x}\, \left (b^{2} x^{2}-4 b x +4\right )}{3 x^{\frac {3}{2}} \sqrt {-b x +2}\, \sqrt {-x \left (b x -2\right )}}\)	\(47\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {2-b x}}{x^{5/2}} \, dx=\frac {{\left (b x - 2\right )} \sqrt {-b x + 2}}{3 \, x^{\frac {3}{2}}} \]

[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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 4.37 \[ \int \frac {\sqrt {2-b x}}{x^{5/2}} \, dx=\begin {cases} \frac {b^{\frac {3}{2}} \sqrt {-1 + \frac {2}{b x}}}{3} - \frac {2 \sqrt {b} \sqrt {-1 + \frac {2}{b x}}}{3 x} & \text {for}\: \frac {1}{\left |{b x}\right |} > \frac {1}{2} \\\frac {i b^{\frac {3}{2}} \sqrt {1 - \frac {2}{b x}}}{3} - \frac {2 i \sqrt {b} \sqrt {1 - \frac {2}{b x}}}{3 x} & \text {otherwise} \end {cases} \]

[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), 1/Abs(b*x) > 1/2), (I*b**(3/2)*
sqrt(1 - 2/(b*x))/3 - 2*I*sqrt(b)*sqrt(1 - 2/(b*x))/(3*x), True))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {2-b x}}{x^{5/2}} \, dx=-\frac {{\left (-b x + 2\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \]

[In]

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

[Out]

-1/3*(-b*x + 2)^(3/2)/x^(3/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (13) = 26\).

Time = 0.30 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.84 \[ \int \frac {\sqrt {2-b x}}{x^{5/2}} \, dx=\frac {{\left (b x - 2\right )} \sqrt {-b x + 2} b^{4}}{3 \, {\left ({\left (b x - 2\right )} b + 2 \, b\right )}^{\frac {3}{2}} {\left | b \right |}} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {2-b x}}{x^{5/2}} \, dx=\frac {\sqrt {2-b\,x}\,\left (\frac {b\,x}{3}-\frac {2}{3}\right )}{x^{3/2}} \]

[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)