\(\int \frac {1}{x^{3/2} (2-b x)^{3/2}} \, dx\) [639]

Optimal result

Integrand size = 16, antiderivative size = 34 \[ \int \frac {1}{x^{3/2} (2-b x)^{3/2}} \, dx=\frac {1}{\sqrt {x} \sqrt {2-b x}}-\frac {\sqrt {2-b x}}{\sqrt {x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {47, 37} \[ \int \frac {1}{x^{3/2} (2-b x)^{3/2}} \, dx=\frac {1}{\sqrt {x} \sqrt {2-b x}}-\frac {\sqrt {2-b x}}{\sqrt {x}} \]

[In]

Int[1/(x^(3/2)*(2 - b*x)^(3/2)),x]

[Out]

1/(Sqrt[x]*Sqrt[2 - b*x]) - Sqrt[2 - b*x]/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*} \text {integral}& = \frac {1}{\sqrt {x} \sqrt {2-b x}}+\int \frac {1}{x^{3/2} \sqrt {2-b x}} \, dx \\ & = \frac {1}{\sqrt {x} \sqrt {2-b x}}-\frac {\sqrt {2-b x}}{\sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^{3/2} (2-b x)^{3/2}} \, dx=\frac {-1+b x}{\sqrt {x} \sqrt {2-b x}} \]

[In]

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

[Out]

(-1 + b*x)/(Sqrt[x]*Sqrt[2 - b*x])

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.53

method	result	size
gosper	\(\frac {b x -1}{\sqrt {x}\, \sqrt {-b x +2}}\)	\(18\)
meijerg	\(-\frac {\sqrt {2}\, \left (-b x +1\right )}{2 \sqrt {x}\, \sqrt {-\frac {b x}{2}+1}}\)	\(23\)
default	\(-\frac {1}{\sqrt {x}\, \sqrt {-b x +2}}+\frac {\sqrt {x}\, b}{\sqrt {-b x +2}}\)	\(28\)
risch	\(\frac {\left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{2 \sqrt {-x \left (b x -2\right )}\, \sqrt {x}\, \sqrt {-b x +2}}+\frac {b \sqrt {x}\, \sqrt {\left (-b x +2\right ) x}}{2 \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}\)	\(74\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^{3/2} (2-b x)^{3/2}} \, dx=-\frac {{\left (b x - 1\right )} \sqrt {-b x + 2} \sqrt {x}}{b x^{2} - 2 \, x} \]

[In]

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

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.60 (sec) , antiderivative size = 90, normalized size of antiderivative = 2.65 \[ \int \frac {1}{x^{3/2} (2-b x)^{3/2}} \, dx=\begin {cases} - \frac {b^{\frac {5}{2}} x \sqrt {-1 + \frac {2}{b x}}}{b^{2} x - 2 b} + \frac {b^{\frac {3}{2}} \sqrt {-1 + \frac {2}{b x}}}{b^{2} x - 2 b} & \text {for}\: \frac {1}{\left |{b x}\right |} > \frac {1}{2} \\- \frac {i \sqrt {b}}{\sqrt {1 - \frac {2}{b x}}} + \frac {i}{\sqrt {b} x \sqrt {1 - \frac {2}{b x}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^{3/2} (2-b x)^{3/2}} \, dx=\frac {b \sqrt {x}}{2 \, \sqrt {-b x + 2}} - \frac {\sqrt {-b x + 2}}{2 \, \sqrt {x}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (26) = 52\).

Time = 0.28 (sec) , antiderivative size = 83, normalized size of antiderivative = 2.44 \[ \int \frac {1}{x^{3/2} (2-b x)^{3/2}} \, dx=-\frac {\sqrt {-b x + 2} b^{2}}{2 \, \sqrt {{\left (b x - 2\right )} b + 2 \, b} {\left | b \right |}} - \frac {2 \, \sqrt {-b} b^{2}}{{\left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b\right )} {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {1}{x^{3/2} (2-b x)^{3/2}} \, dx=\frac {b\,\sqrt {x}}{\sqrt {2-b\,x}}-\frac {1}{\sqrt {x}\,\sqrt {2-b\,x}} \]

[In]

int(1/(x^(3/2)*(2 - b*x)^(3/2)),x)

[Out]

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