\(\int \frac {1}{\sqrt {x} (2+b x)^{3/2}} \, dx\) [619]

Optimal result

Integrand size = 15, antiderivative size = 15 \[ \int \frac {1}{\sqrt {x} (2+b x)^{3/2}} \, dx=\frac {\sqrt {x}}{\sqrt {2+b x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {37} \[ \int \frac {1}{\sqrt {x} (2+b x)^{3/2}} \, dx=\frac {\sqrt {x}}{\sqrt {b x+2}} \]

[In]

Int[1/(Sqrt[x]*(2 + b*x)^(3/2)),x]

[Out]

Sqrt[x]/Sqrt[2 + b*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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {x}}{\sqrt {2+b x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {x} (2+b x)^{3/2}} \, dx=\frac {\sqrt {x}}{\sqrt {2+b x}} \]

[In]

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

[Out]

Sqrt[x]/Sqrt[2 + b*x]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80

method	result	size
gosper	\(\frac {\sqrt {x}}{\sqrt {b x +2}}\)	\(12\)
default	\(\frac {\sqrt {x}}{\sqrt {b x +2}}\)	\(12\)
meijerg	\(\frac {\sqrt {x}\, \sqrt {2}}{2 \sqrt {\frac {b x}{2}+1}}\)	\(17\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\sqrt {x} (2+b x)^{3/2}} \, dx=\frac {\sqrt {x}}{\sqrt {b x + 2}} \]

[In]

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

[Out]

sqrt(x)/sqrt(b*x + 2)

Sympy [A] (verification not implemented)

Time = 0.74 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {x} (2+b x)^{3/2}} \, dx=\frac {1}{\sqrt {b} \sqrt {1 + \frac {2}{b x}}} \]

[In]

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

[Out]

1/(sqrt(b)*sqrt(1 + 2/(b*x)))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\sqrt {x} (2+b x)^{3/2}} \, dx=\frac {\sqrt {x}}{\sqrt {b x + 2}} \]

[In]

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

[Out]

sqrt(x)/sqrt(b*x + 2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (11) = 22\).

Time = 0.32 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.93 \[ \int \frac {1}{\sqrt {x} (2+b x)^{3/2}} \, dx=\frac {4 \, b^{\frac {3}{2}}}{{\left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b\right )} {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\sqrt {x} (2+b x)^{3/2}} \, dx=\frac {\sqrt {x}}{\sqrt {b\,x+2}} \]

[In]

int(1/(x^(1/2)*(b*x + 2)^(3/2)),x)

[Out]

x^(1/2)/(b*x + 2)^(1/2)