\(\int \frac {\sqrt {x}}{(2+b x)^{3/2}} \, dx\) [618]

Optimal result

Integrand size = 15, antiderivative size = 44 \[ \int \frac {\sqrt {x}}{(2+b x)^{3/2}} \, dx=-\frac {2 \sqrt {x}}{b \sqrt {2+b x}}+\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {49, 56, 221} \[ \int \frac {\sqrt {x}}{(2+b x)^{3/2}} \, dx=\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}-\frac {2 \sqrt {x}}{b \sqrt {b x+2}} \]

[In]

Int[Sqrt[x]/(2 + b*x)^(3/2),x]

[Out]

(-2*Sqrt[x])/(b*Sqrt[2 + b*x]) + (2*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(3/2)

Rule 49

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \sqrt {x}}{b \sqrt {2+b x}}+\frac {\int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx}{b} \\ & = -\frac {2 \sqrt {x}}{b \sqrt {2+b x}}+\frac {2 \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right )}{b} \\ & = -\frac {2 \sqrt {x}}{b \sqrt {2+b x}}+\frac {2 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {x}}{(2+b x)^{3/2}} \, dx=-\frac {2 \sqrt {x}}{b \sqrt {2+b x}}-\frac {2 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right )}{b^{3/2}} \]

[In]

Integrate[Sqrt[x]/(2 + b*x)^(3/2),x]

[Out]

(-2*Sqrt[x])/(b*Sqrt[2 + b*x]) - (2*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[2 + b*x]])/b^(3/2)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.09

method	result	size
meijerg	\(\frac {-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}}{\sqrt {\frac {b x}{2}+1}}+2 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {3}{2}} \sqrt {\pi }}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 117, normalized size of antiderivative = 2.66 \[ \int \frac {\sqrt {x}}{(2+b x)^{3/2}} \, dx=\left [\frac {{\left (b x + 2\right )} \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) - 2 \, \sqrt {b x + 2} b \sqrt {x}}{b^{3} x + 2 \, b^{2}}, -\frac {2 \, {\left ({\left (b x + 2\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) + \sqrt {b x + 2} b \sqrt {x}\right )}}{b^{3} x + 2 \, b^{2}}\right ] \]

[In]

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

[Out]

[((b*x + 2)*sqrt(b)*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) - 2*sqrt(b*x + 2)*b*sqrt(x))/(b^3*x + 2*b^2),
-2*((b*x + 2)*sqrt(-b)*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))) + sqrt(b*x + 2)*b*sqrt(x))/(b^3*x + 2*b^2)]

Sympy [A] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {x}}{(2+b x)^{3/2}} \, dx=- \frac {2 \sqrt {x}}{b \sqrt {b x + 2}} + \frac {2 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} \]

[In]

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

[Out]

-2*sqrt(x)/(b*sqrt(b*x + 2)) + 2*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(3/2)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.30 \[ \int \frac {\sqrt {x}}{(2+b x)^{3/2}} \, dx=-\frac {\log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{b^{\frac {3}{2}}} - \frac {2 \, \sqrt {x}}{\sqrt {b x + 2} b} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (33) = 66\).

Time = 1.67 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.86 \[ \int \frac {\sqrt {x}}{(2+b x)^{3/2}} \, dx=-\frac {{\left (\frac {\log \left ({\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2}\right )}{\sqrt {b}} + \frac {8 \, \sqrt {b}}{{\left (\sqrt {b x + 2} \sqrt {b} - \sqrt {{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b}\right )} {\left | b \right |}}{b^{2}} \]

[In]

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

[Out]

-(log((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2)/sqrt(b) + 8*sqrt(b)/((sqrt(b*x + 2)*sqrt(b) - sqrt(
(b*x + 2)*b - 2*b))^2 + 2*b))*abs(b)/b^2

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {x}}{(2+b x)^{3/2}} \, dx=\int \frac {\sqrt {x}}{{\left (b\,x+2\right )}^{3/2}} \,d x \]

[In]

int(x^(1/2)/(b*x + 2)^(3/2),x)

[Out]

int(x^(1/2)/(b*x + 2)^(3/2), x)