\(\int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx\) [611]

Optimal result

Integrand size = 15, antiderivative size = 24 \[ \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx=\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {56, 221} \[ \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx=\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \]

[In]

Int[1/(Sqrt[x]*Sqrt[2 + b*x]),x]

[Out]

(2*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/Sqrt[b]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx=-\frac {2 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right )}{\sqrt {b}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75

method	result	size
meijerg	\(\frac {2 \,\operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{\sqrt {b}}\)	\(18\)
default	\(\frac {\sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{\sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(46\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.29 \[ \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx=\left [\frac {\log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{\sqrt {b}}, -\frac {2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{b}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.68 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx=\frac {2 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} \]

[In]

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

[Out]

2*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/sqrt(b)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (17) = 34\).

Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71 \[ \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx=-\frac {\log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{\sqrt {b}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (17) = 34\).

Time = 6.16 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx=-\frac {2 \, \sqrt {b} \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{{\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx=\frac {4\,\mathrm {atan}\left (\frac {\sqrt {2}-\sqrt {b\,x+2}}{\sqrt {-b}\,\sqrt {x}}\right )}{\sqrt {-b}} \]

[In]

int(1/(x^(1/2)*(b*x + 2)^(1/2)),x)

[Out]

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