\(\int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx\) [508]

Optimal result

Integrand size = 15, antiderivative size = 40 \[ \int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx=\sqrt {x} \sqrt {2+b x}+\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, 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 = {52, 56, 221} \[ \int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx=\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}}+\sqrt {x} \sqrt {b x+2} \]

[In]

Int[Sqrt[2 + b*x]/Sqrt[x],x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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}& = \sqrt {x} \sqrt {2+b x}+\int \frac {1}{\sqrt {x} \sqrt {2+b x}} \, dx \\ & = \sqrt {x} \sqrt {2+b x}+2 \text {Subst}\left (\int \frac {1}{\sqrt {2+b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = \sqrt {x} \sqrt {2+b x}+\frac {2 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx=\sqrt {x} \sqrt {2+b x}-\frac {2 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right )}{\sqrt {b}} \]

[In]

Integrate[Sqrt[2 + b*x]/Sqrt[x],x]

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.22

method	result	size
meijerg	\(-\frac {-\sqrt {\pi }\, \sqrt {b}\, \sqrt {x}\, \sqrt {2}\, \sqrt {\frac {b x}{2}+1}-2 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{\sqrt {b}\, \sqrt {\pi }}\)	\(49\)
default	\(\sqrt {x}\, \sqrt {b x +2}+\frac {\sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{\sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(58\)
risch	\(\sqrt {x}\, \sqrt {b x +2}+\frac {\sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{\sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(58\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 2.15 \[ \int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx=\left [\frac {\sqrt {b x + 2} b \sqrt {x} + \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{b}, \frac {\sqrt {b x + 2} b \sqrt {x} - 2 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{b}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.95 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx=\sqrt {x} \sqrt {b x + 2} + \frac {2 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} \]

[In]

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

[Out]

sqrt(x)*sqrt(b*x + 2) + 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. 68 vs. \(2 (29) = 58\).

Time = 0.31 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.70 \[ \int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx=-\frac {\log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{\sqrt {b}} - \frac {2 \, \sqrt {b x + 2}}{{\left (b - \frac {b x + 2}{x}\right )} \sqrt {x}} \]

[In]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (29) = 58\).

Time = 5.89 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.65 \[ \int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx=-\frac {b {\left (\frac {2 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{\sqrt {b}} - \frac {\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2}}{b}\right )}}{{\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {2+b x}}{\sqrt {x}} \, dx=\sqrt {x}\,\sqrt {b\,x+2}-\frac {4\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {2}-\sqrt {b\,x+2}}\right )}{\sqrt {b}} \]

[In]

int((b*x + 2)^(1/2)/x^(1/2),x)

[Out]

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