3.7.10 \(\int \frac {\sqrt {x}}{\sqrt {2+b x}} \, dx\) [610]

Optimal. Leaf size=43 \[ \frac {\sqrt {x} \sqrt {2+b x}}{b}-\frac {2 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {52, 56, 221} \begin {gather*} \frac {\sqrt {x} \sqrt {b x+2}}{b}-\frac {2 \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/Sqrt[2 + b*x],x]

[Out]

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

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

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Mathematica [A]
time = 0.05, size = 49, normalized size = 1.14 \begin {gather*} \frac {\sqrt {x} \sqrt {2+b x}}{b}+\frac {2 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/Sqrt[2 + b*x],x]

[Out]

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

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Maple [A]
time = 0.12, size = 62, normalized size = 1.44

method	result	size
meijerg	\(\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \sqrt {b}\, \sqrt {\frac {b x}{2}+1}-2 \sqrt {\pi }\, \arcsinh \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {3}{2}} \sqrt {\pi }}\)	\(49\)
default	\(\frac {\sqrt {x}\, \sqrt {b x +2}}{b}-\frac {\sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right )}{b^{\frac {3}{2}} \sqrt {b x +2}\, \sqrt {x}}\)	\(62\)
risch	\(\frac {\sqrt {x}\, \sqrt {b x +2}}{b}-\frac {\sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {x^{2} b +2 x}\right )}{b^{\frac {3}{2}} \sqrt {b x +2}\, \sqrt {x}}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (32) = 64\).
time = 0.51, size = 70, normalized size = 1.63 \begin {gather*} \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 {b x + 2}}{{\left (b^{2} - \frac {{\left (b x + 2\right )} b}{x}\right )} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]
time = 0.54, size = 87, normalized size = 2.02 \begin {gather*} \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^{2}}, \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^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [A]
time = 1.11, size = 54, normalized size = 1.26 \begin {gather*} \frac {x^{\frac {3}{2}}}{\sqrt {b x + 2}} + \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{-4,[1
,1]%%%}+%%%{-4,[1,0]%%%}+%%%{-4,[0,1]%%%}+%%%{-8,[0,0]%%%},0,%%%{6,[2,2]%%%}+%%%{4,[2,1]%%%}+%%%{6,[2,0]%%%}+%
%%{4,[1,2]%%%}+%%%{28

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Mupad [B]
time = 0.59, size = 43, normalized size = 1.00 \begin {gather*} \frac {4\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {2}-\sqrt {b\,x+2}}\right )}{b^{3/2}}+\frac {\sqrt {x}\,\sqrt {b\,x+2}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)/(b*x + 2)^(1/2),x)

[Out]

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

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