\(\int \frac {\sqrt {2+b x}}{x^{3/2}} \, dx\) [509]

Optimal result

Integrand size = 15, antiderivative size = 41 \[ \int \frac {\sqrt {2+b x}}{x^{3/2}} \, dx=-\frac {2 \sqrt {2+b x}}{\sqrt {x}}+2 \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, 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} \[ \int \frac {\sqrt {2+b x}}{x^{3/2}} \, dx=2 \sqrt {b} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )-\frac {2 \sqrt {b x+2}}{\sqrt {x}} \]

[In]

Int[Sqrt[2 + b*x]/x^(3/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.15 \[ \int \frac {\sqrt {2+b x}}{x^{3/2}} \, dx=-\frac {2 \sqrt {2+b x}}{\sqrt {x}}-2 \sqrt {b} \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right ) \]

[In]

Integrate[Sqrt[2 + b*x]/x^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.20

method	result	size
meijerg	\(-\frac {\sqrt {b}\, \left (\frac {4 \sqrt {\pi }\, \sqrt {2}\, \sqrt {\frac {b x}{2}+1}}{\sqrt {x}\, \sqrt {b}}-4 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )\right )}{2 \sqrt {\pi }}\)	\(49\)
risch	\(-\frac {2 \sqrt {b x +2}}{\sqrt {x}}+\frac {\sqrt {b}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right ) \sqrt {x \left (b x +2\right )}}{\sqrt {x}\, \sqrt {b x +2}}\)	\(59\)

[In]

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

[Out]

-1/2*b^(1/2)/Pi^(1/2)*(4*Pi^(1/2)/x^(1/2)*2^(1/2)/b^(1/2)*(1/2*b*x+1)^(1/2)-4*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 = 87, normalized size of antiderivative = 2.12 \[ \int \frac {\sqrt {2+b x}}{x^{3/2}} \, dx=\left [\frac {\sqrt {b} x \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right ) - 2 \, \sqrt {b x + 2} \sqrt {x}}{x}, -\frac {2 \, {\left (\sqrt {-b} x \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right ) + \sqrt {b x + 2} \sqrt {x}\right )}}{x}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.89 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.17 \[ \int \frac {\sqrt {2+b x}}{x^{3/2}} \, dx=- 2 \sqrt {b} \sqrt {1 + \frac {2}{b x}} - \sqrt {b} \log {\left (\frac {1}{b x} \right )} + 2 \sqrt {b} \log {\left (\sqrt {1 + \frac {2}{b x}} + 1 \right )} \]

[In]

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

[Out]

-2*sqrt(b)*sqrt(1 + 2/(b*x)) - sqrt(b)*log(1/(b*x)) + 2*sqrt(b)*log(sqrt(1 + 2/(b*x)) + 1)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.32 \[ \int \frac {\sqrt {2+b x}}{x^{3/2}} \, dx=-\sqrt {b} \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right ) - \frac {2 \, \sqrt {b x + 2}}{\sqrt {x}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (30) = 60\).

Time = 5.82 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.54 \[ \int \frac {\sqrt {2+b x}}{x^{3/2}} \, dx=-\frac {2 \, b^{2} {\left (\frac {\log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{\sqrt {b}} + \frac {\sqrt {b x + 2}}{\sqrt {{\left (b x + 2\right )} b - 2 \, b}}\right )}}{{\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int((b*x + 2)^(1/2)/x^(3/2),x)

[Out]

int((b*x + 2)^(1/2)/x^(3/2), x)