\(\int \frac {(2+b x)^{3/2}}{\sqrt {x}} \, dx\) [536]

Optimal result

Integrand size = 15, antiderivative size = 61 \[ \int \frac {(2+b x)^{3/2}}{\sqrt {x}} \, dx=\frac {3}{2} \sqrt {x} \sqrt {2+b x}+\frac {1}{2} \sqrt {x} (2+b x)^{3/2}+\frac {3 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {52, 56, 221} \[ \int \frac {(2+b x)^{3/2}}{\sqrt {x}} \, dx=\frac {3 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{\sqrt {b}}+\frac {1}{2} \sqrt {x} (b x+2)^{3/2}+\frac {3}{2} \sqrt {x} \sqrt {b x+2} \]

[In]

Int[(2 + b*x)^(3/2)/Sqrt[x],x]

[Out]

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89 \[ \int \frac {(2+b x)^{3/2}}{\sqrt {x}} \, dx=\frac {1}{2} \sqrt {x} \sqrt {2+b x} (5+b x)-\frac {3 \log \left (-\sqrt {b} \sqrt {x}+\sqrt {2+b x}\right )}{\sqrt {b}} \]

[In]

Integrate[(2 + b*x)^(3/2)/Sqrt[x],x]

[Out]

(Sqrt[x]*Sqrt[2 + b*x]*(5 + b*x))/2 - (3*Log[-(Sqrt[b]*Sqrt[x]) + Sqrt[2 + b*x]])/Sqrt[b]

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.89

method	result	size
meijerg	\(\frac {4 \sqrt {\pi }\, \sqrt {b}\, \sqrt {x}\, \sqrt {2}\, \left (\frac {b x}{8}+\frac {5}{8}\right ) \sqrt {\frac {b x}{2}+1}+3 \sqrt {\pi }\, \operatorname {arcsinh}\left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{\sqrt {b}\, \sqrt {\pi }}\)	\(54\)
risch	\(\frac {\left (b x +5\right ) \sqrt {x}\, \sqrt {b x +2}}{2}+\frac {3 \sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(65\)
default	\(\frac {\left (b x +2\right )^{\frac {3}{2}} \sqrt {x}}{2}+\frac {3 \sqrt {x}\, \sqrt {b x +2}}{2}+\frac {3 \sqrt {x \left (b x +2\right )}\, \ln \left (\frac {b x +1}{\sqrt {b}}+\sqrt {b \,x^{2}+2 x}\right )}{2 \sqrt {b x +2}\, \sqrt {x}\, \sqrt {b}}\)	\(72\)

[In]

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

[Out]

3/b^(1/2)/Pi^(1/2)*(4/3*Pi^(1/2)*b^(1/2)*x^(1/2)*2^(1/2)*(1/8*b*x+5/8)*(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 = 0.23 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.72 \[ \int \frac {(2+b x)^{3/2}}{\sqrt {x}} \, dx=\left [\frac {{\left (b^{2} x + 5 \, b\right )} \sqrt {b x + 2} \sqrt {x} + 3 \, \sqrt {b} \log \left (b x + \sqrt {b x + 2} \sqrt {b} \sqrt {x} + 1\right )}{2 \, b}, \frac {{\left (b^{2} x + 5 \, b\right )} \sqrt {b x + 2} \sqrt {x} - 6 \, \sqrt {-b} \arctan \left (\frac {\sqrt {b x + 2} \sqrt {-b}}{b \sqrt {x}}\right )}{2 \, b}\right ] \]

[In]

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

[Out]

[1/2*((b^2*x + 5*b)*sqrt(b*x + 2)*sqrt(x) + 3*sqrt(b)*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1))/b, 1/2*((b
^2*x + 5*b)*sqrt(b*x + 2)*sqrt(x) - 6*sqrt(-b)*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))))/b]

Sympy [A] (verification not implemented)

Time = 2.35 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.25 \[ \int \frac {(2+b x)^{3/2}}{\sqrt {x}} \, dx=\frac {b^{2} x^{\frac {5}{2}}}{2 \sqrt {b x + 2}} + \frac {7 b x^{\frac {3}{2}}}{2 \sqrt {b x + 2}} + \frac {5 \sqrt {x}}{\sqrt {b x + 2}} + \frac {3 \operatorname {asinh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{\sqrt {b}} \]

[In]

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

[Out]

b**2*x**(5/2)/(2*sqrt(b*x + 2)) + 7*b*x**(3/2)/(2*sqrt(b*x + 2)) + 5*sqrt(x)/sqrt(b*x + 2) + 3*asinh(sqrt(2)*s
qrt(b)*sqrt(x)/2)/sqrt(b)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 98 vs. \(2 (42) = 84\).

Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.61 \[ \int \frac {(2+b x)^{3/2}}{\sqrt {x}} \, dx=-\frac {3 \, \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x + 2}}{\sqrt {x}}}{\sqrt {b} + \frac {\sqrt {b x + 2}}{\sqrt {x}}}\right )}{2 \, \sqrt {b}} - \frac {\frac {3 \, \sqrt {b x + 2} b}{\sqrt {x}} - \frac {5 \, {\left (b x + 2\right )}^{\frac {3}{2}}}{x^{\frac {3}{2}}}}{b^{2} - \frac {2 \, {\left (b x + 2\right )} b}{x} + \frac {{\left (b x + 2\right )}^{2}}{x^{2}}} \]

[In]

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

[Out]

-3/2*log(-(sqrt(b) - sqrt(b*x + 2)/sqrt(x))/(sqrt(b) + sqrt(b*x + 2)/sqrt(x)))/sqrt(b) - (3*sqrt(b*x + 2)*b/sq
rt(x) - 5*(b*x + 2)^(3/2)/x^(3/2))/(b^2 - 2*(b*x + 2)*b/x + (b*x + 2)^2/x^2)

Giac [A] (verification not implemented)

none

Time = 6.10 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.26 \[ \int \frac {(2+b x)^{3/2}}{\sqrt {x}} \, dx=\frac {{\left (\sqrt {{\left (b x + 2\right )} b - 2 \, b} \sqrt {b x + 2} {\left (\frac {b x + 2}{b} + \frac {3}{b}\right )} - \frac {6 \, \log \left ({\left | -\sqrt {b x + 2} \sqrt {b} + \sqrt {{\left (b x + 2\right )} b - 2 \, b} \right |}\right )}{\sqrt {b}}\right )} b}{2 \, {\left | b \right |}} \]

[In]

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

[Out]

1/2*(sqrt((b*x + 2)*b - 2*b)*sqrt(b*x + 2)*((b*x + 2)/b + 3/b) - 6*log(abs(-sqrt(b*x + 2)*sqrt(b) + sqrt((b*x
+ 2)*b - 2*b)))/sqrt(b))*b/abs(b)

Mupad [F(-1)]

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

[In]

int((b*x + 2)^(3/2)/x^(1/2),x)

[Out]

int((b*x + 2)^(3/2)/x^(1/2), x)