\(\int (a+\frac {b}{x})^{3/2} \, dx\) [234]

Optimal result

Integrand size = 11, antiderivative size = 54 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \, dx=-3 b \sqrt {a+\frac {b}{x}}+\left (a+\frac {b}{x}\right )^{3/2} x+3 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {248, 43, 52, 65, 214} \[ \int \left (a+\frac {b}{x}\right )^{3/2} \, dx=3 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )+x \left (a+\frac {b}{x}\right )^{3/2}-3 b \sqrt {a+\frac {b}{x}} \]

[In]

Int[(a + b/x)^(3/2),x]

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 248

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \, dx=\sqrt {a+\frac {b}{x}} (-2 b+a x)+3 \sqrt {a} b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right ) \]

[In]

Integrate[(a + b/x)^(3/2),x]

[Out]

Sqrt[a + b/x]*(-2*b + a*x) + 3*Sqrt[a]*b*ArcTanh[Sqrt[a + b/x]/Sqrt[a]]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.44

method	result	size
risch	\(\left (a x -2 b \right ) \sqrt {\frac {a x +b}{x}}+\frac {3 \sqrt {a}\, b \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x \left (a x +b \right )}}{2 \left (a x +b \right )}\)	\(78\)
default	\(\frac {\sqrt {\frac {a x +b}{x}}\, \left (6 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b x}\, x^{2}+3 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a b \,x^{2}-4 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\right )}{2 x \sqrt {x \left (a x +b \right )}\, \sqrt {a}}\)	\(100\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.85 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \, dx=\left [\frac {3}{2} \, \sqrt {a} b \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + {\left (a x - 2 \, b\right )} \sqrt {\frac {a x + b}{x}}, -3 \, \sqrt {-a} b \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (a x - 2 \, b\right )} \sqrt {\frac {a x + b}{x}}\right ] \]

[In]

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

[Out]

[3/2*sqrt(a)*b*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + (a*x - 2*b)*sqrt((a*x + b)/x), -3*sqrt(-a)*b*a
rctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (a*x - 2*b)*sqrt((a*x + b)/x)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 92 vs. \(2 (44) = 88\).

Time = 1.39 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.70 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \, dx=3 \sqrt {a} b \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )} + \frac {a^{2} x^{\frac {3}{2}}}{\sqrt {b} \sqrt {\frac {a x}{b} + 1}} - \frac {a \sqrt {b} \sqrt {x}}{\sqrt {\frac {a x}{b} + 1}} - \frac {2 b^{\frac {3}{2}}}{\sqrt {x} \sqrt {\frac {a x}{b} + 1}} \]

[In]

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

[Out]

3*sqrt(a)*b*asinh(sqrt(a)*sqrt(x)/sqrt(b)) + a**2*x**(3/2)/(sqrt(b)*sqrt(a*x/b + 1)) - a*sqrt(b)*sqrt(x)/sqrt(
a*x/b + 1) - 2*b**(3/2)/(sqrt(x)*sqrt(a*x/b + 1))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.17 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \, dx=\sqrt {a + \frac {b}{x}} a x - \frac {3}{2} \, \sqrt {a} b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right ) - 2 \, \sqrt {a + \frac {b}{x}} b \]

[In]

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

[Out]

sqrt(a + b/x)*a*x - 3/2*sqrt(a)*b*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a))) - 2*sqrt(a + b/x)*b

Giac [F(-2)]

Exception generated. \[ \int \left (a+\frac {b}{x}\right )^{3/2} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value

Mupad [B] (verification not implemented)

Time = 5.60 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.63 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \, dx=-\frac {2\,x\,{\left (a+\frac {b}{x}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {a\,x}{b}\right )}{{\left (\frac {a\,x}{b}+1\right )}^{3/2}} \]

[In]

int((a + b/x)^(3/2),x)

[Out]

-(2*x*(a + b/x)^(3/2)*hypergeom([-3/2, -1/2], 1/2, -(a*x)/b))/((a*x)/b + 1)^(3/2)