\(\int \frac {(a+\frac {b}{x})^{3/2}}{x^{3/2}} \, dx\) [1766]

Optimal result

Integrand size = 17, antiderivative size = 77 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^{3/2}} \, dx=-\frac {3 a \sqrt {a+\frac {b}{x}}}{4 \sqrt {x}}-\frac {\left (a+\frac {b}{x}\right )^{3/2}}{2 \sqrt {x}}-\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 \sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {344, 201, 223, 212} \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^{3/2}} \, dx=-\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{4 \sqrt {b}}-\frac {3 a \sqrt {a+\frac {b}{x}}}{4 \sqrt {x}}-\frac {\left (a+\frac {b}{x}\right )^{3/2}}{2 \sqrt {x}} \]

[In]

Int[(a + b/x)^(3/2)/x^(3/2),x]

[Out]

(-3*a*Sqrt[a + b/x])/(4*Sqrt[x]) - (a + b/x)^(3/2)/(2*Sqrt[x]) - (3*a^2*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x]
)])/(4*Sqrt[b])

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 344

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[-k/c, Subst[
Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n,
0] && FractionQ[m]

Rubi steps \begin{align*} \text {integral}& = -\left (2 \text {Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,\frac {1}{\sqrt {x}}\right )\right ) \\ & = -\frac {\left (a+\frac {b}{x}\right )^{3/2}}{2 \sqrt {x}}-\frac {1}{2} (3 a) \text {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,\frac {1}{\sqrt {x}}\right ) \\ & = -\frac {3 a \sqrt {a+\frac {b}{x}}}{4 \sqrt {x}}-\frac {\left (a+\frac {b}{x}\right )^{3/2}}{2 \sqrt {x}}-\frac {1}{4} \left (3 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right ) \\ & = -\frac {3 a \sqrt {a+\frac {b}{x}}}{4 \sqrt {x}}-\frac {\left (a+\frac {b}{x}\right )^{3/2}}{2 \sqrt {x}}-\frac {1}{4} \left (3 a^2\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right ) \\ & = -\frac {3 a \sqrt {a+\frac {b}{x}}}{4 \sqrt {x}}-\frac {\left (a+\frac {b}{x}\right )^{3/2}}{2 \sqrt {x}}-\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 \sqrt {b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.39 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^{3/2}} \, dx=\frac {\sqrt {a+\frac {b}{x}} \sqrt {x} \left (\frac {(-2 b-5 a x) \sqrt {b+a x}}{4 x^2}-\frac {3 a^2 \text {arctanh}\left (\frac {\sqrt {b+a x}}{\sqrt {b}}\right )}{4 \sqrt {b}}\right )}{\sqrt {b+a x}} \]

[In]

Integrate[(a + b/x)^(3/2)/x^(3/2),x]

[Out]

(Sqrt[a + b/x]*Sqrt[x]*(((-2*b - 5*a*x)*Sqrt[b + a*x])/(4*x^2) - (3*a^2*ArcTanh[Sqrt[b + a*x]/Sqrt[b]])/(4*Sqr
t[b])))/Sqrt[b + a*x]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87

method	result	size
risch	\(-\frac {\left (5 a x +2 b \right ) \sqrt {\frac {a x +b}{x}}}{4 x^{\frac {3}{2}}}-\frac {3 a^{2} \operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) \sqrt {\frac {a x +b}{x}}\, \sqrt {x}}{4 \sqrt {b}\, \sqrt {a x +b}}\)	\(67\)
default	\(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (3 \,\operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right ) a^{2} x^{2}+5 a x \sqrt {a x +b}\, \sqrt {b}+2 b^{\frac {3}{2}} \sqrt {a x +b}\right )}{4 x^{\frac {3}{2}} \sqrt {a x +b}\, \sqrt {b}}\)	\(74\)

[In]

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

[Out]

-1/4*(5*a*x+2*b)/x^(3/2)*((a*x+b)/x)^(1/2)-3/4*a^2/b^(1/2)*arctanh((a*x+b)^(1/2)/b^(1/2))*((a*x+b)/x)^(1/2)/(a
*x+b)^(1/2)*x^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.97 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^{3/2}} \, dx=\left [\frac {3 \, a^{2} \sqrt {b} x^{2} \log \left (\frac {a x - 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) - 2 \, {\left (5 \, a b x + 2 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{8 \, b x^{2}}, \frac {3 \, a^{2} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) - {\left (5 \, a b x + 2 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{4 \, b x^{2}}\right ] \]

[In]

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

[Out]

[1/8*(3*a^2*sqrt(b)*x^2*log((a*x - 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) - 2*(5*a*b*x + 2*b^2)*sqrt(x)
*sqrt((a*x + b)/x))/(b*x^2), 1/4*(3*a^2*sqrt(-b)*x^2*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/b) - (5*a*b*x +
2*b^2)*sqrt(x)*sqrt((a*x + b)/x))/(b*x^2)]

Sympy [A] (verification not implemented)

Time = 2.65 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.99 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^{3/2}} \, dx=- \frac {5 a^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x}}}{4 \sqrt {x}} - \frac {\sqrt {a} b \sqrt {1 + \frac {b}{a x}}}{2 x^{\frac {3}{2}}} - \frac {3 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}} \right )}}{4 \sqrt {b}} \]

[In]

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

[Out]

-5*a**(3/2)*sqrt(1 + b/(a*x))/(4*sqrt(x)) - sqrt(a)*b*sqrt(1 + b/(a*x))/(2*x**(3/2)) - 3*a**2*asinh(sqrt(b)/(s
qrt(a)*sqrt(x)))/(4*sqrt(b))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (55) = 110\).

Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.52 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^{3/2}} \, dx=\frac {3 \, a^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{8 \, \sqrt {b}} - \frac {5 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2} x^{\frac {3}{2}} - 3 \, \sqrt {a + \frac {b}{x}} a^{2} b \sqrt {x}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{2} x^{2} - 2 \, {\left (a + \frac {b}{x}\right )} b x + b^{2}\right )}} \]

[In]

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

[Out]

3/8*a^2*log((sqrt(a + b/x)*sqrt(x) - sqrt(b))/(sqrt(a + b/x)*sqrt(x) + sqrt(b)))/sqrt(b) - 1/4*(5*(a + b/x)^(3
/2)*a^2*x^(3/2) - 3*sqrt(a + b/x)*a^2*b*sqrt(x))/((a + b/x)^2*x^2 - 2*(a + b/x)*b*x + b^2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+\frac {b}{x}\right )^{3/2}}{x^{3/2}} \, dx=\frac {\frac {3 \, a^{3} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} - \frac {5 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{3} - 3 \, \sqrt {a x + b} a^{3} b}{a^{2} x^{2}}}{4 \, a} \]

[In]

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

[Out]

1/4*(3*a^3*arctan(sqrt(a*x + b)/sqrt(-b))/sqrt(-b) - (5*(a*x + b)^(3/2)*a^3 - 3*sqrt(a*x + b)*a^3*b)/(a^2*x^2)
)/a

Mupad [F(-1)]

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

[In]

int((a + b/x)^(3/2)/x^(3/2),x)

[Out]

int((a + b/x)^(3/2)/x^(3/2), x)