\(\int (a+\frac {b}{x})^{3/2} \sqrt {x} \, dx\) [1764]

Optimal result

Integrand size = 17, antiderivative size = 70 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \sqrt {x} \, dx=2 b \sqrt {a+\frac {b}{x}} \sqrt {x}+\frac {2}{3} \left (a+\frac {b}{x}\right )^{3/2} x^{3/2}-2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 70, 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, 283, 223, 212} \[ \int \left (a+\frac {b}{x}\right )^{3/2} \sqrt {x} \, dx=-2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )+\frac {2}{3} x^{3/2} \left (a+\frac {b}{x}\right )^{3/2}+2 b \sqrt {x} \sqrt {a+\frac {b}{x}} \]

[In]

Int[(a + b/x)^(3/2)*Sqrt[x],x]

[Out]

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

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 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [A] (verified)

Time = 6.25 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.00 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \sqrt {x} \, dx=\frac {\sqrt {a+\frac {b}{x}} \sqrt {x} \left (\frac {2}{3} \sqrt {b+a x} (4 b+a x)-2 b^{3/2} \text {arctanh}\left (\frac {\sqrt {b+a x}}{\sqrt {b}}\right )\right )}{\sqrt {b+a x}} \]

[In]

Integrate[(a + b/x)^(3/2)*Sqrt[x],x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.90

method	result	size
default	\(-\frac {2 \sqrt {\frac {a x +b}{x}}\, \sqrt {x}\, \left (3 b^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )-a x \sqrt {a x +b}-4 b \sqrt {a x +b}\right )}{3 \sqrt {a x +b}}\)	\(63\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.66 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \sqrt {x} \, dx=\left [b^{\frac {3}{2}} \log \left (\frac {a x - 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) + \frac {2}{3} \, {\left (a x + 4 \, b\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}, 2 \, \sqrt {-b} b \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) + \frac {2}{3} \, {\left (a x + 4 \, b\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.72 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.01 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \sqrt {x} \, dx=\frac {2 a \sqrt {b} x \sqrt {\frac {a x}{b} + 1}}{3} + \frac {8 b^{\frac {3}{2}} \sqrt {\frac {a x}{b} + 1}}{3} + b^{\frac {3}{2}} \log {\left (\frac {a x}{b} \right )} - 2 b^{\frac {3}{2}} \log {\left (\sqrt {\frac {a x}{b} + 1} + 1 \right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.06 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \sqrt {x} \, dx=\frac {2}{3} \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} x^{\frac {3}{2}} + b^{\frac {3}{2}} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right ) + 2 \, \sqrt {a + \frac {b}{x}} b \sqrt {x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.63 \[ \int \left (a+\frac {b}{x}\right )^{3/2} \sqrt {x} \, dx=\frac {2 \, b^{2} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + \frac {2}{3} \, {\left (a x + b\right )}^{\frac {3}{2}} + 2 \, \sqrt {a x + b} b \]

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(x^(1/2)*(a + b/x)^(3/2),x)

[Out]

int(x^(1/2)*(a + b/x)^(3/2), x)