\(\int \sqrt {a+\frac {b}{x}} x^{3/2} \, dx\) [1754]

Optimal result

Integrand size = 17, antiderivative size = 48 \[ \int \sqrt {a+\frac {b}{x}} x^{3/2} \, dx=-\frac {4 b \left (a+\frac {b}{x}\right )^{3/2} x^{3/2}}{15 a^2}+\frac {2 \left (a+\frac {b}{x}\right )^{3/2} x^{5/2}}{5 a} \]

[Out]

-4/15*b*(a+b/x)^(3/2)*x^(3/2)/a^2+2/5*(a+b/x)^(3/2)*x^(5/2)/a

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {277, 270} \[ \int \sqrt {a+\frac {b}{x}} x^{3/2} \, dx=\frac {2 x^{5/2} \left (a+\frac {b}{x}\right )^{3/2}}{5 a}-\frac {4 b x^{3/2} \left (a+\frac {b}{x}\right )^{3/2}}{15 a^2} \]

[In]

Int[Sqrt[a + b/x]*x^(3/2),x]

[Out]

(-4*b*(a + b/x)^(3/2)*x^(3/2))/(15*a^2) + (2*(a + b/x)^(3/2)*x^(5/2))/(5*a)

Rule 270

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

Rule 277

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 \left (a+\frac {b}{x}\right )^{3/2} x^{5/2}}{5 a}-\frac {(2 b) \int \sqrt {a+\frac {b}{x}} \sqrt {x} \, dx}{5 a} \\ & = -\frac {4 b \left (a+\frac {b}{x}\right )^{3/2} x^{3/2}}{15 a^2}+\frac {2 \left (a+\frac {b}{x}\right )^{3/2} x^{5/2}}{5 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.57 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.85 \[ \int \sqrt {a+\frac {b}{x}} x^{3/2} \, dx=\frac {2 \sqrt {a+\frac {b}{x}} \sqrt {x} \left (-2 b^2+a b x+3 a^2 x^2\right )}{15 a^2} \]

[In]

Integrate[Sqrt[a + b/x]*x^(3/2),x]

[Out]

(2*Sqrt[a + b/x]*Sqrt[x]*(-2*b^2 + a*b*x + 3*a^2*x^2))/(15*a^2)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.69

method	result	size
gosper	\(\frac {2 \left (a x +b \right ) \left (3 a x -2 b \right ) \sqrt {x}\, \sqrt {\frac {a x +b}{x}}}{15 a^{2}}\)	\(33\)
default	\(\frac {2 \left (a x +b \right ) \left (3 a x -2 b \right ) \sqrt {x}\, \sqrt {\frac {a x +b}{x}}}{15 a^{2}}\)	\(33\)
risch	\(\frac {2 \sqrt {\frac {a x +b}{x}}\, \sqrt {x}\, \left (3 a^{2} x^{2}+a b x -2 b^{2}\right )}{15 a^{2}}\)	\(38\)

[In]

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

[Out]

2/15*(a*x+b)*(3*a*x-2*b)*x^(1/2)*((a*x+b)/x)^(1/2)/a^2

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.77 \[ \int \sqrt {a+\frac {b}{x}} x^{3/2} \, dx=\frac {2 \, {\left (3 \, a^{2} x^{2} + a b x - 2 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{15 \, a^{2}} \]

[In]

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

[Out]

2/15*(3*a^2*x^2 + a*b*x - 2*b^2)*sqrt(x)*sqrt((a*x + b)/x)/a^2

Sympy [A] (verification not implemented)

Time = 1.98 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.35 \[ \int \sqrt {a+\frac {b}{x}} x^{3/2} \, dx=\frac {2 \sqrt {b} x^{2} \sqrt {\frac {a x}{b} + 1}}{5} + \frac {2 b^{\frac {3}{2}} x \sqrt {\frac {a x}{b} + 1}}{15 a} - \frac {4 b^{\frac {5}{2}} \sqrt {\frac {a x}{b} + 1}}{15 a^{2}} \]

[In]

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

[Out]

2*sqrt(b)*x**2*sqrt(a*x/b + 1)/5 + 2*b**(3/2)*x*sqrt(a*x/b + 1)/(15*a) - 4*b**(5/2)*sqrt(a*x/b + 1)/(15*a**2)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.73 \[ \int \sqrt {a+\frac {b}{x}} x^{3/2} \, dx=\frac {2 \, {\left (3 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} x^{\frac {5}{2}} - 5 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b x^{\frac {3}{2}}\right )}}{15 \, a^{2}} \]

[In]

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

[Out]

2/15*(3*(a + b/x)^(5/2)*x^(5/2) - 5*(a + b/x)^(3/2)*b*x^(3/2))/a^2

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.77 \[ \int \sqrt {a+\frac {b}{x}} x^{3/2} \, dx=\frac {2}{15} \, {\left (\frac {2 \, b^{\frac {5}{2}}}{a^{2}} + \frac {3 \, {\left (a x + b\right )}^{\frac {5}{2}} - 5 \, {\left (a x + b\right )}^{\frac {3}{2}} b}{a^{2}}\right )} \mathrm {sgn}\left (x\right ) \]

[In]

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

[Out]

2/15*(2*b^(5/2)/a^2 + (3*(a*x + b)^(5/2) - 5*(a*x + b)^(3/2)*b)/a^2)*sgn(x)

Mupad [B] (verification not implemented)

Time = 5.99 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.75 \[ \int \sqrt {a+\frac {b}{x}} x^{3/2} \, dx=\sqrt {a+\frac {b}{x}}\,\left (\frac {2\,x^{5/2}}{5}+\frac {2\,b\,x^{3/2}}{15\,a}-\frac {4\,b^2\,\sqrt {x}}{15\,a^2}\right ) \]

[In]

int(x^(3/2)*(a + b/x)^(1/2),x)

[Out]

(a + b/x)^(1/2)*((2*x^(5/2))/5 + (2*b*x^(3/2))/(15*a) - (4*b^2*x^(1/2))/(15*a^2))