\(\int (a+\frac {b}{x})^{3/2} x^{5/2} \, dx\) [1762]

Optimal result

Integrand size = 17, antiderivative size = 48 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^{5/2} \, dx=-\frac {4 b \left (a+\frac {b}{x}\right )^{5/2} x^{5/2}}{35 a^2}+\frac {2 \left (a+\frac {b}{x}\right )^{5/2} x^{7/2}}{7 a} \]

[Out]

-4/35*b*(a+b/x)^(5/2)*x^(5/2)/a^2+2/7*(a+b/x)^(5/2)*x^(7/2)/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} \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^{5/2} \, dx=\frac {2 x^{7/2} \left (a+\frac {b}{x}\right )^{5/2}}{7 a}-\frac {4 b x^{5/2} \left (a+\frac {b}{x}\right )^{5/2}}{35 a^2} \]

[In]

Int[(a + b/x)^(3/2)*x^(5/2),x]

[Out]

(-4*b*(a + b/x)^(5/2)*x^(5/2))/(35*a^2) + (2*(a + b/x)^(5/2)*x^(7/2))/(7*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 )^{5/2} x^{7/2}}{7 a}-\frac {(2 b) \int \left (a+\frac {b}{x}\right )^{3/2} x^{3/2} \, dx}{7 a} \\ & = -\frac {4 b \left (a+\frac {b}{x}\right )^{5/2} x^{5/2}}{35 a^2}+\frac {2 \left (a+\frac {b}{x}\right )^{5/2} x^{7/2}}{7 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 6.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.79 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^{5/2} \, dx=\frac {2 \sqrt {a+\frac {b}{x}} \sqrt {x} (b+a x)^2 (-2 b+5 a x)}{35 a^2} \]

[In]

Integrate[(a + b/x)^(3/2)*x^(5/2),x]

[Out]

(2*Sqrt[a + b/x]*Sqrt[x]*(b + a*x)^2*(-2*b + 5*a*x))/(35*a^2)

Maple [A] (verified)

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

method	result	size
gosper	\(\frac {2 \left (a x +b \right ) \left (5 a x -2 b \right ) x^{\frac {3}{2}} \left (\frac {a x +b}{x}\right )^{\frac {3}{2}}}{35 a^{2}}\)	\(33\)
default	\(\frac {2 \sqrt {\frac {a x +b}{x}}\, \sqrt {x}\, \left (a x +b \right )^{2} \left (5 a x -2 b \right )}{35 a^{2}}\)	\(35\)
risch	\(\frac {2 \sqrt {\frac {a x +b}{x}}\, \sqrt {x}\, \left (5 a^{3} x^{3}+8 a^{2} b \,x^{2}+a \,b^{2} x -2 b^{3}\right )}{35 a^{2}}\)	\(49\)

[In]

int((a+b/x)^(3/2)*x^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/35*(a*x+b)*(5*a*x-2*b)*x^(3/2)*((a*x+b)/x)^(3/2)/a^2

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^{5/2} \, dx=\frac {2 \, {\left (5 \, a^{3} x^{3} + 8 \, a^{2} b x^{2} + a b^{2} x - 2 \, b^{3}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{35 \, a^{2}} \]

[In]

integrate((a+b/x)^(3/2)*x^(5/2),x, algorithm="fricas")

[Out]

2/35*(5*a^3*x^3 + 8*a^2*b*x^2 + a*b^2*x - 2*b^3)*sqrt(x)*sqrt((a*x + b)/x)/a^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (39) = 78\).

Time = 9.98 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.83 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^{5/2} \, dx=\frac {2 a \sqrt {b} x^{3} \sqrt {\frac {a x}{b} + 1}}{7} + \frac {16 b^{\frac {3}{2}} x^{2} \sqrt {\frac {a x}{b} + 1}}{35} + \frac {2 b^{\frac {5}{2}} x \sqrt {\frac {a x}{b} + 1}}{35 a} - \frac {4 b^{\frac {7}{2}} \sqrt {\frac {a x}{b} + 1}}{35 a^{2}} \]

[In]

integrate((a+b/x)**(3/2)*x**(5/2),x)

[Out]

2*a*sqrt(b)*x**3*sqrt(a*x/b + 1)/7 + 16*b**(3/2)*x**2*sqrt(a*x/b + 1)/35 + 2*b**(5/2)*x*sqrt(a*x/b + 1)/(35*a)
- 4*b**(7/2)*sqrt(a*x/b + 1)/(35*a**2)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.73 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^{5/2} \, dx=\frac {2 \, {\left (5 \, {\left (a + \frac {b}{x}\right )}^{\frac {7}{2}} x^{\frac {7}{2}} - 7 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} b x^{\frac {5}{2}}\right )}}{35 \, a^{2}} \]

[In]

integrate((a+b/x)^(3/2)*x^(5/2),x, algorithm="maxima")

[Out]

2/35*(5*(a + b/x)^(7/2)*x^(7/2) - 7*(a + b/x)^(5/2)*b*x^(5/2))/a^2

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (36) = 72\).

Time = 0.29 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.88 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^{5/2} \, dx=\frac {2}{15} \, b {\left (\frac {2 \, b^{\frac {5}{2}}}{a^{2}} + \frac {3 \, {\left (a x + b\right )}^{\frac {5}{2}} - 5 \, {\left (a x + b\right )}^{\frac {3}{2}} b}{a^{2}}\right )} \mathrm {sgn}\left (x\right ) - \frac {2}{105} \, a {\left (\frac {8 \, b^{\frac {7}{2}}}{a^{3}} - \frac {15 \, {\left (a x + b\right )}^{\frac {7}{2}} - 42 \, {\left (a x + b\right )}^{\frac {5}{2}} b + 35 \, {\left (a x + b\right )}^{\frac {3}{2}} b^{2}}{a^{3}}\right )} \mathrm {sgn}\left (x\right ) \]

[In]

integrate((a+b/x)^(3/2)*x^(5/2),x, algorithm="giac")

[Out]

2/15*b*(2*b^(5/2)/a^2 + (3*(a*x + b)^(5/2) - 5*(a*x + b)^(3/2)*b)/a^2)*sgn(x) - 2/105*a*(8*b^(7/2)/a^3 - (15*(
a*x + b)^(7/2) - 42*(a*x + b)^(5/2)*b + 35*(a*x + b)^(3/2)*b^2)/a^3)*sgn(x)

Mupad [B] (verification not implemented)

Time = 6.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int \left (a+\frac {b}{x}\right )^{3/2} x^{5/2} \, dx=\sqrt {a+\frac {b}{x}}\,\left (\frac {2\,a\,x^{7/2}}{7}+\frac {16\,b\,x^{5/2}}{35}+\frac {2\,b^2\,x^{3/2}}{35\,a}-\frac {4\,b^3\,\sqrt {x}}{35\,a^2}\right ) \]

[In]

int(x^(5/2)*(a + b/x)^(3/2),x)

[Out]

(a + b/x)^(1/2)*((2*a*x^(7/2))/7 + (16*b*x^(5/2))/35 + (2*b^2*x^(3/2))/(35*a) - (4*b^3*x^(1/2))/(35*a^2))