\(\int \frac {(a+\frac {b}{x})^3}{x^{5/2}} \, dx\) [1664]

Optimal result

Integrand size = 15, antiderivative size = 51 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{5/2}} \, dx=-\frac {2 b^3}{9 x^{9/2}}-\frac {6 a b^2}{7 x^{7/2}}-\frac {6 a^2 b}{5 x^{5/2}}-\frac {2 a^3}{3 x^{3/2}} \]

[Out]

-2/9*b^3/x^(9/2)-6/7*a*b^2/x^(7/2)-6/5*a^2*b/x^(5/2)-2/3*a^3/x^(3/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {269, 45} \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{5/2}} \, dx=-\frac {2 a^3}{3 x^{3/2}}-\frac {6 a^2 b}{5 x^{5/2}}-\frac {6 a b^2}{7 x^{7/2}}-\frac {2 b^3}{9 x^{9/2}} \]

[In]

Int[(a + b/x)^3/x^(5/2),x]

[Out]

(-2*b^3)/(9*x^(9/2)) - (6*a*b^2)/(7*x^(7/2)) - (6*a^2*b)/(5*x^(5/2)) - (2*a^3)/(3*x^(3/2))

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(b+a x)^3}{x^{11/2}} \, dx \\ & = \int \left (\frac {b^3}{x^{11/2}}+\frac {3 a b^2}{x^{9/2}}+\frac {3 a^2 b}{x^{7/2}}+\frac {a^3}{x^{5/2}}\right ) \, dx \\ & = -\frac {2 b^3}{9 x^{9/2}}-\frac {6 a b^2}{7 x^{7/2}}-\frac {6 a^2 b}{5 x^{5/2}}-\frac {2 a^3}{3 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.76 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{5/2}} \, dx=-\frac {2 \left (35 b^3+135 a b^2 x+189 a^2 b x^2+105 a^3 x^3\right )}{315 x^{9/2}} \]

[In]

Integrate[(a + b/x)^3/x^(5/2),x]

[Out]

(-2*(35*b^3 + 135*a*b^2*x + 189*a^2*b*x^2 + 105*a^3*x^3))/(315*x^(9/2))

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.71

method	result	size
gosper	\(-\frac {2 \left (105 a^{3} x^{3}+189 a^{2} b \,x^{2}+135 a \,b^{2} x +35 b^{3}\right )}{315 x^{\frac {9}{2}}}\)	\(36\)
derivativedivides	\(-\frac {2 b^{3}}{9 x^{\frac {9}{2}}}-\frac {6 a \,b^{2}}{7 x^{\frac {7}{2}}}-\frac {6 a^{2} b}{5 x^{\frac {5}{2}}}-\frac {2 a^{3}}{3 x^{\frac {3}{2}}}\)	\(36\)
default	\(-\frac {2 b^{3}}{9 x^{\frac {9}{2}}}-\frac {6 a \,b^{2}}{7 x^{\frac {7}{2}}}-\frac {6 a^{2} b}{5 x^{\frac {5}{2}}}-\frac {2 a^{3}}{3 x^{\frac {3}{2}}}\)	\(36\)
trager	\(-\frac {2 \left (105 a^{3} x^{3}+189 a^{2} b \,x^{2}+135 a \,b^{2} x +35 b^{3}\right )}{315 x^{\frac {9}{2}}}\)	\(36\)
risch	\(-\frac {2 \left (105 a^{3} x^{3}+189 a^{2} b \,x^{2}+135 a \,b^{2} x +35 b^{3}\right )}{315 x^{\frac {9}{2}}}\)	\(36\)

[In]

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

[Out]

-2/315*(105*a^3*x^3+189*a^2*b*x^2+135*a*b^2*x+35*b^3)/x^(9/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{5/2}} \, dx=-\frac {2 \, {\left (105 \, a^{3} x^{3} + 189 \, a^{2} b x^{2} + 135 \, a b^{2} x + 35 \, b^{3}\right )}}{315 \, x^{\frac {9}{2}}} \]

[In]

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

[Out]

-2/315*(105*a^3*x^3 + 189*a^2*b*x^2 + 135*a*b^2*x + 35*b^3)/x^(9/2)

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{5/2}} \, dx=- \frac {2 a^{3}}{3 x^{\frac {3}{2}}} - \frac {6 a^{2} b}{5 x^{\frac {5}{2}}} - \frac {6 a b^{2}}{7 x^{\frac {7}{2}}} - \frac {2 b^{3}}{9 x^{\frac {9}{2}}} \]

[In]

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

[Out]

-2*a**3/(3*x**(3/2)) - 6*a**2*b/(5*x**(5/2)) - 6*a*b**2/(7*x**(7/2)) - 2*b**3/(9*x**(9/2))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{5/2}} \, dx=-\frac {2 \, a^{3}}{3 \, x^{\frac {3}{2}}} - \frac {6 \, a^{2} b}{5 \, x^{\frac {5}{2}}} - \frac {6 \, a b^{2}}{7 \, x^{\frac {7}{2}}} - \frac {2 \, b^{3}}{9 \, x^{\frac {9}{2}}} \]

[In]

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

[Out]

-2/3*a^3/x^(3/2) - 6/5*a^2*b/x^(5/2) - 6/7*a*b^2/x^(7/2) - 2/9*b^3/x^(9/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{5/2}} \, dx=-\frac {2 \, {\left (105 \, a^{3} x^{3} + 189 \, a^{2} b x^{2} + 135 \, a b^{2} x + 35 \, b^{3}\right )}}{315 \, x^{\frac {9}{2}}} \]

[In]

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

[Out]

-2/315*(105*a^3*x^3 + 189*a^2*b*x^2 + 135*a*b^2*x + 35*b^3)/x^(9/2)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.69 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{5/2}} \, dx=-\frac {210\,a^3\,x^3+378\,a^2\,b\,x^2+270\,a\,b^2\,x+70\,b^3}{315\,x^{9/2}} \]

[In]

int((a + b/x)^3/x^(5/2),x)

[Out]

-(70*b^3 + 210*a^3*x^3 + 378*a^2*b*x^2 + 270*a*b^2*x)/(315*x^(9/2))