\(\int \frac {(a+\frac {b}{x})^3}{x^{3/2}} \, dx\) [1663]

Optimal result

Integrand size = 15, antiderivative size = 47 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{3/2}} \, dx=-\frac {2 b^3}{7 x^{7/2}}-\frac {6 a b^2}{5 x^{5/2}}-\frac {2 a^2 b}{x^{3/2}}-\frac {2 a^3}{\sqrt {x}} \]

[Out]

-2/7*b^3/x^(7/2)-6/5*a*b^2/x^(5/2)-2*a^2*b/x^(3/2)-2*a^3/x^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, 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} \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{3/2}} \, dx=-\frac {2 a^3}{\sqrt {x}}-\frac {2 a^2 b}{x^{3/2}}-\frac {6 a b^2}{5 x^{5/2}}-\frac {2 b^3}{7 x^{7/2}} \]

[In]

Int[(a + b/x)^3/x^(3/2),x]

[Out]

(-2*b^3)/(7*x^(7/2)) - (6*a*b^2)/(5*x^(5/2)) - (2*a^2*b)/x^(3/2) - (2*a^3)/Sqrt[x]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{3/2}} \, dx=-\frac {2 \left (5 b^3+21 a b^2 x+35 a^2 b x^2+35 a^3 x^3\right )}{35 x^{7/2}} \]

[In]

Integrate[(a + b/x)^3/x^(3/2),x]

[Out]

(-2*(5*b^3 + 21*a*b^2*x + 35*a^2*b*x^2 + 35*a^3*x^3))/(35*x^(7/2))

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.77

method	result	size
gosper	\(-\frac {2 \left (35 a^{3} x^{3}+35 a^{2} b \,x^{2}+21 a \,b^{2} x +5 b^{3}\right )}{35 x^{\frac {7}{2}}}\)	\(36\)
derivativedivides	\(-\frac {2 b^{3}}{7 x^{\frac {7}{2}}}-\frac {6 a \,b^{2}}{5 x^{\frac {5}{2}}}-\frac {2 a^{2} b}{x^{\frac {3}{2}}}-\frac {2 a^{3}}{\sqrt {x}}\)	\(36\)
default	\(-\frac {2 b^{3}}{7 x^{\frac {7}{2}}}-\frac {6 a \,b^{2}}{5 x^{\frac {5}{2}}}-\frac {2 a^{2} b}{x^{\frac {3}{2}}}-\frac {2 a^{3}}{\sqrt {x}}\)	\(36\)
trager	\(-\frac {2 \left (35 a^{3} x^{3}+35 a^{2} b \,x^{2}+21 a \,b^{2} x +5 b^{3}\right )}{35 x^{\frac {7}{2}}}\)	\(36\)
risch	\(-\frac {2 \left (35 a^{3} x^{3}+35 a^{2} b \,x^{2}+21 a \,b^{2} x +5 b^{3}\right )}{35 x^{\frac {7}{2}}}\)	\(36\)

[In]

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

[Out]

-2/35*(35*a^3*x^3+35*a^2*b*x^2+21*a*b^2*x+5*b^3)/x^(7/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{3/2}} \, dx=-\frac {2 \, {\left (35 \, a^{3} x^{3} + 35 \, a^{2} b x^{2} + 21 \, a b^{2} x + 5 \, b^{3}\right )}}{35 \, x^{\frac {7}{2}}} \]

[In]

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

[Out]

-2/35*(35*a^3*x^3 + 35*a^2*b*x^2 + 21*a*b^2*x + 5*b^3)/x^(7/2)

Sympy [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{3/2}} \, dx=- \frac {2 a^{3}}{\sqrt {x}} - \frac {2 a^{2} b}{x^{\frac {3}{2}}} - \frac {6 a b^{2}}{5 x^{\frac {5}{2}}} - \frac {2 b^{3}}{7 x^{\frac {7}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{3/2}} \, dx=-\frac {2 \, a^{3}}{\sqrt {x}} - \frac {2 \, a^{2} b}{x^{\frac {3}{2}}} - \frac {6 \, a b^{2}}{5 \, x^{\frac {5}{2}}} - \frac {2 \, b^{3}}{7 \, x^{\frac {7}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{3/2}} \, dx=-\frac {2 \, {\left (35 \, a^{3} x^{3} + 35 \, a^{2} b x^{2} + 21 \, a b^{2} x + 5 \, b^{3}\right )}}{35 \, x^{\frac {7}{2}}} \]

[In]

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

[Out]

-2/35*(35*a^3*x^3 + 35*a^2*b*x^2 + 21*a*b^2*x + 5*b^3)/x^(7/2)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+\frac {b}{x}\right )^3}{x^{3/2}} \, dx=-\frac {70\,a^3\,x^3+70\,a^2\,b\,x^2+42\,a\,b^2\,x+10\,b^3}{35\,x^{7/2}} \]

[In]

int((a + b/x)^3/x^(3/2),x)

[Out]

-(10*b^3 + 70*a^3*x^3 + 70*a^2*b*x^2 + 42*a*b^2*x)/(35*x^(7/2))