\(\int (a+\frac {b}{x})^3 x^{5/2} \, dx\) [1659]

Optimal result

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

[Out]

2*a*b^2*x^(3/2)+6/5*a^2*b*x^(5/2)+2/7*a^3*x^(7/2)+2*b^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} \[ \int \left (a+\frac {b}{x}\right )^3 x^{5/2} \, dx=\frac {2}{7} a^3 x^{7/2}+\frac {6}{5} a^2 b x^{5/2}+2 a b^2 x^{3/2}+2 b^3 \sqrt {x} \]

[In]

Int[(a + b/x)^3*x^(5/2),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.83 \[ \int \left (a+\frac {b}{x}\right )^3 x^{5/2} \, dx=\frac {2}{35} \sqrt {x} \left (35 b^3+35 a b^2 x+21 a^2 b x^2+5 a^3 x^3\right ) \]

[In]

Integrate[(a + b/x)^3*x^(5/2),x]

[Out]

(2*Sqrt[x]*(35*b^3 + 35*a*b^2*x + 21*a^2*b*x^2 + 5*a^3*x^3))/35

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.98 \[ \int \left (a+\frac {b}{x}\right )^3 x^{5/2} \, dx=\frac {2 a^{3} x^{\frac {7}{2}}}{7} + \frac {6 a^{2} b x^{\frac {5}{2}}}{5} + 2 a b^{2} x^{\frac {3}{2}} + 2 b^{3} \sqrt {x} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.79 \[ \int \left (a+\frac {b}{x}\right )^3 x^{5/2} \, dx=\frac {2}{35} \, {\left (5 \, a^{3} + \frac {21 \, a^{2} b}{x} + \frac {35 \, a b^{2}}{x^{2}} + \frac {35 \, b^{3}}{x^{3}}\right )} x^{\frac {7}{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \left (a+\frac {b}{x}\right )^3 x^{5/2} \, dx=\frac {2}{7} \, a^{3} x^{\frac {7}{2}} + \frac {6}{5} \, a^{2} b x^{\frac {5}{2}} + 2 \, a b^{2} x^{\frac {3}{2}} + 2 \, b^{3} \sqrt {x} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.74 \[ \int \left (a+\frac {b}{x}\right )^3 x^{5/2} \, dx=\frac {2\,a^3\,x^{7/2}}{7}+2\,b^3\,\sqrt {x}+2\,a\,b^2\,x^{3/2}+\frac {6\,a^2\,b\,x^{5/2}}{5} \]

[In]

int(x^(5/2)*(a + b/x)^3,x)

[Out]

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