\(\int (a+\frac {b}{x}) x^{3/2} \, dx\) [1648]

Optimal result

Integrand size = 13, antiderivative size = 21 \[ \int \left (a+\frac {b}{x}\right ) x^{3/2} \, dx=\frac {2}{3} b x^{3/2}+\frac {2}{5} a x^{5/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {14} \[ \int \left (a+\frac {b}{x}\right ) x^{3/2} \, dx=\frac {2}{5} a x^{5/2}+\frac {2}{3} b x^{3/2} \]

[In]

Int[(a + b/x)*x^(3/2),x]

[Out]

(2*b*x^(3/2))/3 + (2*a*x^(5/2))/5

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \left (a+\frac {b}{x}\right ) x^{3/2} \, dx=\frac {2}{15} x^{3/2} (5 b+3 a x) \]

[In]

Integrate[(a + b/x)*x^(3/2),x]

[Out]

(2*x^(3/2)*(5*b + 3*a*x))/15

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

method	result	size
gosper	\(\frac {2 \left (3 a x +5 b \right ) x^{\frac {3}{2}}}{15}\)	\(14\)
derivativedivides	\(\frac {2 b \,x^{\frac {3}{2}}}{3}+\frac {2 a \,x^{\frac {5}{2}}}{5}\)	\(14\)
default	\(\frac {2 b \,x^{\frac {3}{2}}}{3}+\frac {2 a \,x^{\frac {5}{2}}}{5}\)	\(14\)
trager	\(\frac {2 \left (3 a x +5 b \right ) x^{\frac {3}{2}}}{15}\)	\(14\)
risch	\(\frac {2 \left (3 a x +5 b \right ) x^{\frac {3}{2}}}{15}\)	\(14\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int \left (a+\frac {b}{x}\right ) x^{3/2} \, dx=\frac {2}{15} \, {\left (3 \, a x^{2} + 5 \, b x\right )} \sqrt {x} \]

[In]

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

[Out]

2/15*(3*a*x^2 + 5*b*x)*sqrt(x)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \left (a+\frac {b}{x}\right ) x^{3/2} \, dx=\frac {2 a x^{\frac {5}{2}}}{5} + \frac {2 b x^{\frac {3}{2}}}{3} \]

[In]

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

[Out]

2*a*x**(5/2)/5 + 2*b*x**(3/2)/3

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \left (a+\frac {b}{x}\right ) x^{3/2} \, dx=\frac {2}{15} \, {\left (3 \, a + \frac {5 \, b}{x}\right )} x^{\frac {5}{2}} \]

[In]

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

[Out]

2/15*(3*a + 5*b/x)*x^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \left (a+\frac {b}{x}\right ) x^{3/2} \, dx=\frac {2}{5} \, a x^{\frac {5}{2}} + \frac {2}{3} \, b x^{\frac {3}{2}} \]

[In]

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

[Out]

2/5*a*x^(5/2) + 2/3*b*x^(3/2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \left (a+\frac {b}{x}\right ) x^{3/2} \, dx=\frac {2\,x^{3/2}\,\left (5\,b+3\,a\,x\right )}{15} \]

[In]

int(x^(3/2)*(a + b/x),x)

[Out]

(2*x^(3/2)*(5*b + 3*a*x))/15