\(\int \frac {a+\frac {b}{x}}{x^{5/2}} \, dx\) [1652]

Optimal result

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

[Out]

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

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} \[ \int \frac {a+\frac {b}{x}}{x^{5/2}} \, dx=-\frac {2 a}{3 x^{3/2}}-\frac {2 b}{5 x^{5/2}} \]

[In]

Int[(a + b/x)/x^(5/2),x]

[Out]

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

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

Mathematica [A] (verified)

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

[In]

Integrate[(a + b/x)/x^(5/2),x]

[Out]

(-2*(3*b + 5*a*x))/(15*x^(5/2))

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.67

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {a+\frac {b}{x}}{x^{5/2}} \, dx=-\frac {2 \, {\left (5 \, a x + 3 \, b\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

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

[Out]

-2/15*(5*a*x + 3*b)/x^(5/2)

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {a+\frac {b}{x}}{x^{5/2}} \, dx=- \frac {2 a}{3 x^{\frac {3}{2}}} - \frac {2 b}{5 x^{\frac {5}{2}}} \]

[In]

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

[Out]

-2*a/(3*x**(3/2)) - 2*b/(5*x**(5/2))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {a+\frac {b}{x}}{x^{5/2}} \, dx=-\frac {2 \, a}{3 \, x^{\frac {3}{2}}} - \frac {2 \, b}{5 \, x^{\frac {5}{2}}} \]

[In]

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

[Out]

-2/3*a/x^(3/2) - 2/5*b/x^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {a+\frac {b}{x}}{x^{5/2}} \, dx=-\frac {2 \, {\left (5 \, a x + 3 \, b\right )}}{15 \, x^{\frac {5}{2}}} \]

[In]

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

[Out]

-2/15*(5*a*x + 3*b)/x^(5/2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.62 \[ \int \frac {a+\frac {b}{x}}{x^{5/2}} \, dx=-\frac {6\,b+10\,a\,x}{15\,x^{5/2}} \]

[In]

int((a + b/x)/x^(5/2),x)

[Out]

-(6*b + 10*a*x)/(15*x^(5/2))