\(\int \frac {(a+\frac {b}{x})^2}{x^{5/2}} \, dx\) [1658]

Optimal result

Integrand size = 15, antiderivative size = 36 \[ \int \frac {\left (a+\frac {b}{x}\right )^2}{x^{5/2}} \, dx=-\frac {2 b^2}{7 x^{7/2}}-\frac {4 a b}{5 x^{5/2}}-\frac {2 a^2}{3 x^{3/2}} \]

[Out]

-2/7*b^2/x^(7/2)-4/5*a*b/x^(5/2)-2/3*a^2/x^(3/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 36, 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} \[ \int \frac {\left (a+\frac {b}{x}\right )^2}{x^{5/2}} \, dx=-\frac {2 a^2}{3 x^{3/2}}-\frac {4 a b}{5 x^{5/2}}-\frac {2 b^2}{7 x^{7/2}} \]

[In]

Int[(a + b/x)^2/x^(5/2),x]

[Out]

(-2*b^2)/(7*x^(7/2)) - (4*a*b)/(5*x^(5/2)) - (2*a^2)/(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)^2}{x^{9/2}} \, dx \\ & = \int \left (\frac {b^2}{x^{9/2}}+\frac {2 a b}{x^{7/2}}+\frac {a^2}{x^{5/2}}\right ) \, dx \\ & = -\frac {2 b^2}{7 x^{7/2}}-\frac {4 a b}{5 x^{5/2}}-\frac {2 a^2}{3 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.78 \[ \int \frac {\left (a+\frac {b}{x}\right )^2}{x^{5/2}} \, dx=-\frac {2 \left (15 b^2+42 a b x+35 a^2 x^2\right )}{105 x^{7/2}} \]

[In]

Integrate[(a + b/x)^2/x^(5/2),x]

[Out]

(-2*(15*b^2 + 42*a*b*x + 35*a^2*x^2))/(105*x^(7/2))

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.69

method	result	size
gosper	\(-\frac {2 \left (35 a^{2} x^{2}+42 a b x +15 b^{2}\right )}{105 x^{\frac {7}{2}}}\)	\(25\)
derivativedivides	\(-\frac {2 b^{2}}{7 x^{\frac {7}{2}}}-\frac {4 a b}{5 x^{\frac {5}{2}}}-\frac {2 a^{2}}{3 x^{\frac {3}{2}}}\)	\(25\)
default	\(-\frac {2 b^{2}}{7 x^{\frac {7}{2}}}-\frac {4 a b}{5 x^{\frac {5}{2}}}-\frac {2 a^{2}}{3 x^{\frac {3}{2}}}\)	\(25\)
trager	\(-\frac {2 \left (35 a^{2} x^{2}+42 a b x +15 b^{2}\right )}{105 x^{\frac {7}{2}}}\)	\(25\)
risch	\(-\frac {2 \left (35 a^{2} x^{2}+42 a b x +15 b^{2}\right )}{105 x^{\frac {7}{2}}}\)	\(25\)

[In]

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

[Out]

-2/105*(35*a^2*x^2+42*a*b*x+15*b^2)/x^(7/2)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {\left (a+\frac {b}{x}\right )^2}{x^{5/2}} \, dx=-\frac {2 \, {\left (35 \, a^{2} x^{2} + 42 \, a b x + 15 \, b^{2}\right )}}{105 \, x^{\frac {7}{2}}} \]

[In]

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

[Out]

-2/105*(35*a^2*x^2 + 42*a*b*x + 15*b^2)/x^(7/2)

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+\frac {b}{x}\right )^2}{x^{5/2}} \, dx=- \frac {2 a^{2}}{3 x^{\frac {3}{2}}} - \frac {4 a b}{5 x^{\frac {5}{2}}} - \frac {2 b^{2}}{7 x^{\frac {7}{2}}} \]

[In]

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

[Out]

-2*a**2/(3*x**(3/2)) - 4*a*b/(5*x**(5/2)) - 2*b**2/(7*x**(7/2))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {\left (a+\frac {b}{x}\right )^2}{x^{5/2}} \, dx=-\frac {2 \, a^{2}}{3 \, x^{\frac {3}{2}}} - \frac {4 \, a b}{5 \, x^{\frac {5}{2}}} - \frac {2 \, b^{2}}{7 \, x^{\frac {7}{2}}} \]

[In]

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

[Out]

-2/3*a^2/x^(3/2) - 4/5*a*b/x^(5/2) - 2/7*b^2/x^(7/2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {\left (a+\frac {b}{x}\right )^2}{x^{5/2}} \, dx=-\frac {2 \, {\left (35 \, a^{2} x^{2} + 42 \, a b x + 15 \, b^{2}\right )}}{105 \, x^{\frac {7}{2}}} \]

[In]

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

[Out]

-2/105*(35*a^2*x^2 + 42*a*b*x + 15*b^2)/x^(7/2)

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.67 \[ \int \frac {\left (a+\frac {b}{x}\right )^2}{x^{5/2}} \, dx=-\frac {70\,a^2\,x^2+84\,a\,b\,x+30\,b^2}{105\,x^{7/2}} \]

[In]

int((a + b/x)^2/x^(5/2),x)

[Out]

-(30*b^2 + 70*a^2*x^2 + 84*a*b*x)/(105*x^(7/2))