\(\int \frac {(a+b \sqrt {x})^2}{x^3} \, dx\) [2127]

Optimal result

Integrand size = 15, antiderivative size = 30 \[ \int \frac {\left (a+b \sqrt {x}\right )^2}{x^3} \, dx=-\frac {a^2}{2 x^2}-\frac {4 a b}{3 x^{3/2}}-\frac {b^2}{x} \]

[Out]

-1/2*a^2/x^2-4/3*a*b/x^(3/2)-b^2/x

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 30, 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 = {272, 45} \[ \int \frac {\left (a+b \sqrt {x}\right )^2}{x^3} \, dx=-\frac {a^2}{2 x^2}-\frac {4 a b}{3 x^{3/2}}-\frac {b^2}{x} \]

[In]

Int[(a + b*Sqrt[x])^2/x^3,x]

[Out]

-1/2*a^2/x^2 - (4*a*b)/(3*x^(3/2)) - b^2/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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = 2 \text {Subst}\left (\int \frac {(a+b x)^2}{x^5} \, dx,x,\sqrt {x}\right ) \\ & = 2 \text {Subst}\left (\int \left (\frac {a^2}{x^5}+\frac {2 a b}{x^4}+\frac {b^2}{x^3}\right ) \, dx,x,\sqrt {x}\right ) \\ & = -\frac {a^2}{2 x^2}-\frac {4 a b}{3 x^{3/2}}-\frac {b^2}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {\left (a+b \sqrt {x}\right )^2}{x^3} \, dx=\frac {-3 a^2-8 a b \sqrt {x}-6 b^2 x}{6 x^2} \]

[In]

Integrate[(a + b*Sqrt[x])^2/x^3,x]

[Out]

(-3*a^2 - 8*a*b*Sqrt[x] - 6*b^2*x)/(6*x^2)

Maple [A] (verified)

Time = 3.50 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {a^{2}}{2 x^{2}}-\frac {4 a b}{3 x^{\frac {3}{2}}}-\frac {b^{2}}{x}\)	\(25\)
default	\(-\frac {a^{2}}{2 x^{2}}-\frac {4 a b}{3 x^{\frac {3}{2}}}-\frac {b^{2}}{x}\)	\(25\)
trager	\(\frac {\left (-1+x \right ) \left (a^{2} x +2 b^{2} x +a^{2}\right )}{2 x^{2}}-\frac {4 a b}{3 x^{\frac {3}{2}}}\)	\(32\)

[In]

int((a+b*x^(1/2))^2/x^3,x,method=_RETURNVERBOSE)

[Out]

-1/2*a^2/x^2-4/3*a*b/x^(3/2)-b^2/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b \sqrt {x}\right )^2}{x^3} \, dx=-\frac {6 \, b^{2} x + 8 \, a b \sqrt {x} + 3 \, a^{2}}{6 \, x^{2}} \]

[In]

integrate((a+b*x^(1/2))^2/x^3,x, algorithm="fricas")

[Out]

-1/6*(6*b^2*x + 8*a*b*sqrt(x) + 3*a^2)/x^2

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {\left (a+b \sqrt {x}\right )^2}{x^3} \, dx=- \frac {a^{2}}{2 x^{2}} - \frac {4 a b}{3 x^{\frac {3}{2}}} - \frac {b^{2}}{x} \]

[In]

integrate((a+b*x**(1/2))**2/x**3,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b \sqrt {x}\right )^2}{x^3} \, dx=-\frac {6 \, b^{2} x + 8 \, a b \sqrt {x} + 3 \, a^{2}}{6 \, x^{2}} \]

[In]

integrate((a+b*x^(1/2))^2/x^3,x, algorithm="maxima")

[Out]

-1/6*(6*b^2*x + 8*a*b*sqrt(x) + 3*a^2)/x^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b \sqrt {x}\right )^2}{x^3} \, dx=-\frac {6 \, b^{2} x + 8 \, a b \sqrt {x} + 3 \, a^{2}}{6 \, x^{2}} \]

[In]

integrate((a+b*x^(1/2))^2/x^3,x, algorithm="giac")

[Out]

-1/6*(6*b^2*x + 8*a*b*sqrt(x) + 3*a^2)/x^2

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b \sqrt {x}\right )^2}{x^3} \, dx=-\frac {6\,b^2\,x+3\,a^2+8\,a\,b\,\sqrt {x}}{6\,x^2} \]

[In]

int((a + b*x^(1/2))^2/x^3,x)

[Out]

-(6*b^2*x + 3*a^2 + 8*a*b*x^(1/2))/(6*x^2)