3.1657 \(\int \frac {(a+\frac {b}{x})^2}{x^{3/2}} \, dx\)

Optimal. Leaf size=34 \[ -\frac {2 a^2}{\sqrt {x}}-\frac {4 a b}{3 x^{3/2}}-\frac {2 b^2}{5 x^{5/2}} \]

[Out]

-2/5*b^2/x^(5/2)-4/3*a*b/x^(3/2)-2*a^2/x^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {263, 43} \[ -\frac {2 a^2}{\sqrt {x}}-\frac {4 a b}{3 x^{3/2}}-\frac {2 b^2}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^2/x^(3/2),x]

[Out]

(-2*b^2)/(5*x^(5/2)) - (4*a*b)/(3*x^(3/2)) - (2*a^2)/Sqrt[x]

Rule 43

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 263

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

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.82 \[ -\frac {2 \left (15 a^2 x^2+10 a b x+3 b^2\right )}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^2/x^(3/2),x]

[Out]

(-2*(3*b^2 + 10*a*b*x + 15*a^2*x^2))/(15*x^(5/2))

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fricas [A]  time = 0.90, size = 24, normalized size = 0.71 \[ -\frac {2 \, {\left (15 \, a^{2} x^{2} + 10 \, a b x + 3 \, b^{2}\right )}}{15 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*a^2*x^2 + 10*a*b*x + 3*b^2)/x^(5/2)

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giac [A]  time = 0.15, size = 24, normalized size = 0.71 \[ -\frac {2 \, {\left (15 \, a^{2} x^{2} + 10 \, a b x + 3 \, b^{2}\right )}}{15 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(15*a^2*x^2 + 10*a*b*x + 3*b^2)/x^(5/2)

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maple [A]  time = 0.00, size = 25, normalized size = 0.74 \[ -\frac {2 \left (15 a^{2} x^{2}+10 a b x +3 b^{2}\right )}{15 x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^2/x^(3/2),x)

[Out]

-2/15*(15*a^2*x^2+10*a*b*x+3*b^2)/x^(5/2)

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maxima [A]  time = 1.07, size = 24, normalized size = 0.71 \[ -\frac {2 \, a^{2}}{\sqrt {x}} - \frac {4 \, a b}{3 \, x^{\frac {3}{2}}} - \frac {2 \, b^{2}}{5 \, x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 24, normalized size = 0.71 \[ -\frac {30\,a^2\,x^2+20\,a\,b\,x+6\,b^2}{15\,x^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^2/x^(3/2),x)

[Out]

-(6*b^2 + 30*a^2*x^2 + 20*a*b*x)/(15*x^(5/2))

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sympy [A]  time = 0.85, size = 34, normalized size = 1.00 \[ - \frac {2 a^{2}}{\sqrt {x}} - \frac {4 a b}{3 x^{\frac {3}{2}}} - \frac {2 b^{2}}{5 x^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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