3.2128 \(\int \frac {(a+b \sqrt {x})^2}{x^4} \, dx\)

Optimal. Leaf size=32 \[ -\frac {a^2}{3 x^3}-\frac {4 a b}{5 x^{5/2}}-\frac {b^2}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 32, 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 = {266, 43} \[ -\frac {a^2}{3 x^3}-\frac {4 a b}{5 x^{5/2}}-\frac {b^2}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^2/x^4,x]

[Out]

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

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 266

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*} \int \frac {\left (a+b \sqrt {x}\right )^2}{x^4} \, dx &=2 \operatorname {Subst}\left (\int \frac {(a+b x)^2}{x^7} \, dx,x,\sqrt {x}\right )\\ &=2 \operatorname {Subst}\left (\int \left (\frac {a^2}{x^7}+\frac {2 a b}{x^6}+\frac {b^2}{x^5}\right ) \, dx,x,\sqrt {x}\right )\\ &=-\frac {a^2}{3 x^3}-\frac {4 a b}{5 x^{5/2}}-\frac {b^2}{2 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.88 \[ -\frac {10 a^2+24 a b \sqrt {x}+15 b^2 x}{30 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^2/x^4,x]

[Out]

-1/30*(10*a^2 + 24*a*b*Sqrt[x] + 15*b^2*x)/x^3

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fricas [A]  time = 0.84, size = 24, normalized size = 0.75 \[ -\frac {15 \, b^{2} x + 24 \, a b \sqrt {x} + 10 \, a^{2}}{30 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/2))^2/x^4,x, algorithm="fricas")

[Out]

-1/30*(15*b^2*x + 24*a*b*sqrt(x) + 10*a^2)/x^3

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giac [A]  time = 0.16, size = 24, normalized size = 0.75 \[ -\frac {15 \, b^{2} x + 24 \, a b \sqrt {x} + 10 \, a^{2}}{30 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/2))^2/x^4,x, algorithm="giac")

[Out]

-1/30*(15*b^2*x + 24*a*b*sqrt(x) + 10*a^2)/x^3

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maple [A]  time = 0.00, size = 25, normalized size = 0.78 \[ -\frac {b^{2}}{2 x^{2}}-\frac {4 a b}{5 x^{\frac {5}{2}}}-\frac {a^{2}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*x^(1/2))^2/x^4,x)

[Out]

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

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maxima [A]  time = 0.84, size = 24, normalized size = 0.75 \[ -\frac {15 \, b^{2} x + 24 \, a b \sqrt {x} + 10 \, a^{2}}{30 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/2))^2/x^4,x, algorithm="maxima")

[Out]

-1/30*(15*b^2*x + 24*a*b*sqrt(x) + 10*a^2)/x^3

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mupad [B]  time = 0.03, size = 24, normalized size = 0.75 \[ -\frac {15\,b^2\,x+10\,a^2+24\,a\,b\,\sqrt {x}}{30\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^(1/2))^2/x^4,x)

[Out]

-(15*b^2*x + 10*a^2 + 24*a*b*x^(1/2))/(30*x^3)

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sympy [A]  time = 1.06, size = 29, normalized size = 0.91 \[ - \frac {a^{2}}{3 x^{3}} - \frac {4 a b}{5 x^{\frac {5}{2}}} - \frac {b^{2}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x**(1/2))**2/x**4,x)

[Out]

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

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