3.1824 \(\int \frac {(a+\frac {b}{x^2})^2}{x} \, dx\)

Optimal. Leaf size=24 \[ a^2 \log (x)-\frac {a b}{x^2}-\frac {b^2}{4 x^4} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {263, 266, 43} \[ a^2 \log (x)-\frac {a b}{x^2}-\frac {b^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^2/x,x]

[Out]

-b^2/(4*x^4) - (a*b)/x^2 + a^2*Log[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]

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+\frac {b}{x^2}\right )^2}{x} \, dx &=\int \frac {\left (b+a x^2\right )^2}{x^5} \, dx\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {(b+a x)^2}{x^3} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {b^2}{x^3}+\frac {2 a b}{x^2}+\frac {a^2}{x}\right ) \, dx,x,x^2\right )\\ &=-\frac {b^2}{4 x^4}-\frac {a b}{x^2}+a^2 \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 24, normalized size = 1.00 \[ a^2 \log (x)-\frac {a b}{x^2}-\frac {b^2}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^2/x,x]

[Out]

-1/4*b^2/x^4 - (a*b)/x^2 + a^2*Log[x]

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fricas [A]  time = 0.75, size = 28, normalized size = 1.17 \[ \frac {4 \, a^{2} x^{4} \log \relax (x) - 4 \, a b x^{2} - b^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 34, normalized size = 1.42 \[ \frac {1}{2} \, a^{2} \log \left (x^{2}\right ) - \frac {3 \, a^{2} x^{4} + 4 \, a b x^{2} + b^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 23, normalized size = 0.96 \[ a^{2} \ln \relax (x )-\frac {a b}{x^{2}}-\frac {b^{2}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.88, size = 26, normalized size = 1.08 \[ \frac {1}{2} \, a^{2} \log \left (x^{2}\right ) - \frac {4 \, a b x^{2} + b^{2}}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 24, normalized size = 1.00 \[ a^2\,\ln \relax (x)-\frac {\frac {b^2}{4}+a\,b\,x^2}{x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)^2/x,x)

[Out]

a^2*log(x) - (b^2/4 + a*b*x^2)/x^4

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sympy [A]  time = 0.18, size = 24, normalized size = 1.00 \[ a^{2} \log {\relax (x )} + \frac {- 4 a b x^{2} - b^{2}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*log(x) + (-4*a*b*x**2 - b**2)/(4*x**4)

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