3.1822 \(\int (a+\frac {b}{x^2})^2 x \, dx\)

Optimal. Leaf size=27 \[ \frac {a^2 x^2}{2}+2 a b \log (x)-\frac {b^2}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {263, 266, 43} \[ \frac {a^2 x^2}{2}+2 a b \log (x)-\frac {b^2}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^2*x,x]

[Out]

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

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Mathematica [A]  time = 0.00, size = 27, normalized size = 1.00 \[ \frac {a^2 x^2}{2}+2 a b \log (x)-\frac {b^2}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^2*x,x]

[Out]

-1/2*b^2/x^2 + (a^2*x^2)/2 + 2*a*b*Log[x]

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fricas [A]  time = 0.83, size = 27, normalized size = 1.00 \[ \frac {a^{2} x^{4} + 4 \, a b x^{2} \log \relax (x) - b^{2}}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 24, normalized size = 0.89 \[ \frac {1}{2} \, a^{2} x^{2} + 2 \, a b \log \left ({\left | x \right |}\right ) - \frac {b^{2}}{2 \, x^{2}} \]

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

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maple [A]  time = 0.01, size = 24, normalized size = 0.89 \[ \frac {a^{2} x^{2}}{2}+2 a b \ln \relax (x )-\frac {b^{2}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.88, size = 24, normalized size = 0.89 \[ \frac {1}{2} \, a^{2} x^{2} + a b \log \left (x^{2}\right ) - \frac {b^{2}}{2 \, x^{2}} \]

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*x^2 + a*b*log(x^2) - 1/2*b^2/x^2

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mupad [B]  time = 1.06, size = 23, normalized size = 0.85 \[ \frac {a^2\,x^2}{2}-\frac {b^2}{2\,x^2}+2\,a\,b\,\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b/x^2)^2,x)

[Out]

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

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sympy [A]  time = 0.14, size = 24, normalized size = 0.89 \[ \frac {a^{2} x^{2}}{2} + 2 a b \log {\relax (x )} - \frac {b^{2}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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