3.1863 \(\int \frac{x}{(a+\frac{b}{x^2})^2} \, dx\)

Optimal. Leaf size=44 \[ -\frac{b^2}{2 a^3 \left (a x^2+b\right )}-\frac{b \log \left (a x^2+b\right )}{a^3}+\frac{x^2}{2 a^2} \]

[Out]

x^2/(2*a^2) - b^2/(2*a^3*(b + a*x^2)) - (b*Log[b + a*x^2])/a^3

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Rubi [A]  time = 0.0312199, antiderivative size = 44, normalized size of antiderivative = 1., 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} \[ -\frac{b^2}{2 a^3 \left (a x^2+b\right )}-\frac{b \log \left (a x^2+b\right )}{a^3}+\frac{x^2}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b/x^2)^2,x]

[Out]

x^2/(2*a^2) - b^2/(2*a^3*(b + a*x^2)) - (b*Log[b + a*x^2])/a^3

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]]

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])

Rubi steps

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

Mathematica [A]  time = 0.0167486, size = 38, normalized size = 0.86 \[ \frac{-\frac{b^2}{a x^2+b}-2 b \log \left (a x^2+b\right )+a x^2}{2 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b/x^2)^2,x]

[Out]

(a*x^2 - b^2/(b + a*x^2) - 2*b*Log[b + a*x^2])/(2*a^3)

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Maple [A]  time = 0.009, size = 41, normalized size = 0.9 \begin{align*}{\frac{{x}^{2}}{2\,{a}^{2}}}-{\frac{{b}^{2}}{2\,{a}^{3} \left ( a{x}^{2}+b \right ) }}-{\frac{b\ln \left ( a{x}^{2}+b \right ) }{{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.987699, size = 58, normalized size = 1.32 \begin{align*} -\frac{b^{2}}{2 \,{\left (a^{4} x^{2} + a^{3} b\right )}} + \frac{x^{2}}{2 \, a^{2}} - \frac{b \log \left (a x^{2} + b\right )}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.42715, size = 113, normalized size = 2.57 \begin{align*} \frac{a^{2} x^{4} + a b x^{2} - b^{2} - 2 \,{\left (a b x^{2} + b^{2}\right )} \log \left (a x^{2} + b\right )}{2 \,{\left (a^{4} x^{2} + a^{3} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.483175, size = 39, normalized size = 0.89 \begin{align*} - \frac{b^{2}}{2 a^{4} x^{2} + 2 a^{3} b} + \frac{x^{2}}{2 a^{2}} - \frac{b \log{\left (a x^{2} + b \right )}}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22766, size = 55, normalized size = 1.25 \begin{align*} \frac{x^{2}}{2 \, a^{2}} - \frac{b \log \left ({\left | a x^{2} + b \right |}\right )}{a^{3}} - \frac{b^{2}}{2 \,{\left (a x^{2} + b\right )} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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