3.1865 \(\int \frac{1}{(a+\frac{b}{x^2})^2 x} \, dx\)

Optimal. Leaf size=33 \[ \frac{b}{2 a^2 \left (a x^2+b\right )}+\frac{\log \left (a x^2+b\right )}{2 a^2} \]

[Out]

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

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Rubi [A]  time = 0.0241786, antiderivative size = 33, normalized size of antiderivative = 1., 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} \[ \frac{b}{2 a^2 \left (a x^2+b\right )}+\frac{\log \left (a x^2+b\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^2*x),x]

[Out]

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

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

Mathematica [A]  time = 0.007832, size = 27, normalized size = 0.82 \[ \frac{\frac{b}{a x^2+b}+\log \left (a x^2+b\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^2*x),x]

[Out]

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

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Maple [A]  time = 0.007, size = 30, normalized size = 0.9 \begin{align*}{\frac{b}{2\,{a}^{2} \left ( a{x}^{2}+b \right ) }}+{\frac{\ln \left ( a{x}^{2}+b \right ) }{2\,{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.986698, size = 43, normalized size = 1.3 \begin{align*} \frac{b}{2 \,{\left (a^{3} x^{2} + a^{2} b\right )}} + \frac{\log \left (a x^{2} + b\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.43589, size = 76, normalized size = 2.3 \begin{align*} \frac{{\left (a x^{2} + b\right )} \log \left (a x^{2} + b\right ) + b}{2 \,{\left (a^{3} x^{2} + a^{2} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.433492, size = 29, normalized size = 0.88 \begin{align*} \frac{b}{2 a^{3} x^{2} + 2 a^{2} b} + \frac{\log{\left (a x^{2} + b \right )}}{2 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12806, size = 43, normalized size = 1.3 \begin{align*} -\frac{x^{2}}{2 \,{\left (a x^{2} + b\right )} a} + \frac{\log \left ({\left | a x^{2} + b \right |}\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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