3.1881 \(\int \frac{1}{(a+\frac{b}{x^2})^3 x^2} \, dx\)

Optimal. Leaf size=64 \[ -\frac{3 x}{8 a^2 \left (a x^2+b\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 a^{5/2} \sqrt{b}}-\frac{x^3}{4 a \left (a x^2+b\right )^2} \]

[Out]

-x^3/(4*a*(b + a*x^2)^2) - (3*x)/(8*a^2*(b + a*x^2)) + (3*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(8*a^(5/2)*Sqrt[b])

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Rubi [A]  time = 0.0200818, antiderivative size = 64, 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, 288, 205} \[ -\frac{3 x}{8 a^2 \left (a x^2+b\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 a^{5/2} \sqrt{b}}-\frac{x^3}{4 a \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^3*x^2),x]

[Out]

-x^3/(4*a*(b + a*x^2)^2) - (3*x)/(8*a^2*(b + a*x^2)) + (3*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(8*a^(5/2)*Sqrt[b])

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 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+\frac{b}{x^2}\right )^3 x^2} \, dx &=\int \frac{x^4}{\left (b+a x^2\right )^3} \, dx\\ &=-\frac{x^3}{4 a \left (b+a x^2\right )^2}+\frac{3 \int \frac{x^2}{\left (b+a x^2\right )^2} \, dx}{4 a}\\ &=-\frac{x^3}{4 a \left (b+a x^2\right )^2}-\frac{3 x}{8 a^2 \left (b+a x^2\right )}+\frac{3 \int \frac{1}{b+a x^2} \, dx}{8 a^2}\\ &=-\frac{x^3}{4 a \left (b+a x^2\right )^2}-\frac{3 x}{8 a^2 \left (b+a x^2\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 a^{5/2} \sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0403315, size = 55, normalized size = 0.86 \[ \frac{3 \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 a^{5/2} \sqrt{b}}-\frac{5 a x^3+3 b x}{8 a^2 \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^3*x^2),x]

[Out]

-(3*b*x + 5*a*x^3)/(8*a^2*(b + a*x^2)^2) + (3*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(8*a^(5/2)*Sqrt[b])

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Maple [A]  time = 0.007, size = 47, normalized size = 0.7 \begin{align*}{\frac{1}{ \left ( a{x}^{2}+b \right ) ^{2}} \left ( -{\frac{5\,{x}^{3}}{8\,a}}-{\frac{3\,bx}{8\,{a}^{2}}} \right ) }+{\frac{3}{8\,{a}^{2}}\arctan \left ({ax{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-5/8*x^3/a-3/8*b*x/a^2)/(a*x^2+b)^2+3/8/a^2/(a*b)^(1/2)*arctan(a*x/(a*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^3/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.45685, size = 404, normalized size = 6.31 \begin{align*} \left [-\frac{10 \, a^{2} b x^{3} + 6 \, a b^{2} x + 3 \,{\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt{-a b} \log \left (\frac{a x^{2} - 2 \, \sqrt{-a b} x - b}{a x^{2} + b}\right )}{16 \,{\left (a^{5} b x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{3} b^{3}\right )}}, -\frac{5 \, a^{2} b x^{3} + 3 \, a b^{2} x - 3 \,{\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{b}\right )}{8 \,{\left (a^{5} b x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{3} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^3/x^2,x, algorithm="fricas")

[Out]

[-1/16*(10*a^2*b*x^3 + 6*a*b^2*x + 3*(a^2*x^4 + 2*a*b*x^2 + b^2)*sqrt(-a*b)*log((a*x^2 - 2*sqrt(-a*b)*x - b)/(
a*x^2 + b)))/(a^5*b*x^4 + 2*a^4*b^2*x^2 + a^3*b^3), -1/8*(5*a^2*b*x^3 + 3*a*b^2*x - 3*(a^2*x^4 + 2*a*b*x^2 + b
^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/b))/(a^5*b*x^4 + 2*a^4*b^2*x^2 + a^3*b^3)]

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Sympy [A]  time = 0.638077, size = 109, normalized size = 1.7 \begin{align*} - \frac{3 \sqrt{- \frac{1}{a^{5} b}} \log{\left (- a^{2} b \sqrt{- \frac{1}{a^{5} b}} + x \right )}}{16} + \frac{3 \sqrt{- \frac{1}{a^{5} b}} \log{\left (a^{2} b \sqrt{- \frac{1}{a^{5} b}} + x \right )}}{16} - \frac{5 a x^{3} + 3 b x}{8 a^{4} x^{4} + 16 a^{3} b x^{2} + 8 a^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**3/x**2,x)

[Out]

-3*sqrt(-1/(a**5*b))*log(-a**2*b*sqrt(-1/(a**5*b)) + x)/16 + 3*sqrt(-1/(a**5*b))*log(a**2*b*sqrt(-1/(a**5*b))
+ x)/16 - (5*a*x**3 + 3*b*x)/(8*a**4*x**4 + 16*a**3*b*x**2 + 8*a**2*b**2)

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Giac [A]  time = 1.19668, size = 61, normalized size = 0.95 \begin{align*} \frac{3 \, \arctan \left (\frac{a x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{2}} - \frac{5 \, a x^{3} + 3 \, b x}{8 \,{\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)^3/x^2,x, algorithm="giac")

[Out]

3/8*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^2) - 1/8*(5*a*x^3 + 3*b*x)/((a*x^2 + b)^2*a^2)