3.1862 \(\int \frac{x^2}{(a+\frac{b}{x^2})^2} \, dx\)

Optimal. Leaf size=66 \[ \frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{2 a^{7/2}}-\frac{5 b x}{2 a^3}+\frac{5 x^3}{6 a^2}-\frac{x^5}{2 a \left (a x^2+b\right )} \]

[Out]

(-5*b*x)/(2*a^3) + (5*x^3)/(6*a^2) - x^5/(2*a*(b + a*x^2)) + (5*b^(3/2)*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(2*a^(7/2
))

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Rubi [A]  time = 0.0261688, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {263, 288, 302, 205} \[ \frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{2 a^{7/2}}-\frac{5 b x}{2 a^3}+\frac{5 x^3}{6 a^2}-\frac{x^5}{2 a \left (a x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b/x^2)^2,x]

[Out]

(-5*b*x)/(2*a^3) + (5*x^3)/(6*a^2) - x^5/(2*a*(b + a*x^2)) + (5*b^(3/2)*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(2*a^(7/2
))

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 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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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{x^2}{\left (a+\frac{b}{x^2}\right )^2} \, dx &=\int \frac{x^6}{\left (b+a x^2\right )^2} \, dx\\ &=-\frac{x^5}{2 a \left (b+a x^2\right )}+\frac{5 \int \frac{x^4}{b+a x^2} \, dx}{2 a}\\ &=-\frac{x^5}{2 a \left (b+a x^2\right )}+\frac{5 \int \left (-\frac{b}{a^2}+\frac{x^2}{a}+\frac{b^2}{a^2 \left (b+a x^2\right )}\right ) \, dx}{2 a}\\ &=-\frac{5 b x}{2 a^3}+\frac{5 x^3}{6 a^2}-\frac{x^5}{2 a \left (b+a x^2\right )}+\frac{\left (5 b^2\right ) \int \frac{1}{b+a x^2} \, dx}{2 a^3}\\ &=-\frac{5 b x}{2 a^3}+\frac{5 x^3}{6 a^2}-\frac{x^5}{2 a \left (b+a x^2\right )}+\frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{2 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0410853, size = 60, normalized size = 0.91 \[ \frac{x \left (-\frac{3 b^2}{a x^2+b}+2 a x^2-12 b\right )}{6 a^3}+\frac{5 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{2 a^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b/x^2)^2,x]

[Out]

(x*(-12*b + 2*a*x^2 - (3*b^2)/(b + a*x^2)))/(6*a^3) + (5*b^(3/2)*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(2*a^(7/2))

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Maple [A]  time = 0.007, size = 57, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}}{3\,{a}^{2}}}-2\,{\frac{bx}{{a}^{3}}}-{\frac{{b}^{2}x}{2\,{a}^{3} \left ( a{x}^{2}+b \right ) }}+{\frac{5\,{b}^{2}}{2\,{a}^{3}}\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(x^2/(a+1/x^2*b)^2,x)

[Out]

1/3*x^3/a^2-2*b*x/a^3-1/2*b^2/a^3*x/(a*x^2+b)+5/2*b^2/a^3/(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(x^2/(a+b/x^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.46385, size = 348, normalized size = 5.27 \begin{align*} \left [\frac{4 \, a^{2} x^{5} - 20 \, a b x^{3} - 30 \, b^{2} x + 15 \,{\left (a b x^{2} + b^{2}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{a x^{2} + 2 \, a x \sqrt{-\frac{b}{a}} - b}{a x^{2} + b}\right )}{12 \,{\left (a^{4} x^{2} + a^{3} b\right )}}, \frac{2 \, a^{2} x^{5} - 10 \, a b x^{3} - 15 \, b^{2} x + 15 \,{\left (a b x^{2} + b^{2}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a x \sqrt{\frac{b}{a}}}{b}\right )}{6 \,{\left (a^{4} x^{2} + a^{3} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(4*a^2*x^5 - 20*a*b*x^3 - 30*b^2*x + 15*(a*b*x^2 + b^2)*sqrt(-b/a)*log((a*x^2 + 2*a*x*sqrt(-b/a) - b)/(a
*x^2 + b)))/(a^4*x^2 + a^3*b), 1/6*(2*a^2*x^5 - 10*a*b*x^3 - 15*b^2*x + 15*(a*b*x^2 + b^2)*sqrt(b/a)*arctan(a*
x*sqrt(b/a)/b))/(a^4*x^2 + a^3*b)]

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Sympy [A]  time = 0.585575, size = 107, normalized size = 1.62 \begin{align*} - \frac{b^{2} x}{2 a^{4} x^{2} + 2 a^{3} b} - \frac{5 \sqrt{- \frac{b^{3}}{a^{7}}} \log{\left (- \frac{a^{3} \sqrt{- \frac{b^{3}}{a^{7}}}}{b} + x \right )}}{4} + \frac{5 \sqrt{- \frac{b^{3}}{a^{7}}} \log{\left (\frac{a^{3} \sqrt{- \frac{b^{3}}{a^{7}}}}{b} + x \right )}}{4} + \frac{x^{3}}{3 a^{2}} - \frac{2 b x}{a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b**2*x/(2*a**4*x**2 + 2*a**3*b) - 5*sqrt(-b**3/a**7)*log(-a**3*sqrt(-b**3/a**7)/b + x)/4 + 5*sqrt(-b**3/a**7)
*log(a**3*sqrt(-b**3/a**7)/b + x)/4 + x**3/(3*a**2) - 2*b*x/a**3

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Giac [A]  time = 1.15197, size = 82, normalized size = 1.24 \begin{align*} \frac{5 \, b^{2} \arctan \left (\frac{a x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} a^{3}} - \frac{b^{2} x}{2 \,{\left (a x^{2} + b\right )} a^{3}} + \frac{a^{4} x^{3} - 6 \, a^{3} b x}{3 \, a^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/2*b^2*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^3) - 1/2*b^2*x/((a*x^2 + b)*a^3) + 1/3*(a^4*x^3 - 6*a^3*b*x)/a^6