3.19.62 \(\int \frac {x^2}{(a+\frac {b}{x^2})^2} \, dx\) [1862]

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

[Out]

-5/2*b*x/a^3+5/6*x^3/a^2-1/2*x^5/a/(a*x^2+b)+5/2*b^(3/2)*arctan(x*a^(1/2)/b^(1/2))/a^(7/2)

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Rubi [A]
time = 0.02, antiderivative size = 66, normalized size of antiderivative = 1.00, 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 = {269, 294, 308, 211} \begin {gather*} \frac {5 b^{3/2} \text {ArcTan}\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 )} \end {gather*}

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 211

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

Rule 269

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 294

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 308

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]

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*}

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Mathematica [A]
time = 0.03, size = 60, normalized size = 0.91 \begin {gather*} \frac {x \left (-12 b+2 a x^2-\frac {3 b^2}{b+a x^2}\right )}{6 a^3}+\frac {5 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 a^{7/2}} \end {gather*}

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 = 0.04, size = 53, normalized size = 0.80

method	result	size
default	\(\frac {\frac {1}{3} a \,x^{3}-2 b x}{a^{3}}+\frac {b^{2} \left (-\frac {x}{2 \left (a \,x^{2}+b \right )}+\frac {5 \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{3}}\)	\(53\)
risch	\(\frac {x^{3}}{3 a^{2}}-\frac {2 b x}{a^{3}}-\frac {b^{2} x}{2 a^{3} \left (a \,x^{2}+b \right )}+\frac {5 \sqrt {-a b}\, b \ln \left (-\sqrt {-a b}\, x +b \right )}{4 a^{4}}-\frac {5 \sqrt {-a b}\, b \ln \left (\sqrt {-a b}\, x +b \right )}{4 a^{4}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(b/x^2+a)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.51, size = 59, normalized size = 0.89 \begin {gather*} -\frac {b^{2} x}{2 \, {\left (a^{4} x^{2} + a^{3} b\right )}} + \frac {5 \, b^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{3}} + \frac {a x^{3} - 6 \, b x}{3 \, a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 164, normalized size = 2.48 \begin {gather*} \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 {gather*}

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 = 0.13, size = 107, normalized size = 1.62 \begin {gather*} - \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 {gather*}

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 = 0.78, size = 61, normalized size = 0.92 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.07, size = 56, normalized size = 0.85 \begin {gather*} \frac {x^3}{3\,a^2}+\frac {5\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{2\,a^{7/2}}-\frac {b^2\,x}{2\,\left (a^4\,x^2+b\,a^3\right )}-\frac {2\,b\,x}{a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a + b/x^2)^2,x)

[Out]

x^3/(3*a^2) + (5*b^(3/2)*atan((a^(1/2)*x)/b^(1/2)))/(2*a^(7/2)) - (b^2*x)/(2*(a^3*b + a^4*x^2)) - (2*b*x)/a^3

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