3.507 \(\int \frac{x^2}{(a^2+2 a b x^2+b^2 x^4)^2} \, dx\)

Optimal. Leaf size=85 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 a^{5/2} b^{3/2}}+\frac{x}{16 a^2 b \left (a+b x^2\right )}+\frac{x}{24 a b \left (a+b x^2\right )^2}-\frac{x}{6 b \left (a+b x^2\right )^3} \]

[Out]

-x/(6*b*(a + b*x^2)^3) + x/(24*a*b*(a + b*x^2)^2) + x/(16*a^2*b*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(16
*a^(5/2)*b^(3/2))

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Rubi [A]  time = 0.0426251, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {28, 288, 199, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 a^{5/2} b^{3/2}}+\frac{x}{16 a^2 b \left (a+b x^2\right )}+\frac{x}{24 a b \left (a+b x^2\right )^2}-\frac{x}{6 b \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

-x/(6*b*(a + b*x^2)^3) + x/(24*a*b*(a + b*x^2)^2) + x/(16*a^2*b*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(16
*a^(5/2)*b^(3/2))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac{x^2}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=-\frac{x}{6 b \left (a+b x^2\right )^3}+\frac{1}{6} b^2 \int \frac{1}{\left (a b+b^2 x^2\right )^3} \, dx\\ &=-\frac{x}{6 b \left (a+b x^2\right )^3}+\frac{x}{24 a b \left (a+b x^2\right )^2}+\frac{b \int \frac{1}{\left (a b+b^2 x^2\right )^2} \, dx}{8 a}\\ &=-\frac{x}{6 b \left (a+b x^2\right )^3}+\frac{x}{24 a b \left (a+b x^2\right )^2}+\frac{x}{16 a^2 b \left (a+b x^2\right )}+\frac{\int \frac{1}{a b+b^2 x^2} \, dx}{16 a^2}\\ &=-\frac{x}{6 b \left (a+b x^2\right )^3}+\frac{x}{24 a b \left (a+b x^2\right )^2}+\frac{x}{16 a^2 b \left (a+b x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 a^{5/2} b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.037126, size = 69, normalized size = 0.81 \[ \frac{-3 a^2 x+8 a b x^3+3 b^2 x^5}{48 a^2 b \left (a+b x^2\right )^3}+\frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{16 a^{5/2} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(-3*a^2*x + 8*a*b*x^3 + 3*b^2*x^5)/(48*a^2*b*(a + b*x^2)^3) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(16*a^(5/2)*b^(3/2))

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Maple [A]  time = 0.049, size = 58, normalized size = 0.7 \begin{align*}{\frac{1}{ \left ( b{x}^{2}+a \right ) ^{3}} \left ({\frac{b{x}^{5}}{16\,{a}^{2}}}+{\frac{{x}^{3}}{6\,a}}-{\frac{x}{16\,b}} \right ) }+{\frac{1}{16\,b{a}^{2}}\arctan \left ({bx{\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/(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

(1/16*b/a^2*x^5+1/6/a*x^3-1/16*x/b)/(b*x^2+a)^3+1/16/b/a^2/(a*b)^(1/2)*arctan(b*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/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72973, size = 537, normalized size = 6.32 \begin{align*} \left [\frac{6 \, a b^{3} x^{5} + 16 \, a^{2} b^{2} x^{3} - 6 \, a^{3} b x - 3 \,{\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{96 \,{\left (a^{3} b^{5} x^{6} + 3 \, a^{4} b^{4} x^{4} + 3 \, a^{5} b^{3} x^{2} + a^{6} b^{2}\right )}}, \frac{3 \, a b^{3} x^{5} + 8 \, a^{2} b^{2} x^{3} - 3 \, a^{3} b x + 3 \,{\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{48 \,{\left (a^{3} b^{5} x^{6} + 3 \, a^{4} b^{4} x^{4} + 3 \, a^{5} b^{3} x^{2} + a^{6} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

[1/96*(6*a*b^3*x^5 + 16*a^2*b^2*x^3 - 6*a^3*b*x - 3*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*sqrt(-a*b)*log
((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^3*b^5*x^6 + 3*a^4*b^4*x^4 + 3*a^5*b^3*x^2 + a^6*b^2), 1/48*(3*a
*b^3*x^5 + 8*a^2*b^2*x^3 - 3*a^3*b*x + 3*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*sqrt(a*b)*arctan(sqrt(a*b
)*x/a))/(a^3*b^5*x^6 + 3*a^4*b^4*x^4 + 3*a^5*b^3*x^2 + a^6*b^2)]

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Sympy [B]  time = 0.611216, size = 139, normalized size = 1.64 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{5} b^{3}}} \log{\left (- a^{3} b \sqrt{- \frac{1}{a^{5} b^{3}}} + x \right )}}{32} + \frac{\sqrt{- \frac{1}{a^{5} b^{3}}} \log{\left (a^{3} b \sqrt{- \frac{1}{a^{5} b^{3}}} + x \right )}}{32} + \frac{- 3 a^{2} x + 8 a b x^{3} + 3 b^{2} x^{5}}{48 a^{5} b + 144 a^{4} b^{2} x^{2} + 144 a^{3} b^{3} x^{4} + 48 a^{2} b^{4} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

-sqrt(-1/(a**5*b**3))*log(-a**3*b*sqrt(-1/(a**5*b**3)) + x)/32 + sqrt(-1/(a**5*b**3))*log(a**3*b*sqrt(-1/(a**5
*b**3)) + x)/32 + (-3*a**2*x + 8*a*b*x**3 + 3*b**2*x**5)/(48*a**5*b + 144*a**4*b**2*x**2 + 144*a**3*b**3*x**4
+ 48*a**2*b**4*x**6)

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Giac [A]  time = 1.16199, size = 84, normalized size = 0.99 \begin{align*} \frac{\arctan \left (\frac{b x}{\sqrt{a b}}\right )}{16 \, \sqrt{a b} a^{2} b} + \frac{3 \, b^{2} x^{5} + 8 \, a b x^{3} - 3 \, a^{2} x}{48 \,{\left (b x^{2} + a\right )}^{3} a^{2} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

1/16*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b) + 1/48*(3*b^2*x^5 + 8*a*b*x^3 - 3*a^2*x)/((b*x^2 + a)^3*a^2*b)