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

Optimal. Leaf size=65 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{3/2}}+\frac{x}{8 a b \left (a+b x^2\right )}-\frac{x}{4 b \left (a+b x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0183032, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {288, 199, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{3/2}}+\frac{x}{8 a b \left (a+b x^2\right )}-\frac{x}{4 b \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b*x^2)^3,x]

[Out]

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

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

Mathematica [A]  time = 0.0271433, size = 58, normalized size = 0.89 \[ \frac{\frac{\sqrt{a} \sqrt{b} x \left (b x^2-a\right )}{\left (a+b x^2\right )^2}+\tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b*x^2)^3,x]

[Out]

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

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Maple [A]  time = 0.007, size = 49, normalized size = 0.8 \begin{align*}{\frac{1}{ \left ( b{x}^{2}+a \right ) ^{2}} \left ({\frac{{x}^{3}}{8\,a}}-{\frac{x}{8\,b}} \right ) }+{\frac{1}{8\,ab}\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*x^2+a)^3,x)

[Out]

(1/8/a*x^3-1/8*x/b)/(b*x^2+a)^2+1/8/b/a/(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*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.25451, size = 394, normalized size = 6.06 \begin{align*} \left [\frac{2 \, a b^{2} x^{3} - 2 \, a^{2} b x -{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt{-a b} \log \left (\frac{b x^{2} - 2 \, \sqrt{-a b} x - a}{b x^{2} + a}\right )}{16 \,{\left (a^{2} b^{4} x^{4} + 2 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}}, \frac{a b^{2} x^{3} - a^{2} b x +{\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b} x}{a}\right )}{8 \,{\left (a^{2} b^{4} x^{4} + 2 \, a^{3} b^{3} x^{2} + a^{4} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.493554, size = 110, normalized size = 1.69 \begin{align*} - \frac{\sqrt{- \frac{1}{a^{3} b^{3}}} \log{\left (- a^{2} b \sqrt{- \frac{1}{a^{3} b^{3}}} + x \right )}}{16} + \frac{\sqrt{- \frac{1}{a^{3} b^{3}}} \log{\left (a^{2} b \sqrt{- \frac{1}{a^{3} b^{3}}} + x \right )}}{16} + \frac{- a x + b x^{3}}{8 a^{3} b + 16 a^{2} b^{2} x^{2} + 8 a b^{3} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.75238, size = 68, normalized size = 1.05 \begin{align*} \frac{\arctan \left (\frac{b x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a b} + \frac{b x^{3} - a x}{8 \,{\left (b x^{2} + a\right )}^{2} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(b*x^2+a)^3,x, algorithm="giac")

[Out]

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