3.240 \(\int \frac{x^2}{(a-b x^2)^3} \, dx\)

Optimal. Leaf size=67 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{3/2}}-\frac{x}{8 a b \left (a-b x^2\right )}+\frac{x}{4 b \left (a-b x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0197181, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {288, 199, 208} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{3/2}}-\frac{x}{8 a b \left (a-b x^2\right )}+\frac{x}{4 b \left (a-b x^2\right )^2} \]

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)) - ArcTanh[(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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[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{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0293294, size = 56, normalized size = 0.84 \[ \frac{x \left (a+b x^2\right )}{8 a b \left (a-b x^2\right )^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{8 a^{3/2} b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a - b*x^2)^3,x]

[Out]

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

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Maple [A]  time = 0.006, size = 52, 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}{\it Artanh} \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)*arctanh(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 = 1.29872, size = 394, normalized size = 5.88 \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 = 0.47619, size = 104, normalized size = 1.55 \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 = 2.14536, size = 72, normalized size = 1.07 \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)