3.1877 \(\int \frac{x^2}{(a+\frac{b}{x^2})^3} \, dx\)

Optimal. Leaf size=85 \[ \frac{35 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 a^{9/2}}-\frac{7 x^5}{8 a^2 \left (a x^2+b\right )}-\frac{35 b x}{8 a^4}+\frac{35 x^3}{24 a^3}-\frac{x^7}{4 a \left (a x^2+b\right )^2} \]

[Out]

(-35*b*x)/(8*a^4) + (35*x^3)/(24*a^3) - x^7/(4*a*(b + a*x^2)^2) - (7*x^5)/(8*a^2*(b + a*x^2)) + (35*b^(3/2)*Ar
cTan[(Sqrt[a]*x)/Sqrt[b]])/(8*a^(9/2))

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Rubi [A]  time = 0.0358001, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {263, 288, 302, 205} \[ \frac{35 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 a^{9/2}}-\frac{7 x^5}{8 a^2 \left (a x^2+b\right )}-\frac{35 b x}{8 a^4}+\frac{35 x^3}{24 a^3}-\frac{x^7}{4 a \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[x^2/(a + b/x^2)^3,x]

[Out]

(-35*b*x)/(8*a^4) + (35*x^3)/(24*a^3) - x^7/(4*a*(b + a*x^2)^2) - (7*x^5)/(8*a^2*(b + a*x^2)) + (35*b^(3/2)*Ar
cTan[(Sqrt[a]*x)/Sqrt[b]])/(8*a^(9/2))

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 )^3} \, dx &=\int \frac{x^8}{\left (b+a x^2\right )^3} \, dx\\ &=-\frac{x^7}{4 a \left (b+a x^2\right )^2}+\frac{7 \int \frac{x^6}{\left (b+a x^2\right )^2} \, dx}{4 a}\\ &=-\frac{x^7}{4 a \left (b+a x^2\right )^2}-\frac{7 x^5}{8 a^2 \left (b+a x^2\right )}+\frac{35 \int \frac{x^4}{b+a x^2} \, dx}{8 a^2}\\ &=-\frac{x^7}{4 a \left (b+a x^2\right )^2}-\frac{7 x^5}{8 a^2 \left (b+a x^2\right )}+\frac{35 \int \left (-\frac{b}{a^2}+\frac{x^2}{a}+\frac{b^2}{a^2 \left (b+a x^2\right )}\right ) \, dx}{8 a^2}\\ &=-\frac{35 b x}{8 a^4}+\frac{35 x^3}{24 a^3}-\frac{x^7}{4 a \left (b+a x^2\right )^2}-\frac{7 x^5}{8 a^2 \left (b+a x^2\right )}+\frac{\left (35 b^2\right ) \int \frac{1}{b+a x^2} \, dx}{8 a^4}\\ &=-\frac{35 b x}{8 a^4}+\frac{35 x^3}{24 a^3}-\frac{x^7}{4 a \left (b+a x^2\right )^2}-\frac{7 x^5}{8 a^2 \left (b+a x^2\right )}+\frac{35 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 a^{9/2}}\\ \end{align*}

Mathematica [A]  time = 0.0437503, size = 77, normalized size = 0.91 \[ \frac{35 b^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 a^{9/2}}-\frac{56 a^2 b x^5-8 a^3 x^7+175 a b^2 x^3+105 b^3 x}{24 a^4 \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a + b/x^2)^3,x]

[Out]

-(105*b^3*x + 175*a*b^2*x^3 + 56*a^2*b*x^5 - 8*a^3*x^7)/(24*a^4*(b + a*x^2)^2) + (35*b^(3/2)*ArcTan[(Sqrt[a]*x
)/Sqrt[b]])/(8*a^(9/2))

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Maple [A]  time = 0.01, size = 77, normalized size = 0.9 \begin{align*}{\frac{{x}^{3}}{3\,{a}^{3}}}-3\,{\frac{bx}{{a}^{4}}}-{\frac{13\,{b}^{2}{x}^{3}}{8\,{a}^{3} \left ( a{x}^{2}+b \right ) ^{2}}}-{\frac{11\,{b}^{3}x}{8\,{a}^{4} \left ( a{x}^{2}+b \right ) ^{2}}}+{\frac{35\,{b}^{2}}{8\,{a}^{4}}\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)^3,x)

[Out]

1/3*x^3/a^3-3*b*x/a^4-13/8/a^3*b^2/(a*x^2+b)^2*x^3-11/8/a^4*b^3/(a*x^2+b)^2*x+35/8/a^4*b^2/(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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47591, size = 493, normalized size = 5.8 \begin{align*} \left [\frac{16 \, a^{3} x^{7} - 112 \, a^{2} b x^{5} - 350 \, a b^{2} x^{3} - 210 \, b^{3} x + 105 \,{\left (a^{2} b x^{4} + 2 \, a b^{2} x^{2} + b^{3}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{a x^{2} + 2 \, a x \sqrt{-\frac{b}{a}} - b}{a x^{2} + b}\right )}{48 \,{\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}}, \frac{8 \, a^{3} x^{7} - 56 \, a^{2} b x^{5} - 175 \, a b^{2} x^{3} - 105 \, b^{3} x + 105 \,{\left (a^{2} b x^{4} + 2 \, a b^{2} x^{2} + b^{3}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a x \sqrt{\frac{b}{a}}}{b}\right )}{24 \,{\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(16*a^3*x^7 - 112*a^2*b*x^5 - 350*a*b^2*x^3 - 210*b^3*x + 105*(a^2*b*x^4 + 2*a*b^2*x^2 + b^3)*sqrt(-b/a)
*log((a*x^2 + 2*a*x*sqrt(-b/a) - b)/(a*x^2 + b)))/(a^6*x^4 + 2*a^5*b*x^2 + a^4*b^2), 1/24*(8*a^3*x^7 - 56*a^2*
b*x^5 - 175*a*b^2*x^3 - 105*b^3*x + 105*(a^2*b*x^4 + 2*a*b^2*x^2 + b^3)*sqrt(b/a)*arctan(a*x*sqrt(b/a)/b))/(a^
6*x^4 + 2*a^5*b*x^2 + a^4*b^2)]

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Sympy [A]  time = 0.783442, size = 131, normalized size = 1.54 \begin{align*} - \frac{35 \sqrt{- \frac{b^{3}}{a^{9}}} \log{\left (- \frac{a^{4} \sqrt{- \frac{b^{3}}{a^{9}}}}{b} + x \right )}}{16} + \frac{35 \sqrt{- \frac{b^{3}}{a^{9}}} \log{\left (\frac{a^{4} \sqrt{- \frac{b^{3}}{a^{9}}}}{b} + x \right )}}{16} - \frac{13 a b^{2} x^{3} + 11 b^{3} x}{8 a^{6} x^{4} + 16 a^{5} b x^{2} + 8 a^{4} b^{2}} + \frac{x^{3}}{3 a^{3}} - \frac{3 b x}{a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-35*sqrt(-b**3/a**9)*log(-a**4*sqrt(-b**3/a**9)/b + x)/16 + 35*sqrt(-b**3/a**9)*log(a**4*sqrt(-b**3/a**9)/b +
x)/16 - (13*a*b**2*x**3 + 11*b**3*x)/(8*a**6*x**4 + 16*a**5*b*x**2 + 8*a**4*b**2) + x**3/(3*a**3) - 3*b*x/a**4

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Giac [A]  time = 1.12967, size = 99, normalized size = 1.16 \begin{align*} \frac{35 \, b^{2} \arctan \left (\frac{a x}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{4}} - \frac{13 \, a b^{2} x^{3} + 11 \, b^{3} x}{8 \,{\left (a x^{2} + b\right )}^{2} a^{4}} + \frac{a^{6} x^{3} - 9 \, a^{5} b x}{3 \, a^{9}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/8*b^2*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^4) - 1/8*(13*a*b^2*x^3 + 11*b^3*x)/((a*x^2 + b)^2*a^4) + 1/3*(a^6*
x^3 - 9*a^5*b*x)/a^9