3.1872 \(\int \frac{1}{(a+\frac{b}{x^2})^2 x^8} \, dx\)

Optimal. Leaf size=68 \[ \frac{5 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{2 b^{7/2}}+\frac{5 a}{2 b^3 x}+\frac{1}{2 b x^3 \left (a x^2+b\right )}-\frac{5}{6 b^2 x^3} \]

[Out]

-5/(6*b^2*x^3) + (5*a)/(2*b^3*x) + 1/(2*b*x^3*(b + a*x^2)) + (5*a^(3/2)*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(2*b^(7/2
))

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Rubi [A]  time = 0.0249115, antiderivative size = 68, normalized size of antiderivative = 1., 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 = {263, 290, 325, 205} \[ \frac{5 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{2 b^{7/2}}+\frac{5 a}{2 b^3 x}+\frac{1}{2 b x^3 \left (a x^2+b\right )}-\frac{5}{6 b^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^2*x^8),x]

[Out]

-5/(6*b^2*x^3) + (5*a)/(2*b^3*x) + 1/(2*b*x^3*(b + a*x^2)) + (5*a^(3/2)*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(2*b^(7/2
))

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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [A]  time = 0.0384199, size = 67, normalized size = 0.99 \[ \frac{a^2 x}{2 b^3 \left (a x^2+b\right )}+\frac{5 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{2 b^{7/2}}+\frac{2 a}{b^3 x}-\frac{1}{3 b^2 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^2*x^8),x]

[Out]

-1/(3*b^2*x^3) + (2*a)/(b^3*x) + (a^2*x)/(2*b^3*(b + a*x^2)) + (5*a^(3/2)*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(2*b^(7
/2))

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Maple [A]  time = 0.011, size = 59, normalized size = 0.9 \begin{align*}{\frac{{a}^{2}x}{2\,{b}^{3} \left ( a{x}^{2}+b \right ) }}+{\frac{5\,{a}^{2}}{2\,{b}^{3}}\arctan \left ({ax{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{1}{3\,{b}^{2}{x}^{3}}}+2\,{\frac{a}{{b}^{3}x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+1/x^2*b)^2/x^8,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.5536, size = 359, normalized size = 5.28 \begin{align*} \left [\frac{30 \, a^{2} x^{4} + 20 \, a b x^{2} + 15 \,{\left (a^{2} x^{5} + a b x^{3}\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{a x^{2} + 2 \, b x \sqrt{-\frac{a}{b}} - b}{a x^{2} + b}\right ) - 4 \, b^{2}}{12 \,{\left (a b^{3} x^{5} + b^{4} x^{3}\right )}}, \frac{15 \, a^{2} x^{4} + 10 \, a b x^{2} + 15 \,{\left (a^{2} x^{5} + a b x^{3}\right )} \sqrt{\frac{a}{b}} \arctan \left (x \sqrt{\frac{a}{b}}\right ) - 2 \, b^{2}}{6 \,{\left (a b^{3} x^{5} + b^{4} x^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^2/x^8,x, algorithm="fricas")

[Out]

[1/12*(30*a^2*x^4 + 20*a*b*x^2 + 15*(a^2*x^5 + a*b*x^3)*sqrt(-a/b)*log((a*x^2 + 2*b*x*sqrt(-a/b) - b)/(a*x^2 +
 b)) - 4*b^2)/(a*b^3*x^5 + b^4*x^3), 1/6*(15*a^2*x^4 + 10*a*b*x^2 + 15*(a^2*x^5 + a*b*x^3)*sqrt(a/b)*arctan(x*
sqrt(a/b)) - 2*b^2)/(a*b^3*x^5 + b^4*x^3)]

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Sympy [A]  time = 0.686601, size = 114, normalized size = 1.68 \begin{align*} - \frac{5 \sqrt{- \frac{a^{3}}{b^{7}}} \log{\left (x - \frac{b^{4} \sqrt{- \frac{a^{3}}{b^{7}}}}{a^{2}} \right )}}{4} + \frac{5 \sqrt{- \frac{a^{3}}{b^{7}}} \log{\left (x + \frac{b^{4} \sqrt{- \frac{a^{3}}{b^{7}}}}{a^{2}} \right )}}{4} + \frac{15 a^{2} x^{4} + 10 a b x^{2} - 2 b^{2}}{6 a b^{3} x^{5} + 6 b^{4} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**2/x**8,x)

[Out]

-5*sqrt(-a**3/b**7)*log(x - b**4*sqrt(-a**3/b**7)/a**2)/4 + 5*sqrt(-a**3/b**7)*log(x + b**4*sqrt(-a**3/b**7)/a
**2)/4 + (15*a**2*x**4 + 10*a*b*x**2 - 2*b**2)/(6*a*b**3*x**5 + 6*b**4*x**3)

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Giac [A]  time = 1.14466, size = 80, normalized size = 1.18 \begin{align*} \frac{5 \, a^{2} \arctan \left (\frac{a x}{\sqrt{a b}}\right )}{2 \, \sqrt{a b} b^{3}} + \frac{a^{2} x}{2 \,{\left (a x^{2} + b\right )} b^{3}} + \frac{6 \, a x^{2} - b}{3 \, b^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^2/x^8,x, algorithm="giac")

[Out]

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