3.1885 \(\int \frac {1}{(a+\frac {b}{x^2})^3 x^6} \, dx\)

Optimal. Leaf size=62 \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 \sqrt {a} b^{5/2}}+\frac {3 x}{8 b^2 \left (a x^2+b\right )}+\frac {x}{4 b \left (a x^2+b\right )^2} \]

[Out]

1/4*x/b/(a*x^2+b)^2+3/8*x/b^2/(a*x^2+b)+3/8*arctan(x*a^(1/2)/b^(1/2))/b^(5/2)/a^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {263, 199, 205} \[ \frac {3 x}{8 b^2 \left (a x^2+b\right )}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 \sqrt {a} b^{5/2}}+\frac {x}{4 b \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^3*x^6),x]

[Out]

x/(4*b*(b + a*x^2)^2) + (3*x)/(8*b^2*(b + a*x^2)) + (3*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(8*Sqrt[a]*b^(5/2))

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]

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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 55, normalized size = 0.89 \[ \frac {3 \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 \sqrt {a} b^{5/2}}+\frac {3 a x^3+5 b x}{8 b^2 \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^3*x^6),x]

[Out]

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

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fricas [A]  time = 0.65, size = 188, normalized size = 3.03 \[ \left [\frac {6 \, a^{2} b x^{3} + 10 \, a b^{2} x - 3 \, {\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt {-a b} \log \left (\frac {a x^{2} - 2 \, \sqrt {-a b} x - b}{a x^{2} + b}\right )}{16 \, {\left (a^{3} b^{3} x^{4} + 2 \, a^{2} b^{4} x^{2} + a b^{5}\right )}}, \frac {3 \, a^{2} b x^{3} + 5 \, a b^{2} x + 3 \, {\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{b}\right )}{8 \, {\left (a^{3} b^{3} x^{4} + 2 \, a^{2} b^{4} x^{2} + a b^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^3/x^6,x, algorithm="fricas")

[Out]

[1/16*(6*a^2*b*x^3 + 10*a*b^2*x - 3*(a^2*x^4 + 2*a*b*x^2 + b^2)*sqrt(-a*b)*log((a*x^2 - 2*sqrt(-a*b)*x - b)/(a
*x^2 + b)))/(a^3*b^3*x^4 + 2*a^2*b^4*x^2 + a*b^5), 1/8*(3*a^2*b*x^3 + 5*a*b^2*x + 3*(a^2*x^4 + 2*a*b*x^2 + b^2
)*sqrt(a*b)*arctan(sqrt(a*b)*x/b))/(a^3*b^3*x^4 + 2*a^2*b^4*x^2 + a*b^5)]

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giac [A]  time = 0.22, size = 45, normalized size = 0.73 \[ \frac {3 \, \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{2}} + \frac {3 \, a x^{3} + 5 \, b x}{8 \, {\left (a x^{2} + b\right )}^{2} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^3/x^6,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.00, size = 51, normalized size = 0.82 \[ \frac {x}{4 \left (a \,x^{2}+b \right )^{2} b}+\frac {3 x}{8 \left (a \,x^{2}+b \right ) b^{2}}+\frac {3 \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^2)^3/x^6,x)

[Out]

1/4*x/b/(a*x^2+b)^2+3/8*x/b^2/(a*x^2+b)+3/8/b^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*a*x)

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maxima [A]  time = 1.95, size = 58, normalized size = 0.94 \[ \frac {3 \, a x^{3} + 5 \, b x}{8 \, {\left (a^{2} b^{2} x^{4} + 2 \, a b^{3} x^{2} + b^{4}\right )}} + \frac {3 \, \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^3/x^6,x, algorithm="maxima")

[Out]

1/8*(3*a*x^3 + 5*b*x)/(a^2*b^2*x^4 + 2*a*b^3*x^2 + b^4) + 3/8*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*b^2)

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mupad [B]  time = 0.07, size = 55, normalized size = 0.89 \[ \frac {\frac {5\,x}{8\,b}+\frac {3\,a\,x^3}{8\,b^2}}{a^2\,x^4+2\,a\,b\,x^2+b^2}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{8\,\sqrt {a}\,b^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a + b/x^2)^3),x)

[Out]

((5*x)/(8*b) + (3*a*x^3)/(8*b^2))/(b^2 + a^2*x^4 + 2*a*b*x^2) + (3*atan((a^(1/2)*x)/b^(1/2)))/(8*a^(1/2)*b^(5/
2))

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sympy [A]  time = 0.34, size = 105, normalized size = 1.69 \[ - \frac {3 \sqrt {- \frac {1}{a b^{5}}} \log {\left (- b^{3} \sqrt {- \frac {1}{a b^{5}}} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{a b^{5}}} \log {\left (b^{3} \sqrt {- \frac {1}{a b^{5}}} + x \right )}}{16} + \frac {3 a x^{3} + 5 b x}{8 a^{2} b^{2} x^{4} + 16 a b^{3} x^{2} + 8 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**3/x**6,x)

[Out]

-3*sqrt(-1/(a*b**5))*log(-b**3*sqrt(-1/(a*b**5)) + x)/16 + 3*sqrt(-1/(a*b**5))*log(b**3*sqrt(-1/(a*b**5)) + x)
/16 + (3*a*x**3 + 5*b*x)/(8*a**2*b**2*x**4 + 16*a*b**3*x**2 + 8*b**4)

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