3.1879 \(\int \frac {1}{(a+\frac {b}{x^2})^3} \, dx\)

Optimal. Leaf size=74 \[ -\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{7/2}}+\frac {15 x}{8 a^3}-\frac {5 x^3}{8 a^2 \left (a x^2+b\right )}-\frac {x^5}{4 a \left (a x^2+b\right )^2} \]

[Out]

15/8*x/a^3-1/4*x^5/a/(a*x^2+b)^2-5/8*x^3/a^2/(a*x^2+b)-15/8*arctan(x*a^(1/2)/b^(1/2))*b^(1/2)/a^(7/2)

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Rubi [A]  time = 0.03, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {193, 288, 321, 205} \[ -\frac {5 x^3}{8 a^2 \left (a x^2+b\right )}-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{7/2}}+\frac {15 x}{8 a^3}-\frac {x^5}{4 a \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(-3),x]

[Out]

(15*x)/(8*a^3) - x^5/(4*a*(b + a*x^2)^2) - (5*x^3)/(8*a^2*(b + a*x^2)) - (15*Sqrt[b]*ArcTan[(Sqrt[a]*x)/Sqrt[b
]])/(8*a^(7/2))

Rule 193

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[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 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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3} \, dx &=\int \frac {x^6}{\left (b+a x^2\right )^3} \, dx\\ &=-\frac {x^5}{4 a \left (b+a x^2\right )^2}+\frac {5 \int \frac {x^4}{\left (b+a x^2\right )^2} \, dx}{4 a}\\ &=-\frac {x^5}{4 a \left (b+a x^2\right )^2}-\frac {5 x^3}{8 a^2 \left (b+a x^2\right )}+\frac {15 \int \frac {x^2}{b+a x^2} \, dx}{8 a^2}\\ &=\frac {15 x}{8 a^3}-\frac {x^5}{4 a \left (b+a x^2\right )^2}-\frac {5 x^3}{8 a^2 \left (b+a x^2\right )}-\frac {(15 b) \int \frac {1}{b+a x^2} \, dx}{8 a^3}\\ &=\frac {15 x}{8 a^3}-\frac {x^5}{4 a \left (b+a x^2\right )^2}-\frac {5 x^3}{8 a^2 \left (b+a x^2\right )}-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 66, normalized size = 0.89 \[ \frac {8 a^2 x^5+25 a b x^3+15 b^2 x}{8 a^3 \left (a x^2+b\right )^2}-\frac {15 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(-3),x]

[Out]

(15*b^2*x + 25*a*b*x^3 + 8*a^2*x^5)/(8*a^3*(b + a*x^2)^2) - (15*Sqrt[b]*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(8*a^(7/2
))

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fricas [A]  time = 0.83, size = 202, normalized size = 2.73 \[ \left [\frac {16 \, a^{2} x^{5} + 50 \, a b x^{3} + 30 \, b^{2} x + 15 \, {\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - b}{a x^{2} + b}\right )}{16 \, {\left (a^{5} x^{4} + 2 \, a^{4} b x^{2} + a^{3} b^{2}\right )}}, \frac {8 \, a^{2} x^{5} + 25 \, a b x^{3} + 15 \, b^{2} x - 15 \, {\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a x \sqrt {\frac {b}{a}}}{b}\right )}{8 \, {\left (a^{5} x^{4} + 2 \, a^{4} b x^{2} + a^{3} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(16*a^2*x^5 + 50*a*b*x^3 + 30*b^2*x + 15*(a^2*x^4 + 2*a*b*x^2 + b^2)*sqrt(-b/a)*log((a*x^2 - 2*a*x*sqrt(
-b/a) - b)/(a*x^2 + b)))/(a^5*x^4 + 2*a^4*b*x^2 + a^3*b^2), 1/8*(8*a^2*x^5 + 25*a*b*x^3 + 15*b^2*x - 15*(a^2*x
^4 + 2*a*b*x^2 + b^2)*sqrt(b/a)*arctan(a*x*sqrt(b/a)/b))/(a^5*x^4 + 2*a^4*b*x^2 + a^3*b^2)]

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giac [A]  time = 0.15, size = 54, normalized size = 0.73 \[ -\frac {15 \, b \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} + \frac {x}{a^{3}} + \frac {9 \, a b x^{3} + 7 \, b^{2} x}{8 \, {\left (a x^{2} + b\right )}^{2} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/8*b*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^3) + x/a^3 + 1/8*(9*a*b*x^3 + 7*b^2*x)/((a*x^2 + b)^2*a^3)

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maple [A]  time = 0.01, size = 63, normalized size = 0.85 \[ \frac {9 b \,x^{3}}{8 \left (a \,x^{2}+b \right )^{2} a^{2}}+\frac {7 b^{2} x}{8 \left (a \,x^{2}+b \right )^{2} a^{3}}-\frac {15 b \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \sqrt {a b}\, a^{3}}+\frac {x}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^3*x+9/8/a^2*b/(a*x^2+b)^2*x^3+7/8/a^3*b^2/(a*x^2+b)^2*x-15/8/a^3*b/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*a*x)

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maxima [A]  time = 1.97, size = 68, normalized size = 0.92 \[ \frac {9 \, a b x^{3} + 7 \, b^{2} x}{8 \, {\left (a^{5} x^{4} + 2 \, a^{4} b x^{2} + a^{3} b^{2}\right )}} - \frac {15 \, b \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} + \frac {x}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(9*a*b*x^3 + 7*b^2*x)/(a^5*x^4 + 2*a^4*b*x^2 + a^3*b^2) - 15/8*b*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^3) + x
/a^3

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mupad [B]  time = 1.36, size = 64, normalized size = 0.86 \[ \frac {\frac {7\,b^2\,x}{8}+\frac {9\,a\,b\,x^3}{8}}{a^5\,x^4+2\,a^4\,b\,x^2+a^3\,b^2}+\frac {x}{a^3}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{8\,a^{7/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/x^2)^3,x)

[Out]

((7*b^2*x)/8 + (9*a*b*x^3)/8)/(a^3*b^2 + a^5*x^4 + 2*a^4*b*x^2) + x/a^3 - (15*b^(1/2)*atan((a^(1/2)*x)/b^(1/2)
))/(8*a^(7/2))

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sympy [A]  time = 0.43, size = 107, normalized size = 1.45 \[ \frac {15 \sqrt {- \frac {b}{a^{7}}} \log {\left (- a^{3} \sqrt {- \frac {b}{a^{7}}} + x \right )}}{16} - \frac {15 \sqrt {- \frac {b}{a^{7}}} \log {\left (a^{3} \sqrt {- \frac {b}{a^{7}}} + x \right )}}{16} + \frac {9 a b x^{3} + 7 b^{2} x}{8 a^{5} x^{4} + 16 a^{4} b x^{2} + 8 a^{3} b^{2}} + \frac {x}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*sqrt(-b/a**7)*log(-a**3*sqrt(-b/a**7) + x)/16 - 15*sqrt(-b/a**7)*log(a**3*sqrt(-b/a**7) + x)/16 + (9*a*b*x*
*3 + 7*b**2*x)/(8*a**5*x**4 + 16*a**4*b*x**2 + 8*a**3*b**2) + x/a**3

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