3.1949 \(\int \frac {1}{(a+\frac {b}{x^2})^{5/2} x^7} \, dx\)

Optimal. Leaf size=55 \[ \frac {a^2}{3 b^3 \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {2 a}{b^3 \sqrt {a+\frac {b}{x^2}}}-\frac {\sqrt {a+\frac {b}{x^2}}}{b^3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac {a^2}{3 b^3 \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {2 a}{b^3 \sqrt {a+\frac {b}{x^2}}}-\frac {\sqrt {a+\frac {b}{x^2}}}{b^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^(5/2)*x^7),x]

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 48, normalized size = 0.87 \[ -\frac {\sqrt {a+\frac {b}{x^2}} \left (8 a^2 x^4+12 a b x^2+3 b^2\right )}{3 b^3 \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^(5/2)*x^7),x]

[Out]

-1/3*(Sqrt[a + b/x^2]*(3*b^2 + 12*a*b*x^2 + 8*a^2*x^4))/(b^3*(b + a*x^2)^2)

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fricas [A]  time = 1.32, size = 61, normalized size = 1.11 \[ -\frac {{\left (8 \, a^{2} x^{4} + 12 \, a b x^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{3 \, {\left (a^{2} b^{3} x^{4} + 2 \, a b^{4} x^{2} + b^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^7,x, algorithm="fricas")

[Out]

-1/3*(8*a^2*x^4 + 12*a*b*x^2 + 3*b^2)*sqrt((a*x^2 + b)/x^2)/(a^2*b^3*x^4 + 2*a*b^4*x^2 + b^5)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^7,x, algorithm="giac")

[Out]

integrate(1/((a + b/x^2)^(5/2)*x^7), x)

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maple [A]  time = 0.01, size = 50, normalized size = 0.91 \[ -\frac {\left (a \,x^{2}+b \right ) \left (8 a^{2} x^{4}+12 a b \,x^{2}+3 b^{2}\right )}{3 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} b^{3} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^2)^(5/2)/x^7,x)

[Out]

-1/3*(a*x^2+b)*(8*a^2*x^4+12*a*b*x^2+3*b^2)/x^6/b^3/((a*x^2+b)/x^2)^(5/2)

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maxima [A]  time = 0.92, size = 47, normalized size = 0.85 \[ -\frac {\sqrt {a + \frac {b}{x^{2}}}}{b^{3}} - \frac {2 \, a}{\sqrt {a + \frac {b}{x^{2}}} b^{3}} + \frac {a^{2}}{3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^7,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.29, size = 42, normalized size = 0.76 \[ -\frac {\sqrt {a+\frac {b}{x^2}}\,\left (\frac {8\,a^2\,x^4}{3}+4\,a\,b\,x^2+b^2\right )}{b^3\,{\left (a\,x^2+b\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(a + b/x^2)^(5/2)),x)

[Out]

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

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sympy [A]  time = 8.36, size = 153, normalized size = 2.78 \[ \begin {cases} - \frac {8 a^{2} x^{4}}{3 a b^{3} x^{4} \sqrt {a + \frac {b}{x^{2}}} + 3 b^{4} x^{2} \sqrt {a + \frac {b}{x^{2}}}} - \frac {12 a b x^{2}}{3 a b^{3} x^{4} \sqrt {a + \frac {b}{x^{2}}} + 3 b^{4} x^{2} \sqrt {a + \frac {b}{x^{2}}}} - \frac {3 b^{2}}{3 a b^{3} x^{4} \sqrt {a + \frac {b}{x^{2}}} + 3 b^{4} x^{2} \sqrt {a + \frac {b}{x^{2}}}} & \text {for}\: b \neq 0 \\- \frac {1}{6 a^{\frac {5}{2}} x^{6}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**(5/2)/x**7,x)

[Out]

Piecewise((-8*a**2*x**4/(3*a*b**3*x**4*sqrt(a + b/x**2) + 3*b**4*x**2*sqrt(a + b/x**2)) - 12*a*b*x**2/(3*a*b**
3*x**4*sqrt(a + b/x**2) + 3*b**4*x**2*sqrt(a + b/x**2)) - 3*b**2/(3*a*b**3*x**4*sqrt(a + b/x**2) + 3*b**4*x**2
*sqrt(a + b/x**2)), Ne(b, 0)), (-1/(6*a**(5/2)*x**6), True))

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