3.1935 \(\int \frac {1}{(a+\frac {b}{x^2})^{3/2} x^7} \, dx\)

Optimal. Leaf size=54 \[ \frac {a^2}{b^3 \sqrt {a+\frac {b}{x^2}}}+\frac {2 a \sqrt {a+\frac {b}{x^2}}}{b^3}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{3 b^3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 54, 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} \[ \frac {a^2}{b^3 \sqrt {a+\frac {b}{x^2}}}+\frac {2 a \sqrt {a+\frac {b}{x^2}}}{b^3}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^(3/2)*x^7),x]

[Out]

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

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Mathematica [A]  time = 0.01, size = 42, normalized size = 0.78 \[ \frac {8 a^2 x^4+4 a b x^2-b^2}{3 b^3 x^4 \sqrt {a+\frac {b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^(3/2)*x^7),x]

[Out]

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

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fricas [A]  time = 0.84, size = 54, normalized size = 1.00 \[ \frac {{\left (8 \, a^{2} x^{4} + 4 \, a b x^{2} - b^{2}\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{3 \, {\left (a b^{3} x^{4} + b^{4} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(3/2)/x^7,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} x^{7}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(3/2)/x^7,x, algorithm="giac")

[Out]

integrate(1/((a + b/x^2)^(3/2)*x^7), x)

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maple [A]  time = 0.01, size = 50, normalized size = 0.93 \[ \frac {\left (a \,x^{2}+b \right ) \left (8 a^{2} x^{4}+4 a b \,x^{2}-b^{2}\right )}{3 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}} b^{3} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^2)^(3/2)/x^7,x)

[Out]

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

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maxima [A]  time = 0.88, size = 46, normalized size = 0.85 \[ -\frac {{\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}}}{3 \, b^{3}} + \frac {2 \, \sqrt {a + \frac {b}{x^{2}}} a}{b^{3}} + \frac {a^{2}}{\sqrt {a + \frac {b}{x^{2}}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(3/2)/x^7,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.34, size = 47, normalized size = 0.87 \[ \frac {\sqrt {a+\frac {b}{x^2}}\,\left (8\,a^2\,x^4+4\,a\,b\,x^2-b^2\right )}{3\,b^3\,x^2\,\left (a\,x^2+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^7*(a + b/x^2)^(3/2)),x)

[Out]

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

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sympy [B]  time = 2.60, size = 423, normalized size = 7.83 \[ \frac {8 a^{\frac {9}{2}} b^{\frac {7}{2}} x^{6} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{\frac {7}{2}} b^{6} x^{7} + 6 a^{\frac {5}{2}} b^{7} x^{5} + 3 a^{\frac {3}{2}} b^{8} x^{3}} + \frac {12 a^{\frac {7}{2}} b^{\frac {9}{2}} x^{4} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{\frac {7}{2}} b^{6} x^{7} + 6 a^{\frac {5}{2}} b^{7} x^{5} + 3 a^{\frac {3}{2}} b^{8} x^{3}} + \frac {3 a^{\frac {5}{2}} b^{\frac {11}{2}} x^{2} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{\frac {7}{2}} b^{6} x^{7} + 6 a^{\frac {5}{2}} b^{7} x^{5} + 3 a^{\frac {3}{2}} b^{8} x^{3}} - \frac {a^{\frac {3}{2}} b^{\frac {13}{2}} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{\frac {7}{2}} b^{6} x^{7} + 6 a^{\frac {5}{2}} b^{7} x^{5} + 3 a^{\frac {3}{2}} b^{8} x^{3}} - \frac {8 a^{5} b^{3} x^{7}}{3 a^{\frac {7}{2}} b^{6} x^{7} + 6 a^{\frac {5}{2}} b^{7} x^{5} + 3 a^{\frac {3}{2}} b^{8} x^{3}} - \frac {16 a^{4} b^{4} x^{5}}{3 a^{\frac {7}{2}} b^{6} x^{7} + 6 a^{\frac {5}{2}} b^{7} x^{5} + 3 a^{\frac {3}{2}} b^{8} x^{3}} - \frac {8 a^{3} b^{5} x^{3}}{3 a^{\frac {7}{2}} b^{6} x^{7} + 6 a^{\frac {5}{2}} b^{7} x^{5} + 3 a^{\frac {3}{2}} b^{8} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**(3/2)/x**7,x)

[Out]

8*a**(9/2)*b**(7/2)*x**6*sqrt(a*x**2/b + 1)/(3*a**(7/2)*b**6*x**7 + 6*a**(5/2)*b**7*x**5 + 3*a**(3/2)*b**8*x**
3) + 12*a**(7/2)*b**(9/2)*x**4*sqrt(a*x**2/b + 1)/(3*a**(7/2)*b**6*x**7 + 6*a**(5/2)*b**7*x**5 + 3*a**(3/2)*b*
*8*x**3) + 3*a**(5/2)*b**(11/2)*x**2*sqrt(a*x**2/b + 1)/(3*a**(7/2)*b**6*x**7 + 6*a**(5/2)*b**7*x**5 + 3*a**(3
/2)*b**8*x**3) - a**(3/2)*b**(13/2)*sqrt(a*x**2/b + 1)/(3*a**(7/2)*b**6*x**7 + 6*a**(5/2)*b**7*x**5 + 3*a**(3/
2)*b**8*x**3) - 8*a**5*b**3*x**7/(3*a**(7/2)*b**6*x**7 + 6*a**(5/2)*b**7*x**5 + 3*a**(3/2)*b**8*x**3) - 16*a**
4*b**4*x**5/(3*a**(7/2)*b**6*x**7 + 6*a**(5/2)*b**7*x**5 + 3*a**(3/2)*b**8*x**3) - 8*a**3*b**5*x**3/(3*a**(7/2
)*b**6*x**7 + 6*a**(5/2)*b**7*x**5 + 3*a**(3/2)*b**8*x**3)

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