3.366 \(\int \frac {\sqrt {a+b x^2}}{x^6} \, dx\)

Optimal. Leaf size=44 \[ \frac {2 b \left (a+b x^2\right )^{3/2}}{15 a^2 x^3}-\frac {\left (a+b x^2\right )^{3/2}}{5 a x^5} \]

[Out]

-1/5*(b*x^2+a)^(3/2)/a/x^5+2/15*b*(b*x^2+a)^(3/2)/a^2/x^3

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Rubi [A]  time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {271, 264} \[ \frac {2 b \left (a+b x^2\right )^{3/2}}{15 a^2 x^3}-\frac {\left (a+b x^2\right )^{3/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2]/x^6,x]

[Out]

-(a + b*x^2)^(3/2)/(5*a*x^5) + (2*b*(a + b*x^2)^(3/2))/(15*a^2*x^3)

Rule 264

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] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 271

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 31, normalized size = 0.70 \[ \frac {\left (a+b x^2\right )^{3/2} \left (2 b x^2-3 a\right )}{15 a^2 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2]/x^6,x]

[Out]

((a + b*x^2)^(3/2)*(-3*a + 2*b*x^2))/(15*a^2*x^5)

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fricas [A]  time = 0.83, size = 38, normalized size = 0.86 \[ \frac {{\left (2 \, b^{2} x^{4} - a b x^{2} - 3 \, a^{2}\right )} \sqrt {b x^{2} + a}}{15 \, a^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.68, size = 112, normalized size = 2.55 \[ \frac {4 \, {\left (15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} b^{\frac {5}{2}} + 5 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a b^{\frac {5}{2}} + 5 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{2} b^{\frac {5}{2}} - a^{3} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 28, normalized size = 0.64 \[ -\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} \left (-2 b \,x^{2}+3 a \right )}{15 a^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(b*x^2+a)^(3/2)*(-2*b*x^2+3*a)/a^2/x^5

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maxima [A]  time = 1.28, size = 36, normalized size = 0.82 \[ \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b}{15 \, a^{2} x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{5 \, a x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(b*x^2 + a)^(3/2)*b/(a^2*x^3) - 1/5*(b*x^2 + a)^(3/2)/(a*x^5)

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mupad [B]  time = 4.77, size = 37, normalized size = 0.84 \[ -\frac {\sqrt {b\,x^2+a}\,\left (3\,a^2+a\,b\,x^2-2\,b^2\,x^4\right )}{15\,a^2\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/2)/x^6,x)

[Out]

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

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sympy [A]  time = 0.91, size = 68, normalized size = 1.55 \[ - \frac {\sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a x^{2}} + \frac {2 b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(b)*sqrt(a/(b*x**2) + 1)/(5*x**4) - b**(3/2)*sqrt(a/(b*x**2) + 1)/(15*a*x**2) + 2*b**(5/2)*sqrt(a/(b*x**2
) + 1)/(15*a**2)

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