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]

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

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Rubi [A]  time = 0.0106586, antiderivative size = 44, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.0087928, size = 31, normalized size = 0.7 \[ \frac{\left (a+b x^2\right )^{3/2} \left (2 b x^2-3 a\right )}{15 a^2 x^5} \]

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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Maple [A]  time = 0.003, size = 28, normalized size = 0.6 \begin{align*} -{\frac{-2\,b{x}^{2}+3\,a}{15\,{a}^{2}{x}^{5}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}} \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52771, size = 84, normalized size = 1.91 \begin{align*} \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}} \end{align*}

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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Sympy [A]  time = 0.842172, size = 68, normalized size = 1.55 \begin{align*} - \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}} \end{align*}

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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Giac [B]  time = 1.46924, size = 151, normalized size = 3.43 \begin{align*} \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}} \end{align*}

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