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

Optimal. Leaf size=60 \[ -\frac{8 b x}{3 a^3 \sqrt{a+b x^2}}-\frac{4 b x}{3 a^2 \left (a+b x^2\right )^{3/2}}-\frac{1}{a x \left (a+b x^2\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0135271, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {271, 192, 191} \[ -\frac{8 b x}{3 a^3 \sqrt{a+b x^2}}-\frac{4 b x}{3 a^2 \left (a+b x^2\right )^{3/2}}-\frac{1}{a x \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*x^2)^(5/2)),x]

[Out]

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

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 192

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

Rule 191

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

Rubi steps

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

Mathematica [A]  time = 0.0084805, size = 42, normalized size = 0.7 \[ \frac{-3 a^2-12 a b x^2-8 b^2 x^4}{3 a^3 x \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 39, normalized size = 0.7 \begin{align*} -{\frac{8\,{b}^{2}{x}^{4}+12\,ab{x}^{2}+3\,{a}^{2}}{3\,{a}^{3}x} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.34013, size = 123, normalized size = 2.05 \begin{align*} -\frac{{\left (8 \, b^{2} x^{4} + 12 \, a b x^{2} + 3 \, a^{2}\right )} \sqrt{b x^{2} + a}}{3 \,{\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 1.31081, size = 165, normalized size = 2.75 \begin{align*} - \frac{3 a^{2} b^{\frac{9}{2}} \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac{12 a b^{\frac{11}{2}} x^{2} \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac{8 b^{\frac{13}{2}} x^{4} \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.88141, size = 86, normalized size = 1.43 \begin{align*} -\frac{x{\left (\frac{5 \, b^{2} x^{2}}{a^{3}} + \frac{6 \, b}{a^{2}}\right )}}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} + \frac{2 \, \sqrt{b}}{{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x*(5*b^2*x^2/a^3 + 6*b/a^2)/(b*x^2 + a)^(3/2) + 2*sqrt(b)/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)*a^2)