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

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Rubi [A]  time = 0.01, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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 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 {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*}

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Mathematica [A]  time = 0.01, size = 42, normalized size = 0.70 \[ \frac {-3 a^2-12 a b x^2-8 b^2 x^4}{3 a^3 x \left (a+b x^2\right )^{3/2}} \]

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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fricas [A]  time = 0.83, size = 59, normalized size = 0.98 \[ -\frac {{\left (8 \, b^{2} x^{4} + 12 \, a b x^{2} + 3 \, a^{2}\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} \]

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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giac [A]  time = 1.20, size = 64, normalized size = 1.07 \[ -\frac {x {\left (\frac {5 \, b^{2} x^{2}}{a^{3}} + \frac {6 \, b}{a^{2}}\right )}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} + \frac {2 \, \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )} a^{2}} \]

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)

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maple [A]  time = 0.00, size = 39, normalized size = 0.65 \[ -\frac {8 b^{2} x^{4}+12 a b \,x^{2}+3 a^{2}}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{3} x} \]

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 [A]  time = 1.34, size = 50, normalized size = 0.83 \[ -\frac {8 \, b x}{3 \, \sqrt {b x^{2} + a} a^{3}} - \frac {4 \, b x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} a x} \]

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]

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

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mupad [B]  time = 5.18, size = 42, normalized size = 0.70 \[ \frac {4\,a\,\left (b\,x^2+a\right )-8\,{\left (b\,x^2+a\right )}^2+a^2}{3\,a^3\,x\,{\left (b\,x^2+a\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x^2)^(5/2)),x)

[Out]

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

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sympy [B]  time = 1.29, size = 165, normalized size = 2.75 \[ - \frac {3 a^{2} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac {12 a b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac {8 b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} \]

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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