3.247 \(\int \frac{(\frac{c}{a+b x^2})^{3/2}}{x^2} \, dx\)

Optimal. Leaf size=48 \[ -\frac{2 b c x \sqrt{\frac{c}{a+b x^2}}}{a^2}-\frac{c \sqrt{\frac{c}{a+b x^2}}}{a x} \]

[Out]

-((c*Sqrt[c/(a + b*x^2)])/(a*x)) - (2*b*c*x*Sqrt[c/(a + b*x^2)])/a^2

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Rubi [A]  time = 0.114111, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {6720, 271, 191} \[ -\frac{2 b c x \sqrt{\frac{c}{a+b x^2}}}{a^2}-\frac{c \sqrt{\frac{c}{a+b x^2}}}{a x} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x^2))^(3/2)/x^2,x]

[Out]

-((c*Sqrt[c/(a + b*x^2)])/(a*x)) - (2*b*c*x*Sqrt[c/(a + b*x^2)])/a^2

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[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]

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

Mathematica [A]  time = 0.0078871, size = 32, normalized size = 0.67 \[ -\frac{c \left (a+2 b x^2\right ) \sqrt{\frac{c}{a+b x^2}}}{a^2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((c*Sqrt[c/(a + b*x^2)]*(a + 2*b*x^2))/(a^2*x))

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Maple [A]  time = 0.003, size = 37, normalized size = 0.8 \begin{align*} -{\frac{ \left ( b{x}^{2}+a \right ) \left ( 2\,b{x}^{2}+a \right ) }{x{a}^{2}} \left ({\frac{c}{b{x}^{2}+a}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/(b*x^2+a))^(3/2)/x^2,x)

[Out]

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

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Maxima [A]  time = 1.00312, size = 62, normalized size = 1.29 \begin{align*} -\frac{2 \, b^{2} c^{\frac{3}{2}} x^{4} + 3 \, a b c^{\frac{3}{2}} x^{2} + a^{2} c^{\frac{3}{2}}}{{\left (b x^{2} + a\right )}^{\frac{3}{2}} a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.4804, size = 65, normalized size = 1.35 \begin{align*} -\frac{{\left (2 \, b c x^{2} + a c\right )} \sqrt{\frac{c}{b x^{2} + a}}}{a^{2} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (\frac{c}{a + b x^{2}}\right )^{\frac{3}{2}}}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c/(a + b*x**2))**(3/2)/x**2, x)

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Giac [A]  time = 1.18548, size = 109, normalized size = 2.27 \begin{align*} -{\left (\frac{b c x \mathrm{sgn}\left (b x^{2} + a\right )}{\sqrt{b c x^{2} + a c} a^{2}} - \frac{2 \, \sqrt{b c} c \mathrm{sgn}\left (b x^{2} + a\right )}{{\left ({\left (\sqrt{b c} x - \sqrt{b c x^{2} + a c}\right )}^{2} - a c\right )} a}\right )} c \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*c*x*sgn(b*x^2 + a)/(sqrt(b*c*x^2 + a*c)*a^2) - 2*sqrt(b*c)*c*sgn(b*x^2 + a)/(((sqrt(b*c)*x - sqrt(b*c*x^2
+ a*c))^2 - a*c)*a))*c