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

Optimal. Leaf size=21 \[ \frac{c x \sqrt{\frac{c}{a+b x^2}}}{a} \]

[Out]

(c*x*Sqrt[c/(a + b*x^2)])/a

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Rubi [A]  time = 0.017496, antiderivative size = 21, 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 = {6720, 191} \[ \frac{c x \sqrt{\frac{c}{a+b x^2}}}{a} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x^2))^(3/2),x]

[Out]

(c*x*Sqrt[c/(a + b*x^2)])/a

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

Mathematica [A]  time = 0.0047915, size = 21, normalized size = 1. \[ \frac{c x \sqrt{\frac{c}{a+b x^2}}}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x^2))^(3/2),x]

[Out]

(c*x*Sqrt[c/(a + b*x^2)])/a

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Maple [A]  time = 0.003, size = 26, normalized size = 1.2 \begin{align*}{\frac{x \left ( b{x}^{2}+a \right ) }{a} \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)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c/(b*x^2 + a))^(3/2), x)

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Fricas [A]  time = 1.48451, size = 36, normalized size = 1.71 \begin{align*} \frac{c x \sqrt{\frac{c}{b x^{2} + a}}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*x*sqrt(c/(b*x^2 + a))/a

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Sympy [A]  time = 1.46707, size = 66, normalized size = 3.14 \begin{align*} \begin{cases} c^{\frac{3}{2}} x \left (\frac{1}{a + b x^{2}}\right )^{\frac{3}{2}} + \frac{b c^{\frac{3}{2}} x^{3} \left (\frac{1}{a + b x^{2}}\right )^{\frac{3}{2}}}{a} & \text{for}\: a \neq 0 \\- \frac{c^{\frac{3}{2}} x \left (\frac{1}{b}\right )^{\frac{3}{2}} \left (\frac{1}{x^{2}}\right )^{\frac{3}{2}}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((c**(3/2)*x*(1/(a + b*x**2))**(3/2) + b*c**(3/2)*x**3*(1/(a + b*x**2))**(3/2)/a, Ne(a, 0)), (-c**(3/
2)*x*(1/b)**(3/2)*(x**(-2))**(3/2)/2, True))

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Giac [A]  time = 1.2531, size = 38, normalized size = 1.81 \begin{align*} \frac{c^{2} x \mathrm{sgn}\left (b x^{2} + a\right )}{\sqrt{b c x^{2} + a c} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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