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

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

[Out]

-((c*Sqrt[c/(a + b*x^2)])/b)

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Rubi [A]  time = 0.0175141, antiderivative size = 21, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {1591, 15, 30} \[ -\frac{c \sqrt{\frac{c}{a+b x^2}}}{b} \]

Antiderivative was successfully verified.

[In]

Int[x*(c/(a + b*x^2))^(3/2),x]

[Out]

-((c*Sqrt[c/(a + b*x^2)])/b)

Rule 1591

Int[((a_.) + (b_.)*(Pq_)^(n_.))^(p_.)*(Qr_), x_Symbol] :> With[{q = Expon[Pq, x], r = Expon[Qr, x]}, Dist[Coef
f[Qr, x, r]/(q*Coeff[Pq, x, q]), Subst[Int[(a + b*x^n)^p, x], x, Pq], x] /; EqQ[r, q - 1] && EqQ[Coeff[Qr, x,
r]*D[Pq, x], q*Coeff[Pq, x, q]*Qr]] /; FreeQ[{a, b, n, p}, x] && PolyQ[Pq, x] && PolyQ[Qr, x]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0048062, size = 21, normalized size = 1. \[ -\frac{c \sqrt{\frac{c}{a+b x^2}}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(c/(a + b*x^2))^(3/2),x]

[Out]

-((c*Sqrt[c/(a + b*x^2)])/b)

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Maple [A]  time = 0.004, size = 26, normalized size = 1.2 \begin{align*} -{\frac{b{x}^{2}+a}{b} \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(x*(c/(b*x^2+a))^(3/2),x)

[Out]

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

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Maxima [A]  time = 0.967719, size = 26, normalized size = 1.24 \begin{align*} -\frac{c \sqrt{\frac{c}{b x^{2} + a}}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*sqrt(c/(b*x^2 + a))/b

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Fricas [A]  time = 1.49485, size = 35, normalized size = 1.67 \begin{align*} -\frac{c \sqrt{\frac{c}{b x^{2} + a}}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*sqrt(c/(b*x^2 + a))/b

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Sympy [A]  time = 1.4509, size = 53, normalized size = 2.52 \begin{align*} \begin{cases} - \frac{a c^{\frac{3}{2}} \left (\frac{1}{a + b x^{2}}\right )^{\frac{3}{2}}}{b} - c^{\frac{3}{2}} x^{2} \left (\frac{1}{a + b x^{2}}\right )^{\frac{3}{2}} & \text{for}\: b \neq 0 \\\frac{x^{2} \left (\frac{c}{a}\right )^{\frac{3}{2}}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19169, size = 38, normalized size = 1.81 \begin{align*} -\frac{c^{2} \mathrm{sgn}\left (b x^{2} + a\right )}{\sqrt{b c x^{2} + a c} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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