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*(c/(b*x^2+a))^(1/2)/b

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Rubi [A]  time = 0.02, antiderivative size = 21, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.00 \[ -\frac {c \sqrt {\frac {c}{a+b x^2}}}{b} \]

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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fricas [A]  time = 0.44, size = 19, normalized size = 0.90 \[ -\frac {c \sqrt {\frac {c}{b x^{2} + a}}}{b} \]

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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giac [A]  time = 0.39, size = 28, normalized size = 1.33 \[ -\frac {c^{2} \mathrm {sgn}\left (b x^{2} + a\right )}{\sqrt {b c x^{2} + a c} b} \]

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)

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maple [A]  time = 0.00, size = 26, normalized size = 1.24 \[ -\frac {\left (b \,x^{2}+a \right ) \left (\frac {c}{b \,x^{2}+a}\right )^{\frac {3}{2}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(1/(b*x^2+a)*c)^(3/2),x)

[Out]

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

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maxima [A]  time = 1.06, size = 19, normalized size = 0.90 \[ -\frac {c \sqrt {\frac {c}{b x^{2} + a}}}{b} \]

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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mupad [B]  time = 2.64, size = 19, normalized size = 0.90 \[ -\frac {c\,\sqrt {\frac {c}{b\,x^2+a}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c/(a + b*x^2))^(3/2),x)

[Out]

-(c*(c/(a + b*x^2))^(1/2))/b

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sympy [A]  time = 1.51, size = 53, normalized size = 2.52 \[ \begin {cases} - \frac {a c^{\frac {3}{2}} \left (\frac {1}{a + b x^{2}}\right )^{\frac {3}{2}}}{b} - c^{\frac {3}{2}} x^{2} \left (\frac {1}{a + b x^{2}}\right )^{\frac {3}{2}} & \text {for}\: b \neq 0 \\\frac {x^{2} \left (\frac {c}{a}\right )^{\frac {3}{2}}}{2} & \text {otherwise} \end {cases} \]

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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