3.149 \(\int \frac{x (A+B x^2)}{(b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=37 \[ -\frac{A b-x^2 (b B-2 A c)}{b^2 \sqrt{b x^2+c x^4}} \]

[Out]

-((A*b - (b*B - 2*A*c)*x^2)/(b^2*Sqrt[b*x^2 + c*x^4]))

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Rubi [A]  time = 0.121296, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {2034, 636} \[ -\frac{A b-x^2 (b B-2 A c)}{b^2 \sqrt{b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

-((A*b - (b*B - 2*A*c)*x^2)/(b^2*Sqrt[b*x^2 + c*x^4]))

Rule 2034

Int[(x_)^(m_.)*((b_.)*(x_)^(k_.) + (a_.)*(x_)^(j_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n
, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a*x^Simplify[j/n] + b*x^Simplify[k/n])^p*(c + d*x)^q, x], x, x^n], x]
/; FreeQ[{a, b, c, d, j, k, m, n, p, q}, x] &&  !IntegerQ[p] && NeQ[k, j] && IntegerQ[Simplify[j/n]] && Integ
erQ[Simplify[k/n]] && IntegerQ[Simplify[(m + 1)/n]] && NeQ[n^2, 1]

Rule 636

Int[((d_.) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-2*(b*d - 2*a*e + (2*c*
d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]), x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] &&
NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{x \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{A+B x}{\left (b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{A b-(b B-2 A c) x^2}{b^2 \sqrt{b x^2+c x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0187079, size = 37, normalized size = 1. \[ \frac{b B x^2-A \left (b+2 c x^2\right )}{b^2 \sqrt{x^2 \left (b+c x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(A + B*x^2))/(b*x^2 + c*x^4)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.004, size = 47, normalized size = 1.3 \begin{align*} -{\frac{ \left ( c{x}^{2}+b \right ){x}^{2} \left ( 2\,A{x}^{2}c-B{x}^{2}b+Ab \right ) }{{b}^{2}} \left ( c{x}^{4}+b{x}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.1736, size = 88, normalized size = 2.38 \begin{align*} -A{\left (\frac{2 \, c x^{2}}{\sqrt{c x^{4} + b x^{2}} b^{2}} + \frac{1}{\sqrt{c x^{4} + b x^{2}} b}\right )} + \frac{B x^{2}}{\sqrt{c x^{4} + b x^{2}} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")

[Out]

-A*(2*c*x^2/(sqrt(c*x^4 + b*x^2)*b^2) + 1/(sqrt(c*x^4 + b*x^2)*b)) + B*x^2/(sqrt(c*x^4 + b*x^2)*b)

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Fricas [A]  time = 1.32803, size = 93, normalized size = 2.51 \begin{align*} \frac{\sqrt{c x^{4} + b x^{2}}{\left ({\left (B b - 2 \, A c\right )} x^{2} - A b\right )}}{b^{2} c x^{4} + b^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")

[Out]

sqrt(c*x^4 + b*x^2)*((B*b - 2*A*c)*x^2 - A*b)/(b^2*c*x^4 + b^3*x^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x**2+A)/(c*x**4+b*x**2)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.18676, size = 49, normalized size = 1.32 \begin{align*} \frac{\frac{{\left (B b - 2 \, A c\right )} x^{2}}{b^{2}} - \frac{A}{b}}{\sqrt{c x^{4} + b x^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x^2+A)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")

[Out]

((B*b - 2*A*c)*x^2/b^2 - A/b)/sqrt(c*x^4 + b*x^2)