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

Optimal. Leaf size=108 \[ -\frac{b \left (c+d x^2\right )^{3/2} (3 b c-2 a d)}{3 d^4}+\frac{\sqrt{c+d x^2} (b c-a d) (3 b c-a d)}{d^4}+\frac{c (b c-a d)^2}{d^4 \sqrt{c+d x^2}}+\frac{b^2 \left (c+d x^2\right )^{5/2}}{5 d^4} \]

[Out]

(c*(b*c - a*d)^2)/(d^4*Sqrt[c + d*x^2]) + ((b*c - a*d)*(3*b*c - a*d)*Sqrt[c + d*x^2])/d^4 - (b*(3*b*c - 2*a*d)
*(c + d*x^2)^(3/2))/(3*d^4) + (b^2*(c + d*x^2)^(5/2))/(5*d^4)

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Rubi [A]  time = 0.0859802, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {446, 77} \[ -\frac{b \left (c+d x^2\right )^{3/2} (3 b c-2 a d)}{3 d^4}+\frac{\sqrt{c+d x^2} (b c-a d) (3 b c-a d)}{d^4}+\frac{c (b c-a d)^2}{d^4 \sqrt{c+d x^2}}+\frac{b^2 \left (c+d x^2\right )^{5/2}}{5 d^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(b*c - a*d)^2)/(d^4*Sqrt[c + d*x^2]) + ((b*c - a*d)*(3*b*c - a*d)*Sqrt[c + d*x^2])/d^4 - (b*(3*b*c - 2*a*d)
*(c + d*x^2)^(3/2))/(3*d^4) + (b^2*(c + d*x^2)^(5/2))/(5*d^4)

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b x^2\right )^2}{\left (c+d x^2\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x (a+b x)^2}{(c+d x)^{3/2}} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{c (b c-a d)^2}{d^3 (c+d x)^{3/2}}+\frac{(b c-a d) (3 b c-a d)}{d^3 \sqrt{c+d x}}-\frac{b (3 b c-2 a d) \sqrt{c+d x}}{d^3}+\frac{b^2 (c+d x)^{3/2}}{d^3}\right ) \, dx,x,x^2\right )\\ &=\frac{c (b c-a d)^2}{d^4 \sqrt{c+d x^2}}+\frac{(b c-a d) (3 b c-a d) \sqrt{c+d x^2}}{d^4}-\frac{b (3 b c-2 a d) \left (c+d x^2\right )^{3/2}}{3 d^4}+\frac{b^2 \left (c+d x^2\right )^{5/2}}{5 d^4}\\ \end{align*}

Mathematica [A]  time = 0.0608721, size = 97, normalized size = 0.9 \[ \frac{15 a^2 d^2 \left (2 c+d x^2\right )+10 a b d \left (-8 c^2-4 c d x^2+d^2 x^4\right )+3 b^2 \left (8 c^2 d x^2+16 c^3-2 c d^2 x^4+d^3 x^6\right )}{15 d^4 \sqrt{c+d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*a^2*d^2*(2*c + d*x^2) + 10*a*b*d*(-8*c^2 - 4*c*d*x^2 + d^2*x^4) + 3*b^2*(16*c^3 + 8*c^2*d*x^2 - 2*c*d^2*x^
4 + d^3*x^6))/(15*d^4*Sqrt[c + d*x^2])

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Maple [A]  time = 0.007, size = 108, normalized size = 1. \begin{align*}{\frac{3\,{b}^{2}{x}^{6}{d}^{3}+10\,ab{d}^{3}{x}^{4}-6\,{b}^{2}c{d}^{2}{x}^{4}+15\,{a}^{2}{d}^{3}{x}^{2}-40\,abc{d}^{2}{x}^{2}+24\,{b}^{2}{c}^{2}d{x}^{2}+30\,{a}^{2}c{d}^{2}-80\,ab{c}^{2}d+48\,{b}^{2}{c}^{3}}{15\,{d}^{4}}{\frac{1}{\sqrt{d{x}^{2}+c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^2*d^3*x^6+10*a*b*d^3*x^4-6*b^2*c*d^2*x^4+15*a^2*d^3*x^2-40*a*b*c*d^2*x^2+24*b^2*c^2*d*x^2+30*a^2*c*d
^2-80*a*b*c^2*d+48*b^2*c^3)/(d*x^2+c)^(1/2)/d^4

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.3194, size = 246, normalized size = 2.28 \begin{align*} \frac{{\left (3 \, b^{2} d^{3} x^{6} + 48 \, b^{2} c^{3} - 80 \, a b c^{2} d + 30 \, a^{2} c d^{2} - 2 \,{\left (3 \, b^{2} c d^{2} - 5 \, a b d^{3}\right )} x^{4} +{\left (24 \, b^{2} c^{2} d - 40 \, a b c d^{2} + 15 \, a^{2} d^{3}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{15 \,{\left (d^{5} x^{2} + c d^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*b^2*d^3*x^6 + 48*b^2*c^3 - 80*a*b*c^2*d + 30*a^2*c*d^2 - 2*(3*b^2*c*d^2 - 5*a*b*d^3)*x^4 + (24*b^2*c^2
*d - 40*a*b*c*d^2 + 15*a^2*d^3)*x^2)*sqrt(d*x^2 + c)/(d^5*x^2 + c*d^4)

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Sympy [A]  time = 1.74068, size = 236, normalized size = 2.19 \begin{align*} \begin{cases} \frac{2 a^{2} c}{d^{2} \sqrt{c + d x^{2}}} + \frac{a^{2} x^{2}}{d \sqrt{c + d x^{2}}} - \frac{16 a b c^{2}}{3 d^{3} \sqrt{c + d x^{2}}} - \frac{8 a b c x^{2}}{3 d^{2} \sqrt{c + d x^{2}}} + \frac{2 a b x^{4}}{3 d \sqrt{c + d x^{2}}} + \frac{16 b^{2} c^{3}}{5 d^{4} \sqrt{c + d x^{2}}} + \frac{8 b^{2} c^{2} x^{2}}{5 d^{3} \sqrt{c + d x^{2}}} - \frac{2 b^{2} c x^{4}}{5 d^{2} \sqrt{c + d x^{2}}} + \frac{b^{2} x^{6}}{5 d \sqrt{c + d x^{2}}} & \text{for}\: d \neq 0 \\\frac{\frac{a^{2} x^{4}}{4} + \frac{a b x^{6}}{3} + \frac{b^{2} x^{8}}{8}}{c^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*a**2*c/(d**2*sqrt(c + d*x**2)) + a**2*x**2/(d*sqrt(c + d*x**2)) - 16*a*b*c**2/(3*d**3*sqrt(c + d*
x**2)) - 8*a*b*c*x**2/(3*d**2*sqrt(c + d*x**2)) + 2*a*b*x**4/(3*d*sqrt(c + d*x**2)) + 16*b**2*c**3/(5*d**4*sqr
t(c + d*x**2)) + 8*b**2*c**2*x**2/(5*d**3*sqrt(c + d*x**2)) - 2*b**2*c*x**4/(5*d**2*sqrt(c + d*x**2)) + b**2*x
**6/(5*d*sqrt(c + d*x**2)), Ne(d, 0)), ((a**2*x**4/4 + a*b*x**6/3 + b**2*x**8/8)/c**(3/2), True))

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Giac [A]  time = 1.13426, size = 180, normalized size = 1.67 \begin{align*} \frac{3 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} b^{2} - 15 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} b^{2} c + 45 \, \sqrt{d x^{2} + c} b^{2} c^{2} + 10 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a b d - 60 \, \sqrt{d x^{2} + c} a b c d + 15 \, \sqrt{d x^{2} + c} a^{2} d^{2} + \frac{15 \,{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2}\right )}}{\sqrt{d x^{2} + c}}}{15 \, d^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*(d*x^2 + c)^(5/2)*b^2 - 15*(d*x^2 + c)^(3/2)*b^2*c + 45*sqrt(d*x^2 + c)*b^2*c^2 + 10*(d*x^2 + c)^(3/2)
*a*b*d - 60*sqrt(d*x^2 + c)*a*b*c*d + 15*sqrt(d*x^2 + c)*a^2*d^2 + 15*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2)/sqrt
(d*x^2 + c))/d^4