3.151 \(\int x^3 (a+b x^2)^2 (c+d x^2)^2 \, dx\)

Optimal. Leaf size=87 \[ \frac{1}{8} x^8 \left (a^2 d^2+4 a b c d+b^2 c^2\right )+\frac{1}{4} a^2 c^2 x^4+\frac{1}{5} b d x^{10} (a d+b c)+\frac{1}{3} a c x^6 (a d+b c)+\frac{1}{12} b^2 d^2 x^{12} \]

[Out]

(a^2*c^2*x^4)/4 + (a*c*(b*c + a*d)*x^6)/3 + ((b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^8)/8 + (b*d*(b*c + a*d)*x^10)/5
+ (b^2*d^2*x^12)/12

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Rubi [A]  time = 0.10562, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {446, 77} \[ \frac{1}{8} x^8 \left (a^2 d^2+4 a b c d+b^2 c^2\right )+\frac{1}{4} a^2 c^2 x^4+\frac{1}{5} b d x^{10} (a d+b c)+\frac{1}{3} a c x^6 (a d+b c)+\frac{1}{12} b^2 d^2 x^{12} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c^2*x^4)/4 + (a*c*(b*c + a*d)*x^6)/3 + ((b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^8)/8 + (b*d*(b*c + a*d)*x^10)/5
+ (b^2*d^2*x^12)/12

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 x^3 \left (a+b x^2\right )^2 \left (c+d x^2\right )^2 \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int x (a+b x)^2 (c+d x)^2 \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (a^2 c^2 x+2 a c (b c+a d) x^2+\left (b^2 c^2+4 a b c d+a^2 d^2\right ) x^3+2 b d (b c+a d) x^4+b^2 d^2 x^5\right ) \, dx,x,x^2\right )\\ &=\frac{1}{4} a^2 c^2 x^4+\frac{1}{3} a c (b c+a d) x^6+\frac{1}{8} \left (b^2 c^2+4 a b c d+a^2 d^2\right ) x^8+\frac{1}{5} b d (b c+a d) x^{10}+\frac{1}{12} b^2 d^2 x^{12}\\ \end{align*}

Mathematica [A]  time = 0.0249023, size = 81, normalized size = 0.93 \[ \frac{1}{120} x^4 \left (15 x^4 \left (a^2 d^2+4 a b c d+b^2 c^2\right )+30 a^2 c^2+24 b d x^6 (a d+b c)+40 a c x^2 (a d+b c)+10 b^2 d^2 x^8\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*(30*a^2*c^2 + 40*a*c*(b*c + a*d)*x^2 + 15*(b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^4 + 24*b*d*(b*c + a*d)*x^6 +
10*b^2*d^2*x^8))/120

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Maple [A]  time = 0., size = 90, normalized size = 1. \begin{align*}{\frac{{b}^{2}{d}^{2}{x}^{12}}{12}}+{\frac{ \left ( 2\,ab{d}^{2}+2\,{b}^{2}cd \right ){x}^{10}}{10}}+{\frac{ \left ({a}^{2}{d}^{2}+4\,cabd+{b}^{2}{c}^{2} \right ){x}^{8}}{8}}+{\frac{ \left ( 2\,{a}^{2}cd+2\,ab{c}^{2} \right ){x}^{6}}{6}}+{\frac{{a}^{2}{c}^{2}{x}^{4}}{4}} \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)^2,x)

[Out]

1/12*b^2*d^2*x^12+1/10*(2*a*b*d^2+2*b^2*c*d)*x^10+1/8*(a^2*d^2+4*a*b*c*d+b^2*c^2)*x^8+1/6*(2*a^2*c*d+2*a*b*c^2
)*x^6+1/4*a^2*c^2*x^4

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Maxima [A]  time = 0.990316, size = 115, normalized size = 1.32 \begin{align*} \frac{1}{12} \, b^{2} d^{2} x^{12} + \frac{1}{5} \,{\left (b^{2} c d + a b d^{2}\right )} x^{10} + \frac{1}{8} \,{\left (b^{2} c^{2} + 4 \, a b c d + a^{2} d^{2}\right )} x^{8} + \frac{1}{4} \, a^{2} c^{2} x^{4} + \frac{1}{3} \,{\left (a b c^{2} + a^{2} c d\right )} x^{6} \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)^2,x, algorithm="maxima")

[Out]

1/12*b^2*d^2*x^12 + 1/5*(b^2*c*d + a*b*d^2)*x^10 + 1/8*(b^2*c^2 + 4*a*b*c*d + a^2*d^2)*x^8 + 1/4*a^2*c^2*x^4 +
1/3*(a*b*c^2 + a^2*c*d)*x^6

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Fricas [A]  time = 1.14923, size = 223, normalized size = 2.56 \begin{align*} \frac{1}{12} x^{12} d^{2} b^{2} + \frac{1}{5} x^{10} d c b^{2} + \frac{1}{5} x^{10} d^{2} b a + \frac{1}{8} x^{8} c^{2} b^{2} + \frac{1}{2} x^{8} d c b a + \frac{1}{8} x^{8} d^{2} a^{2} + \frac{1}{3} x^{6} c^{2} b a + \frac{1}{3} x^{6} d c a^{2} + \frac{1}{4} x^{4} c^{2} a^{2} \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)^2,x, algorithm="fricas")

[Out]

1/12*x^12*d^2*b^2 + 1/5*x^10*d*c*b^2 + 1/5*x^10*d^2*b*a + 1/8*x^8*c^2*b^2 + 1/2*x^8*d*c*b*a + 1/8*x^8*d^2*a^2
+ 1/3*x^6*c^2*b*a + 1/3*x^6*d*c*a^2 + 1/4*x^4*c^2*a^2

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Sympy [A]  time = 0.076072, size = 92, normalized size = 1.06 \begin{align*} \frac{a^{2} c^{2} x^{4}}{4} + \frac{b^{2} d^{2} x^{12}}{12} + x^{10} \left (\frac{a b d^{2}}{5} + \frac{b^{2} c d}{5}\right ) + x^{8} \left (\frac{a^{2} d^{2}}{8} + \frac{a b c d}{2} + \frac{b^{2} c^{2}}{8}\right ) + x^{6} \left (\frac{a^{2} c d}{3} + \frac{a b c^{2}}{3}\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)**2,x)

[Out]

a**2*c**2*x**4/4 + b**2*d**2*x**12/12 + x**10*(a*b*d**2/5 + b**2*c*d/5) + x**8*(a**2*d**2/8 + a*b*c*d/2 + b**2
*c**2/8) + x**6*(a**2*c*d/3 + a*b*c**2/3)

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Giac [A]  time = 1.18127, size = 127, normalized size = 1.46 \begin{align*} \frac{1}{12} \, b^{2} d^{2} x^{12} + \frac{1}{5} \, b^{2} c d x^{10} + \frac{1}{5} \, a b d^{2} x^{10} + \frac{1}{8} \, b^{2} c^{2} x^{8} + \frac{1}{2} \, a b c d x^{8} + \frac{1}{8} \, a^{2} d^{2} x^{8} + \frac{1}{3} \, a b c^{2} x^{6} + \frac{1}{3} \, a^{2} c d x^{6} + \frac{1}{4} \, a^{2} c^{2} x^{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)^2,x, algorithm="giac")

[Out]

1/12*b^2*d^2*x^12 + 1/5*b^2*c*d*x^10 + 1/5*a*b*d^2*x^10 + 1/8*b^2*c^2*x^8 + 1/2*a*b*c*d*x^8 + 1/8*a^2*d^2*x^8
+ 1/3*a*b*c^2*x^6 + 1/3*a^2*c*d*x^6 + 1/4*a^2*c^2*x^4