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

Optimal. Leaf size=55 \[ \frac {1}{4} a^2 c x^4+\frac {1}{8} b x^8 (2 a d+b c)+\frac {1}{6} a x^6 (a d+2 b c)+\frac {1}{10} b^2 d x^{10} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {446, 76} \[ \frac {1}{4} a^2 c x^4+\frac {1}{8} b x^8 (2 a d+b c)+\frac {1}{6} a x^6 (a d+2 b c)+\frac {1}{10} b^2 d x^{10} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*x^2)^2*(c + d*x^2),x]

[Out]

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

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

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]]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 55, normalized size = 1.00 \[ \frac {1}{4} a^2 c x^4+\frac {1}{8} b x^8 (2 a d+b c)+\frac {1}{6} a x^6 (a d+2 b c)+\frac {1}{10} b^2 d x^{10} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 53, normalized size = 0.96 \[ \frac {1}{10} x^{10} d b^{2} + \frac {1}{8} x^{8} c b^{2} + \frac {1}{4} x^{8} d b a + \frac {1}{3} x^{6} c b a + \frac {1}{6} x^{6} d a^{2} + \frac {1}{4} x^{4} c a^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.32, size = 53, normalized size = 0.96 \[ \frac {1}{10} \, b^{2} d x^{10} + \frac {1}{8} \, b^{2} c x^{8} + \frac {1}{4} \, a b d x^{8} + \frac {1}{3} \, a b c x^{6} + \frac {1}{6} \, a^{2} d x^{6} + \frac {1}{4} \, a^{2} c x^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 52, normalized size = 0.95 \[ \frac {b^{2} d \,x^{10}}{10}+\frac {\left (2 a b d +b^{2} c \right ) x^{8}}{8}+\frac {a^{2} c \,x^{4}}{4}+\frac {\left (a^{2} d +2 a b c \right ) x^{6}}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.99, size = 51, normalized size = 0.93 \[ \frac {1}{10} \, b^{2} d x^{10} + \frac {1}{8} \, {\left (b^{2} c + 2 \, a b d\right )} x^{8} + \frac {1}{4} \, a^{2} c x^{4} + \frac {1}{6} \, {\left (2 \, a b c + a^{2} d\right )} x^{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 51, normalized size = 0.93 \[ x^6\,\left (\frac {d\,a^2}{6}+\frac {b\,c\,a}{3}\right )+x^8\,\left (\frac {c\,b^2}{8}+\frac {a\,d\,b}{4}\right )+\frac {a^2\,c\,x^4}{4}+\frac {b^2\,d\,x^{10}}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*x^2)^2*(c + d*x^2),x)

[Out]

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

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sympy [A]  time = 0.07, size = 53, normalized size = 0.96 \[ \frac {a^{2} c x^{4}}{4} + \frac {b^{2} d x^{10}}{10} + x^{8} \left (\frac {a b d}{4} + \frac {b^{2} c}{8}\right ) + x^{6} \left (\frac {a^{2} d}{6} + \frac {a b c}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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