3.1.12 \(\int x (a+b x^2)^2 (A+B x^2) \, dx\)

Optimal. Leaf size=42 \[ \frac {\left (a+b x^2\right )^3 (A b-a B)}{6 b^2}+\frac {B \left (a+b x^2\right )^4}{8 b^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {444, 43} \begin {gather*} \frac {\left (a+b x^2\right )^3 (A b-a B)}{6 b^2}+\frac {B \left (a+b x^2\right )^4}{8 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*x^2)^2*(A + B*x^2),x]

[Out]

((A*b - a*B)*(a + b*x^2)^3)/(6*b^2) + (B*(a + b*x^2)^4)/(8*b^2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 51, normalized size = 1.21 \begin {gather*} \frac {1}{24} x^2 \left (12 a^2 A+4 b x^4 (2 a B+A b)+6 a x^2 (a B+2 A b)+3 b^2 B x^6\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*x^2)^2*(A + B*x^2),x]

[Out]

(x^2*(12*a^2*A + 6*a*(2*A*b + a*B)*x^2 + 4*b*(A*b + 2*a*B)*x^4 + 3*b^2*B*x^6))/24

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \left (a+b x^2\right )^2 \left (A+B x^2\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x*(a + b*x^2)^2*(A + B*x^2),x]

[Out]

IntegrateAlgebraic[x*(a + b*x^2)^2*(A + B*x^2), x]

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fricas [A]  time = 0.38, size = 53, normalized size = 1.26 \begin {gather*} \frac {1}{8} x^{8} b^{2} B + \frac {1}{3} x^{6} b a B + \frac {1}{6} x^{6} b^{2} A + \frac {1}{4} x^{4} a^{2} B + \frac {1}{2} x^{4} b a A + \frac {1}{2} x^{2} a^{2} A \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)^2*(B*x^2+A),x, algorithm="fricas")

[Out]

1/8*x^8*b^2*B + 1/3*x^6*b*a*B + 1/6*x^6*b^2*A + 1/4*x^4*a^2*B + 1/2*x^4*b*a*A + 1/2*x^2*a^2*A

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giac [A]  time = 0.28, size = 53, normalized size = 1.26 \begin {gather*} \frac {1}{8} \, B b^{2} x^{8} + \frac {1}{3} \, B a b x^{6} + \frac {1}{6} \, A b^{2} x^{6} + \frac {1}{4} \, B a^{2} x^{4} + \frac {1}{2} \, A a b x^{4} + \frac {1}{2} \, A a^{2} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)^2*(B*x^2+A),x, algorithm="giac")

[Out]

1/8*B*b^2*x^8 + 1/3*B*a*b*x^6 + 1/6*A*b^2*x^6 + 1/4*B*a^2*x^4 + 1/2*A*a*b*x^4 + 1/2*A*a^2*x^2

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maple [A]  time = 0.00, size = 52, normalized size = 1.24 \begin {gather*} \frac {B \,b^{2} x^{8}}{8}+\frac {\left (b^{2} A +2 a b B \right ) x^{6}}{6}+\frac {A \,a^{2} x^{2}}{2}+\frac {\left (2 a b A +a^{2} B \right ) x^{4}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*x^2+a)^2*(B*x^2+A),x)

[Out]

1/8*b^2*B*x^8+1/6*(A*b^2+2*B*a*b)*x^6+1/4*(2*A*a*b+B*a^2)*x^4+1/2*a^2*A*x^2

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maxima [A]  time = 1.37, size = 51, normalized size = 1.21 \begin {gather*} \frac {1}{8} \, B b^{2} x^{8} + \frac {1}{6} \, {\left (2 \, B a b + A b^{2}\right )} x^{6} + \frac {1}{2} \, A a^{2} x^{2} + \frac {1}{4} \, {\left (B a^{2} + 2 \, A a b\right )} x^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^2+a)^2*(B*x^2+A),x, algorithm="maxima")

[Out]

1/8*B*b^2*x^8 + 1/6*(2*B*a*b + A*b^2)*x^6 + 1/2*A*a^2*x^2 + 1/4*(B*a^2 + 2*A*a*b)*x^4

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mupad [B]  time = 0.04, size = 51, normalized size = 1.21 \begin {gather*} x^4\,\left (\frac {B\,a^2}{4}+\frac {A\,b\,a}{2}\right )+x^6\,\left (\frac {A\,b^2}{6}+\frac {B\,a\,b}{3}\right )+\frac {A\,a^2\,x^2}{2}+\frac {B\,b^2\,x^8}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(A + B*x^2)*(a + b*x^2)^2,x)

[Out]

x^4*((B*a^2)/4 + (A*a*b)/2) + x^6*((A*b^2)/6 + (B*a*b)/3) + (A*a^2*x^2)/2 + (B*b^2*x^8)/8

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sympy [A]  time = 0.08, size = 53, normalized size = 1.26 \begin {gather*} \frac {A a^{2} x^{2}}{2} + \frac {B b^{2} x^{8}}{8} + x^{6} \left (\frac {A b^{2}}{6} + \frac {B a b}{3}\right ) + x^{4} \left (\frac {A a b}{2} + \frac {B a^{2}}{4}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x**2+a)**2*(B*x**2+A),x)

[Out]

A*a**2*x**2/2 + B*b**2*x**8/8 + x**6*(A*b**2/6 + B*a*b/3) + x**4*(A*a*b/2 + B*a**2/4)

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