3.3.52 \(\int x^2 (A+B x) (a+c x^2) \, dx\)

Optimal. Leaf size=37 \[ \frac {1}{3} a A x^3+\frac {1}{4} a B x^4+\frac {1}{5} A c x^5+\frac {1}{6} B c x^6 \]

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Rubi [A]  time = 0.03, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {766} \begin {gather*} \frac {1}{3} a A x^3+\frac {1}{4} a B x^4+\frac {1}{5} A c x^5+\frac {1}{6} B c x^6 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*(A + B*x)*(a + c*x^2),x]

[Out]

(a*A*x^3)/3 + (a*B*x^4)/4 + (A*c*x^5)/5 + (B*c*x^6)/6

Rule 766

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int x^2 (A+B x) \left (a+c x^2\right ) \, dx &=\int \left (a A x^2+a B x^3+A c x^4+B c x^5\right ) \, dx\\ &=\frac {1}{3} a A x^3+\frac {1}{4} a B x^4+\frac {1}{5} A c x^5+\frac {1}{6} B c x^6\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 37, normalized size = 1.00 \begin {gather*} \frac {1}{3} a A x^3+\frac {1}{4} a B x^4+\frac {1}{5} A c x^5+\frac {1}{6} B c x^6 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*A*x^3)/3 + (a*B*x^4)/4 + (A*c*x^5)/5 + (B*c*x^6)/6

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.35, size = 29, normalized size = 0.78 \begin {gather*} \frac {1}{6} x^{6} c B + \frac {1}{5} x^{5} c A + \frac {1}{4} x^{4} a B + \frac {1}{3} x^{3} a A \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^6*c*B + 1/5*x^5*c*A + 1/4*x^4*a*B + 1/3*x^3*a*A

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giac [A]  time = 0.17, size = 29, normalized size = 0.78 \begin {gather*} \frac {1}{6} \, B c x^{6} + \frac {1}{5} \, A c x^{5} + \frac {1}{4} \, B a x^{4} + \frac {1}{3} \, A a x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*c*x^6 + 1/5*A*c*x^5 + 1/4*B*a*x^4 + 1/3*A*a*x^3

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maple [A]  time = 0.05, size = 30, normalized size = 0.81 \begin {gather*} \frac {1}{6} B c \,x^{6}+\frac {1}{5} A c \,x^{5}+\frac {1}{4} B a \,x^{4}+\frac {1}{3} A a \,x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*A*a*x^3+1/4*B*a*x^4+1/5*A*c*x^5+1/6*B*c*x^6

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maxima [A]  time = 0.50, size = 29, normalized size = 0.78 \begin {gather*} \frac {1}{6} \, B c x^{6} + \frac {1}{5} \, A c x^{5} + \frac {1}{4} \, B a x^{4} + \frac {1}{3} \, A a x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*B*c*x^6 + 1/5*A*c*x^5 + 1/4*B*a*x^4 + 1/3*A*a*x^3

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mupad [B]  time = 0.04, size = 29, normalized size = 0.78 \begin {gather*} \frac {B\,c\,x^6}{6}+\frac {A\,c\,x^5}{5}+\frac {B\,a\,x^4}{4}+\frac {A\,a\,x^3}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + c*x^2)*(A + B*x),x)

[Out]

(A*a*x^3)/3 + (B*a*x^4)/4 + (A*c*x^5)/5 + (B*c*x^6)/6

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sympy [A]  time = 0.07, size = 32, normalized size = 0.86 \begin {gather*} \frac {A a x^{3}}{3} + \frac {A c x^{5}}{5} + \frac {B a x^{4}}{4} + \frac {B c x^{6}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a*x**3/3 + A*c*x**5/5 + B*a*x**4/4 + B*c*x**6/6

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