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

Optimal. Leaf size=46 \[ \frac {1}{3} x^3 (a C+A c)+a A x+\frac {1}{2} a B x^2+\frac {1}{4} B c x^4+\frac {1}{5} c C x^5 \]

[Out]

a*A*x+1/2*a*B*x^2+1/3*(A*c+C*a)*x^3+1/4*B*c*x^4+1/5*c*C*x^5

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Rubi [A]  time = 0.03, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1657} \[ \frac {1}{3} x^3 (a C+A c)+a A x+\frac {1}{2} a B x^2+\frac {1}{4} B c x^4+\frac {1}{5} c C x^5 \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)*(A + B*x + C*x^2),x]

[Out]

a*A*x + (a*B*x^2)/2 + ((A*c + a*C)*x^3)/3 + (B*c*x^4)/4 + (c*C*x^5)/5

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 46, normalized size = 1.00 \[ \frac {1}{3} x^3 (a C+A c)+a A x+\frac {1}{2} a B x^2+\frac {1}{4} B c x^4+\frac {1}{5} c C x^5 \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*A*x + (a*B*x^2)/2 + ((A*c + a*C)*x^3)/3 + (B*c*x^4)/4 + (c*C*x^5)/5

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fricas [A]  time = 0.84, size = 40, normalized size = 0.87 \[ \frac {1}{5} x^{5} c C + \frac {1}{4} x^{4} c B + \frac {1}{3} x^{3} a C + \frac {1}{3} x^{3} c A + \frac {1}{2} x^{2} a B + x a A \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*c*C + 1/4*x^4*c*B + 1/3*x^3*a*C + 1/3*x^3*c*A + 1/2*x^2*a*B + x*a*A

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giac [A]  time = 0.16, size = 40, normalized size = 0.87 \[ \frac {1}{5} \, C c x^{5} + \frac {1}{4} \, B c x^{4} + \frac {1}{3} \, C a x^{3} + \frac {1}{3} \, A c x^{3} + \frac {1}{2} \, B a x^{2} + A a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*C*c*x^5 + 1/4*B*c*x^4 + 1/3*C*a*x^3 + 1/3*A*c*x^3 + 1/2*B*a*x^2 + A*a*x

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maple [A]  time = 0.00, size = 39, normalized size = 0.85 \[ \frac {C c \,x^{5}}{5}+\frac {B c \,x^{4}}{4}+\frac {B a \,x^{2}}{2}+A a x +\frac {\left (A c +a C \right ) x^{3}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*A*x+1/2*a*B*x^2+1/3*(A*c+C*a)*x^3+1/4*B*c*x^4+1/5*c*C*x^5

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maxima [A]  time = 0.44, size = 38, normalized size = 0.83 \[ \frac {1}{5} \, C c x^{5} + \frac {1}{4} \, B c x^{4} + \frac {1}{2} \, B a x^{2} + \frac {1}{3} \, {\left (C a + A c\right )} x^{3} + A a x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*C*c*x^5 + 1/4*B*c*x^4 + 1/2*B*a*x^2 + 1/3*(C*a + A*c)*x^3 + A*a*x

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mupad [B]  time = 0.03, size = 39, normalized size = 0.85 \[ \frac {C\,c\,x^5}{5}+\frac {B\,c\,x^4}{4}+\left (\frac {A\,c}{3}+\frac {C\,a}{3}\right )\,x^3+\frac {B\,a\,x^2}{2}+A\,a\,x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^2)*(A + B*x + C*x^2),x)

[Out]

x^3*((A*c)/3 + (C*a)/3) + A*a*x + (B*a*x^2)/2 + (B*c*x^4)/4 + (C*c*x^5)/5

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sympy [A]  time = 0.07, size = 42, normalized size = 0.91 \[ A a x + \frac {B a x^{2}}{2} + \frac {B c x^{4}}{4} + \frac {C c x^{5}}{5} + x^{3} \left (\frac {A c}{3} + \frac {C a}{3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*a*x + B*a*x**2/2 + B*c*x**4/4 + C*c*x**5/5 + x**3*(A*c/3 + C*a/3)

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