3.143 $$\int (a+b x+c x^2) (A+C x^2) \, dx$$

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 + (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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Rubi [A]  time = 0.0295376, antiderivative size = 46, normalized size of antiderivative = 1., 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 + b*x + c*x^2)*(A + 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+b x+c x^2\right ) \left (A+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*}

Mathematica [A]  time = 0.0093153, size = 46, normalized size = 1. $\frac{1}{3} x^3 (a C+A c)+a A x+\frac{1}{2} A b x^2+\frac{1}{4} b C x^4+\frac{1}{5} c C x^5$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a + b*x + c*x^2)*(A + 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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Maple [A]  time = 0.05, size = 39, normalized size = 0.9 \begin{align*} aAx+{\frac{Ab{x}^{2}}{2}}+{\frac{ \left ( Ac+aC \right ){x}^{3}}{3}}+{\frac{bC{x}^{4}}{4}}+{\frac{cC{x}^{5}}{5}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)*(C*x^2+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 = 0.983255, size = 51, normalized size = 1.11 \begin{align*} \frac{1}{5} \, C c x^{5} + \frac{1}{4} \, C b x^{4} + \frac{1}{2} \, A b x^{2} + \frac{1}{3} \,{\left (C a + A c\right )} x^{3} + A a x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.31628, size = 104, normalized size = 2.26 \begin{align*} \frac{1}{5} x^{5} c C + \frac{1}{4} x^{4} b C + \frac{1}{3} x^{3} a C + \frac{1}{3} x^{3} c A + \frac{1}{2} x^{2} b A + x a A \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.063846, size = 42, normalized size = 0.91 \begin{align*} A a x + \frac{A b x^{2}}{2} + \frac{C b x^{4}}{4} + \frac{C c x^{5}}{5} + x^{3} \left (\frac{A c}{3} + \frac{C a}{3}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16908, size = 54, normalized size = 1.17 \begin{align*} \frac{1}{5} \, C c x^{5} + \frac{1}{4} \, C b x^{4} + \frac{1}{3} \, C a x^{3} + \frac{1}{3} \, A c x^{3} + \frac{1}{2} \, A b x^{2} + A a x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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