\(\int (a+b x+c x^2) (A+C x^2) \, dx\) [143]

Optimal result

Integrand size = 18, antiderivative size = 46 \[ \int \left (a+b x+c x^2\right ) \left (A+C x^2\right ) \, dx=a A x+\frac {1}{2} A b x^2+\frac {1}{3} (A c+a C) x^3+\frac {1}{4} b C x^4+\frac {1}{5} c C x^5 \]

[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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {1671} \[ \int \left (a+b x+c x^2\right ) \left (A+C x^2\right ) \, dx=\frac {1}{3} x^3 (a C+A c)+a A x+\frac {1}{2} A b x^2+\frac {1}{4} b C x^4+\frac {1}{5} c C x^5 \]

[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 1671

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*} \text {integral}& = \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] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \left (a+b x+c x^2\right ) \left (A+C x^2\right ) \, dx=a A x+\frac {1}{2} A b x^2+\frac {1}{3} (A c+a C) x^3+\frac {1}{4} b C x^4+\frac {1}{5} c C x^5 \]

[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

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85

method	result	size
default	\(a A x +\frac {A b \,x^{2}}{2}+\frac {\left (A c +C a \right ) x^{3}}{3}+\frac {b C \,x^{4}}{4}+\frac {c C \,x^{5}}{5}\)	\(39\)
norman	\(\frac {c C \,x^{5}}{5}+\frac {b C \,x^{4}}{4}+\left (\frac {A c}{3}+\frac {C a}{3}\right ) x^{3}+\frac {A b \,x^{2}}{2}+a A x\)	\(40\)
gosper	\(\frac {1}{5} c C \,x^{5}+\frac {1}{4} b C \,x^{4}+\frac {1}{3} A c \,x^{3}+\frac {1}{3} x^{3} C a +\frac {1}{2} A b \,x^{2}+a A x\)	\(41\)
risch	\(\frac {1}{5} c C \,x^{5}+\frac {1}{4} b C \,x^{4}+\frac {1}{3} A c \,x^{3}+\frac {1}{3} x^{3} C a +\frac {1}{2} A b \,x^{2}+a A x\)	\(41\)
parallelrisch	\(\frac {1}{5} c C \,x^{5}+\frac {1}{4} b C \,x^{4}+\frac {1}{3} A c \,x^{3}+\frac {1}{3} x^{3} C a +\frac {1}{2} A b \,x^{2}+a A x\)	\(41\)

[In]

int((c*x^2+b*x+a)*(C*x^2+A),x,method=_RETURNVERBOSE)

[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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \left (a+b x+c x^2\right ) \left (A+C x^2\right ) \, dx=\frac {1}{5} \, C c x^{5} + \frac {1}{4} \, C b x^{4} + \frac {1}{2} \, A b x^{2} + \frac {1}{3} \, {\left (C a + A c\right )} x^{3} + A a x \]

[In]

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

[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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91 \[ \int \left (a+b x+c x^2\right ) \left (A+C x^2\right ) \, dx=A a x + \frac {A b x^{2}}{2} + \frac {C b x^{4}}{4} + \frac {C c x^{5}}{5} + x^{3} \left (\frac {A c}{3} + \frac {C a}{3}\right ) \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \left (a+b x+c x^2\right ) \left (A+C x^2\right ) \, dx=\frac {1}{5} \, C c x^{5} + \frac {1}{4} \, C b x^{4} + \frac {1}{2} \, A b x^{2} + \frac {1}{3} \, {\left (C a + A c\right )} x^{3} + A a x \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int \left (a+b x+c x^2\right ) \left (A+C x^2\right ) \, dx=\frac {1}{5} \, C c x^{5} + \frac {1}{4} \, C b x^{4} + \frac {1}{3} \, C a x^{3} + \frac {1}{3} \, A c x^{3} + \frac {1}{2} \, A b x^{2} + A a x \]

[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

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.85 \[ \int \left (a+b x+c x^2\right ) \left (A+C x^2\right ) \, dx=\frac {C\,c\,x^5}{5}+\frac {C\,b\,x^4}{4}+\left (\frac {A\,c}{3}+\frac {C\,a}{3}\right )\,x^3+\frac {A\,b\,x^2}{2}+A\,a\,x \]

[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 + (A*b*x^2)/2 + (C*b*x^4)/4 + (C*c*x^5)/5