\(\int (a+b x^2) (A+B x+C x^2+D x^3) \, dx\) [65]

Optimal result

Integrand size = 23, antiderivative size = 60 \[ \int \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=a A x+\frac {1}{2} a B x^2+\frac {1}{3} (A b+a C) x^3+\frac {1}{4} (b B+a D) x^4+\frac {1}{5} b C x^5+\frac {1}{6} b D x^6 \]

[Out]

a*A*x+1/2*a*B*x^2+1/3*(A*b+C*a)*x^3+1/4*(B*b+D*a)*x^4+1/5*b*C*x^5+1/6*b*D*x^6

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {1824} \[ \int \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{3} x^3 (a C+A b)+a A x+\frac {1}{4} x^4 (a D+b B)+\frac {1}{2} a B x^2+\frac {1}{5} b C x^5+\frac {1}{6} b D x^6 \]

[In]

Int[(a + b*x^2)*(A + B*x + C*x^2 + D*x^3),x]

[Out]

a*A*x + (a*B*x^2)/2 + ((A*b + a*C)*x^3)/3 + ((b*B + a*D)*x^4)/4 + (b*C*x^5)/5 + (b*D*x^6)/6

Rule 1824

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a A+a B x+(A b+a C) x^2+(b B+a D) x^3+b C x^4+b D x^5\right ) \, dx \\ & = a A x+\frac {1}{2} a B x^2+\frac {1}{3} (A b+a C) x^3+\frac {1}{4} (b B+a D) x^4+\frac {1}{5} b C x^5+\frac {1}{6} b D x^6 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00 \[ \int \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=a A x+\frac {1}{2} a B x^2+\frac {1}{3} (A b+a C) x^3+\frac {1}{4} (b B+a D) x^4+\frac {1}{5} b C x^5+\frac {1}{6} b D x^6 \]

[In]

Integrate[(a + b*x^2)*(A + B*x + C*x^2 + D*x^3),x]

[Out]

a*A*x + (a*B*x^2)/2 + ((A*b + a*C)*x^3)/3 + ((b*B + a*D)*x^4)/4 + (b*C*x^5)/5 + (b*D*x^6)/6

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.85

method	result	size
default	\(a A x +\frac {B a \,x^{2}}{2}+\frac {\left (A b +C a \right ) x^{3}}{3}+\frac {\left (B b +D a \right ) x^{4}}{4}+\frac {b C \,x^{5}}{5}+\frac {b D x^{6}}{6}\)	\(51\)
norman	\(\frac {b D x^{6}}{6}+\frac {b C \,x^{5}}{5}+\left (\frac {B b}{4}+\frac {D a}{4}\right ) x^{4}+\left (\frac {A b}{3}+\frac {C a}{3}\right ) x^{3}+\frac {B a \,x^{2}}{2}+a A x\)	\(53\)
gosper	\(\frac {1}{6} b D x^{6}+\frac {1}{5} b C \,x^{5}+\frac {1}{4} b B \,x^{4}+\frac {1}{4} x^{4} D a +\frac {1}{3} A b \,x^{3}+\frac {1}{3} x^{3} C a +\frac {1}{2} B a \,x^{2}+a A x\)	\(55\)
parallelrisch	\(\frac {1}{6} b D x^{6}+\frac {1}{5} b C \,x^{5}+\frac {1}{4} b B \,x^{4}+\frac {1}{4} x^{4} D a +\frac {1}{3} A b \,x^{3}+\frac {1}{3} x^{3} C a +\frac {1}{2} B a \,x^{2}+a A x\)	\(55\)

[In]

int((b*x^2+a)*(D*x^3+C*x^2+B*x+A),x,method=_RETURNVERBOSE)

[Out]

a*A*x+1/2*B*a*x^2+1/3*(A*b+C*a)*x^3+1/4*(B*b+D*a)*x^4+1/5*b*C*x^5+1/6*b*D*x^6

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.83 \[ \int \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{6} \, D b x^{6} + \frac {1}{5} \, C b x^{5} + \frac {1}{4} \, {\left (D a + B b\right )} x^{4} + \frac {1}{2} \, B a x^{2} + \frac {1}{3} \, {\left (C a + A b\right )} x^{3} + A a x \]

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A),x, algorithm="fricas")

[Out]

1/6*D*b*x^6 + 1/5*C*b*x^5 + 1/4*(D*a + B*b)*x^4 + 1/2*B*a*x^2 + 1/3*(C*a + A*b)*x^3 + A*a*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.93 \[ \int \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=A a x + \frac {B a x^{2}}{2} + \frac {C b x^{5}}{5} + \frac {D b x^{6}}{6} + x^{4} \left (\frac {B b}{4} + \frac {D a}{4}\right ) + x^{3} \left (\frac {A b}{3} + \frac {C a}{3}\right ) \]

[In]

integrate((b*x**2+a)*(D*x**3+C*x**2+B*x+A),x)

[Out]

A*a*x + B*a*x**2/2 + C*b*x**5/5 + D*b*x**6/6 + x**4*(B*b/4 + D*a/4) + x**3*(A*b/3 + C*a/3)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.83 \[ \int \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{6} \, D b x^{6} + \frac {1}{5} \, C b x^{5} + \frac {1}{4} \, {\left (D a + B b\right )} x^{4} + \frac {1}{2} \, B a x^{2} + \frac {1}{3} \, {\left (C a + A b\right )} x^{3} + A a x \]

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A),x, algorithm="maxima")

[Out]

1/6*D*b*x^6 + 1/5*C*b*x^5 + 1/4*(D*a + B*b)*x^4 + 1/2*B*a*x^2 + 1/3*(C*a + A*b)*x^3 + A*a*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{6} \, D b x^{6} + \frac {1}{5} \, C b x^{5} + \frac {1}{4} \, D a x^{4} + \frac {1}{4} \, B b x^{4} + \frac {1}{3} \, C a x^{3} + \frac {1}{3} \, A b x^{3} + \frac {1}{2} \, B a x^{2} + A a x \]

[In]

integrate((b*x^2+a)*(D*x^3+C*x^2+B*x+A),x, algorithm="giac")

[Out]

1/6*D*b*x^6 + 1/5*C*b*x^5 + 1/4*D*a*x^4 + 1/4*B*b*x^4 + 1/3*C*a*x^3 + 1/3*A*b*x^3 + 1/2*B*a*x^2 + A*a*x

Mupad [B] (verification not implemented)

Time = 5.88 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.90 \[ \int \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {a\,x^4\,D}{4}+\frac {b\,x^6\,D}{6}+A\,a\,x+\frac {B\,a\,x^2}{2}+\frac {A\,b\,x^3}{3}+\frac {C\,a\,x^3}{3}+\frac {B\,b\,x^4}{4}+\frac {C\,b\,x^5}{5} \]

[In]

int((a + b*x^2)*(A + B*x + C*x^2 + x^3*D),x)

[Out]

(a*x^4*D)/4 + (b*x^6*D)/6 + A*a*x + (B*a*x^2)/2 + (A*b*x^3)/3 + (C*a*x^3)/3 + (B*b*x^4)/4 + (C*b*x^5)/5