\(\int x^2 (a+b x^2) (A+B x+C x^2+D x^3) \, dx\) [63]

Optimal result

Integrand size = 26, antiderivative size = 65 \[ \int x^2 \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{3} a A x^3+\frac {1}{4} a B x^4+\frac {1}{5} (A b+a C) x^5+\frac {1}{6} (b B+a D) x^6+\frac {1}{7} b C x^7+\frac {1}{8} b D x^8 \]

[Out]

1/3*a*A*x^3+1/4*a*B*x^4+1/5*(A*b+C*a)*x^5+1/6*(B*b+D*a)*x^6+1/7*b*C*x^7+1/8*b*D*x^8

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, Rules used = {1816} \[ \int x^2 \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{5} x^5 (a C+A b)+\frac {1}{3} a A x^3+\frac {1}{6} x^6 (a D+b B)+\frac {1}{4} a B x^4+\frac {1}{7} b C x^7+\frac {1}{8} b D x^8 \]

[In]

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

[Out]

(a*A*x^3)/3 + (a*B*x^4)/4 + ((A*b + a*C)*x^5)/5 + ((b*B + a*D)*x^6)/6 + (b*C*x^7)/7 + (b*D*x^8)/8

Rule 1816

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

Rubi steps \begin{align*} \text {integral}& = \int \left (a A x^2+a B x^3+(A b+a C) x^4+(b B+a D) x^5+b C x^6+b D x^7\right ) \, dx \\ & = \frac {1}{3} a A x^3+\frac {1}{4} a B x^4+\frac {1}{5} (A b+a C) x^5+\frac {1}{6} (b B+a D) x^6+\frac {1}{7} b C x^7+\frac {1}{8} b D x^8 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{3} a A x^3+\frac {1}{4} a B x^4+\frac {1}{5} (A b+a C) x^5+\frac {1}{6} (b B+a D) x^6+\frac {1}{7} b C x^7+\frac {1}{8} b D x^8 \]

[In]

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

[Out]

(a*A*x^3)/3 + (a*B*x^4)/4 + ((A*b + a*C)*x^5)/5 + ((b*B + a*D)*x^6)/6 + (b*C*x^7)/7 + (b*D*x^8)/8

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {a A \,x^{3}}{3}+\frac {B a \,x^{4}}{4}+\frac {\left (A b +C a \right ) x^{5}}{5}+\frac {\left (B b +D a \right ) x^{6}}{6}+\frac {b C \,x^{7}}{7}+\frac {b D x^{8}}{8}\)	\(54\)
norman	\(\frac {b D x^{8}}{8}+\frac {b C \,x^{7}}{7}+\left (\frac {B b}{6}+\frac {D a}{6}\right ) x^{6}+\left (\frac {A b}{5}+\frac {C a}{5}\right ) x^{5}+\frac {B a \,x^{4}}{4}+\frac {a A \,x^{3}}{3}\)	\(56\)
gosper	\(\frac {1}{8} b D x^{8}+\frac {1}{7} b C \,x^{7}+\frac {1}{6} b B \,x^{6}+\frac {1}{6} x^{6} D a +\frac {1}{5} x^{5} A b +\frac {1}{5} x^{5} C a +\frac {1}{4} B a \,x^{4}+\frac {1}{3} a A \,x^{3}\)	\(58\)
parallelrisch	\(\frac {1}{8} b D x^{8}+\frac {1}{7} b C \,x^{7}+\frac {1}{6} b B \,x^{6}+\frac {1}{6} x^{6} D a +\frac {1}{5} x^{5} A b +\frac {1}{5} x^{5} C a +\frac {1}{4} B a \,x^{4}+\frac {1}{3} a A \,x^{3}\)	\(58\)

[In]

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

[Out]

1/3*a*A*x^3+1/4*B*a*x^4+1/5*(A*b+C*a)*x^5+1/6*(B*b+D*a)*x^6+1/7*b*C*x^7+1/8*b*D*x^8

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82 \[ \int x^2 \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{8} \, D b x^{8} + \frac {1}{7} \, C b x^{7} + \frac {1}{6} \, {\left (D a + B b\right )} x^{6} + \frac {1}{4} \, B a x^{4} + \frac {1}{5} \, {\left (C a + A b\right )} x^{5} + \frac {1}{3} \, A a x^{3} \]

[In]

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

[Out]

1/8*D*b*x^8 + 1/7*C*b*x^7 + 1/6*(D*a + B*b)*x^6 + 1/4*B*a*x^4 + 1/5*(C*a + A*b)*x^5 + 1/3*A*a*x^3

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int x^2 \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {A a x^{3}}{3} + \frac {B a x^{4}}{4} + \frac {C b x^{7}}{7} + \frac {D b x^{8}}{8} + x^{6} \left (\frac {B b}{6} + \frac {D a}{6}\right ) + x^{5} \left (\frac {A b}{5} + \frac {C a}{5}\right ) \]

[In]

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

[Out]

A*a*x**3/3 + B*a*x**4/4 + C*b*x**7/7 + D*b*x**8/8 + x**6*(B*b/6 + D*a/6) + x**5*(A*b/5 + C*a/5)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82 \[ \int x^2 \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{8} \, D b x^{8} + \frac {1}{7} \, C b x^{7} + \frac {1}{6} \, {\left (D a + B b\right )} x^{6} + \frac {1}{4} \, B a x^{4} + \frac {1}{5} \, {\left (C a + A b\right )} x^{5} + \frac {1}{3} \, A a x^{3} \]

[In]

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

[Out]

1/8*D*b*x^8 + 1/7*C*b*x^7 + 1/6*(D*a + B*b)*x^6 + 1/4*B*a*x^4 + 1/5*(C*a + A*b)*x^5 + 1/3*A*a*x^3

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int x^2 \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{8} \, D b x^{8} + \frac {1}{7} \, C b x^{7} + \frac {1}{6} \, D a x^{6} + \frac {1}{6} \, B b x^{6} + \frac {1}{5} \, C a x^{5} + \frac {1}{5} \, A b x^{5} + \frac {1}{4} \, B a x^{4} + \frac {1}{3} \, A a x^{3} \]

[In]

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

[Out]

1/8*D*b*x^8 + 1/7*C*b*x^7 + 1/6*D*a*x^6 + 1/6*B*b*x^6 + 1/5*C*a*x^5 + 1/5*A*b*x^5 + 1/4*B*a*x^4 + 1/3*A*a*x^3

Mupad [B] (verification not implemented)

Time = 6.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.88 \[ \int x^2 \left (a+b x^2\right ) \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {a\,x^6\,D}{6}+\frac {b\,x^8\,D}{8}+\frac {A\,a\,x^3}{3}+\frac {B\,a\,x^4}{4}+\frac {A\,b\,x^5}{5}+\frac {C\,a\,x^5}{5}+\frac {B\,b\,x^6}{6}+\frac {C\,b\,x^7}{7} \]

[In]

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

[Out]

(a*x^6*D)/6 + (b*x^8*D)/8 + (A*a*x^3)/3 + (B*a*x^4)/4 + (A*b*x^5)/5 + (C*a*x^5)/5 + (B*b*x^6)/6 + (C*b*x^7)/7