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

Optimal result

Integrand size = 16, antiderivative size = 33 \[ \int x \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {1}{2} a A x^2+\frac {1}{4} (A b+a B) x^4+\frac {1}{6} b B x^6 \]

[Out]

1/2*a*A*x^2+1/4*(A*b+B*a)*x^4+1/6*b*B*x^6

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {455, 45} \[ \int x \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {1}{4} x^4 (a B+A b)+\frac {1}{2} a A x^2+\frac {1}{6} b B x^6 \]

[In]

Int[x*(a + b*x^2)*(A + B*x^2),x]

[Out]

(a*A*x^2)/2 + ((A*b + a*B)*x^4)/4 + (b*B*x^6)/6

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int x \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {1}{2} a A x^2+\frac {1}{4} (A b+a B) x^4+\frac {1}{6} b B x^6 \]

[In]

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

[Out]

(a*A*x^2)/2 + ((A*b + a*B)*x^4)/4 + (b*B*x^6)/6

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {a A \,x^{2}}{2}+\frac {\left (A b +B a \right ) x^{4}}{4}+\frac {b B \,x^{6}}{6}\)	\(28\)
norman	\(\frac {b B \,x^{6}}{6}+\left (\frac {A b}{4}+\frac {B a}{4}\right ) x^{4}+\frac {a A \,x^{2}}{2}\)	\(29\)
gosper	\(\frac {1}{6} b B \,x^{6}+\frac {1}{4} x^{4} A b +\frac {1}{4} x^{4} B a +\frac {1}{2} a A \,x^{2}\)	\(30\)
risch	\(\frac {1}{6} b B \,x^{6}+\frac {1}{4} x^{4} A b +\frac {1}{4} x^{4} B a +\frac {1}{2} a A \,x^{2}\)	\(30\)
parallelrisch	\(\frac {1}{6} b B \,x^{6}+\frac {1}{4} x^{4} A b +\frac {1}{4} x^{4} B a +\frac {1}{2} a A \,x^{2}\)	\(30\)

[In]

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

[Out]

1/2*a*A*x^2+1/4*(A*b+B*a)*x^4+1/6*b*B*x^6

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int x \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {1}{6} \, B b x^{6} + \frac {1}{4} \, {\left (B a + A b\right )} x^{4} + \frac {1}{2} \, A a x^{2} \]

[In]

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

[Out]

1/6*B*b*x^6 + 1/4*(B*a + A*b)*x^4 + 1/2*A*a*x^2

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int x \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {A a x^{2}}{2} + \frac {B b x^{6}}{6} + x^{4} \left (\frac {A b}{4} + \frac {B a}{4}\right ) \]

[In]

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

[Out]

A*a*x**2/2 + B*b*x**6/6 + x**4*(A*b/4 + B*a/4)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int x \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {1}{6} \, B b x^{6} + \frac {1}{4} \, {\left (B a + A b\right )} x^{4} + \frac {1}{2} \, A a x^{2} \]

[In]

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

[Out]

1/6*B*b*x^6 + 1/4*(B*a + A*b)*x^4 + 1/2*A*a*x^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int x \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {1}{6} \, B b x^{6} + \frac {1}{4} \, B a x^{4} + \frac {1}{4} \, A b x^{4} + \frac {1}{2} \, A a x^{2} \]

[In]

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

[Out]

1/6*B*b*x^6 + 1/4*B*a*x^4 + 1/4*A*b*x^4 + 1/2*A*a*x^2

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int x \left (a+b x^2\right ) \left (A+B x^2\right ) \, dx=\frac {B\,b\,x^6}{6}+\left (\frac {A\,b}{4}+\frac {B\,a}{4}\right )\,x^4+\frac {A\,a\,x^2}{2} \]

[In]

int(x*(A + B*x^2)*(a + b*x^2),x)

[Out]

x^4*((A*b)/4 + (B*a)/4) + (A*a*x^2)/2 + (B*b*x^6)/6