\(\int x^2 (A+B x) (a+c x^2) \, dx\) [252]

Optimal result

Integrand size = 16, antiderivative size = 37 \[ \int x^2 (A+B x) \left (a+c x^2\right ) \, dx=\frac {1}{3} a A x^3+\frac {1}{4} a B x^4+\frac {1}{5} A c x^5+\frac {1}{6} B c x^6 \]

[Out]

1/3*a*A*x^3+1/4*a*B*x^4+1/5*A*c*x^5+1/6*B*c*x^6

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {780} \[ \int x^2 (A+B x) \left (a+c x^2\right ) \, dx=\frac {1}{3} a A x^3+\frac {1}{4} a B x^4+\frac {1}{5} A c x^5+\frac {1}{6} B c x^6 \]

[In]

Int[x^2*(A + B*x)*(a + c*x^2),x]

[Out]

(a*A*x^3)/3 + (a*B*x^4)/4 + (A*c*x^5)/5 + (B*c*x^6)/6

Rule 780

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int x^2 (A+B x) \left (a+c x^2\right ) \, dx=\frac {1}{3} a A x^3+\frac {1}{4} a B x^4+\frac {1}{5} A c x^5+\frac {1}{6} B c x^6 \]

[In]

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

[Out]

(a*A*x^3)/3 + (a*B*x^4)/4 + (A*c*x^5)/5 + (B*c*x^6)/6

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81

method	result	size
gosper	\(\frac {1}{3} a A \,x^{3}+\frac {1}{4} a B \,x^{4}+\frac {1}{5} A c \,x^{5}+\frac {1}{6} B c \,x^{6}\)	\(30\)
default	\(\frac {1}{3} a A \,x^{3}+\frac {1}{4} a B \,x^{4}+\frac {1}{5} A c \,x^{5}+\frac {1}{6} B c \,x^{6}\)	\(30\)
norman	\(\frac {1}{3} a A \,x^{3}+\frac {1}{4} a B \,x^{4}+\frac {1}{5} A c \,x^{5}+\frac {1}{6} B c \,x^{6}\)	\(30\)
risch	\(\frac {1}{3} a A \,x^{3}+\frac {1}{4} a B \,x^{4}+\frac {1}{5} A c \,x^{5}+\frac {1}{6} B c \,x^{6}\)	\(30\)
parallelrisch	\(\frac {1}{3} a A \,x^{3}+\frac {1}{4} a B \,x^{4}+\frac {1}{5} A c \,x^{5}+\frac {1}{6} B c \,x^{6}\)	\(30\)

[In]

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

[Out]

1/3*a*A*x^3+1/4*a*B*x^4+1/5*A*c*x^5+1/6*B*c*x^6

Fricas [A] (verification not implemented)

none

Time = 0.54 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int x^2 (A+B x) \left (a+c x^2\right ) \, dx=\frac {1}{6} \, B c x^{6} + \frac {1}{5} \, A c x^{5} + \frac {1}{4} \, B a x^{4} + \frac {1}{3} \, A a x^{3} \]

[In]

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

[Out]

1/6*B*c*x^6 + 1/5*A*c*x^5 + 1/4*B*a*x^4 + 1/3*A*a*x^3

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int x^2 (A+B x) \left (a+c x^2\right ) \, dx=\frac {A a x^{3}}{3} + \frac {A c x^{5}}{5} + \frac {B a x^{4}}{4} + \frac {B c x^{6}}{6} \]

[In]

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

[Out]

A*a*x**3/3 + A*c*x**5/5 + B*a*x**4/4 + B*c*x**6/6

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int x^2 (A+B x) \left (a+c x^2\right ) \, dx=\frac {1}{6} \, B c x^{6} + \frac {1}{5} \, A c x^{5} + \frac {1}{4} \, B a x^{4} + \frac {1}{3} \, A a x^{3} \]

[In]

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

[Out]

1/6*B*c*x^6 + 1/5*A*c*x^5 + 1/4*B*a*x^4 + 1/3*A*a*x^3

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int x^2 (A+B x) \left (a+c x^2\right ) \, dx=\frac {1}{6} \, B c x^{6} + \frac {1}{5} \, A c x^{5} + \frac {1}{4} \, B a x^{4} + \frac {1}{3} \, A a x^{3} \]

[In]

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

[Out]

1/6*B*c*x^6 + 1/5*A*c*x^5 + 1/4*B*a*x^4 + 1/3*A*a*x^3

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.78 \[ \int x^2 (A+B x) \left (a+c x^2\right ) \, dx=\frac {B\,c\,x^6}{6}+\frac {A\,c\,x^5}{5}+\frac {B\,a\,x^4}{4}+\frac {A\,a\,x^3}{3} \]

[In]

int(x^2*(a + c*x^2)*(A + B*x),x)

[Out]

(A*a*x^3)/3 + (B*a*x^4)/4 + (A*c*x^5)/5 + (B*c*x^6)/6