\(\int x^2 (a+b x) (A+B x) \, dx\) [82]

Optimal result

Integrand size = 14, antiderivative size = 33 \[ \int x^2 (a+b x) (A+B x) \, dx=\frac {1}{3} a A x^3+\frac {1}{4} (A b+a B) x^4+\frac {1}{5} b B x^5 \]

[Out]

1/3*a*A*x^3+1/4*(A*b+B*a)*x^4+1/5*b*B*x^5

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {77} \[ \int x^2 (a+b x) (A+B x) \, dx=\frac {1}{4} x^4 (a B+A b)+\frac {1}{3} a A x^3+\frac {1}{5} b B x^5 \]

[In]

Int[x^2*(a + b*x)*(A + B*x),x]

[Out]

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

Rule 77

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00 \[ \int x^2 (a+b x) (A+B x) \, dx=\frac {1}{3} a A x^3+\frac {1}{4} (A b+a B) x^4+\frac {1}{5} b B x^5 \]

[In]

Integrate[x^2*(a + b*x)*(A + B*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {a A \,x^{3}}{3}+\frac {\left (A b +B a \right ) x^{4}}{4}+\frac {b B \,x^{5}}{5}\)	\(28\)
norman	\(\frac {b B \,x^{5}}{5}+\left (\frac {A b}{4}+\frac {B a}{4}\right ) x^{4}+\frac {a A \,x^{3}}{3}\)	\(29\)
gosper	\(\frac {1}{5} b B \,x^{5}+\frac {1}{4} x^{4} A b +\frac {1}{4} x^{4} B a +\frac {1}{3} a A \,x^{3}\)	\(30\)
risch	\(\frac {1}{5} b B \,x^{5}+\frac {1}{4} x^{4} A b +\frac {1}{4} x^{4} B a +\frac {1}{3} a A \,x^{3}\)	\(30\)
parallelrisch	\(\frac {1}{5} b B \,x^{5}+\frac {1}{4} x^{4} A b +\frac {1}{4} x^{4} B a +\frac {1}{3} a A \,x^{3}\)	\(30\)

[In]

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

[Out]

1/3*a*A*x^3+1/4*(A*b+B*a)*x^4+1/5*b*B*x^5

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int x^2 (a+b x) (A+B x) \, dx=\frac {1}{5} \, B b x^{5} + \frac {1}{3} \, A a x^{3} + \frac {1}{4} \, {\left (B a + A b\right )} x^{4} \]

[In]

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

[Out]

1/5*B*b*x^5 + 1/3*A*a*x^3 + 1/4*(B*a + A*b)*x^4

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int x^2 (a+b x) (A+B x) \, dx=\frac {A a x^{3}}{3} + \frac {B b x^{5}}{5} + x^{4} \left (\frac {A b}{4} + \frac {B a}{4}\right ) \]

[In]

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

[Out]

A*a*x**3/3 + B*b*x**5/5 + x**4*(A*b/4 + B*a/4)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.82 \[ \int x^2 (a+b x) (A+B x) \, dx=\frac {1}{5} \, B b x^{5} + \frac {1}{3} \, A a x^{3} + \frac {1}{4} \, {\left (B a + A b\right )} x^{4} \]

[In]

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

[Out]

1/5*B*b*x^5 + 1/3*A*a*x^3 + 1/4*(B*a + A*b)*x^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int x^2 (a+b x) (A+B x) \, dx=\frac {1}{5} \, B b x^{5} + \frac {1}{4} \, B a x^{4} + \frac {1}{4} \, A b x^{4} + \frac {1}{3} \, A a x^{3} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int x^2 (a+b x) (A+B x) \, dx=\frac {B\,b\,x^5}{5}+\left (\frac {A\,b}{4}+\frac {B\,a}{4}\right )\,x^4+\frac {A\,a\,x^3}{3} \]

[In]

int(x^2*(A + B*x)*(a + b*x),x)

[Out]

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