\(\int x (a+b x)^2 (A+B x) \, dx\) [94]

Optimal result

Integrand size = 14, antiderivative size = 55 \[ \int x (a+b x)^2 (A+B x) \, dx=\frac {1}{2} a^2 A x^2+\frac {1}{3} a (2 A b+a B) x^3+\frac {1}{4} b (A b+2 a B) x^4+\frac {1}{5} b^2 B x^5 \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, 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} \[ \int x (a+b x)^2 (A+B x) \, dx=\frac {1}{2} a^2 A x^2+\frac {1}{4} b x^4 (2 a B+A b)+\frac {1}{3} a x^3 (a B+2 A b)+\frac {1}{5} b^2 B x^5 \]

[In]

Int[x*(a + b*x)^2*(A + B*x),x]

[Out]

(a^2*A*x^2)/2 + (a*(2*A*b + a*B)*x^3)/3 + (b*(A*b + 2*a*B)*x^4)/4 + (b^2*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^2 A x+a (2 A b+a B) x^2+b (A b+2 a B) x^3+b^2 B x^4\right ) \, dx \\ & = \frac {1}{2} a^2 A x^2+\frac {1}{3} a (2 A b+a B) x^3+\frac {1}{4} b (A b+2 a B) x^4+\frac {1}{5} b^2 B x^5 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91 \[ \int x (a+b x)^2 (A+B x) \, dx=\frac {1}{60} x^2 \left (10 a^2 (3 A+2 B x)+10 a b x (4 A+3 B x)+3 b^2 x^2 (5 A+4 B x)\right ) \]

[In]

Integrate[x*(a + b*x)^2*(A + B*x),x]

[Out]

(x^2*(10*a^2*(3*A + 2*B*x) + 10*a*b*x*(4*A + 3*B*x) + 3*b^2*x^2*(5*A + 4*B*x)))/60

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {b^{2} B \,x^{5}}{5}+\frac {\left (b^{2} A +2 a b B \right ) x^{4}}{4}+\frac {\left (2 a b A +a^{2} B \right ) x^{3}}{3}+\frac {a^{2} A \,x^{2}}{2}\)	\(52\)
norman	\(\frac {b^{2} B \,x^{5}}{5}+\left (\frac {1}{4} b^{2} A +\frac {1}{2} a b B \right ) x^{4}+\left (\frac {2}{3} a b A +\frac {1}{3} a^{2} B \right ) x^{3}+\frac {a^{2} A \,x^{2}}{2}\)	\(52\)
gosper	\(\frac {1}{5} b^{2} B \,x^{5}+\frac {1}{4} x^{4} b^{2} A +\frac {1}{2} x^{4} a b B +\frac {2}{3} x^{3} a b A +\frac {1}{3} x^{3} a^{2} B +\frac {1}{2} a^{2} A \,x^{2}\)	\(54\)
risch	\(\frac {1}{5} b^{2} B \,x^{5}+\frac {1}{4} x^{4} b^{2} A +\frac {1}{2} x^{4} a b B +\frac {2}{3} x^{3} a b A +\frac {1}{3} x^{3} a^{2} B +\frac {1}{2} a^{2} A \,x^{2}\)	\(54\)
parallelrisch	\(\frac {1}{5} b^{2} B \,x^{5}+\frac {1}{4} x^{4} b^{2} A +\frac {1}{2} x^{4} a b B +\frac {2}{3} x^{3} a b A +\frac {1}{3} x^{3} a^{2} B +\frac {1}{2} a^{2} A \,x^{2}\)	\(54\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int x (a+b x)^2 (A+B x) \, dx=\frac {1}{5} \, B b^{2} x^{5} + \frac {1}{2} \, A a^{2} x^{2} + \frac {1}{4} \, {\left (2 \, B a b + A b^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.98 \[ \int x (a+b x)^2 (A+B x) \, dx=\frac {A a^{2} x^{2}}{2} + \frac {B b^{2} x^{5}}{5} + x^{4} \left (\frac {A b^{2}}{4} + \frac {B a b}{2}\right ) + x^{3} \cdot \left (\frac {2 A a b}{3} + \frac {B a^{2}}{3}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int x (a+b x)^2 (A+B x) \, dx=\frac {1}{5} \, B b^{2} x^{5} + \frac {1}{2} \, A a^{2} x^{2} + \frac {1}{4} \, {\left (2 \, B a b + A b^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B a^{2} + 2 \, A a b\right )} x^{3} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.96 \[ \int x (a+b x)^2 (A+B x) \, dx=\frac {1}{5} \, B b^{2} x^{5} + \frac {1}{2} \, B a b x^{4} + \frac {1}{4} \, A b^{2} x^{4} + \frac {1}{3} \, B a^{2} x^{3} + \frac {2}{3} \, A a b x^{3} + \frac {1}{2} \, A a^{2} x^{2} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.93 \[ \int x (a+b x)^2 (A+B x) \, dx=x^3\,\left (\frac {B\,a^2}{3}+\frac {2\,A\,b\,a}{3}\right )+x^4\,\left (\frac {A\,b^2}{4}+\frac {B\,a\,b}{2}\right )+\frac {A\,a^2\,x^2}{2}+\frac {B\,b^2\,x^5}{5} \]

[In]

int(x*(A + B*x)*(a + b*x)^2,x)

[Out]

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