\(\int (a+b x)^2 (A+B x) \, dx\) [95]

Optimal result

Integrand size = 13, antiderivative size = 38 \[ \int (a+b x)^2 (A+B x) \, dx=\frac {(A b-a B) (a+b x)^3}{3 b^2}+\frac {B (a+b x)^4}{4 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \[ \int (a+b x)^2 (A+B x) \, dx=\frac {(a+b x)^3 (A b-a B)}{3 b^2}+\frac {B (a+b x)^4}{4 b^2} \]

[In]

Int[(a + b*x)^2*(A + B*x),x]

[Out]

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

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])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21 \[ \int (a+b x)^2 (A+B x) \, dx=\frac {1}{12} x \left (6 a^2 (2 A+B x)+4 a b x (3 A+2 B x)+b^2 x^2 (4 A+3 B x)\right ) \]

[In]

Integrate[(a + b*x)^2*(A + B*x),x]

[Out]

(x*(6*a^2*(2*A + B*x) + 4*a*b*x*(3*A + 2*B*x) + b^2*x^2*(4*A + 3*B*x)))/12

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26

method	result	size
norman	\(\frac {b^{2} B \,x^{4}}{4}+\left (\frac {1}{3} b^{2} A +\frac {2}{3} a b B \right ) x^{3}+\left (a b A +\frac {1}{2} a^{2} B \right ) x^{2}+a^{2} A x\)	\(48\)
default	\(\frac {b^{2} B \,x^{4}}{4}+\frac {\left (b^{2} A +2 a b B \right ) x^{3}}{3}+\frac {\left (2 a b A +a^{2} B \right ) x^{2}}{2}+a^{2} A x\)	\(49\)
gosper	\(\frac {1}{4} b^{2} B \,x^{4}+\frac {1}{3} x^{3} b^{2} A +\frac {2}{3} x^{3} a b B +x^{2} a b A +\frac {1}{2} x^{2} a^{2} B +a^{2} A x\)	\(50\)
risch	\(\frac {1}{4} b^{2} B \,x^{4}+\frac {1}{3} x^{3} b^{2} A +\frac {2}{3} x^{3} a b B +x^{2} a b A +\frac {1}{2} x^{2} a^{2} B +a^{2} A x\)	\(50\)
parallelrisch	\(\frac {1}{4} b^{2} B \,x^{4}+\frac {1}{3} x^{3} b^{2} A +\frac {2}{3} x^{3} a b B +x^{2} a b A +\frac {1}{2} x^{2} a^{2} B +a^{2} A x\)	\(50\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int (a+b x)^2 (A+B x) \, dx=\frac {1}{4} \, B b^{2} x^{4} + A a^{2} x + \frac {1}{3} \, {\left (2 \, B a b + A b^{2}\right )} x^{3} + \frac {1}{2} \, {\left (B a^{2} + 2 \, A a b\right )} x^{2} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.29 \[ \int (a+b x)^2 (A+B x) \, dx=A a^{2} x + \frac {B b^{2} x^{4}}{4} + x^{3} \left (\frac {A b^{2}}{3} + \frac {2 B a b}{3}\right ) + x^{2} \left (A a b + \frac {B a^{2}}{2}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int (a+b x)^2 (A+B x) \, dx=\frac {1}{4} \, B b^{2} x^{4} + A a^{2} x + \frac {1}{3} \, {\left (2 \, B a b + A b^{2}\right )} x^{3} + \frac {1}{2} \, {\left (B a^{2} + 2 \, A a b\right )} x^{2} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.29 \[ \int (a+b x)^2 (A+B x) \, dx=\frac {1}{4} \, B b^{2} x^{4} + \frac {2}{3} \, B a b x^{3} + \frac {1}{3} \, A b^{2} x^{3} + \frac {1}{2} \, B a^{2} x^{2} + A a b x^{2} + A a^{2} x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.24 \[ \int (a+b x)^2 (A+B x) \, dx=x^2\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )+x^3\,\left (\frac {A\,b^2}{3}+\frac {2\,B\,a\,b}{3}\right )+\frac {B\,b^2\,x^4}{4}+A\,a^2\,x \]

[In]

int((A + B*x)*(a + b*x)^2,x)

[Out]

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