\(\int (a+b x) (A+B x) \, dx\) [1015]

Optimal result

Integrand size = 11, antiderivative size = 28 \[ \int (a+b x) (A+B x) \, dx=a A x+\frac {1}{2} (A b+a B) x^2+\frac {1}{3} b B x^3 \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \[ \int (a+b x) (A+B x) \, dx=\frac {1}{2} x^2 (a B+A b)+a A x+\frac {1}{3} b B x^3 \]

[In]

Int[(a + b*x)*(A + B*x),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int (a+b x) (A+B x) \, dx=a A x+\frac {1}{2} (A b+a B) x^2+\frac {1}{3} b B x^3 \]

[In]

Integrate[(a + b*x)*(A + B*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method	result	size
default	\(a A x +\frac {\left (A b +B a \right ) x^{2}}{2}+\frac {b B \,x^{3}}{3}\)	\(25\)
norman	\(\frac {b B \,x^{3}}{3}+\left (\frac {A b}{2}+\frac {B a}{2}\right ) x^{2}+a A x\)	\(26\)
gosper	\(\frac {1}{3} b B \,x^{3}+\frac {1}{2} x^{2} A b +\frac {1}{2} x^{2} B a +a A x\)	\(27\)
risch	\(\frac {1}{3} b B \,x^{3}+\frac {1}{2} x^{2} A b +\frac {1}{2} x^{2} B a +a A x\)	\(27\)
parallelrisch	\(\frac {1}{3} b B \,x^{3}+\frac {1}{2} x^{2} A b +\frac {1}{2} x^{2} B a +a A x\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int (a+b x) (A+B x) \, dx=\frac {1}{3} x^{3} b B + \frac {1}{2} x^{2} a B + \frac {1}{2} x^{2} b A + x a A \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int (a+b x) (A+B x) \, dx=A a x + \frac {B b x^{3}}{3} + x^{2} \left (\frac {A b}{2} + \frac {B a}{2}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int (a+b x) (A+B x) \, dx=\frac {1}{3} \, B b x^{3} + A a x + \frac {1}{2} \, {\left (B a + A b\right )} x^{2} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int (a+b x) (A+B x) \, dx=\frac {1}{3} \, B b x^{3} + \frac {1}{2} \, B a x^{2} + \frac {1}{2} \, A b x^{2} + A a x \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int (a+b x) (A+B x) \, dx=\frac {B\,b\,x^3}{3}+\left (\frac {A\,b}{2}+\frac {B\,a}{2}\right )\,x^2+A\,a\,x \]

[In]

int((A + B*x)*(a + b*x),x)

[Out]

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