\(\int (a+b x) (c+d x) \, dx\) [1239]

Optimal result

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

[Out]

a*c*x+1/2*(a*d+b*c)*x^2+1/3*b*d*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} \[ \int (a+b x) (c+d x) \, dx=\frac {1}{2} x^2 (a d+b c)+a c x+\frac {1}{3} b d x^3 \]

[In]

Int[(a + b*x)*(c + d*x),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int (a+b x) (c+d x) \, dx=a c x+\frac {1}{2} (b c+a d) x^2+\frac {1}{3} b d x^3 \]

[In]

Integrate[(a + b*x)*(c + d*x),x]

[Out]

a*c*x + ((b*c + a*d)*x^2)/2 + (b*d*x^3)/3

Maple [A] (verified)

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

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

[In]

int((b*x+a)*(d*x+c),x,method=_RETURNVERBOSE)

[Out]

a*c*x+1/2*(a*d+b*c)*x^2+1/3*b*d*x^3

Fricas [A] (verification not implemented)

none

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

[In]

integrate((b*x+a)*(d*x+c),x, algorithm="fricas")

[Out]

1/3*x^3*d*b + 1/2*x^2*c*b + 1/2*x^2*d*a + x*c*a

Sympy [A] (verification not implemented)

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

[In]

integrate((b*x+a)*(d*x+c),x)

[Out]

a*c*x + b*d*x**3/3 + x**2*(a*d/2 + b*c/2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int (a+b x) (c+d x) \, dx=\frac {1}{3} \, b d x^{3} + a c x + \frac {1}{2} \, {\left (b c + a d\right )} x^{2} \]

[In]

integrate((b*x+a)*(d*x+c),x, algorithm="maxima")

[Out]

1/3*b*d*x^3 + a*c*x + 1/2*(b*c + a*d)*x^2

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int (a+b x) (c+d x) \, dx=\frac {1}{3} \, b d x^{3} + \frac {1}{2} \, b c x^{2} + \frac {1}{2} \, a d x^{2} + a c x \]

[In]

integrate((b*x+a)*(d*x+c),x, algorithm="giac")

[Out]

1/3*b*d*x^3 + 1/2*b*c*x^2 + 1/2*a*d*x^2 + a*c*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int (a+b x) (c+d x) \, dx=\frac {b\,d\,x^3}{3}+\left (\frac {a\,d}{2}+\frac {b\,c}{2}\right )\,x^2+a\,c\,x \]

[In]

int((a + b*x)*(c + d*x),x)

[Out]

x^2*((a*d)/2 + (b*c)/2) + a*c*x + (b*d*x^3)/3