\(\int (a+b x) (a c-b c x) \, dx\) [1030]

Optimal result

Integrand size = 15, antiderivative size = 18 \[ \int (a+b x) (a c-b c x) \, dx=a^2 c x-\frac {1}{3} b^2 c x^3 \]

[Out]

a^2*c*x-1/3*b^2*c*x^3

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {41} \[ \int (a+b x) (a c-b c x) \, dx=a^2 c x-\frac {1}{3} b^2 c x^3 \]

[In]

Int[(a + b*x)*(a*c - b*c*x),x]

[Out]

a^2*c*x - (b^2*c*x^3)/3

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int (a+b x) (a c-b c x) \, dx=c \left (a^2 x-\frac {b^2 x^3}{3}\right ) \]

[In]

Integrate[(a + b*x)*(a*c - b*c*x),x]

[Out]

c*(a^2*x - (b^2*x^3)/3)

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
default	\(a^{2} c x -\frac {1}{3} b^{2} c \,x^{3}\)	\(17\)
norman	\(a^{2} c x -\frac {1}{3} b^{2} c \,x^{3}\)	\(17\)
risch	\(a^{2} c x -\frac {1}{3} b^{2} c \,x^{3}\)	\(17\)
parallelrisch	\(a^{2} c x -\frac {1}{3} b^{2} c \,x^{3}\)	\(17\)
gosper	\(\frac {c x \left (-b^{2} x^{2}+3 a^{2}\right )}{3}\)	\(19\)

[In]

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

[Out]

a^2*c*x-1/3*b^2*c*x^3

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int (a+b x) (a c-b c x) \, dx=-\frac {1}{3} \, b^{2} c x^{3} + a^{2} c x \]

[In]

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

[Out]

-1/3*b^2*c*x^3 + a^2*c*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int (a+b x) (a c-b c x) \, dx=a^{2} c x - \frac {b^{2} c x^{3}}{3} \]

[In]

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

[Out]

a**2*c*x - b**2*c*x**3/3

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int (a+b x) (a c-b c x) \, dx=-\frac {1}{3} \, b^{2} c x^{3} + a^{2} c x \]

[In]

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

[Out]

-1/3*b^2*c*x^3 + a^2*c*x

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int (a+b x) (a c-b c x) \, dx=-\frac {1}{3} \, b^{2} c x^{3} + a^{2} c x \]

[In]

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

[Out]

-1/3*b^2*c*x^3 + a^2*c*x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int (a+b x) (a c-b c x) \, dx=\frac {c\,x\,\left (3\,a^2-b^2\,x^2\right )}{3} \]

[In]

int((a*c - b*c*x)*(a + b*x),x)

[Out]

(c*x*(3*a^2 - b^2*x^2))/3