\(\int (a+b x)^2 (a c-b c x) \, dx\) [1040]

Optimal result

Integrand size = 17, antiderivative size = 32 \[ \int (a+b x)^2 (a c-b c x) \, dx=\frac {2 a c (a+b x)^3}{3 b}-\frac {c (a+b x)^4}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {45} \[ \int (a+b x)^2 (a c-b c x) \, dx=\frac {2 a c (a+b x)^3}{3 b}-\frac {c (a+b x)^4}{4 b} \]

[In]

Int[(a + b*x)^2*(a*c - b*c*x),x]

[Out]

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

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 (2 a c (a+b x)^2-c (a+b x)^3\right ) \, dx \\ & = \frac {2 a c (a+b x)^3}{3 b}-\frac {c (a+b x)^4}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.25 \[ \int (a+b x)^2 (a c-b c x) \, dx=c \left (a^3 x+\frac {1}{2} a^2 b x^2-\frac {1}{3} a b^2 x^3-\frac {b^3 x^4}{4}\right ) \]

[In]

Integrate[(a + b*x)^2*(a*c - b*c*x),x]

[Out]

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

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09

method	result	size
gosper	\(\frac {c x \left (-3 b^{3} x^{3}-4 a \,b^{2} x^{2}+6 a^{2} b x +12 a^{3}\right )}{12}\)	\(35\)
default	\(-\frac {1}{4} b^{3} c \,x^{4}-\frac {1}{3} a \,b^{2} c \,x^{3}+\frac {1}{2} a^{2} b c \,x^{2}+a^{3} c x\)	\(37\)
norman	\(-\frac {1}{4} b^{3} c \,x^{4}-\frac {1}{3} a \,b^{2} c \,x^{3}+\frac {1}{2} a^{2} b c \,x^{2}+a^{3} c x\)	\(37\)
risch	\(-\frac {1}{4} b^{3} c \,x^{4}-\frac {1}{3} a \,b^{2} c \,x^{3}+\frac {1}{2} a^{2} b c \,x^{2}+a^{3} c x\)	\(37\)
parallelrisch	\(-\frac {1}{4} b^{3} c \,x^{4}-\frac {1}{3} a \,b^{2} c \,x^{3}+\frac {1}{2} a^{2} b c \,x^{2}+a^{3} c x\)	\(37\)

[In]

int((b*x+a)^2*(-b*c*x+a*c),x,method=_RETURNVERBOSE)

[Out]

1/12*c*x*(-3*b^3*x^3-4*a*b^2*x^2+6*a^2*b*x+12*a^3)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int (a+b x)^2 (a c-b c x) \, dx=-\frac {1}{4} \, b^{3} c x^{4} - \frac {1}{3} \, a b^{2} c x^{3} + \frac {1}{2} \, a^{2} b c x^{2} + a^{3} c x \]

[In]

integrate((b*x+a)^2*(-b*c*x+a*c),x, algorithm="fricas")

[Out]

-1/4*b^3*c*x^4 - 1/3*a*b^2*c*x^3 + 1/2*a^2*b*c*x^2 + a^3*c*x

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int (a+b x)^2 (a c-b c x) \, dx=a^{3} c x + \frac {a^{2} b c x^{2}}{2} - \frac {a b^{2} c x^{3}}{3} - \frac {b^{3} c x^{4}}{4} \]

[In]

integrate((b*x+a)**2*(-b*c*x+a*c),x)

[Out]

a**3*c*x + a**2*b*c*x**2/2 - a*b**2*c*x**3/3 - b**3*c*x**4/4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int (a+b x)^2 (a c-b c x) \, dx=-\frac {1}{4} \, b^{3} c x^{4} - \frac {1}{3} \, a b^{2} c x^{3} + \frac {1}{2} \, a^{2} b c x^{2} + a^{3} c x \]

[In]

integrate((b*x+a)^2*(-b*c*x+a*c),x, algorithm="maxima")

[Out]

-1/4*b^3*c*x^4 - 1/3*a*b^2*c*x^3 + 1/2*a^2*b*c*x^2 + a^3*c*x

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int (a+b x)^2 (a c-b c x) \, dx=-\frac {1}{4} \, b^{3} c x^{4} - \frac {1}{3} \, a b^{2} c x^{3} + \frac {1}{2} \, a^{2} b c x^{2} + a^{3} c x \]

[In]

integrate((b*x+a)^2*(-b*c*x+a*c),x, algorithm="giac")

[Out]

-1/4*b^3*c*x^4 - 1/3*a*b^2*c*x^3 + 1/2*a^2*b*c*x^2 + a^3*c*x

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int (a+b x)^2 (a c-b c x) \, dx=c\,a^3\,x+\frac {c\,a^2\,b\,x^2}{2}-\frac {c\,a\,b^2\,x^3}{3}-\frac {c\,b^3\,x^4}{4} \]

[In]

int((a*c - b*c*x)*(a + b*x)^2,x)

[Out]

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