\(\int (a+b x) (a c-b c x)^3 \, dx\) [4]

Optimal result

Integrand size = 17, antiderivative size = 38 \[ \int (a+b x) (a c-b c x)^3 \, dx=-\frac {a c^3 (a-b x)^4}{2 b}+\frac {c^3 (a-b x)^5}{5 b} \]

[Out]

-1/2*a*c^3*(-b*x+a)^4/b+1/5*c^3*(-b*x+a)^5/b

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {45} \[ \int (a+b x) (a c-b c x)^3 \, dx=\frac {c^3 (a-b x)^5}{5 b}-\frac {a c^3 (a-b x)^4}{2 b} \]

[In]

Int[(a + b*x)*(a*c - b*c*x)^3,x]

[Out]

-1/2*(a*c^3*(a - b*x)^4)/b + (c^3*(a - b*x)^5)/(5*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 (a c-b c x)^3-\frac {(a c-b c x)^4}{c}\right ) \, dx \\ & = -\frac {a c^3 (a-b x)^4}{2 b}+\frac {c^3 (a-b x)^5}{5 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.05 \[ \int (a+b x) (a c-b c x)^3 \, dx=c^3 \left (a^4 x-a^3 b x^2+\frac {1}{2} a b^3 x^4-\frac {b^4 x^5}{5}\right ) \]

[In]

Integrate[(a + b*x)*(a*c - b*c*x)^3,x]

[Out]

c^3*(a^4*x - a^3*b*x^2 + (a*b^3*x^4)/2 - (b^4*x^5)/5)

Maple [A] (verified)

Time = 0.35 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.97

method	result	size
gosper	\(\frac {x \left (-2 b^{4} x^{4}+5 a \,b^{3} x^{3}-10 a^{3} b x +10 a^{4}\right ) c^{3}}{10}\)	\(37\)
default	\(-\frac {1}{5} b^{4} c^{3} x^{5}+\frac {1}{2} a \,b^{3} c^{3} x^{4}-a^{3} c^{3} b \,x^{2}+a^{4} c^{3} x\)	\(45\)
norman	\(-\frac {1}{5} b^{4} c^{3} x^{5}+\frac {1}{2} a \,b^{3} c^{3} x^{4}-a^{3} c^{3} b \,x^{2}+a^{4} c^{3} x\)	\(45\)
risch	\(-\frac {1}{5} b^{4} c^{3} x^{5}+\frac {1}{2} a \,b^{3} c^{3} x^{4}-a^{3} c^{3} b \,x^{2}+a^{4} c^{3} x\)	\(45\)
parallelrisch	\(-\frac {1}{5} b^{4} c^{3} x^{5}+\frac {1}{2} a \,b^{3} c^{3} x^{4}-a^{3} c^{3} b \,x^{2}+a^{4} c^{3} x\)	\(45\)

[In]

int((b*x+a)*(-b*c*x+a*c)^3,x,method=_RETURNVERBOSE)

[Out]

1/10*x*(-2*b^4*x^4+5*a*b^3*x^3-10*a^3*b*x+10*a^4)*c^3

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int (a+b x) (a c-b c x)^3 \, dx=-\frac {1}{5} \, b^{4} c^{3} x^{5} + \frac {1}{2} \, a b^{3} c^{3} x^{4} - a^{3} b c^{3} x^{2} + a^{4} c^{3} x \]

[In]

integrate((b*x+a)*(-b*c*x+a*c)^3,x, algorithm="fricas")

[Out]

-1/5*b^4*c^3*x^5 + 1/2*a*b^3*c^3*x^4 - a^3*b*c^3*x^2 + a^4*c^3*x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int (a+b x) (a c-b c x)^3 \, dx=a^{4} c^{3} x - a^{3} b c^{3} x^{2} + \frac {a b^{3} c^{3} x^{4}}{2} - \frac {b^{4} c^{3} x^{5}}{5} \]

[In]

integrate((b*x+a)*(-b*c*x+a*c)**3,x)

[Out]

a**4*c**3*x - a**3*b*c**3*x**2 + a*b**3*c**3*x**4/2 - b**4*c**3*x**5/5

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int (a+b x) (a c-b c x)^3 \, dx=-\frac {1}{5} \, b^{4} c^{3} x^{5} + \frac {1}{2} \, a b^{3} c^{3} x^{4} - a^{3} b c^{3} x^{2} + a^{4} c^{3} x \]

[In]

integrate((b*x+a)*(-b*c*x+a*c)^3,x, algorithm="maxima")

[Out]

-1/5*b^4*c^3*x^5 + 1/2*a*b^3*c^3*x^4 - a^3*b*c^3*x^2 + a^4*c^3*x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int (a+b x) (a c-b c x)^3 \, dx=-\frac {1}{5} \, b^{4} c^{3} x^{5} + \frac {1}{2} \, a b^{3} c^{3} x^{4} - a^{3} b c^{3} x^{2} + a^{4} c^{3} x \]

[In]

integrate((b*x+a)*(-b*c*x+a*c)^3,x, algorithm="giac")

[Out]

-1/5*b^4*c^3*x^5 + 1/2*a*b^3*c^3*x^4 - a^3*b*c^3*x^2 + a^4*c^3*x

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.16 \[ \int (a+b x) (a c-b c x)^3 \, dx=a^4\,c^3\,x-a^3\,b\,c^3\,x^2+\frac {a\,b^3\,c^3\,x^4}{2}-\frac {b^4\,c^3\,x^5}{5} \]

[In]

int((a*c - b*c*x)^3*(a + b*x),x)

[Out]

a^4*c^3*x - (b^4*c^3*x^5)/5 - a^3*b*c^3*x^2 + (a*b^3*c^3*x^4)/2