\(\int (a+b x)^2 (a c-b c x)^3 \, dx\) [1038]

Optimal result

Integrand size = 19, antiderivative size = 57 \[ \int (a+b x)^2 (a c-b c x)^3 \, dx=-\frac {a^2 c^3 (a-b x)^4}{b}+\frac {4 a c^3 (a-b x)^5}{5 b}-\frac {c^3 (a-b x)^6}{6 b} \]

[Out]

-a^2*c^3*(-b*x+a)^4/b+4/5*a*c^3*(-b*x+a)^5/b-1/6*c^3*(-b*x+a)^6/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {45} \[ \int (a+b x)^2 (a c-b c x)^3 \, dx=-\frac {a^2 c^3 (a-b x)^4}{b}-\frac {c^3 (a-b x)^6}{6 b}+\frac {4 a c^3 (a-b x)^5}{5 b} \]

[In]

Int[(a + b*x)^2*(a*c - b*c*x)^3,x]

[Out]

-((a^2*c^3*(a - b*x)^4)/b) + (4*a*c^3*(a - b*x)^5)/(5*b) - (c^3*(a - b*x)^6)/(6*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 (4 a^2 (a c-b c x)^3-\frac {4 a (a c-b c x)^4}{c}+\frac {(a c-b c x)^5}{c^2}\right ) \, dx \\ & = -\frac {a^2 c^3 (a-b x)^4}{b}+\frac {4 a c^3 (a-b x)^5}{5 b}-\frac {c^3 (a-b x)^6}{6 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19 \[ \int (a+b x)^2 (a c-b c x)^3 \, dx=c^3 \left (a^5 x-\frac {1}{2} a^4 b x^2-\frac {2}{3} a^3 b^2 x^3+\frac {1}{2} a^2 b^3 x^4+\frac {1}{5} a b^4 x^5-\frac {b^5 x^6}{6}\right ) \]

[In]

Integrate[(a + b*x)^2*(a*c - b*c*x)^3,x]

[Out]

c^3*(a^5*x - (a^4*b*x^2)/2 - (2*a^3*b^2*x^3)/3 + (a^2*b^3*x^4)/2 + (a*b^4*x^5)/5 - (b^5*x^6)/6)

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.04

method	result	size
gosper	\(\frac {x \left (-5 b^{5} x^{5}+6 a \,b^{4} x^{4}+15 a^{2} b^{3} x^{3}-20 a^{3} b^{2} x^{2}-15 a^{4} b x +30 a^{5}\right ) c^{3}}{30}\)	\(59\)
default	\(-\frac {1}{6} b^{5} c^{3} x^{6}+\frac {1}{5} a \,b^{4} c^{3} x^{5}+\frac {1}{2} a^{2} b^{3} c^{3} x^{4}-\frac {2}{3} a^{3} c^{3} b^{2} x^{3}-\frac {1}{2} a^{4} c^{3} b \,x^{2}+a^{5} c^{3} x\)	\(73\)
norman	\(-\frac {1}{6} b^{5} c^{3} x^{6}+\frac {1}{5} a \,b^{4} c^{3} x^{5}+\frac {1}{2} a^{2} b^{3} c^{3} x^{4}-\frac {2}{3} a^{3} c^{3} b^{2} x^{3}-\frac {1}{2} a^{4} c^{3} b \,x^{2}+a^{5} c^{3} x\)	\(73\)
risch	\(-\frac {1}{6} b^{5} c^{3} x^{6}+\frac {1}{5} a \,b^{4} c^{3} x^{5}+\frac {1}{2} a^{2} b^{3} c^{3} x^{4}-\frac {2}{3} a^{3} c^{3} b^{2} x^{3}-\frac {1}{2} a^{4} c^{3} b \,x^{2}+a^{5} c^{3} x\)	\(73\)
parallelrisch	\(-\frac {1}{6} b^{5} c^{3} x^{6}+\frac {1}{5} a \,b^{4} c^{3} x^{5}+\frac {1}{2} a^{2} b^{3} c^{3} x^{4}-\frac {2}{3} a^{3} c^{3} b^{2} x^{3}-\frac {1}{2} a^{4} c^{3} b \,x^{2}+a^{5} c^{3} x\)	\(73\)

[In]

int((b*x+a)^2*(-b*c*x+a*c)^3,x,method=_RETURNVERBOSE)

[Out]

1/30*x*(-5*b^5*x^5+6*a*b^4*x^4+15*a^2*b^3*x^3-20*a^3*b^2*x^2-15*a^4*b*x+30*a^5)*c^3

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.26 \[ \int (a+b x)^2 (a c-b c x)^3 \, dx=-\frac {1}{6} \, b^{5} c^{3} x^{6} + \frac {1}{5} \, a b^{4} c^{3} x^{5} + \frac {1}{2} \, a^{2} b^{3} c^{3} x^{4} - \frac {2}{3} \, a^{3} b^{2} c^{3} x^{3} - \frac {1}{2} \, a^{4} b c^{3} x^{2} + a^{5} c^{3} x \]

[In]

integrate((b*x+a)^2*(-b*c*x+a*c)^3,x, algorithm="fricas")

[Out]

-1/6*b^5*c^3*x^6 + 1/5*a*b^4*c^3*x^5 + 1/2*a^2*b^3*c^3*x^4 - 2/3*a^3*b^2*c^3*x^3 - 1/2*a^4*b*c^3*x^2 + a^5*c^3
*x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.37 \[ \int (a+b x)^2 (a c-b c x)^3 \, dx=a^{5} c^{3} x - \frac {a^{4} b c^{3} x^{2}}{2} - \frac {2 a^{3} b^{2} c^{3} x^{3}}{3} + \frac {a^{2} b^{3} c^{3} x^{4}}{2} + \frac {a b^{4} c^{3} x^{5}}{5} - \frac {b^{5} c^{3} x^{6}}{6} \]

[In]

integrate((b*x+a)**2*(-b*c*x+a*c)**3,x)

[Out]

a**5*c**3*x - a**4*b*c**3*x**2/2 - 2*a**3*b**2*c**3*x**3/3 + a**2*b**3*c**3*x**4/2 + a*b**4*c**3*x**5/5 - b**5
*c**3*x**6/6

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.26 \[ \int (a+b x)^2 (a c-b c x)^3 \, dx=-\frac {1}{6} \, b^{5} c^{3} x^{6} + \frac {1}{5} \, a b^{4} c^{3} x^{5} + \frac {1}{2} \, a^{2} b^{3} c^{3} x^{4} - \frac {2}{3} \, a^{3} b^{2} c^{3} x^{3} - \frac {1}{2} \, a^{4} b c^{3} x^{2} + a^{5} c^{3} x \]

[In]

integrate((b*x+a)^2*(-b*c*x+a*c)^3,x, algorithm="maxima")

[Out]

-1/6*b^5*c^3*x^6 + 1/5*a*b^4*c^3*x^5 + 1/2*a^2*b^3*c^3*x^4 - 2/3*a^3*b^2*c^3*x^3 - 1/2*a^4*b*c^3*x^2 + a^5*c^3
*x

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.26 \[ \int (a+b x)^2 (a c-b c x)^3 \, dx=-\frac {1}{6} \, b^{5} c^{3} x^{6} + \frac {1}{5} \, a b^{4} c^{3} x^{5} + \frac {1}{2} \, a^{2} b^{3} c^{3} x^{4} - \frac {2}{3} \, a^{3} b^{2} c^{3} x^{3} - \frac {1}{2} \, a^{4} b c^{3} x^{2} + a^{5} c^{3} x \]

[In]

integrate((b*x+a)^2*(-b*c*x+a*c)^3,x, algorithm="giac")

[Out]

-1/6*b^5*c^3*x^6 + 1/5*a*b^4*c^3*x^5 + 1/2*a^2*b^3*c^3*x^4 - 2/3*a^3*b^2*c^3*x^3 - 1/2*a^4*b*c^3*x^2 + a^5*c^3
*x

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.26 \[ \int (a+b x)^2 (a c-b c x)^3 \, dx=a^5\,c^3\,x-\frac {a^4\,b\,c^3\,x^2}{2}-\frac {2\,a^3\,b^2\,c^3\,x^3}{3}+\frac {a^2\,b^3\,c^3\,x^4}{2}+\frac {a\,b^4\,c^3\,x^5}{5}-\frac {b^5\,c^3\,x^6}{6} \]

[In]

int((a*c - b*c*x)^3*(a + b*x)^2,x)

[Out]

a^5*c^3*x - (b^5*c^3*x^6)/6 - (a^4*b*c^3*x^2)/2 + (a*b^4*c^3*x^5)/5 - (2*a^3*b^2*c^3*x^3)/3 + (a^2*b^3*c^3*x^4
)/2