\(\int \frac {(a c-b c x)^3}{a+b x} \, dx\) [1049]

Optimal result

Integrand size = 19, antiderivative size = 61 \[ \int \frac {(a c-b c x)^3}{a+b x} \, dx=-4 a^2 c^3 x+\frac {a c^3 (a-b x)^2}{b}+\frac {c^3 (a-b x)^3}{3 b}+\frac {8 a^3 c^3 \log (a+b x)}{b} \]

[Out]

-4*a^2*c^3*x+a*c^3*(-b*x+a)^2/b+1/3*c^3*(-b*x+a)^3/b+8*a^3*c^3*ln(b*x+a)/b

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 61, 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} \[ \int \frac {(a c-b c x)^3}{a+b x} \, dx=\frac {8 a^3 c^3 \log (a+b x)}{b}-4 a^2 c^3 x+\frac {c^3 (a-b x)^3}{3 b}+\frac {a c^3 (a-b x)^2}{b} \]

[In]

Int[(a*c - b*c*x)^3/(a + b*x),x]

[Out]

-4*a^2*c^3*x + (a*c^3*(a - b*x)^2)/b + (c^3*(a - b*x)^3)/(3*b) + (8*a^3*c^3*Log[a + b*x])/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 c^3+\frac {8 a^3 c^3}{a+b x}-2 a c^2 (a c-b c x)-c (a c-b c x)^2\right ) \, dx \\ & = -4 a^2 c^3 x+\frac {a c^3 (a-b x)^2}{b}+\frac {c^3 (a-b x)^3}{3 b}+\frac {8 a^3 c^3 \log (a+b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.69 \[ \int \frac {(a c-b c x)^3}{a+b x} \, dx=c^3 \left (-7 a^2 x+2 a b x^2-\frac {b^2 x^3}{3}+\frac {8 a^3 \log (a+b x)}{b}\right ) \]

[In]

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

[Out]

c^3*(-7*a^2*x + 2*a*b*x^2 - (b^2*x^3)/3 + (8*a^3*Log[a + b*x])/b)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.67

method	result	size
default	\(c^{3} \left (-\frac {b^{2} x^{3}}{3}+2 a b \,x^{2}-7 a^{2} x +\frac {8 a^{3} \ln \left (b x +a \right )}{b}\right )\)	\(41\)
norman	\(-7 a^{2} c^{3} x -\frac {b^{2} c^{3} x^{3}}{3}+2 a \,c^{3} b \,x^{2}+\frac {8 a^{3} c^{3} \ln \left (b x +a \right )}{b}\)	\(49\)
risch	\(-7 a^{2} c^{3} x -\frac {b^{2} c^{3} x^{3}}{3}+2 a \,c^{3} b \,x^{2}+\frac {8 a^{3} c^{3} \ln \left (b x +a \right )}{b}\)	\(49\)
parallelrisch	\(\frac {-b^{3} c^{3} x^{3}+6 a \,c^{3} b^{2} x^{2}+24 a^{3} c^{3} \ln \left (b x +a \right )-21 a^{2} c^{3} x b}{3 b}\)	\(54\)

[In]

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

[Out]

c^3*(-1/3*b^2*x^3+2*a*b*x^2-7*a^2*x+8/b*a^3*ln(b*x+a))

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.85 \[ \int \frac {(a c-b c x)^3}{a+b x} \, dx=-\frac {b^{3} c^{3} x^{3} - 6 \, a b^{2} c^{3} x^{2} + 21 \, a^{2} b c^{3} x - 24 \, a^{3} c^{3} \log \left (b x + a\right )}{3 \, b} \]

[In]

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

[Out]

-1/3*(b^3*c^3*x^3 - 6*a*b^2*c^3*x^2 + 21*a^2*b*c^3*x - 24*a^3*c^3*log(b*x + a))/b

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.80 \[ \int \frac {(a c-b c x)^3}{a+b x} \, dx=\frac {8 a^{3} c^{3} \log {\left (a + b x \right )}}{b} - 7 a^{2} c^{3} x + 2 a b c^{3} x^{2} - \frac {b^{2} c^{3} x^{3}}{3} \]

[In]

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

[Out]

8*a**3*c**3*log(a + b*x)/b - 7*a**2*c**3*x + 2*a*b*c**3*x**2 - b**2*c**3*x**3/3

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.79 \[ \int \frac {(a c-b c x)^3}{a+b x} \, dx=-\frac {1}{3} \, b^{2} c^{3} x^{3} + 2 \, a b c^{3} x^{2} - 7 \, a^{2} c^{3} x + \frac {8 \, a^{3} c^{3} \log \left (b x + a\right )}{b} \]

[In]

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

[Out]

-1/3*b^2*c^3*x^3 + 2*a*b*c^3*x^2 - 7*a^2*c^3*x + 8*a^3*c^3*log(b*x + a)/b

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.97 \[ \int \frac {(a c-b c x)^3}{a+b x} \, dx=\frac {8 \, a^{3} c^{3} \log \left ({\left | b x + a \right |}\right )}{b} - \frac {b^{5} c^{3} x^{3} - 6 \, a b^{4} c^{3} x^{2} + 21 \, a^{2} b^{3} c^{3} x}{3 \, b^{3}} \]

[In]

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

[Out]

8*a^3*c^3*log(abs(b*x + a))/b - 1/3*(b^5*c^3*x^3 - 6*a*b^4*c^3*x^2 + 21*a^2*b^3*c^3*x)/b^3

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.79 \[ \int \frac {(a c-b c x)^3}{a+b x} \, dx=\frac {8\,a^3\,c^3\,\ln \left (a+b\,x\right )}{b}-\frac {b^2\,c^3\,x^3}{3}-7\,a^2\,c^3\,x+2\,a\,b\,c^3\,x^2 \]

[In]

int((a*c - b*c*x)^3/(a + b*x),x)

[Out]

(8*a^3*c^3*log(a + b*x))/b - (b^2*c^3*x^3)/3 - 7*a^2*c^3*x + 2*a*b*c^3*x^2