\(\int \frac {(a c-b c x)^3}{(a+b x)^2} \, dx\) [1056]

Optimal result

Integrand size = 19, antiderivative size = 54 \[ \int \frac {(a c-b c x)^3}{(a+b x)^2} \, dx=5 a c^3 x-\frac {1}{2} b c^3 x^2-\frac {8 a^3 c^3}{b (a+b x)}-\frac {12 a^2 c^3 \log (a+b x)}{b} \]

[Out]

5*a*c^3*x-1/2*b*c^3*x^2-8*a^3*c^3/b/(b*x+a)-12*a^2*c^3*ln(b*x+a)/b

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 54, 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} \[ \int \frac {(a c-b c x)^3}{(a+b x)^2} \, dx=-\frac {8 a^3 c^3}{b (a+b x)}-\frac {12 a^2 c^3 \log (a+b x)}{b}+5 a c^3 x-\frac {1}{2} b c^3 x^2 \]

[In]

Int[(a*c - b*c*x)^3/(a + b*x)^2,x]

[Out]

5*a*c^3*x - (b*c^3*x^2)/2 - (8*a^3*c^3)/(b*(a + b*x)) - (12*a^2*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 (5 a c^3-b c^3 x+\frac {8 a^3 c^3}{(a+b x)^2}-\frac {12 a^2 c^3}{a+b x}\right ) \, dx \\ & = 5 a c^3 x-\frac {1}{2} b c^3 x^2-\frac {8 a^3 c^3}{b (a+b x)}-\frac {12 a^2 c^3 \log (a+b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {(a c-b c x)^3}{(a+b x)^2} \, dx=c^3 \left (5 a x-\frac {b x^2}{2}-\frac {8 a^3}{b (a+b x)}-\frac {12 a^2 \log (a+b x)}{b}\right ) \]

[In]

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

[Out]

c^3*(5*a*x - (b*x^2)/2 - (8*a^3)/(b*(a + b*x)) - (12*a^2*Log[a + b*x])/b)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83

method	result	size
default	\(c^{3} \left (-\frac {b \,x^{2}}{2}+5 a x -\frac {12 a^{2} \ln \left (b x +a \right )}{b}-\frac {8 a^{3}}{b \left (b x +a \right )}\right )\)	\(45\)
risch	\(5 a \,c^{3} x -\frac {b \,c^{3} x^{2}}{2}-\frac {8 a^{3} c^{3}}{b \left (b x +a \right )}-\frac {12 a^{2} c^{3} \ln \left (b x +a \right )}{b}\)	\(53\)
norman	\(\frac {13 a^{2} c^{3} x -\frac {1}{2} b^{2} c^{3} x^{3}+\frac {9}{2} a \,c^{3} b \,x^{2}}{b x +a}-\frac {12 a^{2} c^{3} \ln \left (b x +a \right )}{b}\)	\(58\)
parallelrisch	\(-\frac {x^{3} a \,b^{3} c^{3}+24 \ln \left (b x +a \right ) x \,a^{3} b \,c^{3}-9 x^{2} a^{2} b^{2} c^{3}+24 \ln \left (b x +a \right ) a^{4} c^{3}-26 x \,a^{3} b \,c^{3}}{2 a \left (b x +a \right ) b}\)	\(82\)

[In]

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

[Out]

c^3*(-1/2*b*x^2+5*a*x-12*a^2/b*ln(b*x+a)-8/b*a^3/(b*x+a))

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.46 \[ \int \frac {(a c-b c x)^3}{(a+b x)^2} \, dx=-\frac {b^{3} c^{3} x^{3} - 9 \, a b^{2} c^{3} x^{2} - 10 \, a^{2} b c^{3} x + 16 \, a^{3} c^{3} + 24 \, {\left (a^{2} b c^{3} x + a^{3} c^{3}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{2} x + a b\right )}} \]

[In]

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

[Out]

-1/2*(b^3*c^3*x^3 - 9*a*b^2*c^3*x^2 - 10*a^2*b*c^3*x + 16*a^3*c^3 + 24*(a^2*b*c^3*x + a^3*c^3)*log(b*x + a))/(
b^2*x + a*b)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int \frac {(a c-b c x)^3}{(a+b x)^2} \, dx=- \frac {8 a^{3} c^{3}}{a b + b^{2} x} - \frac {12 a^{2} c^{3} \log {\left (a + b x \right )}}{b} + 5 a c^{3} x - \frac {b c^{3} x^{2}}{2} \]

[In]

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

[Out]

-8*a**3*c**3/(a*b + b**2*x) - 12*a**2*c**3*log(a + b*x)/b + 5*a*c**3*x - b*c**3*x**2/2

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int \frac {(a c-b c x)^3}{(a+b x)^2} \, dx=-\frac {1}{2} \, b c^{3} x^{2} - \frac {8 \, a^{3} c^{3}}{b^{2} x + a b} + 5 \, a c^{3} x - \frac {12 \, a^{2} c^{3} \log \left (b x + a\right )}{b} \]

[In]

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

[Out]

-1/2*b*c^3*x^2 - 8*a^3*c^3/(b^2*x + a*b) + 5*a*c^3*x - 12*a^2*c^3*log(b*x + a)/b

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.48 \[ \int \frac {(a c-b c x)^3}{(a+b x)^2} \, dx=\frac {12 \, a^{2} c^{3} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b} - \frac {8 \, a^{3} c^{3}}{{\left (b x + a\right )} b} + \frac {{\left (\frac {12 \, a c^{3}}{b x + a} - c^{3}\right )} {\left (b x + a\right )}^{2}}{2 \, b} \]

[In]

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

[Out]

12*a^2*c^3*log(abs(b*x + a)/((b*x + a)^2*abs(b)))/b - 8*a^3*c^3/((b*x + a)*b) + 1/2*(12*a*c^3/(b*x + a) - c^3)
*(b*x + a)^2/b

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.96 \[ \int \frac {(a c-b c x)^3}{(a+b x)^2} \, dx=5\,a\,c^3\,x-\frac {b\,c^3\,x^2}{2}-\frac {12\,a^2\,c^3\,\ln \left (a+b\,x\right )}{b}-\frac {8\,a^3\,c^3}{b\,\left (a+b\,x\right )} \]

[In]

int((a*c - b*c*x)^3/(a + b*x)^2,x)

[Out]

5*a*c^3*x - (b*c^3*x^2)/2 - (12*a^2*c^3*log(a + b*x))/b - (8*a^3*c^3)/(b*(a + b*x))