\(\int \frac {(a c-b c x)^2}{(a+b x)^2} \, dx\) [1057]

Optimal result

Integrand size = 19, antiderivative size = 39 \[ \int \frac {(a c-b c x)^2}{(a+b x)^2} \, dx=c^2 x-\frac {4 a^2 c^2}{b (a+b x)}-\frac {4 a c^2 \log (a+b x)}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 39, 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} \[ \int \frac {(a c-b c x)^2}{(a+b x)^2} \, dx=-\frac {4 a^2 c^2}{b (a+b x)}-\frac {4 a c^2 \log (a+b x)}{b}+c^2 x \]

[In]

Int[(a*c - b*c*x)^2/(a + b*x)^2,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {(a c-b c x)^2}{(a+b x)^2} \, dx=c^2 \left (x-\frac {4 a^2}{b (a+b x)}-\frac {4 a \log (a+b x)}{b}\right ) \]

[In]

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

[Out]

c^2*(x - (4*a^2)/(b*(a + b*x)) - (4*a*Log[a + b*x])/b)

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.87

method	result	size
default	\(c^{2} \left (x -\frac {4 a \ln \left (b x +a \right )}{b}-\frac {4 a^{2}}{b \left (b x +a \right )}\right )\)	\(34\)
risch	\(c^{2} x -\frac {4 a^{2} c^{2}}{b \left (b x +a \right )}-\frac {4 a \,c^{2} \ln \left (b x +a \right )}{b}\)	\(40\)
norman	\(\frac {b \,c^{2} x^{2}+5 a \,c^{2} x}{b x +a}-\frac {4 a \,c^{2} \ln \left (b x +a \right )}{b}\)	\(41\)
parallelrisch	\(-\frac {4 \ln \left (b x +a \right ) x a b \,c^{2}-b^{2} c^{2} x^{2}+4 a^{2} c^{2} \ln \left (b x +a \right )+5 a^{2} c^{2}}{\left (b x +a \right ) b}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.56 \[ \int \frac {(a c-b c x)^2}{(a+b x)^2} \, dx=\frac {b^{2} c^{2} x^{2} + a b c^{2} x - 4 \, a^{2} c^{2} - 4 \, {\left (a b c^{2} x + a^{2} c^{2}\right )} \log \left (b x + a\right )}{b^{2} x + a b} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \frac {(a c-b c x)^2}{(a+b x)^2} \, dx=- \frac {4 a^{2} c^{2}}{a b + b^{2} x} - \frac {4 a c^{2} \log {\left (a + b x \right )}}{b} + c^{2} x \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.03 \[ \int \frac {(a c-b c x)^2}{(a+b x)^2} \, dx=-\frac {4 \, a^{2} c^{2}}{b^{2} x + a b} + c^{2} x - \frac {4 \, a c^{2} \log \left (b x + a\right )}{b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.51 \[ \int \frac {(a c-b c x)^2}{(a+b x)^2} \, dx=\frac {4 \, a c^{2} \log \left (\frac {{\left | b x + a \right |}}{{\left (b x + a\right )}^{2} {\left | b \right |}}\right )}{b} + \frac {{\left (b x + a\right )} c^{2}}{b} - \frac {4 \, a^{2} c^{2}}{{\left (b x + a\right )} b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00 \[ \int \frac {(a c-b c x)^2}{(a+b x)^2} \, dx=c^2\,x-\frac {4\,a\,c^2\,\ln \left (a+b\,x\right )}{b}-\frac {4\,a^2\,c^2}{b\,\left (a+b\,x\right )} \]

[In]

int((a*c - b*c*x)^2/(a + b*x)^2,x)

[Out]

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