3.1057 \(\int \frac{(a c-b c x)^2}{(a+b x)^2} \, dx\)

Optimal. Leaf size=39 \[ -\frac{4 a^2 c^2}{b (a+b x)}-\frac{4 a c^2 \log (a+b x)}{b}+c^2 x \]

[Out]

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

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Rubi [A]  time = 0.0209701, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.053, Rules used = {43} \[ -\frac{4 a^2 c^2}{b (a+b x)}-\frac{4 a c^2 \log (a+b x)}{b}+c^2 x \]

Antiderivative was successfully verified.

[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 43

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*} \int \frac{(a c-b c x)^2}{(a+b x)^2} \, dx &=\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]  time = 0.0179474, size = 33, normalized size = 0.85 \[ c^2 \left (-\frac{4 a^2}{b (a+b x)}-\frac{4 a \log (a+b x)}{b}+x\right ) \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.005, size = 40, normalized size = 1. \begin{align*}{c}^{2}x-4\,{\frac{{a}^{2}{c}^{2}}{b \left ( bx+a \right ) }}-4\,{\frac{a{c}^{2}\ln \left ( bx+a \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.04708, size = 54, normalized size = 1.38 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.5316, size = 124, normalized size = 3.18 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]  time = 0.344966, size = 36, normalized size = 0.92 \begin{align*} - \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.07997, size = 80, normalized size = 2.05 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)