3.1061 \(\int \frac{1}{(a+b x)^2 (a c-b c x)^2} \, dx\)

Optimal. Leaf size=46 \[ \frac{x}{2 a^2 c^2 \left (a^2-b^2 x^2\right )}+\frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{2 a^3 b c^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0166867, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {41, 199, 208} \[ \frac{x}{2 a^2 c^2 \left (a^2-b^2 x^2\right )}+\frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{2 a^3 b c^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x)^2*(a*c - b*c*x)^2),x]

[Out]

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

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{(a+b x)^2 (a c-b c x)^2} \, dx &=\int \frac{1}{\left (a^2 c-b^2 c x^2\right )^2} \, dx\\ &=\frac{x}{2 a^2 c^2 \left (a^2-b^2 x^2\right )}+\frac{\int \frac{1}{a^2 c-b^2 c x^2} \, dx}{2 a^2 c}\\ &=\frac{x}{2 a^2 c^2 \left (a^2-b^2 x^2\right )}+\frac{\tanh ^{-1}\left (\frac{b x}{a}\right )}{2 a^3 b c^2}\\ \end{align*}

Mathematica [A]  time = 0.0232392, size = 74, normalized size = 1.61 \[ \frac{\left (b^2 x^2-a^2\right ) \log (a-b x)+\left (a^2-b^2 x^2\right ) \log (a+b x)+2 a b x}{4 a^3 b c^2 (a-b x) (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*x)^2*(a*c - b*c*x)^2),x]

[Out]

(2*a*b*x + (-a^2 + b^2*x^2)*Log[a - b*x] + (a^2 - b^2*x^2)*Log[a + b*x])/(4*a^3*b*c^2*(a - b*x)*(a + b*x))

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 76, normalized size = 1.7 \begin{align*}{\frac{\ln \left ( bx+a \right ) }{4\,{c}^{2}{a}^{3}b}}-{\frac{1}{4\,{c}^{2}{a}^{2}b \left ( bx+a \right ) }}-{\frac{\ln \left ( bx-a \right ) }{4\,{c}^{2}{a}^{3}b}}-{\frac{1}{4\,{c}^{2}{a}^{2}b \left ( bx-a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x+a)^2/(-b*c*x+a*c)^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.03254, size = 86, normalized size = 1.87 \begin{align*} -\frac{x}{2 \,{\left (a^{2} b^{2} c^{2} x^{2} - a^{4} c^{2}\right )}} + \frac{\log \left (b x + a\right )}{4 \, a^{3} b c^{2}} - \frac{\log \left (b x - a\right )}{4 \, a^{3} b c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^2/(-b*c*x+a*c)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.64226, size = 146, normalized size = 3.17 \begin{align*} -\frac{2 \, a b x -{\left (b^{2} x^{2} - a^{2}\right )} \log \left (b x + a\right ) +{\left (b^{2} x^{2} - a^{2}\right )} \log \left (b x - a\right )}{4 \,{\left (a^{3} b^{3} c^{2} x^{2} - a^{5} b c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^2/(-b*c*x+a*c)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.407274, size = 49, normalized size = 1.07 \begin{align*} - \frac{x}{- 2 a^{4} c^{2} + 2 a^{2} b^{2} c^{2} x^{2}} + \frac{- \frac{\log{\left (- \frac{a}{b} + x \right )}}{4} + \frac{\log{\left (\frac{a}{b} + x \right )}}{4}}{a^{3} b c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)**2/(-b*c*x+a*c)**2,x)

[Out]

-x/(-2*a**4*c**2 + 2*a**2*b**2*c**2*x**2) + (-log(-a/b + x)/4 + log(a/b + x)/4)/(a**3*b*c**2)

________________________________________________________________________________________

Giac [A]  time = 1.058, size = 112, normalized size = 2.43 \begin{align*} -\frac{1}{4 \,{\left (b c x - a c\right )} a^{2} b c} + \frac{\log \left ({\left | -\frac{2 \, a c}{b c x - a c} - 1 \right |}\right )}{4 \, a^{3} b c^{2}} + \frac{1}{8 \, a^{3} b{\left (\frac{2 \, a c}{b c x - a c} + 1\right )} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x+a)^2/(-b*c*x+a*c)^2,x, algorithm="giac")

[Out]

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